Black–Scholes model, Geometric Brownian motion, Model risk｜星空物语聿

1. Formula
   Observation

   The formula written at the top of the page is $C = S_0 N(d_1) - X e$^{-rT} $N(d_2)$, and it is handwritten again later.

C

call option price

option value

1. Formula
   Observation

   The formula written at the top of the page is $P = X e$^{-rT} $N(-d_2) - S_0 N(-d_1)$, and it is handwritten again later.

P

put option price

option value

1. Formula
   Observation

   appears in the formulas for C, P, and $d_1$.

2. Audio
   Observation

   referred to as 'stock price' in the explanation.

$S_0$

current price of the underlying asset

positive price quantity

1. Formula
   Observation

   appears in the formulas for C, P, and $d_1$.

2. Audio
   Observation

   referred to as 'exercise price' in the explanation.

X

option strike price

positive price quantity

1. Formula
   Observation

   appears in the exponential term e^{-rT} and in the numerator of $d_1$.

2. Audio
   Observation

   In the explanation, r is read as an interest-rate-related parameter.

r

risk-free interest rate

annualized interest rate

1. Formula
   Observation

   appears in e^{-rT}, in the numerator and denominator $\sigma\sqrt{T}$ of $d_1$.

2. Audio
   Observation

   In the explanation, it says 'this maturity is four months'.

T

time to maturity

time length measured in years

1. Formula
   Observation

   appears in the numerator of $d_1$ sigma to the power of2 over2 and in the denominator sigma times the square root of T.

2. Audio
   Observation

   In the explanation, it says 'this becomes the variance', referring to $\sigma^2$.

$\sigma$

volatility / standard deviation

non-negative parameter

1. Formula
   Observation

   written in the formula as $N(d_1)$, $N(d_2)$, $N(-d_1)$, and $N(-d_2)$.

2. Audio
   Observation

   referred to in the explanation as the standard normal distribution.

$N(\cdot)$

cumulative distribution function of the standard normal distribution

real-number input, taking values in (0, 1)

1. Formula
   Observation

   the page gives $d_1 = [\ln (S_0/X) + (r + \sigma^2/2)T]/(\sigma\sqrt{T})$.

$d_1$

intermediate variable in the Black-Scholes-Merton formula

real number

1. Formula
   Observation

   the page gives $d_2 = d_1 - \sigma\sqrt{T}$.

$d_2$

intermediate variable in the Black-Scholes-Merton formula

real number

1. Formula
   Observation

   The numerator of $d_1$ contains $\ln (S_0/X)$.

2. Audio
   Observation

   Pronounced as 'ln' in the explanation.

$\ln$

natural logarithm

positive real number

1. Formula
   Observation

   The current price of the option equals the current price of the underlying asset multiplied by the cumulative distribution function value of the standard normal distribution at d1, minus the strike price multiplied by e raised to the power of the negative product of the interest rate and the time to maturity, then multiplied by the cumulative distribution function value of the standard normal distribution at d2.

$C_0$

Current value of a European call option

real number

1. Diagram
   Observation

   The green title reads 'III. Black-Scholes-Merton'.

2. Audio
   Observation

   The instructor says, 'Then there is one final method for option pricing: the Black-Scholes-Merton pricing formula.'

This segment focuses on the Black-Scholes-Merton formula in option pricing, first stating the formula's name and then requiring memorization of the formula itself without derivation.

1. Applicable to the topic of option pricing

2. The video emphasizes memorizing the formula only.

1. Formula
   Observation

   At the top of the page, $C = S_0 N(d_1) - X e$^{-rT} $N(d_2)$ is already written.

2. Audio
   Observation

   While writing, the instructor says, 'The call option price equals the stock price $S_0$ multiplied by the standard normal distribution... minus the discounted strike price multiplied by the value of the standard normal distribution at $d_2$.'

The call option price consists of the current stock price term minus the discounted strike price term, with both terms multiplied by the standard normal distribution evaluated at different intermediate variables.

1. Using the standard normal distribution function $N(\cdot)$

2. Requires knowing $S_0$, X, r, T, $\sigma$ and first computing $d_1$ and $d_2$

1. C
2. $S_0$
3. X
4. r
5. T
6. $\sigma$
7. $N(\cdot)$
8. $d_1$
9. $d_2$

1. Formula
   Observation

   At the top of the page, $P = X e$^{-rT} $N(-d_2) - S_0 N(-d_1)$ is already written.

