# Covariance of integrated Brownian motion

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 $$s. Show that the covariance of the integrated brownian motion is given by $$\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?

• I think that the problem may be in the fact that you are using as density the product of the densities, which holds if the RV are independent. Since you are (implicitly ) assuming that for sure you'll obtain a zero covariance Commented Feb 3, 2020 at 21:38

Recall that $$\mathbb{E} W_u W_v = \min(u,v)$$. By Fubini,
$$\mathbb{E} Z_t Z_s = \int_0^t\int_0^s \mathbb{E}W_u W_v dvdu = \int_0^t\int_0^s \min(u , v) dv du$$ $$=\int_s^t \int_0^s \min(u , v) dv du + \int_0^s\int_0^s \min(u , v) dv du$$ $$= \int_s^t \int_0^s v dv du + \int_0^s\big[\int_u^s udv + \int_0^u vdv\big]du$$ $$=\frac{s^2}{2}(t-s) + \frac{s^3}{2}- \frac{s^3}{6} = s^2\Big(\frac{t}{2}-\frac{s}{6}\Big).$$
• Well for the MGF, you appear to be trying to make use of the fact that $(Z_t,Z_s)$ is jointly Gaussian; computation of its MGF would require knowledge of the joint distribution of $(Z_t,Z_s)$, which is what you're trying to obtain by finding the covariance; I don't see how it could be a fruitful approach. Commented Feb 4, 2020 at 18:04