Consider the following exercise (ex. 3.3.5 from Calin's book An Informal Introduction to stochastic Calculus with Applications):
Let $Z_t = \int_0^t W_u du$, where $W$ is the standard Brownian
motion. Let $s<t$. Show that the covariance of the integrated brownian
motion is given by $$ \mathrm{Cov}\left(Z_s Z_t\right) =
s^2\left(\frac{t}{2} – \frac{s}{6}\right). $$
I tried to do this exercise by computing the moment generating function of $Z_s Z_t$, as follows:
$$
\begin{array}\\
M_{Z_s Z_t}(u) &= \mathbb{E}\left[e^{uZ_sZ_t}\right] \\
&=\int_{\mathbb{R}^2} e^{uxy} \frac{3}{2\pi\sqrt{s^3t^3}}\exp\left\{-\frac12 \frac{x^2}{s^3/3}\right\} \exp\left\{-\frac12 \frac{y^2}{t^3/3}\right\} dxdy \\
&= \int_{\mathbb{R}^2} \frac{3}{2\pi\sqrt{s^3t^3}} \exp\left\{-\frac12\frac{x^2}{s^2/3}\right\} \exp\left\{-\frac12 \frac{(y-\frac{t^3}{3} ux)^2}{t^3/3}\right\} \exp\left\{\frac12 \frac{t^3}{3} u^2x^2\right\} dxdy \\
&= \int_{\mathbb{R}} \frac{1}{\sqrt{2\pi s^3/3}}\exp\{-\frac12 \frac{1-s^3t^3 u^2/9}{s^3/3}x^2\} dx \\
&= \left(1 – \frac{s^3t^3}{9} u^2\right)^{-\frac12}.
\end{array}
$$
However,
$$
\mathbb{E}\left[Z_sZ_t\right] = M_{Z_sZ_t}'(0) = 0.
$$
Since $\mathbb{E}[Z_t]=0$, surely I cannot recover the correct value for the covariance.
Can somebody spot my error?
Best Answer
Recall that $\mathbb{E} W_u W_v = \min(u,v)$. By Fubini,
$$ \mathbb{E} Z_t Z_s = \int_0^t\int_0^s \mathbb{E}W_u W_v dvdu = \int_0^t\int_0^s \min(u , v) dv du $$ $$ =\int_s^t \int_0^s \min(u , v) dv du + \int_0^s\int_0^s \min(u , v) dv du $$ $$ = \int_s^t \int_0^s v dv du + \int_0^s\big[\int_u^s udv + \int_0^u vdv\big]du $$ $$ =\frac{s^2}{2}(t-s) + \frac{s^3}{2}- \frac{s^3}{6} = s^2\Big(\frac{t}{2}-\frac{s}{6}\Big). $$