Geometric Brownian Motion

1. Audio
   Observation

   The speaker says, "We say that a process S follows geometric Brownian motion".

2. Formula
   Observation

   The letter S is written on the board as part of the definition and later in the SDE.

S

A stochastic process that follows geometric Brownian motion.

Stochastic process

1. Formula
   Observation

   The variable t appears in the differential dt in the SDE and in Itô's lemma.

t

Time variable.

Real number

1. Formula
   Observation

   The variable W appears in the differential dW in the SDE and as an argument in $f(W,t)$ in Itô's lemma.

W

Standard Brownian motion (Wiener process).

Stochastic process

1. Audio
   Observation

   The speaker says, "for constants mu and sigma".

2. Formula
   Observation

   The Greek letter mu appears in the drift term of the SDE.

$\mu$

Constant drift parameter in the geometric Brownian motion SDE.

Real constant

1. Audio
   Observation

   The speaker says, "for constants mu and sigma".

2. Formula
   Observation

   The Greek letter sigma appears in the diffusion term of the SDE.

$\sigma$

Constant volatility parameter in the geometric Brownian motion SDE.

Real constant

1. Audio
   Observation

   The speaker says, "if we consider f of W and t".

2. Formula
   Observation

   The function f is written with arguments (W,t) in Itô's lemma.

$f(W,t)$

A function of the Brownian motion W and time t to which Itô's lemma is applied.

Function of two variables

1. Formula
   Observation

   The partial derivative $f_t$ appears in the first term of the Itô's lemma expansion.

$f_t$

Partial derivative of f with respect to time t.

Function

1. Formula
   Observation

   The second partial derivative f_{WW} appears multiplied by $1/2$ in the dt term of Itô's lemma.

f_{WW}

Second partial derivative of f with respect to W.

Function

1. Formula
   Observation

   The first partial derivative $f_W$ appears multiplied by dW in Itô's lemma.

$f_W$

First partial derivative of f with respect to W.

Function

1. Formula
   Observation

   The board shows the process symbol S in the definition and in the ansatz $S = f(W,t)$.

S

A stochastic process satisfying the geometric Brownian motion SDE.

Process value; the video does not state an explicit range.

1. Formula
   Observation

   The board shows dW in the SDE and W as an argument of $f(W,t)$.

W

Brownian motion driving the stochastic differential equation.

The video does not state an explicit domain.

1. Formula
   Observation

   The board shows dt in the SDE and t as the second argument of $f(W,t)$.

t

Time variable in the SDE and in the trial solution $f(W,t)$.

The video does not state an explicit time interval.

1. Audio
   Observation

   The speaker states the definition of geometric Brownian motion and the corresponding SDE.

2. Formula
   Observation

   The text 'We say that a process S follows geometric Brownian motion if it satisfies the SDE: dS = $\mu S dt + \sigma S$ dW for $\mu$ and $\sigma$ constant.' is written on the board.

A stochastic process S is said to follow geometric Brownian motion if it satisfies the specified stochastic differential equation involving constants for drift and volatility.

1. $\mu$ and $\sigma$ are constants

1. S
2. $\mu$
3. $\sigma$
4. t
5. W

1. Audio
   Observation

   The speaker recalls Itô's lemma for a function $f(W,t)$.

2. Formula
   Observation

   The text 'Ito Lemma: $f(W,t) \Rightarrow$ df(W,t) = [$f_t + \frac{1}{2} f$_{WW}] $dt + f_W$ dW.' is written on the board.

Itô's lemma provides the differential of a function f that depends on a standard Brownian motion W and time t, incorporating both first and second-order partial derivatives due to the properties of Brownian motion.

1. f is a sufficiently smooth function of W and t

1. $f(W,t)$
2. $f_t$
3. f_{WW}
4. $f_W$
5. W
6. t

1. Formula
   Observation

   Top line on the board: We say that a process S follows geometric Brownian motion if it satisfies the SDE: dS = $\mu S dt + \sigma S$ dW for $\mu$ and $\sigma$ constant.

The video defines geometric Brownian motion by specifying the stochastic differential equation that the process S must satisfy, with constant coefficients $\mu$ and $\sigma$.

1. $\mu$ is constant

2. $\sigma$ is constant

1. Formula
   Observation

   Second line on the board: Itô Lemma: $f(W,t) \Rightarrow$ df(W,t) = [$f_t + \frac{1}{2} f$_{WW}] $dt + f_W$ dW.