2. Audio
   Observation

   The instructor says, "Then for P, there is actually a pattern following the one above... overall negative signs, and then inside them, the arguments also become negative."

The put option formula mirrors the call option formula structurally, but the two terms are swapped and the arguments of N are negated.

1. Memorize by comparing with the call formula

2. Requires knowing $d_1$, $d_2$ and their negatives to substitute into $N(\cdot)$

1. Call option Black-Scholes-Merton formula
2. P
3. $N(\cdot)$
4. $d_1$
5. $d_2$

1. Formula
   Observation

   The page gives $d_1 = [\ln (S_0/X) + (r + \sigma^2/2)T]/(\sigma\sqrt{T})$.

2. Audio
   Observation

   The instructor reads term by term: "The numerator is ln of S0 divided by X, the strike price, plus r plus one-half sigma squared, times T, then divided by sigma times the square root of T."

$d_1$ is determined jointly by the logarithmic price ratio, the drift term, and the volatility term; it is the core intermediate quantity used before evaluating $N(d_1)$ and $N(-d_1)$.

1. $S_0$ is greater than0 and X is greater than0 in order to take the natural log of $S_0$ over X

2. $\sigma > 0$ is required for it to serve as a denominator.

3. $\sqrt{T}$ is meaningful only when $T > 0$.

1. $d_1$
2. $\ln$
3. $S_0$
4. X
5. r
6. $\sigma$
7. T

1. Formula
   Observation

   The page gives $d_2 = d_1 - \sigma\sqrt{T}$.

2. Audio
   Observation

   The instructor says, 'Then $d_2$ equals $d_1$ minus this value in the denominator.'

$d_2$ is not recomputed separately; instead, it is obtained by subtracting $\sigma\sqrt{T}$ from $d_1$.

1. $d_1$ must be computed first.

1. $d_2$
2. Definition of the intermediate variable $d_1$
3. $\sigma$
4. T

1. Formula
   Observation

   The current price of the option equals the current price of the underlying asset multiplied by the value of the standard normal cumulative distribution function corresponding to d1, minus the strike price multiplied by e raised to the power of the negative product of the interest rate and the expiration time, multiplied by the value of the standard normal cumulative distribution function corresponding to d2.

Used to compute the theoretical value of a European call option without dividends.

1. No stock dividends considered

2. European option

1. Formula
   Observation

   The current price of the put option equals the strike price multiplied by e raised to the power of the negative product of the interest rate and the expiration time, multiplied by the value of the standard normal cumulative distribution function corresponding to negative d2, minus the current price of the underlying asset multiplied by the value of the standard normal cumulative distribution function corresponding to negative d1.

Used to compute the theoretical value of a European put option without dividends.

1. No dividend on the underlying stock

2. European option

1. Formula
   Observation

   d1 equals the natural logarithm of the ratio of the underlying asset's current price to the strike price, plus the sum of the interest rate and half the square of the volatility, all multiplied by the time to maturity, and the result is divided by the product of the volatility and the square root of the time to maturity

Compute the intermediate variable $d_1$ required for the Black-Scholes formula.

1. Formula
   Observation

   d2 equals d1 minus the product of the volatility and the square root of the time to maturity

using $d_1$ another intermediate variable d_ required for the Black-Scholes formula2.

1. Formula for computing the intermediate variable $d_1$

1. Audio
   Observation

   The instructor says, 'Because its derivation is relatively complex, you do not need to master the derivation—just memorize this formula.'

The video states that, at this stage, learners do not need to master the derivation of the Black-Scholes-Merton formula; they only need to memorize the formula.

1. For the current learning segment

2. Focuses on memorizing the formula rather than proving its derivation

1. Audio
   Observation

   The instructor says, "Then for P, there is actually a pattern consistent with the one above... overall negative, and then inside, the argument must also become negative."

2. Formula
   Observation

   The screen simultaneously displays both the C and P formulas, with the two terms in P swapped and the argument of N negated.

By comparing the forms of C and P, the video explains that the put option formula can be memorized from the call formula following the rule: "swap the two terms and negate the argument of N."

1. The call option formula form is known.

2. This is presented solely as a mnemonic rule, not a rigorous proof.

1. Formula
   Observation

   In the example solution area, $d_1$ = [$\ln (60/65)$ + (11% + 45%^$2/2$) * $4/12$]/(45% * sqrt($4/12$)) ≈ -0.0371.

