Problem 18.12 Easy Free 3 variations

Killing the Drift

Let $W_t$ be a standard Brownian motion, and fix a constant $c$ . Define $X_t = W_t^3 - ctW_t$ . Find the value of $c$ for which $X_t$ is a martingale.

Work it out, then compare with the answer.

SolutionShow the worked solution

Write $X_t=f(t,W_t)$ with $f(t,w)=w^3-ctw$ . Then

f_t=-cw,\qquad f_w=3w^2-ct,\qquad f_{ww}=6w .

Itô’s lemma gives

dX_t=\Bigl(f_t+\tfrac12 f_{ww}\Bigr)dt+f_w\,dW_t=(3-c)W_t\,dt+(3W_t^2-ct)\,dW_t .

The integrand $3W_t^2-ct$ is square-integrable on every $[0,T]$ . So the $dW_t$ part is a true martingale, and $X_t$ is a martingale exactly when the drift $(3-c)W_t$ vanishes. That happens precisely when $c=3$ .

To see that the condition is also necessary, compute the conditional expectation directly. For $s<t$ , write $W_t=W_s+(W_t-W_s)$ , where the increment is independent of $\mathcal F_s$ and is $\mathcal N(0,t-s)$ . This gives

\mathbb{E}[W_t^3\mid\mathcal F_s]=W_s^3+3(t-s)W_s ,

and hence

\mathbb{E}[X_t\mid\mathcal F_s]=X_s+(3-c)(t-s)W_s .

The extra term is zero almost surely only when $c=3$ .

\boxed{c=3}

Variations

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

Variation 18.12a

Let $W_t$ be a standard Brownian motion. For a non-negative integer $n$ , consider $X_t = W_t^n$ . Find the sum of $n!$ over all values of $n$ for which $X_t$ is a martingale.

Work it out, then compare with the answer.

Solution

For $f(w)=w^n$ , Itô’s lemma gives

dX_t=nW_t^{n-1}\,dW_t+\tfrac12 n(n-1)W_t^{n-2}\,dt .

The drift vanishes only for $n=0$ and $n=1$ . In those cases $X_t\equiv1$ and $X_t=W_t$ , both of which are martingales.

For $n\ge2$ the process is not a martingale. For $s<t$ ,

\mathbb{E}[X_t-X_s\mid\mathcal F_s]=\tfrac12 n(n-1)\int_s^t\mathbb{E}[W_u^{n-2}\mid\mathcal F_s]\,du .

The conditional expectation inside is a polynomial in $W_s$ with leading term $W_s^{n-2}$ , so the right-hand side is not identically zero. For example, when $n=2$ it equals $t-s$ .

The sum is therefore $0!+1!$ :

\boxed{2}

Variation 18.12b

Let $W_t$ be a standard Brownian motion, and let $\{a_t\}_{t\ge0}$ be a process adapted to the natural filtration of $W$ . Define $X_t = W_t^{10} - \int_0^t a_s\,ds$ . There are constants $A$ and $B$ such that the choice $a_t = A W_t^B$ makes $X_t$ a martingale. Find $AB$ .

Work it out, then compare with the answer.

Solution

The integral term has finite variation, so it contributes no second-order term. Itô’s lemma gives

dX_t=10W_t^9\,dW_t+\bigl(\tfrac12\cdot 90\,W_t^8-a_t\bigr)dt=10W_t^9\,dW_t+(45W_t^8-a_t)\,dt .

The drift is unique, because a continuous finite-variation martingale is constant. So the drift must vanish, which forces $a_t=45W_t^8$ . This gives $A=45$ and $B=8$ .

\boxed{AB=360}

Variation 18.12c

Let $W_t$ be a standard Brownian motion and define the Itô integral $Y_t = \int_0^t s\,dW_s$ . Let $\{a_t\}_{t\ge0}$ be a process of the form $a_t = k\int_0^t Y_s\,ds$ for a constant $k$ , chosen so that $Z_t = tY_t - a_t$ is a martingale. Find $k$ .

Work it out, then compare with the answer.

Solution

We have $dY_t=t\,dW_t$ . Since $t$ has finite variation, the product rule has no covariation term:

d(tY_t)=Y_t\,dt+t\,dY_t=Y_t\,dt+t^2\,dW_t .

With $a_t=k\int_0^tY_s\,ds$ we get $da_t=kY_t\,dt$ , and therefore

dZ_t=(1-k)Y_t\,dt+t^2\,dW_t .

The integral $\int_0^t s^2\,dW_s$ is an $L^2$ martingale. So $Z_t$ is a martingale exactly when the drift vanishes, that is, when $k=1$ .

\boxed{k=1}

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