The board states the differential of $f(W,t)$ under Itô's lemma as a sum of a dt term involving $f_t$ and one half of f_{WW}, plus a dW term involving $f_W$.

1. Applied to a function $f(W,t)$ of Brownian motion and time

2. The video does not explicitly list smoothness assumptions

1. Formula
   Observation

   Third line on the board: Seek solutions $S = f(W,t)$.

2. Audio
   Observation

   The lecturer says they will look only at functions of Brownian motion and time as the potential solution class of this SDE.

To solve the SDE, the lecturer restricts attention to processes expressible as a function of Brownian motion and time.

1. The method searches within functions of (W,t) only

1. Definition of geometric Brownian motion via an SDE

1. Formula
   Observation

   The board writes By Itô, $f_W = \sigma S = f \cdot \sigma$ and $f_t + \frac{1}{2} f$_{WW} = $\mu S = f \cdot \mu$.

2. Audio
   Observation

   The lecturer says they consider this system of partial differential equations.

Substituting the ansatz into Itô's lemma and comparing with the SDE yields two coefficient-matching equations for f.

1. Uses $S = f(W,t)$

2. Uses the SDE dS = $\mu S dt + \sigma S$ dW

3. Uses Itô's lemma

1. Definition of geometric Brownian motion via an SDE
2. Itô's lemma for a function of Brownian motion and time
3. Trial solution class $S = f(W,t)$

1. Formula
   Observation

   The board rewrites the first equation as $f_W = \sigma f$.

2. Audio
   Observation

   The lecturer notes that t does not appear in this PDE, which implies it is actually just an ordinary differential equation.

Because the equation $f_W = \sigma f$ contains no t-derivative and treats t only as a parameter, it can be solved as an ordinary differential equation in W.

1. $\sigma$ is constant

2. t appears only parametrically

1. PDE system obtained by matching Itô coefficients

1. Formula
   Observation

   The board writes $f(W,t) = \varphi(t) e^{\sigma W}$.

2. Audio
   Observation

   The lecturer says this tells me that $f(W,t)$ is a constant function $\varphi(t) e^{\sigma W}$.

Solving $f_W = \sigma f$ gives an exponential dependence on W multiplied by an arbitrary function of t.

1. Follows from $f_W = \sigma f$

2. $\varphi(t)$ is arbitrary at this stage

1. First PDE reduces to an ODE in W

1. Formula
   Observation

   The board writes $f_t = \varphi$'(t)$e^{\sigma W}$, $f_W = \varphi(t)\cdot \sigma e^{\sigma W}$, and f_{WW} = $\varphi(t)\cdot \sigma^2 e^{\sigma W}$.

2. Audio
   Observation

   The lecturer says they will take the form of f and plug it into the second PDE, then compute $f_t$, $f_W$, and f_{WW}.

After obtaining $f(W,t)=\varphi(t)e^{\sigma W}$, the lecturer computes the derivatives needed for the second PDE.

1. Uses the product rule for $f_t$

2. Uses repeated differentiation in W for $f_W$ and f_{WW}

1. General form of f from the first PDE

1. Formula
   Observation

   Top line on the board reads: We say that a process S follows geometric Brownian motion if it satisfies the SDE: dS = mu S dt + sigma S dW for mu and sigma constant.

The video defines a process S as geometric Brownian motion when its stochastic differential equation has proportional drift and diffusion terms, with constant coefficients mu and sigma.

1. $\mu$ and $\sigma$ are constant

1. Formula
   Observation

   Second line on the board reads: Itô Lemma: $f(W,t)$ => df(W,t) = [$f_t + 1/2 f$_{WW}]$dt + f_W$ dW.

For a function f of Brownian motion and time, the differential df(W,t) consists of a dt term containing the partial time derivative plus one half of the second partial with respect to W, and a dW term containing the first partial with respect to W.

1. f is sufficiently smooth in (W,t) as assumed by the lecture

1. Definition of geometric Brownian motion

1. Formula
   Observation

   Third line on the board reads: Seek solution $S = f(W,t)$.

To solve the SDE, the lecturer seeks the unknown process S in the form of a function f of the Brownian motion and time.

1. Definition of geometric Brownian motion

1. Formula
   Observation

   The board writes $f_W = \sigma S = f \cdot \sigma$ and $f_t + \frac{1}{2} f$_{WW} = $\mu S = \mu f$.