2. Formula
   Observation

   Then $d_2 = d_1 - 45$% * sqrt($4/12$) = -0.1871.

3. Audio
   Observation

   The instructor says, "First compute d1... substitute these numbers in, and then it approximates to this value... when looking up these two values in the table, round to the nearest hundredth."

1. In the audio, r is pronounced as '11', but the on-screen example problem displays it as 11%; follow the on-screen percentage sign.

1. Expression

   $$d_1 = \frac{\ln(60/65) + (11\% + 45\%^2/2)\times 4/12}{45\%\times \sqrt{4/12}}$$
   Explanation

   Substitute the given values from the example problem—$S_0=60$, $X=65$, $r=11$%, $\sigma=45$%, and $T=4/12$—into the definition of $d_1$.

   Justification

   Apply the ki-d1-formula to substitute parameters.

   Shown in the video
2. Expression

   $$d_1 \approx -0.0371$$
   Explanation

   After completing the numerical computation, obtain an approximate value for $d_1$.

   Justification

   The screen directly displays the computational result.

   Shown in the video
3. Expression

   $$d_2 = d_1 - 45\%\times \sqrt{4/12}$$
   Explanation

   Substitute the already-computed $d_1$, along with volatility and time to maturity, into the expression defining $d_2$.

   Justification

   Apply the ki-d2-formula.

   Shown in the video
4. Expression

   $$d_2 = -0.1871$$
   Explanation

   Obtain the numerical value of $d_2$.

   Justification

   The screen directly displays the computational result.

   Shown in the video

The example problem first computes $d_1$≈-0.0371, then uses $d_2=d_1-\sigma\sqrt{T}$ to obtain $d_2=-0.1871$.

1. Audio
   Observation

   The instructor verbally summarizes: 'They are both negative overall, and then inside, they must become negative again.'

2. Formula
   Observation

   In the image, C uses $N(d_1)$ and $N(d_2)$, while P uses $N(-d_2)$ and $N(-d_1)$, with the two terms swapped in order.

1. Expression

   $$C = S_0 N(d_1) - X e^{-rT} N(d_2)$$
   Explanation

   First, fix the call option formula as the reference.

   Justification

   The image already displays this formula.

   Shown in the video
2. Expression

   $$P = X e^{-rT} N(-d_2) - S_0 N(-d_1)$$
   Explanation

   Swap the positions of the two terms and change the arguments of all N functions to their opposites.

   Justification

   This is a memory aid summarized verbally in the video, not a formal proof.

   Shown in the video

The video explains how the put formula can be remembered from the call formula using patterns of sign changes and term ordering, via structural comparison.

1. Formula
   Observation

   d1 equals the natural logarithm of sixty-five over sixty, plus the sum of eleven percent and half the square of forty-five percent, all multiplied by four over twelve, and the result is divided by the product of forty-five percent and the square root of four over twelve, approximately equal to negative zero point zero three seven one

1. Expression

   $$\ln(60/65) + (11\% + 45\%^2/2) \times 4/12$$
   Explanation

   Substitute the known parameters into the numerator of the $d_1$ formula.

   Justification

   Substitution method

   Shown in the video
2. Expression

   $$45\% \times \sqrt{4/12}$$
   Explanation

   Compute the denominator of the $d_1$ formula.

   Justification

   Substitution method

   Shown in the video
3. Expression

   $$\frac{-0.07918 + 0.04275}{0.2598} \approx -0.0371$$
   Explanation

   Obtain an approximate value for $d_1$.

   Justification

   Numerical computation

   Shown in the video

d1 is approximately negative zero point zero three seven one

1. Formula
   Observation

   d2 equals negative zero point zero three seven one minus the product of forty-five percent and the square root of four twelfths, resulting in negative zero point one eight seven one

1. Expression

   $$-0.0371 - 45\% \times \sqrt{4/12}$$
   Explanation

   Substitute the computed $d_1$ and the known parameters into the $d_2$ formula.

   Justification

   Substitution method

   Shown in the video
2. Expression

   $$-0.0371 - 0.15 = -0.1871$$
   Explanation

   Obtain the value of $d_2$.

   Justification

   Numerical computation

   Shown in the video

d2 equals negative zero point one eight seven one

1. Formula
   Observation

   The example text states: "maturity of 4 months; standard deviation of 45%; strike price of 65; stock price of 60; interest rate of 11%".

