Problem 18.8 Easy Free 2 variations

Variance by the Itô Isometry

Let $W_t$ be a standard Brownian motion. The variance of $\int_0^t W_s\,dW_s$ has the form $kt^2$ for a constant $k$ . Find $k$ .

Work it out, then compare with the answer.

SolutionShow the worked solution

Key idea. For an adapted integrand $f$ with $\mathbb{E}\int_0^t f_s^2\,ds<\infty$ , the Itô integral $\int_0^t f_s\,dW_s$ has mean $0$ . The Itô isometry then gives

\operatorname{Var}\Big(\int_0^t f_s\,dW_s\Big)=\mathbb{E}\Big[\Big(\int_0^t f_s\,dW_s\Big)^2\Big]=\int_0^t \mathbb{E}[f_s^2]\,ds .

Here $f_s=W_s$ is adapted. It is square-integrable because $\mathbb{E}\int_0^t W_s^2\,ds=\int_0^t s\,ds<\infty$ . So the integral has mean zero, and

\operatorname{Var}\Big(\int_0^t W_s\,dW_s\Big)=\int_0^t \mathbb{E}[W_s^2]\,ds=\int_0^t s\,ds=\frac{t^2}{2}.

Alternative. Itô’s formula gives $d(W_t^2)=2W_t\,dW_t+dt$ . Hence

\int_0^t W_s\,dW_s=\tfrac12\big(W_t^2-t\big).

Write $W_t=\sqrt t\,Z$ with $Z\sim\mathcal{N}(0,1)$ , so that $\operatorname{Var}(W_t^2)=t^2\big(\mathbb{E}[Z^4]-1\big)=2t^2$ . The variance is therefore $\tfrac14\cdot 2t^2=t^2/2$ .

\boxed{k=\tfrac12}

Variations

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

Variation 18.8a

Let $W_t$ be a standard Brownian motion. The variance of $\int_0^t W_s^2\,dW_s$ has the form $kt^3$ for a constant $k$ . Find $k$ .

Work it out, then compare with the answer.

Solution

The integrand $W_s^2$ is adapted. It is square-integrable because $\mathbb{E}[W_s^4]=3s^2$ . Write $W_s=\sqrt s\,Z$ with $\mathbb{E}[Z^4]=3$ . The integral has mean $0$ , and the isometry gives

\operatorname{Var}\Big(\int_0^t W_s^2\,dW_s\Big)=\int_0^t \mathbb{E}[W_s^4]\,ds=\int_0^t 3s^2\,ds=t^3.

\boxed{k=1}

Variation 18.8b

Let $W_t$ be a standard Brownian motion, and let $X_t = \int_0^t s\,dW_s$ . The variance of $X_t$ has the form $kt^3$ for a constant $k$ . Find $k$ .

Work it out, then compare with the answer.

Solution

The integrand $s$ is deterministic, so $X_t\sim\mathcal{N}\big(0,\int_0^t s^2\,ds\big)$ . Its variance is

\operatorname{Var}(X_t)=\int_0^t s^2\,ds=\frac{t^3}{3}.

\boxed{k=\tfrac13}

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