Krzysztof Ostaszewski (Pass Actuarial Exams) · YouTube · 10:32
This 120-second clip presents a worked exercise on a one-period securities market with three states, a risk-free bank account earning 10%, and two risky securities X and Y each priced at 100 today. The visible payoff table gives and . The lecturer explains that the existing price structure implies a linear system in the state prices , , , and the document displays the solution [, , ]. Near the end, the lecturer defines the Arrow-Debreu security as the security paying 1 in state 1 and 0 otherwise, with price . This 120-second clip works through a one-period securities market exercise with a risk-free bank account and two risky securities. The lecturer translates the payoff table and the common current price of 100 into three linear pricing equations for the state prices . The displayed solution is , showing that a state price vector exists for the model. This 120-second clip works through a one-period securities market exercise with two risky securities and a risk-free bank account. From the given payoffs and current prices, the lecturer forms a 3x3 linear system for the Arrow-Debreu state prices , argues that the coefficient matrix has nonzero determinant so the solution is unique, and obtains . He then uses with to convert these state prices into risk-neutral probabilities , explicitly contrasting the fact that the -vector is not a probability vector while the -vector sums to 1. This 120-second clip presents a typed lecture note on arbitrage-free pricing in finite-state securities markets. It first recalls a three-state example where a known state-price vector prices any security, illustrating the rule with a payoff vector (0,1,2) valued at 0.8727. The document then scrolls to a two-dimensional exercise with and , asking for the arbitrage-free price of a European put on the second asset with strike 1.10. The solution derives the Arrow-Debreu state prices from the replication equations, solves them by Cramer's rule to obtain psi_1= and psi_2=, notes that the risk-free rate is zero so q=psi here, decomposes the put payoff as [0,0.20]^T, and prices it as . This 120-second clip is a screen-recorded lecture over a typed document on one-period two-state securities pricing. It finishes a previous put-pricing example by computing ₁+₂= from state prices [, ]. The lecturer then works two exercises. In the first, with and , he forms the state-price equations ψ₁+ψ₂=1 and 1.5ψ₁+0.90ψ₂=1, solves [ψ₁ ψ₂]=[, ], and notes that because the risk-free rate is 0, the risk-neutral probabilities equal the state prices. He spends time explaining why the rate is zero: the first security has the same payoff in every state. In the second exercise, with and , he identifies the risk-free rate as 10%, solves 1.1ψ₁+1.1ψ₂=1 and 1.2ψ₁+0.9ψ₂=1 to obtain [, ], and converts these to risk-neutral probabilities [, ] using [q₁ q₂]=[ψ₁() ψ₂()]. This video segment covers the calculation of the state-price vector and risk-neutral probabilities in a binomial model. The instructor demonstrates how to set up and solve the system of equations for the state-price vector using Cramer's rule, and then uses the resulting state prices to compute the risk-neutral probabilities. The segment concludes with an exercise on determining if a given securities market is arbitrage-free and complete.
Use the learning inspector for key ideas and moments, or open the reading tabs for the complete notes.
Generated from the video's visuals and explanation; not verbatim speech.
The clip opens on a written exercise in a Word document. The problem defines a one-period securities market with three ingredients: a risk-free bank account, Security X, and Security Y. The bank account earns 10% per year, and every security costs 100 today.
The next visual element is the state table. There are exactly three scenarios, and the lecturer remarks that this is only a simplified classroom model because real markets would contain many more states. The table assigns payoffs at time one: Security X pays 220, 55, and 0 across the three states, while Security Y pays 0, 0, and 250.
The task stated on screen is to compute the state price vector if it exists, and otherwise explain why not. This frames the mathematical question as a test of whether the observed prices can be represented by nonnegative state prices attached to each scenario.
The solution section then displays a linear system said to be implied by the existing price structure. Each equation corresponds to one traded asset: the left-hand side multiplies that asset's state payoffs by the unknown state prices , , , and the right-hand side is the asset's current price 100.
The lecturer begins unpacking the first equation. Because the bank account is risk-free, investing 100 returns 110 in every state. That is why the first row contains 110 in all three positions. The speaker also emphasizes the ordering of assets: first the bank account, then Security X, then Security Y.
Finally, the lecture introduces the conceptual object behind the unknowns. An Arrow-Debreu security is described as a claim that pays 1 in state 1 and 0 in the other states, and its price is denoted . By extension, the displayed vector [, , ] is the collection of prices of the three state-contingent claims, and the document reports the solved values [, , ].
The clip opens on a worked exercise in a one-period securities market. The visible data are a risk-free bank account paying 110 in every scenario, Security X with payoffs 220, 55, 0, Security Y with payoffs 0, 0, 250, and a current price of 100 for each asset. The lecturer begins with the bank account and says that a payment of 110 in scenario 1 is worth today.
He continues state by state: the bank account also pays 110 in state 2 and state 3, so those contributions are and . Because the amount invested today is 100, the sum of these three state-price-weighted payoffs must equal 100. This gives the first displayed equation, .
Next he turns to Security X. Its payoff in state 1 is 220, so today that component is worth . Its payoff in state 2 is 55, contributing . Its payoff in state 3 is zero, contributing . Since Security X also costs 100 today, the second equation is .
Then he writes the equation for Security Y. The payoff is zero in states 1 and 2, so those terms are and . The payoff in state 3 is 250, giving the term . With current price 100, the third equation becomes . The visual emphasis here is that zero-payoff states are still shown explicitly in the algebra.
With all three pricing equations assembled, the document displays the solution , , , and therefore the state price vector . The lecturer starts to summarize that once all the price information is written out, the model leads to this conclusion.
The clip opens on a worked exercise in a one-period securities market model with two risky securities and a risk-free bank account. The bank account pays 10% interest, each security costs 100 today, and there are three possible future states. The lecturer immediately frames the problem algebraically: there are three equations in three unknowns, and those unknowns are the Arrow-Debreu state prices .
He then explains why the solution is unique. The relevant object is the coefficient matrix of the displayed linear system, namely . His spoken criterion is that the solution is unique when this matrix has nonzero determinant. To see this quickly, he notes that switching the second and third rows makes the determinant easy to compute, and he concludes that it is indeed nonzero.
With uniqueness established, he reads off the solution shown on the page: , , and . Thus the state price vector is .
Next he connects state prices to risk-neutral probabilities using the formula . Since the interest rate is , he multiplies each state price by . This gives , , and , so .
He pauses to stress a conceptual distinction: the vector is not a probability vector, because its entries do not sum to 1. They are prices of Arrow-Debreu securities, not probabilities. By contrast, after the multiplication by , the new vector does behave like probabilities.
To verify this, he adds the entries of the -vector: , and therefore . This confirms that is a valid probability vector. The highlighted row on screen corresponds exactly to this converted risk-neutral probability vector.
At the end of the clip, he begins to say that once these probabilities are known, one can calculate the price of another security in the market. The sentence is cut off before the next example is fully developed.
The clip opens on a worked three-state example. The document states that once the state-price vector is known, the price of any security in the market can be calculated.
The displayed vector is [,,]=[,,], obtained from probabilities multiplied by 1.10. The example security pays 0, 1, and 2 in the three states.
Using the state-price rule, the price is computed as ()()+()()+()(), which the document evaluates to 0.8727.