2. Formula
   Observation

   The solution area writes: $C_0 = 60N(d_1) - 65e$^{-11%\times $4/12$}$N(d_2)$, and $P_0 = 65e$^{-11%\times $4/12$}$N(-d_2) - 60N(-d_1)$.

3. Formula
   Observation

   The solution area further gives $d_1$ ≈ -0.0371 and $d_2 = -0.1871$.

4. Audio
   Observation

   The instructor says, "This maturity is four months… first compute $d_1$… substitute this number… when looking up the table, round to the nearest value."

1. In the audio, r is pronounced as "11", while the on-screen text shows 11%; interpret r as a percentage according to the on-screen text.

2. At the end of the clip, the final option price value has not yet appeared; only $d_1$, $d_2$, and the formula setup are completed.

Write down the Black-Scholes option pricing formula and use it to compute the European option value for the following stock (assuming no dividend payments).

1. Time to maturity: 4 months

2. Standard deviation: 45 percent

3. Option strike price: 65

4. Stock price: 60

5. Interest rate: 11 percent

Write the call and put option value expressions, and first compute the intermediate variables $d_1$ and $d_2$.

1. Expression

   $$T = 4/12$$
   Explanation

   Convert "4 months" into annualized time to maturity.

   Justification

   In the formula, T is measured in years; the on-screen solution directly uses $4/12$.

   Shown in the video
2. Expression

   $$C_0 = 60N(d_1) - 65e^{-11\%\times 4/12}N(d_2)$$
   Explanation

   Substitute the known parameters into the call option formula.

   Justification

   Use ki-call-formula.

   Shown in the video
3. Expression

   $$P_0 = 65e^{-11\%\times 4/12}N(-d_2) - 60N(-d_1)$$
   Explanation

   Substitute the known parameters into the put option formula.

   Justification

   Use ki-put-formula.

   Shown in the video
4. Expression

   $$d_1 = \frac{\ln(60/65) + (11\% + 45\%^2/2)\times 4/12}{45\%\times \sqrt{4/12}} \approx -0.0371$$
   Explanation

   First compute $d_1$.

   Justification

   Use ki-d1-formula and substitute the example data.

   Shown in the video
5. Expression

   $$d_2 = d_1 - 45\%\times \sqrt{4/12} = -0.1871$$
   Explanation

   then use $d_1$ to compute $d_2$.

   Justification

   Use ki-d2-formula.

   Shown in the video

Within this segment, the visible results are $d_1$≈-0.0371 and $d_2=-0.1871$, and expressions for $C_0$ and $P_0$ are listed; the final numerical option price is not given in this segment.

The video notes that when looking up values in the table later, $d_1$ and $d_2$ should first be rounded to two decimal places before evaluating $N(\cdot)$.

1. Formula
   Observation

   Write the Black-Scholes option pricing formula and use it to compute the value of a European option on the following stock (assuming no dividends): maturity 4 months, volatility 45 percent, strike price 65, stock price 60, interest rate 11 percent.

Compute the value of a European option given the parameters.

1. Maturity T equals 4 divided by 12 years

2. Volatility sigma equals 45 percent

3. Strike price X equals 65

4. Stock price S subscript 0 equals 60

5. Interest rate r equals 11 percent

Derive expressions for C subscript 0 and P subscript 0.

1. Expression

   $$C_0 = 60 N(d_1) - 65 e^{-11\% \times 4/12} N(d_2)$$
   Explanation

   Substitute the parameters into the call option formula.

   Justification

   Formula substitution

   Shown in the video
2. Expression

   $$P_0 = 65 e^{-11\% \times 4/12} N(-d_2) - 60 N(-d_1)$$
   Explanation

   Substitute the parameters into the put option formula.

   Justification

   Formula substitution

   Shown in the video
3. Expression

   $$d_1 \approx -0.0371$$
   Explanation

   Compute the intermediate variable $d_1$.

   Justification

   Numerical calculation

   Shown in the video
4. Expression

   $$d_2 = -0.1871$$
   Explanation

   Compute the intermediate variable $d_2$.

   Justification

   Numerical calculation

   Shown in the video

The current price of the call option equals sixty times the cumulative distribution function value of the standard normal distribution at negative zero point zero three seven one, minus sixty-five times e raised to the power of negative eleven percent multiplied by four twelfths, multiplied by the cumulative distribution function value of the standard normal distribution at negative zero point one eight seven one; the current price of the put option equals sixty-five times e raised to the power of negative eleven percent multiplied by four twelfths, multiplied by the cumulative distribution function value of the standard normal distribution at zero point one eight seven one, minus sixty times the cumulative distribution function value of the standard normal distribution at zero point zero three seven one

Since the N value is not provided, only the expression is written; the N value will be given in subsequent questions.

