Problem 18.7 Easy Free 3 variations

Log-Price Dynamics of GBM

A price process satisfies $dS_t = 3S_t\,dt + 4S_t\,dW_t$ for a standard Brownian motion $W_t$ . Writing the dynamics of $X_t=\ln S_t$ as $dX_t = a\,dt + b\,dW_t$ , find $a^2+b^2$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. Apply Itô’s lemma to $f(x)=\ln x$ . Because $f''<0$ , the second-order term lowers the drift by $\tfrac12\sigma^2$ .

We have $f'(x)=1/x$ and $f''(x)=-1/x^2$ . The quadratic variation of $S$ is $d[S,S]_t=16S_t^2\,dt$ . Itô’s lemma gives

dX_t=\frac{1}{S_t}\,dS_t-\frac12\cdot\frac{1}{S_t^2}\,d[S,S]_t .

Substituting $dS_t$ and $d[S,S]_t$ :

dX_t=(3\,dt+4\,dW_t)-8\,dt=-5\,dt+4\,dW_t .

Hence $a=3-\tfrac12\cdot 4^2=-5$ and $b=4$ , so

a^2+b^2=25+16=41.

\boxed{41}

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 18.7a

Consider the process $M_t$ solving $dM_t = \sigma M_t\,dW_t$ . Two attempts below compute $d\ln(M_t^2)$ and reach different answers, so exactly one contains an error. Let $a=1$ if Method 1 is the flawed one and $a=2$ if Method 2 is; let $l$ be the index of the first incorrect displayed line in that method’s derivation (lines numbered from $1$ within each method). Find $100a+l$ .

Method 1

\begin{align*} dM^2 &= (M^2)'\,dM + \tfrac12 (M^2)''\,d[M,M] \\ dM^2 &= 2M\,dM + d[M,M] \\ dM^2 &= 2\sigma M^2\,dW_t + \sigma^2 M^2\,dt \\ \frac{dM^2}{M^2} &= 2\sigma\,dW_t + \sigma^2\,dt \\ d\ln M^2 &= 2\sigma\,dW_t + \sigma^2\,dt \end{align*}

Method 2

\begin{align*} d\ln M^2 &= (\ln M^2)'\,dM + \tfrac12(\ln M^2)''\,d[M,M] \\ d\ln M^2 &= \frac{2M}{M^2}\,dM + \tfrac12\left(\frac{2M}{M^2}\right)'\,d[M,M] \\ d\ln M^2 &= \frac{2}{M}\,dM + \tfrac12\left(\frac{2}{M}\right)'\,d[M,M] \\ d\ln M^2 &= \frac{2}{M}\,dM - \frac12\cdot\frac{2}{M^2}\,d[M,M] \\ d\ln M^2 &= \frac{2}{M}\,dM - \frac{1}{M^2}\,d[M,M] \\ d\ln M^2 &= 2\sigma\,dW_t - \sigma^2\,dt \end{align*}

Work it out, then compare with the answer.

Solution

First find the correct dynamics. Since $\ln M^2=2\ln M$ , and Itô’s lemma gives $d\ln M=\sigma\,dW_t-\tfrac12\sigma^2\,dt$ , we get

d\ln M^2=2\sigma\,dW_t-\sigma^2\,dt .

Method 2 is correct. Take $f(M)=\ln M^2$ , so $f'=2/M$ and $f''=-2/M^2$ . Every line follows from the one before. With $d[M,M]=\sigma^2M^2\,dt$ , the last line is $2\sigma\,dW_t-\sigma^2\,dt$ , which matches.

Method 1 is flawed. Lines 1 to 4 are valid: Itô for $x^2$ , then division by $M^2$ . Line 5 sets $d\ln M^2=dM^2/M^2$ , which is false. The correct relation is

d\ln Y=\frac{dY}{Y}-\frac{d[Y,Y]}{2Y^2}.

Here $Y=M^2$ , so the omitted term is $-\tfrac12(2\sigma)^2\,dt=-2\sigma^2\,dt$ . Adding it to line 4 recovers $2\sigma\,dW_t-\sigma^2\,dt$ .

So $a=1$ and $l=5$ , giving $100a+l=105$ .

\boxed{105}

Variation 18.7b

Let $W_t$ be a standard Brownian motion and let $X_t$ be a process such that $Y_t = 1/X_t$ satisfies $dY_t = Y_t(2\,dt - dW_t)$ . Writing the resulting dynamics of $X_t$ as $dX_t = X_t(a\,dt + b\,dW_t)$ for integers $a,b$ , find $ab$ .

Work it out, then compare with the answer.

Solution

Take $X=f(Y)=1/Y$ , so $f'(y)=-1/y^2$ and $f''(y)=2/y^3$ . The quadratic variation is $d[Y,Y]_t=Y_t^2\,dt$ . Itô’s lemma gives

dX_t=-\frac{1}{Y_t^2}\,Y_t(2\,dt-dW_t)+\frac{1}{Y_t^3}\,Y_t^2\,dt =\frac{1}{Y_t}(-dt+dW_t)=X_t(-dt+dW_t).

So $a=-1$ and $b=1$ , giving $ab=-1$ .

\boxed{-1}

Variation 18.7c

Let $S_t$ be a geometric Brownian motion with both drift and volatility parameters equal to $1$ , and $S_0=1$ . Then $\mathbb{E}[S_2^2] = e^k$ for some constant $k$ . Find $k$ .

Work it out, then compare with the answer.

Solution

The process satisfies $dS_t=S_t(dt+dW_t)$ , so $d[S,S]_t=S_t^2\,dt$ . Itô’s lemma for $x^2$ gives

dS_t^2=2S_t\,dS_t+d[S,S]_t=S_t^2(3\,dt+2\,dW_t).

So $S_t^2$ is a geometric Brownian motion with drift $3$ and initial value $1$ . Hence $\mathbb{E}[S_t^2]=e^{3t}$ .

Check: $S_t=e^{t/2+W_t}$ , so $S_t^2=e^{t+2W_t}$ . Then $\mathbb{E}[S_t^2]=e^{t+2t}=e^{3t}$ .

At $t=2$ this is $e^6$ , so $k=6$ .

\boxed{6}

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