The lecturer then introduces a simpler two-dimensional market exercise. The given data are and .
The task is to find the arbitrage-free price of a European put on the second asset, expiring at time 1, with strike .
To obtain the state prices, the solution writes the replication equations for the two traded assets: psi_1+psi_2=1 and 1.2psi_1+0.9psi_2=1.
These equations say that the cost of reproducing each asset's statewise payoff must equal its observed time-0 price.
Applying Cramer's rule gives psi_1=(-0.1)/(-0.3)= and psi_2=(-0.2)/(-0.3)=, so the state-price vector is [,].
The note then compares q and psi through =psi_s(). Because the risk-free asset has price 1 and payoff 1, the implied risk-free rate is , hence [,]=[psi_1,psi_2].
For the put itself, the second asset pays 1.20 in state 1 and 0.90 in state 2. With strike 1.10, the payoff is 0 in state 1 and 0.20 in state 2.
The document writes this as [0,0.20]^, expressing the option payoff in the Arrow-Debreu basis.
By linearity of pricing, the put price is 0·psi_1+0.2·psi_2=.
Thus the final arbitrage-free value of the European put in this two-state market is .
The clip opens on the last lines of a completed example. The page shows the state price vector [ψ₁ ψ₂]=[, ] and the put payoff vector [0; 0.20]. Using state-price weighting, the price is computed as ₁+₂=. The lecturer verbally confirms the value as two fifteenths and remarks that this is the standard calculation in a two-state market.
The document scrolls to a new exercise. The market is specified by and , and the task is to find a state price vector. The method is to impose that each security's current price equals the sum of its state payoffs weighted by the state prices.
From the first security, whose payoffs are 1 in both states and whose price is 1, we get ψ₁+ψ₂=1. From the second security, whose payoffs are 1.50 and 0.90 and whose price is 1, we get 1.5ψ₁+0.90ψ₂=1. Solving this linear system gives ψ₂= and ψ₁=, so the state price vector is [, ].
The page then converts state prices into risk-neutral probabilities via [q₁ q₂]=[ψ₁() ψ₂()]. Because this particular market has , the factor equals 1, and therefore [q₁ q₂]=[, ] numerically matches the state price vector.
The lecturer pauses to explain why . The first security pays 1 in state 1 and 1 in state 2, so its payoff is constant across states; that makes it the risk-free security. Since one unit invested today returns tomorrow, observing a return of exactly 1 implies , hence .
A second exercise appears with and . Here the first security again has a constant payoff across states, but now the payoff is 1.10. Therefore the risk-free gross return is 1.10 and the risk-free rate is %.
To find the state prices, the same pricing logic gives 1.1ψ₁+1.1ψ₂=1 from the risk-free security and 1.2ψ₁+0.9ψ₂=1 from the second security. The written solution applies Cramer's rule and obtains ψ₁= and ψ₂=.
Finally, the risk-neutral probabilities are computed by multiplying each state price by . This yields q₁= and q₂=, so [q₁ q₂]=[, ]. The clip ends while the lecturer is still speaking, but the full written solution remains visible on screen.
We can see that the risk-free rate is 10%. To find the state-price vector, we write the system of equations: and . The solution for is , and for is .
Therefore, we have the following risk-neutral probabilities: and . Using the notation: = value of a unit invested in the risky instrument in the "up" state, = value of a unit invested in the risky instrument in the "down" state, risk-neutral probabilities are calculated as shown.
Exercise: You are given a securities market described by: , . Is this market arbitrage-free? Is it complete?
A finite-state one-period market is specified with a risk-free bank account and two risky securities X and Y. Each asset costs 100 today. The bank account pays 110 in every state because it earns 10% interest and is risk-free.
The risky payoffs are given state by state: Security X pays 220, 55, 0 in states 1, 2, 3; Security Y pays 0, 0, 250 in states 1, 2, 3.
The unknowns , , are the prices of the state-contingent claims for the three states. The exercise asks whether such a vector exists for the observed market prices.
Each traded asset generates one equation: current price equals the sum of its state payoffs weighted by the state prices. For this example the displayed system is , , and .
The document states that the system solves to , , , so the state price vector is [, , ]. Intermediate algebra is not shown in this clip.
An Arrow-Debreu security is a claim paying 1 in one chosen state and 0 elsewhere. In the clip, pays 1 in state 1 and 0 otherwise, and its price is .
The problem specifies three traded assets and three scenarios. The bank account is risk-free and pays 110 in every scenario because it earns 10% interest on an initial price of 100. Security X pays 220, 55, and 0 across scenarios 1, 2, and 3. Security Y pays 0, 0, and 250. Each asset has today's price 100.
The current value of the bank account is obtained by multiplying each scenario payoff by the corresponding state price and summing. Since the payoff is 110 in all three states and the initial price is 100, the pricing equation is .
Security X has payoffs 220, 55, and 0 in the three scenarios. Applying the same state-price valuation rule and using its current price 100 gives .
Security Y has payoffs 0, 0, and 250. Only the third scenario contributes, but the zero terms are still written explicitly in the displayed solution. With current price 100, the equation is .
Solving the three displayed linear equations yields , , and . Therefore the state price vector is , which answers the exercise prompt by showing that a state price vector exists.
The worked example considers a one-period securities market with two risky securities and one risk-free bank account. The bank account earns 10% per period, each security has current price 100, and there are three possible future states. These data are used to infer state prices.
The unknowns are the prices of Arrow-Debreu securities, i.e. the state prices for the three future states. Each traded asset's current price is matched to the sum of its state-contingent payoffs weighted by these state prices.
From the market data, the document writes the system , , and . These correspond respectively to pricing the bank account, Security X, and Security Y.
The lecturer argues that the solution is unique because the coefficient matrix of the system has nonzero determinant. He points out that swapping the second and third rows makes the determinant easy to check, and concludes that the matrix is nonsingular.
Solving the displayed system gives the unique state price vector . This is the main numerical answer of the exercise.
The document uses the relation to convert state prices into risk-neutral probabilities. With , multiplying each by yields .
The lecturer explicitly warns that should not be mistaken for probabilities. Their sum is not 1, because they are Arrow-Debreu security prices rather than state probabilities.
After conversion, the entries of add to . This is the check the lecturer uses to confirm that the transformed vector is indeed a probability vector.
If a finite-state market has known state prices , then any security with statewise payoffs is priced by summing q_sX_s over states. The opening example applies this to payoffs (0,1,2) and obtains 0.8727.
With [,,]=[,,], a security paying 0, 1, and 2 in the three states has displayed price ()()+()()+()()=0.8727.
The exercise market has two assets and two states, with time-0 prices and time-1 payoff matrix . The target instrument is a European put on the second asset with strike 1.10.
The state prices psi_1 and psi_2 are determined by requiring that the cost of replicating each traded asset equals its observed price. This yields psi_1+psi_2=1 for the first asset and 1.2psi_1+0.9psi_2=1 for the second asset.
Solving the 2x2 linear system by Cramer's rule gives psi_1= and psi_2=. The displayed numerator determinants are -0.1 and -0.2, and the coefficient determinant is -0.3.
Explore conditions, steps and evidence. Supplementary explanations are labeled separately from content shown in the video.