2. Audio
   Observation

   The lecturer says, according to Ito's lemma, we have that $f_W$ is sigma S and $f_t$ plus one half f_{WW} is mu S.

If $S=f(W,t)$ solves dS=$\mu S dt+\sigma S$ dW, then the coefficients in df(W,t) must satisfy $f_W=\sigma f$ and $f_t+\frac12 f$_{WW}=$\mu f$.

1. S follows the geometric Brownian motion SDE

2. S is sought in the form $S=f(W,t)$

3. Itô's lemma applies to $f(W,t)$

For the trial solution class $S=f(W,t)$, the displayed identities hold coefficientwise.

1. Audio
   Observation

   The lecturer says notice how the variable t does not appear in this PDE, which implies that this is actually just an ordinary differential equation.

2. Formula
   Observation

   The board shows $f_W = \sigma f$.

The equation $f_W=\sigma f$ can be treated as an ordinary differential equation in W because t does not enter the differentiation and only acts as a parameter.

1. $\sigma$ is constant

2. The equation is considered with respect to W while t is held fixed

For each fixed t, the relation is an ODE in W.

1. Formula
   Observation

   The board writes $f(W,t) = \varphi(t)e^{\sigma W}$.

2. Audio
   Observation

   The lecturer says this tells me that $f(W,t)$ is a constant function $\varphi(t) e^{\sigma W}$.

The general solution of $f_W=\sigma f$ has the form $f(W,t)=\varphi(t)e^{\sigma W}$, where $\varphi$ depends only on t.

1. $f_W=\sigma f$ holds

2. $\sigma$ is constant

There exists a function $\varphi(t)$ such that $f(W,t)=\varphi(t)e^{\sigma W}$.

1. Formula
   Observation

   Top line on the board states the defining SDE for geometric Brownian motion.

A process S follows geometric Brownian motion if it satisfies dS = mu S dt + sigma S dW with mu and sigma constant.

1. mu and sigma are constant

For a process S and constants mu, sigma.

1. Formula
   Observation

   Second line on the board states Itô's lemma in the form used in the derivation.

For $f(W,t)$, df(W,t) = [$f_t + 1/2 f$_{WW}]$dt + f_W$ dW.

1. The lecture assumes the usual smoothness conditions needed to apply Itô's lemma, but does not state them explicitly.

For a sufficiently smooth function f of W and t.

1. Formula
   Observation

   Right side of the board shows the system obtained by matching coefficients after applying Itô's lemma.

If $S=f(W,t)$ solves the SDE, then f must satisfy $f_W$ = sigma f and $f_t + 1/2 f$_{WW} = mu f.

1. Use the ansatz $S=f(W,t)$ and Itô's lemma.

For the sought solution f.

1. Formula
   Observation

   Center-right derivation reduces the second PDE to an ODE after canceling e^{sigma W}.

Substituting $f(W,t)$=phi(t)e^{sigma W} into the second PDE yields phi'(t) + ($1/2$ sigma^2 - mu)phi(t)=0, equivalently phi'(t)=(mu - $1/2$ sigma^2)phi(t).

1. $f(W,t)$=phi(t)e^{sigma W}

2. The derivative formulas for $f_t$, $f_W$, f_{WW} hold.

For all t in the domain considered.

1. Formula
   Observation

   Bottom line on the board gives the final explicit expression for S.

The solution produced by the derivation is $S = c_1 \exp$(sigma W + (mu - $1/2$ sigma^2)t), and $c_1 = S_0$ when $t=0$.

1. mu and sigma are constant

2. The ansatz $S=f(W,t)$ is valid

3. The ODE solution for phi(t) is used

For the process S constructed in the lecture.

1. Formula
   Observation

   The board moves from $S=f(W,t)$ to By Itô, then writes $f_W=\sigma S=f\cdot\sigma$ and $f_t+\frac12 f$_{WW}=$\mu S=f\cdot\mu$.

2. Audio
   Observation

   The lecturer says, Then according to Ito's lemma, so by Ito, we have that $f_W$ is sigma S and $f_t$ plus one half f_{WW} is mu S.

1. Expression

   $$S = f(W,t)$$
   Explanation

   Start with the proposed solution class for the process S.

   Justification

   Stated on the board as the search ansatz.

   Shown in the video
2. Expression

   $$df(W,t) = \left[f_t + \frac{1}{2} f_{WW}\right]dt + f_W dW$$
   Explanation

   Apply Itô's lemma to the trial function $f(W,t)$.