1. Diagram
   Observation

   The upper part of the screen statically displays a green title and three blue formulas: C, P, $d_1$, and $d_2$.

1. Green title: "Three, Black-Scholes-Merton"

2. The blue formula $C = S_0 N(d_1) - X e$^{-rT} $N(d_2)$

3. The blue formula $P = X e$^{-rT} $N(-d_2) - S_0 N(-d_1)$

4. The blue formula $d_1 = [\ln (S_0/X) + (r + \sigma^2/2)T]/(\sigma\sqrt{T})$

5. The blue formula $d_2 = d_1 - \sigma\sqrt{T}$

1. No writing action; the screen remains static as a formula overview.

1. The relative positions of the four formulas remain unchanged.

2. The notation system remains consistent.

This segment first presents the complete formula framework using static board writing, establishing a reference for subsequent memorization and example substitution.

1. Animation
   Observation

   The instructor handwrites the C formula and then the P formula in blank areas.

2. Audio
   Observation

   While handwriting, the instructor verbally explains the meaning of each term.

1. The newly written C formula

2. Newly written put formula

3. Original formula above

1. First write $C = S_0 N(d_1) - X e$^{-rT} $N(d_2)$

2. Then write $P = X e$^{-rT} $N(-d_2) - S_0 N(-d_1)$

3. The writing order reflects the presentation sequence: call first, then put

1. The meanings of the symbols $S_0$, X, r, T, N, $d_1$, and $d_2$ remain unchanged

By handwriting from memory, abstract formulas are broken down into visible term sequences and symbol changes, highlighting the contrast between the put formula and the call formula.

1. Animation
   Observation

   The presenter continues handwriting the full fractional expression for $d_1$, then writes $d_2 = d_1 - \sigma\sqrt{T}$.

1. Fractional expression for $d_1$

2. Relationship for $d_2$

1. First write the numerator of $d_1$: $\ln (S_0/X)$ and ($r+\sigma^2/2$)T

2. Next, write the denominator $\sigma\sqrt{T}$.

3. Finally, write $d_2$ by subtracting $\sigma\sqrt{T}$ from $d_1$.

1. $d_2$ always depends on $d_1$ and is not recalculated independently.

Visually fix the computation order as 'compute $d_1$ first, then compute $d_2$'.

1. Animation
   Observation

   Scroll the page down to the example section, where the problem statement and solution appear, with $d_1$, $d_2$, and other locations circled in red.

2. Formula
   Observation

   The solution area displays parameter substitution and approximate results: $d_1$≈-0.0371, $d_2=-0.1871$.

1. Example problem statement.

2. Solution area: expressions for $C_0$ and $P_0$.

3. Expression for computing $d_1$.

4. Expression for computing $d_2$.

5. Red circles and arrows.

1. The screen scrolls from the formula memorization section to the example problem section.

2. A red pen circles $d_1$, $d_2$, and the locations of related parameters.

3. The numerical result is highlighted.

1. Problem parameters remain unchanged: $S_0=60$, $X=65$, $r=11$%, $\sigma=45$%, $T=4/12$.

This segment applies the earlier general formula to concrete numbers, demonstrating the process of 'plugging into the formula—substituting parameters—computing intermediate variables.'

1. Animation
   Observation

   Circle the $N(d_1)$ and $N(d_2)$ parts of the formula in red ink and underline them below.

1. Black-Scholes formula

1. Add red highlighting and circled annotations.

1. The mathematical structure of the formula itself remains unchanged.

Emphasize that $N(d_1)$ and $N(d_2)$ are key components when computing option value; typically, problems either provide these values or require looking them up in a table.

1. Animation
   Observation

   The screen scrolls upward from a printed example solution to reveal a handwritten summary of the Black-Scholes formula.

1. Printed example solution.

2. Handwritten formula notes.

1. The viewport shifts to show upper content.

1. The page content itself remains unchanged.

Transition from a specific example to memorizing and summarizing the general formula.

1. Audio
   Observation

   The instructor explicitly states, 'Because its derivation is relatively complex, you do not need to master the derivation—just memorize this formula.'

When learning the Black-Scholes-Merton formula, one must fully understand the complete derivation before using it.