The solution system uses the coefficient 110 in each equation for the bank account payoff.
The lecturer says that if you put 100 in the bank account, it will pay you 110 no matter what happens at time one.
110
One-year payoff of the risk-free bank account per unit invested today, corresponding to 10% interest.
Monetary payoff in the three states.
The exercise table lists Security X as 220, 55, and 0 across Scenario 1, 2, and 3.
The lecturer reads scenario 1 as Security X pays 220, scenario 2 as Security X pays 55, and scenario 3 as Security X pays 0.
Security X
Risky security with one-year payoffs 220 in state 1, 55 in state 2, and 0 in state 3.
Payoff vector across three states.
The exercise table lists Security Y as 0, 0, and 250 across Scenario 1, 2, and 3.
The lecturer reads scenario 1 as Security Y pays 0, scenario 2 as Security Y pays 0, and scenario 3 as Security Y pays 250.
Security Y
Risky security with one-year payoffs 0 in state 1, 0 in state 2, and 250 in state 3.
Payoff vector across three states.
The exercise text states that the price of each security today is 100.
The lecturer says, 'The price of each security today is 100.'
100
Today's price of each traded security in the model.
Current monetary price.
The displayed system uses , , as unknowns.
The lecturer says a payment of one in state one and zero otherwise is called the Arrow-Debreu security , and the price of it is psi_1.
State price (price of the Arrow-Debreu security) for state 1.
Real number; solved value shown as .
The displayed system uses , , as unknowns.
The solution line gives .
State price (price of the Arrow-Debreu security) for state 2.
Real number; solved value shown as .
The displayed system uses , , as unknowns.
The solution line gives .
State price (price of the Arrow-Debreu security) for state 3.
Real number; solved value shown as .
The document writes [, , ] = [, , ].
The exercise asks to calculate the state price vector for this securities market model, if one exists.
[, , ]
Vector of prices of the three Arrow-Debreu state securities.
Three-component vector over the three states.
The lecturer says, 'A payment of one in state one and zero otherwise is called the Arrow-Debreu security .'
Arrow-Debreu security paying 1 in state 1 and 0 in the other states.
Payoff vector (1,0,0).
The displayed solution uses in the first equation and in the final state-price vector.
The lecturer repeatedly says the value today is the payment times psi-one, psi-two, or psi-three.
State price for scenario 1 (also called state 1).
Scalar unknown solved from the pricing system; displayed value .
The displayed solution uses in the second equation and in the final state-price vector.
The lecturer says the value of the payment in state two is 110 times psi-two and later 55 times psi-two.
State price for scenario 2 (also called state 2).
Scalar unknown solved from the pricing system; displayed value .
The displayed solution uses in the third equation and in the final state-price vector.
The lecturer says the value of the payment in state three is 110 times psi-three and later 250 psi-three.
State price for scenario 3 (also called state 3).
Scalar unknown solved from the pricing system; displayed value .
Exercise text: 'The one-period securities market model consists of 2 securities and a bank account ... The bank account pays interest of 10% per year and is risk-free. The price of each security today is 100.'
The lecturer reads the same setup aloud.
The clip defines a one-period market with three traded assets: a risk-free bank account and two risky securities X and Y. Each asset costs 100 today. The bank account returns 110 in every state because it pays 10% interest and is risk-free. Securities X and Y have state-dependent payoffs listed in a table.
One-period model
Three states of the world
Bank account is risk-free
Each security has current price 100
Table shows Scenario 1: Security X 220, Security Y 0; Scenario 2: Security X 55, Security Y 0; Scenario 3: Security X 0, Security Y 250.
The lecturer reads the three scenarios and their payoffs.
The risky securities are specified by their one-year payoffs in three possible states. Security X pays 220, 55, and 0 in states 1, 2, and 3 respectively. Security Y pays 0, 0, and 250 in states 1, 2, and 3 respectively.
States are labeled Scenario 1, Scenario 2, Scenario 3
The lecturer says, 'A payment of one in state one and zero otherwise is called the Arrow-Debreu security , and the price of it is psi_1.'
Only is explicitly named in the audible portion; and are implied by the displayed vector but not separately spoken in this clip.
An Arrow-Debreu security is defined here as a security that pays 1 in one particular state and 0 in all other states. Its price is the corresponding state price . In this clip, the lecturer explicitly names with price .
Finite set of states
Payoff normalized to 1 in one state and 0 elsewhere
Exercise asks: 'Calculate the state price vector for this securities market model, if one exists. If there is no state price vector, explain why.'
Solution displays [, , ] = [, , ].
The state price vector collects the prices of the Arrow-Debreu securities for all states. For this three-state example it is written as [, , ], and the displayed solution gives the numerical vector [, , ].
One-period finite-state market
Prices consistent with no-arbitrage linear pricing
Document states: 'We have the following system of equations implied by the existing price structure' followed by three equations.
The lecturer says, 'You have the following system of equations implied by the existing price structure.'
The method shown is to impose that each traded asset's current price equals the sum of its state payoffs weighted by the state prices. This produces one linear equation per traded asset. Here the three assets give three equations in the three unknowns , , .
Asset prices are known
State payoffs are known
Unknowns are state prices
Visible text: 'The one-period securities market model consists of 2 securities and a bank account (a bank account is assumed to be risk-free). The bank account pays interest of 10% per year and is risk-free. The price of each security today is 100. There are three possible scenarios for the prices of the risky securities in one year:'
Visible table rows: Scenario 1: Security X 220, Security Y 0; Scenario 2: Security X 55, Security Y 0; Scenario 3: Security X 0, Security Y 250.
The lecturer explains the bank account pays 110 in each state and that each risky security also has today's price 100.
The exercise defines a one-period market with three traded assets: a risk-free bank account and two risky securities X and Y. Each asset costs 100 today. Over one year there are three possible scenarios. The bank account pays 110 in every scenario because it earns 10% interest. Security X pays 220, 55, and 0 in scenarios 1, 2, and 3 respectively. Security Y pays 0, 0, and 250 in scenarios 1, 2, and 3 respectively.
One-period model
Three scenarios
Bank account is risk-free
Bank account interest rate is 10% per year
Price of each security today is 100
Visible text: 'We have the following system of equations implied by the existing price structure:' followed by , , and .
The lecturer derives each equation term by term: 110 times psi-one plus 110 times psi-two plus 110 times psi-three equals 100; then 220 times psi-one plus 55 times psi-two plus zero times psi-three equals 100; then zero times psi-one plus zero times psi-two plus 250 psi-three equals 100.
The state price vector is found by requiring that the current price of each traded asset equal the sum over scenarios of that asset's payoff multiplied by the corresponding state price. This produces one linear equation per asset. In this exercise the three equations correspond respectively to the bank account, Security X, and Security Y.
Prices are known today
Payoffs in each scenario are known
State prices are constant across assets within a scenario
Visible text: 'This solves to: , , . The state price vector is:' followed by .
Solving the displayed 3-by-3 linear system yields the unique state prices shown in the document. The result is presented as the state price vector for the market model.
Use the three pricing equations from the market data
The visible exercise text describes a one-period securities market model with 2 securities and a bank account.
The speaker discusses solving for the prices of the Arrow-Debreu securities implied by this market.