   Justification

   Itô's lemma is written on the board.

   Shown in the video
3. Expression

   $$dS = \mu S dt + \sigma S dW$$
   Explanation

   Compare with the defining SDE for geometric Brownian motion.

   Justification

   Given by the top line of the board.

   Shown in the video
4. Expression

   $$f_W = \sigma S = f \cdot \sigma$$
   Explanation

   Match the coefficients of dW in the two differentials and substitute $S=f(W,t)$.

   Justification

   Coefficient comparison after substitution.

   Shown in the video
5. Expression

   $$f_t + \frac{1}{2} f_{WW} = \mu S = f \cdot \mu$$
   Explanation

   Match the coefficients of dt in the two differentials and substitute $S=f(W,t)$.

   Justification

   Coefficient comparison after substitution.

   Shown in the video

The SDE together with the ansatz $S=f(W,t)$ yields the system $f_W=\sigma f$ and $f_t+\frac12 f$_{WW}=$\mu f$.

1. Formula
   Observation

   The board writes $f_W = \sigma f \Rightarrow f(W,t) = \varphi(t)e^{\sigma W}$.

2. Audio
   Observation

   The lecturer says the first PDE tells me that $f_W$ is sigma f, and because t does not appear this is just an ordinary differential equation, which gives $f(W,t)=\varphi(t)e^{\sigma W}$.

1. Expression

   $$f_W = \sigma f$$
   Explanation

   Take the first equation from the PDE system.

   Justification

   Derived earlier by coefficient matching.

   Shown in the video
2. Expression

   $$\text{treat } t \text{ as a parameter}$$
   Explanation

   Observe that the equation differentiates only with respect to W, so t enters only parametrically.

   Justification

   Explicitly noted in the lecture audio.

   Shown in the video
3. Expression

   $$f(W,t) = \varphi(t)e^{\sigma W}$$
   Explanation

   Solve the resulting ODE in W; the integration “constant” may depend on t.

   Justification

   Standard solution of y'=$\sigma y$ with parameter t.

   Shown in the video

The first PDE forces the candidate solution to have the separated form $f(W,t)=\varphi(t)e^{\sigma W}$.

1. Formula
   Observation

   The board writes $f_t = \varphi$'(t)$e^{\sigma W}$, $f_W = \varphi(t)\cdot \sigma e^{\sigma W}$, and f_{WW} = $\varphi(t)\cdot \sigma^2 e^{\sigma W}$.

2. Audio
   Observation

   The lecturer says they will take the form of f and plug it into the second PDE, computing $f_t$, $f_W$, and f_{WW}.

1. Expression

   $$f(W,t) = \varphi(t)e^{\sigma W}$$
   Explanation

   Start from the general form obtained from the first PDE.

   Justification

   Previous derivation step.

   Shown in the video
2. Expression

   $$f_t = \varphi'(t)e^{\sigma W}$$
   Explanation

   Differentiate with respect to t using the product rule; $e^{\sigma W}$ is constant in t.

   Justification

   Ordinary differentiation rules.

   Shown in the video
3. Expression

   $$f_W = \varphi(t)\sigma e^{\sigma W}$$
   Explanation

   Differentiate with respect to W; $\varphi(t)$ is constant in W.

   Justification

   Derivative of $e^{\sigma W}$ with respect to W.

   Shown in the video
4. Expression

   $$f_{WW} = \varphi(t)\sigma^2 e^{\sigma W}$$
   Explanation

   Differentiate $f_W$ again with respect to W.

   Justification

   Repeated differentiation of the exponential factor.

   Shown in the video

These three derivative formulas are the quantities to be substituted into the second PDE.

1. Formula
   Observation

   The board shows the ansatz $S=f(W,t)$, Itô's lemma, and the resulting system { $f_W$ = sigma $S = f$ sigma ; $f_t + 1/2 f$_{WW} = mu $S = f$ mu }.

1. Expression

   $$S=f(W,t)$$
   Explanation

   Assume the solution can be written as a function of Brownian motion and time.

   Justification

   Stated on the board as the search ansatz.

   Shown in the video
2. Expression

   $$df(W,t)=\left[f_t+\frac{1}{2}f_{WW}\right]dt+f_W\,dW$$
   Explanation

   Apply Itô's lemma to the assumed form.

   Justification

   Itô's lemma is written on the board.