In this segment, the video explicitly emphasizes memorizing the formula and substituting values, without requiring mastery of the derivation.

1. Audio
   Observation

   The instructor emphasizes that the put formula 'follows a pattern with the one above' and says 'inside it, the arguments must become negative.'

2. Formula
   Observation

   The visual shows that the arguments of N in P are -d_2 and -d_1, not $d_2$ and $d_1$.

When writing the P formula, retain the notation $N(d_1)$ and $N(d_2)$ without changing the signs.

The put formula shown in the video is $P = X e$^{-rT} $N(-d_2) - S_0 N(-d_1)$; the arguments inside N must be negated, and the order of the two terms is also swapped.

1. Audio
   Observation

   The instructor says, "When looking up these two values in the table, round them to the nearest hundredth."

Directly look up the unrounded $d_1$ and $d_2$ in the standard normal distribution table.

The video reminds viewers to first round $d_1$ and $d_2$ to the nearest hundredth, then look up the corresponding $N(\cdot)$ values.

1. Audio
   Observation

   This value is usually given, or at least known; just substitute it in and compute the result.

2. Formula
   Observation

   If N values are not provided, write only the expression; later questions will supply the N values.

Students may think they must compute cumulative probabilities of the standard normal distribution themselves.

The video points out that, in exams or exercises, values of $N(d_1)$ and $N(d_2)$ are usually provided directly, or it is acceptable to leave expressions containing $N(d)$ unevaluated.

1. Formula
   Observation

   The screen displays the C and P formulas side by side.

2. Audio
   Observation

   The instructor says that P "follows a pattern visible above."

The put formula contrasts with the call formula in term order and in the sign of the argument of N; the video uses this contrast to aid memorization.

1. Formula
   Observation

   The page first writes the definition of $d_1$, then writes $d_2 = d_1 - \sigma\sqrt{T}$.

2. Audio
   Observation

   The instructor says, "$d_2$ equals $d_1$ minus this denominator value."

Computing $d_2$ depends on first computing $d_1$; the video explicitly reflects this dependency order in the formulas.

1. Formula
   Observation

   In the example solution section, directly substitute the previously written formulas for C, P, $d_1$, and $d_2$.

The example applies the general Black-Scholes-Merton formula to a specific set of parameters.

1. Formula
   Observation

   In the example, $d_1$ is calculated first, then $d_2$.

The first step in the numerical solution of the example is substituting into the definition of $d_1$.

1. Formula
   Observation

   In the example, $d_2 = d_1 - 45$% $\times \sqrt{4/12}$.

The second step of the example uses the relationship between $d_2$ and $d_1$ to complete the intermediate variable calculation.

1. Formula
   Observation

   d2 equals d1 minus the product of volatility and the square root of time to maturity.

Computing $d_2$ requires first finding the value of $d_1$.

1. Formula
   Observation

   The example problem directly substitutes into the Black-Scholes formula for calculation.

The example problem demonstrates how to apply the Black-Scholes formula to solve a concrete option pricing problem.

1. Audio
   Observation

   At the beginning, it directly names 'this Black-Scholes-Merton pricing formula'.

1. Introduction to the Black-Scholes-Merton pricing formula
2. Call option Black-Scholes-Merton formula
3. Put option Black-Scholes-Merton formula

1. Formula
   Observation

   Handwrite the replication of the C formula.

1. Call option Black-Scholes-Merton formula

1. Formula
   Observation

   Handwrite the replication of the P formula.

1. Put option Black-Scholes-Merton formula
2. Incorrectly writing the put formula with the same positive arguments as the call formula.

1. Formula
   Observation

   The page gives the definition formulas for $d_1$ and $d_2$.

1. Definition of the intermediate variable $d_1$
2. Definition of the intermediate variable $d_2$

1. Formula
   Observation

   The example solution area shows the parameter substitution process.

1. Substituting into the Black-Scholes-Merton formula
2. Substituting numerical values for $d_1$ and $d_2$ in the example problem.

1. Audio
   Observation

   The instructor reminds viewers to round to four decimal places when using tables.

1. Do not round decimal places before consulting the table.

1. Formula
   Observation

   The formula for computing $d_1$ and the process of substituting numerical values.

1. Formula for computing the intermediate variable $d_1$
2. Numerical derivation of $d_1$

1. Audio
   Observation

   The instructor mentions that N-values are usually provided.

1. Obtaining values of $N(d)$