The clip works with a one-period market containing two risky securities and one risk-free bank account. The bank account earns 10% per period, and all three assets have current price 100. There are three possible future states, so the model is used to infer state prices from the given asset prices and payoffs.
One-period setting
Two risky securities plus one risk-free bank account
Three possible future states
Current price of each security is 100
Bank account interest rate is 10%
The document writes the solution as and displays .
The speaker says the unknowns are the prices of the Arrow-Debreu securities and then reads off the unique solution.
The unknowns are the prices of Arrow-Debreu securities, i.e. the state prices for the three future states. They are solved from the requirement that each traded asset's current price equals the sum of its state-contingent payoffs weighted by the corresponding state prices.
Defined for each future state
Used to price any contingent payoff by state-by-state weighting
The document explicitly states and computes .
The speaker says the risk-neutral probabilities are equal to times the corresponding .
Once the state prices are known, the risk-neutral probabilities are obtained by multiplying each state price by . In this example, with , the resulting vector sums to 1, confirming that it is a probability vector.
Requires already-computed state prices
Uses the same one-period interest rate as the bank account
The document states that knowing [,,] allows calculation of the price of any security in the market.
The example computes .
The speaker explains each term as probability of a state times discounted payoff in that state.
Once the state-price vector is known, the arbitrage-free price of a security is obtained by summing, over all states, the state price multiplied by the payoff in that state. In the displayed three-state example, the state prices are already written with the discount factor included, so the formula shown divides the raw payoffs by while multiplying by the numerical state-price components , , and .
A finite set of states of the world is used.
The state-price vector is known.
is the payoff of the security in state s.
The displayed example uses a one-period discount factor .
The displayed solution says 'This solves to: , , .'
The next line says 'The state price vector is: [, , ] = [, , ].'
The algebraic derivation between the system and the solution is not shown step by step in the visible portion.
For the given one-period market with bank account payoff 110 in all states, Security X payoffs (220,55,0), Security Y payoffs (0,0,250), and current price 100 for each asset, the displayed solution asserts that a state price vector exists and equals [, , ].
The three pricing equations are implied by the existing price structure.
The unknowns are the state prices , , .
For the specific numerical market model shown on screen.
The lecturer says, 'The bank account, if you put 100 in it, will pay you 110 no matter what happens at time one.'
The first equation uses 110 in each state.
In this model, investing 100 in the bank account yields 110 in every state at time one, so the bank account contributes the coefficient 110 to each state-price equation.
Bank account pays 10% interest per year.
Bank account is risk-free.
For all three states in the example.
Visible prompt: 'Calculate the state price vector for this securities market model, if one exists. If there is no state price vector, explain why.'
Visible solution line: 'This solves to: , , . The state price vector is: .'
The clip does not verbally justify uniqueness beyond displaying the solved vector.
For the displayed one-period market with payoffs (110,110,110), (220,55,0), and (0,0,250) and today's prices all equal to 100, the associated linear pricing system has the solution , , , so a state price vector exists.
One-period securities market model
Three scenarios
Risk-free bank account with 10% interest
Current price of each security is 100
Payoffs as listed in the table
There exists a vector satisfying all three pricing equations simultaneously.
The speaker says the solution will be unique if the matrix of the system of equations is non-zero, then checks that it is not zero by writing out the matrix and noting that switching rows makes the determinant easy to compute.
The displayed system has coefficient matrix .
The spoken phrase 'non-zero' refers to the determinant being nonzero; the video does not separately display the determinant value.
For the displayed 3-equation, 3-unknown linear system, the solution for is unique because the coefficient matrix has nonzero determinant.
The system is the displayed 3x3 linear system in .
The coefficient matrix is nonsingular.
For the specific system shown in the exercise.
The speaker says the sum of the three entries in is not equal to one because these are not probabilities; they are prices of Arrow-Debreu securities.
The displayed values are .
The vector is not a probability vector because its entries do not sum to 1.
Use the computed state prices from the exercise.
For the displayed solution vector.
The speaker adds , divides by 50 to get 1, and concludes these are probabilities.
The document displays .
The vector is a probability vector because its entries sum to 1.
Define with .
Use the computed values.
For the specific numerical vector shown.
The text says Knowing this, we can calculate the price of any security in this market.
An explicit example payoff vector (0,1,2) is priced as 0.8727.
If the state-price vector [,,] is known for a market, then the arbitrage-free price of any security in that market is obtained by summing state prices times statewise payoffs.
The market has three states.
The state-price vector is known.
The security has payoffs defined in each state.
For any security with statewise payoffs ,,, price = .
The note displays [,]=[psi_1(),psi_2()]=[psi_1,psi_2].
The speaker says the risk-free interest rate is actually zero.
In the two-dimensional exercise market, the risk-free interest rate is , so the Arrow-Debreu prices psi_s coincide with the state prices .
The first asset has time-0 price 1 and time-1 payoff 1 in both states.
The relation =psi_s() is being used.
For both states ,2, =psi_s because .
Document line: ₁ + ₂ = .
The lecturer says, "and that's point four over three or two fifteenth."
Using the previously obtained state prices ψ₁= and ψ₂=, the price of the put with payoff vector [0;0.20] is .
The put payoff is 0 in state 1 and 0.20 in state 2.
ψ₁= and ψ₂= from the preceding example.
Pricing is by state-price weighting.
For the specific two-state market and specific put shown on screen.
The lecturer says, "This is exactly the standard calculation in the two dimensional market."
The lecturer does not define "two dimensional market" within this clip; the phrase is taken from the audio only.
The displayed computation of the put price is presented as the standard calculation in a two-dimensional (two-state) market.
The market has two states.
State prices have already been computed.
Refers to the class of two-state one-period markets discussed in the lecture.
The lecturer says the setup is telling you that the risk-free interest rate is zero.
first column is [1;1] while first entry is 1.
In the market with and , the risk-free interest rate is 0.
The first security is risk-free.
Its initial price is 1 and its payoff is 1 in both states.
For the specific securities market shown in the first exercise.
Document text: We can see that the risk-free rate is 10%.
The lecturer says, "Now the risk free interest rate is 10%."
with .
In the market with and , the risk-free interest rate is 10%.
The first security is risk-free.
Its initial price is 1 and its payoff is 1.10 in both states.
For the specific securities market shown in the second exercise.
The document explicitly lists the three equations under 'We have the following system of equations implied by the existing price structure'.
The lecturer introduces the system and begins explaining the bank account row.
The clip does not show a fully written intermediate derivation for each asset; it presents the final system directly.
The bank account pays 110 in each of the three states, and its current price is 100, so the weighted sum of state prices must equal 100.
Observed from the displayed first equation and the lecturer's statement that 100 in the bank account pays 110 no matter what happens at time one.
Security X pays 220 in state 1, 55 in state 2, and 0 in state 3, with current price 100.
Observed from the payoff table and the displayed second equation.
Security Y pays 0 in states 1 and 2 and 250 in state 3, with current price 100.
Observed from the payoff table and the displayed third equation.
The existing price structure implies a 3-by-3 linear system in the unknown state prices , , .
The document states 'This solves to: , , .'
The third equation immediately gives .
The intermediate algebra for and is not shown in the visible frames, so only the final solved values are directly evidenced.
From the Security Y equation, only appears, so it can be read off directly.