   Shown in the video
3. Expression

   $$dS=\mu S\,dt+\sigma S\,dW$$
   Explanation

   Compare with the defining SDE for geometric Brownian motion.

   Justification

   Given by the top line of the board.

   Shown in the video
4. Expression

   $$f_W=\sigma f,\qquad f_t+\frac{1}{2}f_{WW}=\mu f$$
   Explanation

   Match the coefficients of dW and dt to obtain two equations for f.

   Justification

   Coefficient comparison between the Itô expansion and the SDE.

   Shown in the video

The SDE is converted into the PDE system $f_W=\sigma f$ and $f_t+\frac{1}{2}f$_{WW}=$\mu f$.

1. Formula
   Observation

   Left column shows $f_W$ = sigma f => $f(W,t)$ = phi(t)e^{sigma W} and the derivative list.

1. Expression

   $$f_W=\sigma f$$
   Explanation

   Start from the first PDE.

   Justification

   Obtained from coefficient matching.

   Shown in the video
2. Expression

   $$f(W,t)=\phi(t)e^{\sigma W}$$
   Explanation

   Solve the first-order equation in W, allowing the integration “constant” to depend on t.

   Justification

   Standard solution of an ODE with respect to W.

   Shown in the video
3. Expression

   $$f_t=\phi'(t)e^{\sigma W}$$
   Explanation

   Differentiate with respect to t.

   Justification

   Direct differentiation of the separated form.

   Shown in the video
4. Expression

   $$f_W=\phi(t)\sigma e^{\sigma W}$$
   Explanation

   Differentiate with respect to W once.

   Justification

   Direct differentiation of the separated form.

   Shown in the video
5. Expression

   $$f_{WW}=\phi(t)\sigma^2 e^{\sigma W}$$
   Explanation

   Differentiate with respect to W twice.

   Justification

   Direct differentiation of the separated form.

   Shown in the video

The separated representation supplies the exact derivative expressions needed for the second PDE.

1. Formula
   Observation

   Center-right derivation shows phi'(t)e^{sigma W} + $1/2$ sigma^2 phi(t)e^{sigma W} = phi(t) mu e^{sigma W}, then cancellation and rearrangement.

2. Audio
   Observation

   The lecturer says the second ODE now implies, putting all this together, phi'(t)e^{sigma W} + $1/2$ sigma^2 phi(t)e^{sigma W} must equal phi(t) mu e^{sigma W}; then he cancels e^{sigma W} and rearranges.

1. Expression

   $$\phi'(t)e^{\sigma W}+\frac{1}{2}\sigma^2\phi(t)e^{\sigma W}=\phi(t)\mu e^{\sigma W}$$
   Explanation

   Substitute the derivative formulas into the second PDE $f_t+\frac{1}{2}f$_{WW}=$\mu f$.

   Justification

   Uses the previously derived expressions for $f_t$ and f_{WW}.

   Shown in the video
2. Expression

   $$\phi'(t)+\left(\frac{1}{2}\sigma^2-\mu\right)\phi(t)=0$$
   Explanation

   Cancel the common nonzero factor $e^{\sigma W}$ from every term.

   Justification

   Algebraic simplification shown on the board and spoken aloud.

   Shown in the video
3. Expression

   $$\phi'(t)=\left(\mu-\frac{1}{2}\sigma^2\right)\phi(t)$$
   Explanation

   Rearrange the linear ODE into standard form.

   Justification

   Move the phi term to the right-hand side.

   Shown in the video

The second PDE reduces to the first-order linear ODE $\phi$'(t)=($\mu-\frac{1}{2}\sigma^2$)$\phi(t)$.

1. Formula
   Observation

   Board writes phi(t)=$c_1 \exp$((mu - $1/2$ sigma^2)t), then $S=f(W,t)=c_1 \exp$(sigma W + (mu - $1/2$ sigma^2)t).

2. Audio
   Observation

   The lecturer says the solution of this ODE is phi(t)=$c_1 e$^{(mu - $1/2$ sigma^2)t}, then writes the solution of the SDE and notes $c_1=S_0$ at $t=0$.

1. Expression

   $$\phi(t)=c_1 e^{\left(\mu-\frac{1}{2}\sigma^2\right)t}$$
   Explanation

   Solve the ODE for phi(t).

   Justification

   Standard exponential solution of a first-order linear ODE.

   Shown in the video
2. Expression

   $$S=f(W,t)=\phi(t)e^{\sigma W}$$
   Explanation

   Return to the separated form of f.

