# Covariance of exponential integrated Brownian Motion

I have difficulties finding the covariance of the exponential integrated Brownian motion and would appreciate any guidance on how to move on.

The problem is as such,

Denote the integrated Brownian motion as $$Z_t = \int_{0}^{t} \, W_s \, ds$$ then the exponential integrated Brownian motion is $$V_t = e^{Z_t}$$

Show that $$cov(V_s, V_t) = e^\frac{t+3s}{2}$$

I tried using the approach of $$cov(X,Y) = E[XY] - E[X]E[Y]$$ but I am unsure what $$E[V_tV_s] = E[e^{\int_{0}^{t} \, W_u \, du}e^{\int_{0}^{s} \, W_u \, du}]$$ is.

I have these properties at hand

1. $$E[V_t] = e^\frac{t^3}{6}$$
2. The MGF with parameter $$u$$ of $$Z_t = E[e^{uZ_t}] = e^{u^2t^3/6}$$

3. For $$X_t = e^{W_t}$$, $$cov(X_s, X_t) = e^\frac{t+3s}{2} - e^\frac{t+s}{2}$$ and $$E[e^{W_s}e^{W_t}] = e^\frac{t+3s}{2}$$

Could there be a possibility this exercise be flawed (possible typo by the author whose exercise I took from)?

If the exercise is indeed correct, then I believe it is sufficient to show $$E[V_tV_s] = e^{\frac{t+3s}{2}} + e^{\frac{t^3}{6}}e^{\frac{s^3}{6}}$$ of which I am clueless.

Thanks

For $$0 , $$E(e^{\int_0^{t} W_u du}e^{\int_0^{s} W_u du}|\mathcal F_s)=e^{2\int_0^{s} W_u du}E(e^{\int_s^{t} (W_u-W_s) du}|\mathcal F_s) e^{(t-s)W_s}$$. The conditioning in the last expression is unnecessary by independence. Also, $$\int_s^{t} (W_u-W_s) du$$ has the same distribution as $$\int_0^{t-s} W_u du$$ (because $$(W_{r+s}-W_s)_{r \geq 0}$$ is also a Brownian motion). With these hints I believe you can compute the covariance of $$V_s$$ and $$V_t$$.
• By conditioning on $\mathcal F_s$, we obtain $E[e^{\int_{0}^{t} W_u \, du}e^{\int_{0}^{s} W_u \, du}|\mathcal F_s] = e^{W_s(t-s)}e^{\int_{0}^{s} W_u \,du}e^{\frac{(t-s)^3}{6}}$. I am thinking of using the property $E[X] = E\left[E[X|\mathcal F_s]\right]$ so $E[e^{\int_{0}^{t} W_u \, du}e^{\int_{0}^{s} W_u \, du}] = E\left [ E[e^{\int_{0}^{t} W_u \, du}e^{\int_{0}^{s} W_u \, du}|\mathcal F_s] \right ] = E\left[e^{W_s(t-s)}e^{\int_{0}^{s} W_u \,du}e^{\frac{(t-s)^3}{6}}\right]$ $= e^{\frac{(t-s)^3}{6}}E\left[e^{W_s(t-s)}e^{\int_{0}^{s} W_u \,du}\right]$ Am I on the right track? Commented Jul 9, 2019 at 8:57
• @KaviRamaMurthy May I ask for another hint on how to proceed from $E[e^{W_s(t-s)}e^{\int_{0}^{s} W_u du}]$? I know we can't split the expectation since $W_s$ is not independent of $\int_{0}^{s} W_udu$ since $cov(Z_s,W_s) = \frac{s^2}{2}$ Commented Jul 9, 2019 at 9:13