Derived from the displayed equation .
Substituting the known into the Security X equation leaves a relation between and .
Derived from the displayed second equation after removing the zero term.
Substituting into the bank-account equation leaves another relation between and .
Derived from the displayed first equation.
The document reports these as the solution values for the remaining two unknowns.
Observed directly from the displayed solution line; the exact elimination steps are not shown on screen.
The displayed solution is [, , ] = [, , ].
From 00:00 to 00:34 the lecturer builds the bank-account equation by saying the value today is 110 times psi-one, plus 110 times psi-two, plus 110 times psi-three, and since we put a hundred in it, that has to be equal to 100.
From 00:35 to 01:06 the lecturer builds the Security X equation from payoffs 220, 55, and 0 and sets the sum equal to 100.
From 01:07 to 01:46 the lecturer builds the Security Y equation from payoffs 0, 0, and 250 and sets the sum equal to 100.
The completed displayed system is , , .
The algebraic solving steps after writing the system are not spoken in this clip; only the final solution is shown on screen.
For the risk-free bank account, each scenario pays 110, so its today's price is the sum of 110 multiplied by each state price.
State-price pricing rule applied to the bank account payoff vector (110,110,110) and current price 100.
For Security X, the scenario payoffs are 220, 55, and 0, so the current price is the weighted sum of those payoffs by the state prices.
State-price pricing rule applied to Security X payoff vector (220,55,0) and current price 100.
For Security Y, the scenario payoffs are 0, 0, and 250, so only the third state price contributes to its current price.
State-price pricing rule applied to Security Y payoff vector (0,0,250) and current price 100.
Solving the three displayed linear equations gives the state prices shown in the document.
Direct solution of the 3-equation linear system displayed on screen.
The market data imply the linear system whose solution is .
The document displays the system , , .
The speaker says there are three equations with three unknowns, and the unknowns are the prices of the Arrow-Debreu securities.
The bank account pays 110 in every state, and its current price is 100, so pricing by state prices gives this equation.
Observed directly from the displayed exercise and system.
Security X pays 220 in state 1, 55 in state 2, and 0 in state 3, with current price 100.
Observed from the payoff table and displayed equation.
Security Y pays 0 in states 1 and 2 and 250 in state 3, with current price 100.
Observed from the payoff table and displayed equation.
The market data imply a 3x3 linear system whose unknowns are the state prices .
The speaker says the solution is unique if the matrix of the system is non-zero, then writes out the matrix and notes that switching the second and third rows makes the determinant calculation simple.
The coefficient matrix is visible as .
The exact determinant value is not spoken or displayed; only the nonzero conclusion is given.
Write down the coefficient matrix of the displayed system.
Observed from the document and the speaker's verbal check.
After swapping the second and third rows, the matrix becomes easier to inspect for triangular structure.
Stated by the speaker.
The speaker concludes the determinant is nonzero, hence the system has a unique solution.
Audio explanation; the determinant value itself is not shown.
Because the coefficient matrix is nonsingular, the state-price vector is uniquely determined.
The document states: This solves to .
The speaker reads off the solution as 24 over 55, 4 over 55, and 2 over 5.
The intermediate algebraic steps are not shown on screen.
From the third equation , one obtains .
Derived from the displayed third equation.
Substituting into the remaining two equations yields the displayed values for and .
The final values are shown in the document; the full substitution steps are omitted on screen.
The unique state price vector is .
The document shows and then .
The speaker says to multiply all three by 1.10, obtaining , , and , then checks that they sum to 1.
Use the stated relation between risk-neutral probabilities and state prices.
Observed formula in the document and spoken by the lecturer.
Multiply the first state price by .
Shown in the document and read aloud.
Multiply the second state price by .
Shown in the document and read aloud.
Multiply the third state price by .
Shown in the document and read aloud.
Add the three entries to verify that the result is a probability vector.
Explicitly checked by the speaker.
The risk-neutral probability vector is , and it sums to 1.
The displayed vector [,,]=[,,] and the price expression ending in 0.8727 are visible.
The speaker reads the three terms and identifies them as probabilities times discounted payoffs.
The document does not restate within this clip how the probabilities , , and were obtained.
The state-price vector is displayed in factored form and simplified numerically.
Directly shown in the document.
The security with payoffs 0,1,2 is priced by multiplying each state price by the corresponding discounted payoff representation shown on screen.
The text says this is an example of calculating the price of a security paying 0,1,2 in the three states.
The arithmetic simplifies to the displayed final price.
Shown in the document and read aloud by the speaker.
The example security has price 0.8727 under the displayed state-price calculation.
The system 1·psi_1+1·psi_2=1 and 1.2·psi_1+0.9·psi_2=1 is displayed.
Cramer's rule expressions and the solution [psi_1,psi_2]=[,] are displayed.
The speaker says we set the system of equations and solve it.
Replicating the first asset, which pays 1 in both states, must cost its observed time-0 price 1.
Asset-price consistency using the state-price representation.
Replicating the second asset, which pays 1.2 in state 1 and 0.9 in state 2, must cost its observed time-0 price 1.
Asset-price consistency using the state-price representation.
Cramer's rule replaces the first column by the right-hand side and divides by the coefficient determinant.
Explicitly labeled using Cramer's Rule in the document.
Cramer's rule replaces the second column by the right-hand side and divides by the coefficient determinant.
Explicitly labeled using Cramer's Rule in the document.
The unique state-price vector for the exercise is [psi_1,psi_2]=[,].
The put payoff is displayed as [0,0.20]^.
The price line is displayed as 0·psi_1+0.2·psi_2=.
The speaker describes multiplying each basis payoff by its corresponding state price.
In state 1 the second asset pays 1.20, above the strike 1.10, so the put payoff is zero.
Definition of a European put payoff.
In state 2 the second asset pays 0.90, below the strike 1.10, so the put payoff is 0.20.
Definition of a European put payoff.
The payoff vector is written as a linear combination of the two Arrow-Debreu basis payoffs.
Shown directly in the document.
Linearity of pricing gives the option price as the same linear combination of state prices.
State-price pricing rule plus the solved values psi_1=, psi_2=.
The arbitrage-free price of the European put is .
The document shows the put cash flows [0;0.20] and the price calculation ₁+₂=.
The lecturer reads the final value as two fifteenth.
Price the put by weighting its state payoffs by the state prices.
State-price pricing rule shown in the document.
Substitute the previously obtained state prices ψ₁= and ψ₂=.
Values are visible earlier on the same page.
Compute the product and add the zero term.
Arithmetic simplification.
Rewrite as .
Equivalent fraction simplification shown on screen.
The put price in the preceding example is .
Document shows the system ψ₁+ψ₂=1 and 1.5ψ₁+0.90ψ₂=1, then states Hence 0.60ψ₂=0.50, so that ψ₂= and ψ₁=.
The lecturer says, "And we solve it, psi 1 is 1 6, psi 2 is 1 6," although the displayed solution is ψ₁= and ψ₂=.
Audio wording around the solved values is imprecise; the displayed formulas are clear and are used as the authoritative evidence.
Use the first security, whose payoffs are 1 in both states and whose price is 1.
State-price equation from and .
Use the second security, whose payoffs are 1.50 and 0.90 and whose price is 1.