   Justification

   From the earlier solution of $f_W=\sigma f$.

   Shown in the video
3. Expression

   $$S=c_1 e^{\sigma W+\left(\mu-\frac{1}{2}\sigma^2\right)t}$$
   Explanation

   Combine the factors into one exponential.

   Justification

   Algebraic multiplication of exponentials.

   Shown in the video
4. Expression

   $$c_1=S_0$$
   Explanation

   Evaluate at $t=0$ to identify the constant with the initial value.

   Justification

   Spoken conclusion at the end of the clip.

   Shown in the video

The explicit solution is $S=c_1 e^{\sigma W+(\mu-\frac{1}{2}\sigma^2)t}$, with $c_1$ fixed by the initial condition as $S_0$.

1. Formula
   Observation

   Seek solutions $S = f(W,t)$. By Itô's, {$f_w = \sigma S = f \cdot \sigma$; $f_t + \frac{1}{2}f$_{ww} = $\mu S = f \cdot \mu$}.

2. Formula
   Observation

   $f_w = \sigma f \Rightarrow f(W,t) = \phi(t)e^{\sigma W}$.

3. Formula
   Observation

   $\phi$'(t)$e^{\sigma W} + \frac{1}{2}\sigma^2 \phi(t)e^{\sigma W} = \phi(t)\mu e^{\sigma W}$.

4. Formula
   Observation

   $\phi$'(t) + ($\frac{1}{2}\sigma^2 - \mu$)$\phi(t) = 0 \Rightarrow \phi$'(t) = ($\mu - \frac{1}{2}\sigma^2$)$\phi(t)$.

5. Formula
   Observation

   $\phi(t) = c_1 \exp((\mu - \frac{1}{2}\sigma^2)t)$.

6. Formula
   Observation

   $S = f(W,t) = c_1 \exp(\sigma W + (\mu - \frac{1}{2}\sigma^2)t)$.

1. Expression

   $$S = f(W,t)$$
   Explanation

   Assume the solution S can be written as a function f of Brownian motion W and time t.

   Justification

   Method of undetermined coefficients / ansatz

   Shown in the video
2. Expression

   $$df(W,t) = \left[f_t + \frac{1}{2}f_{ww}\right]dt + f_w dW$$
   Explanation

   Apply Itô's lemma to the assumed function $f(W,t)$.

   Justification

   Itô's Lemma

   Shown in the video
3. Expression

   $$f_w = \sigma f \quad \text{and} \quad f_t + \frac{1}{2}f_{ww} = \mu f$$
   Explanation

   Equate the coefficients of dW and dt from Itô's expansion with those in the original SDE dS = $\mu S dt + \sigma S$ dW.

   Justification

   Matching coefficients of the stochastic differential equation

   Shown in the video
4. Expression

   $$f(W,t) = \phi(t)e^{\sigma W}$$
   Explanation

   Solve the partial differential equation $f_w = \sigma f$ with respect to W.

   Justification

   Separation of variables / solving first-order ODE in W

   Shown in the video
5. Expression

   $$\phi'(t)e^{\sigma W} + \frac{1}{2}\sigma^2 \phi(t)e^{\sigma W} = \mu \phi(t)e^{\sigma W}$$
   Explanation

   Substitute $f(W,t) = \phi(t)e^{\sigma W}$ and its derivatives into the second PDE.

   Justification

   Substitution

   Shown in the video
6. Expression

   $$\phi'(t) + \left(\frac{1}{2}\sigma^2 - \mu\right)\phi(t) = 0$$
   Explanation

   Divide by the non-zero term $e^{\sigma W}$ and rearrange to form an ordinary differential equation for $\phi(t)$.

   Justification

   Algebraic manipulation

   Shown in the video
7. Expression

   $$\phi(t) = c_1 \exp\left(\left(\mu - \frac{1}{2}\sigma^2\right)t\right)$$
   Explanation

   Solve the first-order linear ordinary differential equation for $\phi(t)$.

   Justification

   Standard ODE solution technique

   Shown in the video
8. Expression

   $$S = f(W,t) = c_1 \exp\left(\sigma W + \left(\mu - \frac{1}{2}\sigma^2\right)t\right)$$
   Explanation

   Combine the results for $\phi(t)$ and the exponential term in W to get the final explicit solution for S.

   Justification

   Substitution back into the ansatz

   Shown in the video