State-price equation from and .
Eliminate ψ₁ using ψ₁=1-ψ₂ from the first equation.
Substitution from the first linear equation.
Simplify to isolate ψ₂.
Algebraic rearrangement shown on screen.
Solve for ψ₂ and then back-substitute for ψ₁.
Linear system solution displayed in the document.
For the first displayed exercise, the state price vector is [ψ₁ ψ₂]=[, ].
Full exercise and solution are visible on the Word document.
The lecturer reads the exercise and then introduces the solution system.
The clip ends before any further application of the state price vector beyond the displayed solution.
Given a one-period securities market with a risk-free bank account paying 10% per year, current price 100 for each asset, Security X payoffs (220,55,0), and Security Y payoffs (0,0,250), calculate the state price vector if it exists.
Bank account payoff is 110 in every state.
Current price of each security is 100.
There are three states.
Security X payoffs: 220, 55, 0.
Security Y payoffs: 0, 0, 250.
Find [, , ].
Write the pricing equation for the bank account.
Observed from the displayed system and explained by the lecturer's statement about the bank account.
Write the pricing equation for Security X.
Observed from the payoff table and displayed system.
Write the pricing equation for Security Y.
Observed from the payoff table and displayed system.
Solve the linear system for the three state prices.
Observed from the displayed solution line; intermediate algebra is not shown.
[, , ] = [, , ].
Substitution back into the three displayed equations is not performed on screen in this clip, so verification is limited to consistency with the shown solution line.
Visible exercise statement and full worked solution on the Word document.
The lecturer verbally walks through how each equation is formed from the table and the price 100.
The clip does not include a separate numerical check after obtaining the solution.
In a one-period securities market model with a risk-free bank account paying 10% per year and two risky securities, the price of each security today is 100. The one-year scenario payoffs are: bank account (110,110,110), Security , Security . Calculate the state price vector if one exists; otherwise explain why not.
Risk-free bank account interest rate = 10% per year
Current price of each security = 100
Scenario 1 payoffs: bank account 110, Security X 220, Security Y 0
Scenario 2 payoffs: bank account 110, Security X 55, Security Y 0
Scenario 3 payoffs: bank account 110, Security X 0, Security Y 250
Find such that each asset's current price equals the sum of its scenario payoffs times the corresponding state prices.
Write the pricing equation for the bank account using its payoff 110 in every scenario.
Apply the state-price valuation formula to the bank account.
Write the pricing equation for Security X using payoffs 220, 55, and 0.
Apply the state-price valuation formula to Security X.
Write the pricing equation for Security Y using payoffs 0, 0, and 250.
Apply the state-price valuation formula to Security Y.
Solve the resulting linear system.
Algebraic solution of the three displayed equations.
The state price vector is .
Substitute the displayed values back into the three equations: , , and .
The visible exercise asks to calculate the state price vector for the given one-period securities market model.
The lecturer solves the system, comments on uniqueness, and then converts the solution to risk-neutral probabilities.
The clip ends before the lecturer fully develops the next pricing application mentioned at the end.
In a one-period securities market model with 2 securities and a bank account, the bank account pays 10% interest and is risk-free. The price of each security today is 100. The payoffs are: Scenario 1: , ; Scenario 2: , ; Scenario 3: , . Calculate the state price vector if it exists; if not, explain why.
Two risky securities X and Y plus one risk-free bank account
Current price of each security is 100
Bank account interest rate is 10%
Three possible future states
Payoffs: (220,0), (55,0), (0,250)
Find the state price vector and determine whether it is unique.
Price the bank account using state prices.
Displayed in the solution section.
Price Security X using state prices.
Displayed in the solution section.
Price Security Y using state prices.
Displayed in the solution section.
Check that the coefficient matrix is nonsingular, so the solution is unique.
Explained verbally by the lecturer; the determinant value is not written out.
Solve the linear system to obtain the state prices.
Displayed as the solution in the document.
Write the answer as the state price vector.
Displayed in the document.
.
The lecturer verifies uniqueness by checking that the coefficient matrix has nonzero determinant, and later verifies that the transformed vector sums to 1.
The text gives an example security that pays 0 in the first state, 1 in the second state, and 2 in the third state.
The displayed price is 0.8727.
The prior derivation of the probabilities , , and is outside this clip.
Given [,,]=[,,], calculate the price of a security paying 0 in state 1, 1 in state 2, and 2 in state 3.
Three states of the world.
State-price vector [,,]=[,,].
Security payoffs are 0,1,2 in states 1,2,3 respectively.
The displayed formula uses division by .
Compute the security price.
Multiply each state price by the corresponding payoff term as displayed.
State-price pricing rule shown in the document.
Evaluate the sum.
Displayed result and spoken by the lecturer.
0.8727
The value matches the displayed final equality in the document.
The Exercise section asks for the arbitrage-free price of a European put on the second asset with strike 1.10.
The solution ends with price .
In a market with and , calculate the arbitrage-free price of a European put on the second asset expiring at time 1 with strike 1.10.
Two assets and two states.
.
.
Strike .
The option is a European put on the second asset.
Find the arbitrage-free price of the put.
Set up replication equations for the two traded assets.
No-arbitrage state-price equations shown in the solution.
Solve the linear system using Cramer's rule.
Explicitly shown and stated in the document.
Write the put payoff in the Arrow-Debreu basis.
Shown in the document after computing the state prices.
Price the put by applying state prices to its payoff components.
State-price pricing rule plus linearity.
The final displayed equality in the document is .
Top of page shows [ψ₁ ψ₂]=[, ], the put cash flows [0;0.20], and the price ₁+₂=.
The lecturer comments that this is the standard calculation in the two dimensional market.
Given state prices ψ₁= and ψ₂=, price a put whose time-1 cash flows are 0 in state 1 and 0.20 in state 2.
[ψ₁ ψ₂]=[, ].
Put payoff vector is [0;0.20].
Compute the put price at time 0.
Weight each state payoff by the corresponding state price.
State-price pricing rule.
Substitute the given state prices.
Direct substitution.
Simplify the arithmetic.
Shown on screen and read aloud.
The result matches the displayed equality chain ₁+₂=.
Exercise text: You are given a securities market described by , . Find a state price vector.
Solution text derives ψ₁=, ψ₂= and then [q₁ q₂]=[, ].
The lecturer reads the exercise and explains forming the system of equations.
You are given a securities market described by and . Find a state price vector.
.
.
Two states and two securities.
Determine [ψ₁ ψ₂], and then identify the associated risk-neutral probabilities.
Use the first security, which pays 1 in both states and costs 1.
State-price equation from the first column of .
Use the second security, which pays 1.50 and 0.90 and costs 1.
State-price equation from the second column of .
Solve the linear system.
Displayed solution on the page.
Because the risk-free rate is 0, the risk-neutral probabilities equal the state prices numerically.
Document formula plus lecturer's explanation that .
State price vector: [ψ₁ ψ₂]=[, ]. Risk-neutral probabilities: [q₁ q₂]=[, ].
Substituting ψ₁= and ψ₂= into the two displayed equations gives and .
Exercise text: You are given securities market described by , . Find the risk-neutral probabilities.
Solution text states the risk-free rate is 10%, solves ψ₁= and ψ₂=, then computes [q₁ q₂]=[, ].
The lecturer says, "Now the risk free interest rate is 10%" and begins reading the solution.
The clip ends while the lecturer is still speaking, but the full written solution is visible on screen.
You are given a securities market described by and . Find the risk-neutral probabilities.
.
.
First security is risk-free with payoff 1.10 in both states.
Determine [q₁ q₂].
Read the risk-free rate from the constant payoff 1.10 of the first security.
Document statement and market data.
State-price equation from the risk-free security.
Pricing by state prices.
State-price equation from the second security.
Pricing by state prices.
Solve the linear system using the displayed Cramer's rule determinants.
Written solution on screen.
Convert state prices to risk-neutral probabilities by multiplying by .
Displayed formula and arithmetic.
[q₁ q₂]=[, ].
The displayed determinant calculations yield ψ₁= and ψ₂=, and multiplying each by 1.1 gives and respectively.
,
Given a securities market described by and , determine if the market is arbitrage-free and complete.
Risk-free rate is 10%
Determine if the market is arbitrage-free and complete.
Set up the system of equations for the state-price vector.
Definition of state-price vector
Solve the system of equations to find the state-price vector.
Cramer's rule
Calculate the risk-neutral probabilities using the state-price vector.
Definition of risk-neutral probability
The market is arbitrage-free and complete because there exists a unique strictly positive state-price vector.
Check that the sum of the risk-neutral probabilities equals 1 and that they are strictly positive.
A Microsoft Word document titled 'Exercises' fills the screen, showing the exercise statement, payoff table, and beginning of the solution.
Word document
Exercise heading
Payoff table
Beginning of solution text
No major visual change in the first minute beyond cursor presence.
The market data and displayed equations remain fixed on screen.
The visual presentation anchors the lecture in a written worked example rather than a live derivation board.
Around 01:01 the page scrolls downward to reveal more of the solution, including the solved state price vector and additional text below.
Around 01:12 the page scrolls back upward toward the exercise statement.
Exact frame-level scroll timing is approximate.
Solution equations
State price vector line
Additional lower text partially visible
The viewport moves down to show the solved vector and then moves back up.
The underlying document content does not change; only the visible region changes.
The scroll exposes the final answer before the lecturer continues explaining the setup and meaning of the equations.
The mouse cursor moves near the payoff table and later near the first equation while the lecturer explains the bank account and Arrow-Debreu security.
Cursor path is visible but not always precisely aligned with the spoken phrase.
Mouse cursor
Scenario table
First pricing equation
Cursor shifts attention from the payoff table to the system of equations.
The numerical contents of the table and equations stay unchanged.
The pointing supports the verbal explanation that each row of the system corresponds to one traded asset.
A Microsoft Word document titled 'Exercises' fills the screen, showing the problem statement, a payoff table, the sentence introducing the system of equations, the three displayed equations, and the final state price vector.
The mouse cursor moves among the table entries and equation terms while the lecturer explains how each term is formed.
Word document page
Exercise heading
Payoff table for Scenarios 1-3
System of three linear equations
Final state price vector
Mouse cursor
Cursor points to different payoff entries and equation terms as the lecturer maps table values to equation coefficients.
No scene change occurs; the same document remains visible throughout.
The displayed market data do not change.
The three equations and final solution remain visible on the page.
The visual layout supports a direct translation from scenario payoffs and current prices into a linear system for the state price vector.
The screen scrolls downward within the Word document to reveal more of the solution area.
Microsoft Word document
Exercise statement
System of equations
Solution lines
The visible portion shifts downward
More of the solution text and matrices come into view
The same exercise remains on screen
The mathematical content does not change during the scroll
The scroll is purely navigational and exposes the already-written solution for discussion.
A yellow highlight appears over the line containing .
Document text
Row vector
Yellow highlight
The -row becomes visually emphasized
The numerical values themselves do not change
The highlight directs attention to the converted probability vector while the lecturer checks that its entries sum to 1.
The mouse cursor moves among the displayed equations, the solution values, and the highlighted -row while the lecturer speaks.
Exact cursor path is not always fully legible frame by frame.
Mouse cursor
Equations
Solution vector
Highlighted probability row
Cursor shifts from the system of equations to the solution and then to the -row
The underlying document content stays fixed
The pointing helps map the spoken explanation to the specific formulas on screen.
A Microsoft Word document titled 06-NoArbitrage-ExtendedNotes.docx is scrolled downward through several pages.
The visible content moves from a three-state example to an Exercise section, then to the solution and a new Exercise heading.
Word document page
three-state example paragraph
Exercise heading
solution equations
final put-price line
next Exercise heading
The view begins on the three-state pricing example.
Around 41 seconds it scrolls to the two-dimensional exercise statement.
Around 66-84 seconds the solution system and Cramer's-rule computations become centered.
Around 96-120 seconds the put payoff decomposition and final price are shown, followed by the start of another exercise.
The material remains a static typed document rather than a live board derivation.
The mathematical content is presented as printed formulas and prose.
The visual progression mirrors the lecture order: first a completed three-state pricing example, then a new two-state exercise, then its solution and final option price.
A yellow highlight appears over the line The solution is [psi_1,psi_2]=[,].
The speaker is discussing the solved state prices and the zero risk-free rate at this moment.
Solution line
yellow highlight
cursor
The solution line becomes highlighted.
The cursor moves near the highlighted expression.
The surrounding equations remain unchanged on screen.
The emphasis marks the key intermediate result needed for the subsequent put pricing.
The screen recording scrolls downward through a Word document, revealing the exercise statement, then its solution, then a second exercise and solution.
Mouse cursor moves over formulas and text while the page position changes.
Word document page
Exercise headings
Matrix formulas
Mouse cursor
The view moves from the tail end of a previous put-pricing example to a new exercise.
Further scrolling reveals the solution to the first exercise.
Additional scrolling reveals a second exercise and its full written solution.
The content remains a static typed document rather than live handwriting.
The mathematical notation stays in matrix and equation form throughout.
The visual motion organizes the lesson as a sequence of worked examples: finish one pricing calculation, then inspect a new market, solve for state prices, and convert them to risk-neutral probabilities.
The cursor hovers around the first exercise' and while the lecturer explains why the interest rate is zero.
Mouse cursor
Attention shifts from the solution lines back to the given market matrices.
The displayed numerical entries do not change.
The visual emphasis supports the verbal argument that the first column of is constant across states, identifying the risk-free security and hence .
The document scrolls down to reveal the next section.
Document
Scroll position
Content of the document
The instructor is moving to the next part of the lecture.
The lecturer says, 'In the real world it's not three, it's thousands, but if you can write them out, you should be able to calculate things.'
One might think a realistic securities market has only three states because this example does.
The video explicitly notes that real markets have thousands of states; the three-state example is a tractable teaching model.
The lecturer says, 'That's our first security in this market, the bank account. The second is security X, the third is security Y.'
One might treat only the risky securities X and Y as assets in the pricing system.
The video counts the bank account as the first security, so its payoff contributes the first pricing equation.
The lecturer explicitly says 'zero times psi-three' and 'zero times psi-one plus zero times psi-two plus 250 psi-three' when forming the equations.
The displayed equations retain the zero coefficients as and .
One might omit scenarios with zero payoff entirely when writing the asset pricing equation.
In this worked example, zero-payoff scenarios are still represented in the equations as terms like or , making explicit that those states contribute nothing to the asset's current price.
The speaker explicitly warns that the sum of the -entries is not 1 because these are not probabilities; they are prices of Arrow-Debreu securities.
One might think is already a probability vector.
The lecturer states that are Arrow-Debreu security prices, not probabilities, and their sum need not be 1.
The speaker says the solution is unique if the matrix of the system is non-zero.
The intended meaning is almost certainly that the determinant is nonzero, but the spoken wording is imprecise.
Saying the matrix itself is 'non-zero' could be misread as merely not the zero matrix.
In context, the lecturer is using the coefficient matrix test for uniqueness, i.e. nonsingularity / nonzero determinant.
The document separately displays [,] and [psi_1,psi_2] and then relates them by multiplication by ().
The speaker notes that in this case the risk-free interest rate is zero, so the two coincide.
One may think and psi_s are always the same object without qualification.
The document distinguishes them by =psi_s(). They coincide in this exercise only because the implied risk-free rate is .
The put payoff is shown as [0,0.20]^T even though the second asset falls from 1.20 to 0.90 across states.
The speaker states the cash flows are zero and point twenty.
A learner might incorrectly record a negative cash flow in the low state instead of a positive option payoff.
A put payoff is max(K-S,0), so when and the payoff is +0.20, not -0.20. This clarification is an analyst explanation of the displayed result.
The lecturer says, "And we solve it, psi 1 is 1 6, psi 2 is 1 6."
The displayed solution on the same page reads ψ₂= and ψ₁=.
One might infer from the audio alone that both ψ₁ and ψ₂ equal .
The written solution clearly gives ψ₁= and ψ₂=; the spoken line appears to be a slip, and the displayed algebra should be used as the authoritative result.
First exercise shows [q₁ q₂]=[ψ₁() ψ₂()] and then equals [] because .
Second exercise shows [q₁ q₂]=[, ]=[], not equal to [].
Because the first example has q₁=ψ₁ and q₂=ψ₂, one may think state prices and risk-neutral probabilities are always numerically the same.
They coincide only when , i.e. when the risk-free rate is zero. In the second example with %, the probabilities are obtained by multiplying the state prices by 1.1.
The payoff table is followed by the system of equations built from those payoffs and the common price 100.
The statewise payoff numbers for X and Y are directly used as coefficients in the linear pricing equations.
The lecturer defines as the security paying 1 in state 1 and 0 otherwise, with price psi_1.
The solution writes the vector [psi_1, psi_2, psi_3].
The state price vector is assembled from the prices of the Arrow-Debreu securities for each state.
The exercise statement defines the market and the solution applies that definition numerically.
The general one-period market description is instantiated in the numerical example with specific payoffs and prices.
The payoff table and current prices are immediately followed by the displayed system of equations.
The lecturer derives each equation directly from the table entries and the statement that each price today is 100.
The market setup supplies the payoffs and current prices that are inserted into the state-price valuation formula to produce the linear system.
The document states 'This solves to:' after the system and then lists the three state prices and the vector.
The computed state price vector depends on first deriving the three pricing equations from the market data.
The existence question in the prompt is answered by the displayed solved vector.
The explicit solution demonstrates that a state price vector exists for the given market model.
The exercise data are used to build the displayed linear system for .
The one-period market model supplies asset prices and state-contingent payoffs, which are then used to solve for the state price vector.
The document explicitly defines .
Risk-neutral probabilities are obtained from state prices by multiplying by ; thus the probability vector is derived from the state-price vector.
The lecturer ties uniqueness of the solution to the coefficient matrix having nonzero determinant.
The claim about uniqueness is what justifies treating the displayed -vector as the unique state price vector for the model.
The first part states that knowing the state-price vector lets us price any security, and the next Exercise applies the same logic in a two-state market.
The general state-price pricing rule introduced in the three-state example is applied to the two-dimensional exercise.
The exercise data , are immediately converted into the system for psi_1 and psi_2.
The numerical asset prices and payoffs in the exercise are the inputs needed to write the state-price equations.
The document says This gives (using Cramer's Rule) directly after displaying the system.
Cramer's rule is the method used to solve the displayed linear system.
Exercise asks to calculate the state price vector; solution displays [psi_1, psi_2, psi_3].
The document shows three equations implied by the existing price structure.
The lecturer explains that putting 100 in the bank account pays 110 no matter what happens at time one.
The lecturer defines as paying 1 in state 1 and 0 otherwise, with price psi_1.
The lecturer spends most of the clip explaining how each coefficient in the three equations comes from the payoff table and the price 100.
The final displayed vector is .
The displayed equations include terms such as and .
The speaker identifies the unknowns as the prices of the Arrow-Debreu securities.
The lecturer explains uniqueness by checking the coefficient matrix determinant.
The speaker contrasts the -vector with the -vector and checks that only the latter sums to 1.
The document displays .
The displayed system comes from pricing the bank account and the two securities.
Covered · The exercise statement, payoff table, and initial reading of the problem are visible and audible.
Covered · The page scrolls to reveal the displayed system and solved state price vector.
Covered · The lecturer explains the bank account row, introduces Arrow-Debreu security , and the displayed solution values remain visible.
Covered · Lecturer explains the bank account payoff 110 in each state and forms the first pricing equation.
Covered · Lecturer forms the Security X equation from payoffs 220, 55, and 0 with current price 100.
Covered · Lecturer forms the Security Y equation from payoffs 0, 0, and 250 with current price 100.
Covered · The completed system and the solved state price vector remain on screen while the lecturer begins a summary sentence.
Covered · The exercise statement and the displayed 3-equation system are introduced.
Covered · The lecturer explains uniqueness by inspecting the coefficient matrix and its determinant.
Covered · The unique solution is read off.
Covered · The lecturer converts state prices to risk-neutral probabilities and checks that they sum to 1.
Covered · The lecturer begins to say that this information can be used to calculate a price, but the clip ends before the next application is developed.
Covered · Opening three-state example shows the state-price vector and prices a security with payoffs 0,1,2 as 0.8727.
Covered · The document scrolls to the Exercise section stating the two-asset, two-state market and the European put with strike 1.10.
Covered · The solution sets up the Arrow-Debreu price equations and solves them by Cramer's rule to get psi_1= and psi_2=.
Covered · A note relates q and psi through the factor and observes that here , so the two vectors coincide.
Covered · The put payoff is decomposed into Arrow-Debreu basis payoffs and priced as 0·psi_1+0.2·psi_2=; the next Exercise heading begins to appear at the end.
Covered · Tail end of the previous example: put payoff vector, state prices [, ], price , and the lecturer's comment that this is the standard two-state calculation.
Covered · First displayed exercise: read and , form the state-price equations, solve for [, ], and compute risk-neutral probabilities when .
Covered · Conceptual explanation of why the first exercise has zero risk-free rate: the first security pays the same amount in every state.
Covered · Second displayed exercise: identify %, solve the state-price system by Cramer's rule to get [, ], then convert to risk-neutral probabilities [, ].
Covered · Calculation of the state-price vector.
Covered · Calculation of the risk-neutral probabilities.
Reviewed application connection to No-arbitrage.
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