# Covariance of the product of log normal process and normal procces

I tried to compute the following covariance : $$Cov(e^{\int_{t}^{T}W^1_sds},\int_{t}^{t+1}W^2_sds)$$

where $$W^1_t$$ and $$W^2_t$$ are Brownian motions such that $$dW_t^1dW_t^2=\rho dt$$

My idea was to use Ito‘s lemma on the process $$(e^{\int_{u}^{T}W^1_sds}\int_{u}^{t+1}W^2_sds)_u$$ with $$u, and then try to find the expectation from the resulting EDP when we take the expectation on both sides of the Ito equation.

But I am not sure if this is working, since the resulting EDP does not have a solution.

Can anyone help me in this regards ? Do you have other ideas on how to compute this covariance ?

I'll give it a try, but am not yet 100% sure that it's the way to go.

## Ansatz:

Let's find the distribution of the integral of a Brownian motion with respect to time (call it $$x$$) and find the expectation of the product of two such integrals $$x$$ and $$y$$. Then, calculate the covariance as $$Cov(x,y)=E(xy)-E(x)E(y)$$.

## 1. Distribution of $$I(t,T)\equiv\int\limits_{s=t}^{T}W_s\mathrm{d}s$$

From this answer, we know that $$I(t,T)\equiv \int\limits_{s=t}^{T}W_s\mathrm{d}s=\int\limits_{s=t}^{T}(T-s)dW_s$$

and that $$I(t,T)$$ is normally distributed with $$I(t,T)\sim \mathrm{N}\left(0,\frac{1}{3}(T-t)^3\right)$$

## 2. Expectation of $$I(t,T)I(t,U)$$

By the same way of reasoning (and some weak recollection of Iso isometry, I'd argue:

\begin{align} \mathrm{E}\left(I(t,T)I(t,U)\right)&=\mathrm{E}\left(\int\limits_{s=t}^{T}(T-s)\mathrm{d}W_s\int\limits_{x=t}^{U}(U-x)\mathrm{d}W_x\right)\\ &=\mathrm{E}\left(\int\limits_{s=t}^{T}\int\limits_{x=t}^{U}(T-s)(U-x)\mathrm{d}W_s\mathrm{d}W_x\right)\\ &=\int\limits_{s=t}^{T}\int\limits_{x=t}^{U}(T-s)(U-x)\mathrm{E}\left(\mathrm{d}W_s\mathrm{d}W_x\right)\\ &=\int\limits_{s=t}^{U}(U-s)^2\rho\mathrm{d}t\\ &=\frac{1}{3}(U-t)^3 \end{align} N.B.: we assume $$U\leq T$$.

## 3. $$I(t,T)$$ and $$I(t,U)$$ are bivariate normally distributed

Let's simplify and let $$x_1\equiv I(t,T)$$, $$x_2\equiv I(t,U)$$ and $$\mathbf{x}=\left(x_1,x_2\right)^T$$, also let $$\sigma_1^2=\frac{1}{3}(T-t)^3$$, $$\sigma_2^2=\frac{1}{3}(U-t)^3$$ and $$\sigma_{1,2}=\frac{1}{3}\rho (U-t)^3$$. Then $$\mathbf{x}$$ is bivariate normally distributed as

$$\mathbf{x}\equiv\begin{pmatrix}x_1\\x_2\end{pmatrix}\sim \mathrm{N}\left(\mathbf{0},\begin{pmatrix}\sigma_1^2 & \sigma_{1,2}\\ \sigma_{1,2} & \sigma_2^2\end{pmatrix}\right)$$

## 4. Apply the moment generating function (MGF) trick:

Now let's use a little trick I learned just recently. Given a real vector $$\mathbf{t}$$, the MGF of the multivariate normal distribution is defined as $$\varphi_X(t)\equiv\mathrm{E}\left(e^{\mathbf{t}^T\mathbf{x}}\right)=e^{\mathbf{t}^T\mathbf{\mu}+\frac{1}{2}\mathbf{t}^T\mathbf{\Sigma}\mathbf{t}}$$ and, in our case, this is

$$\varphi(t_1,t_2)=\mathrm{E}\left(e^{t_1x_1+t_2x_2}\right)=e^{\frac{1}{2}t_1^2\sigma_1^2+\frac{1}{2}t_2^2\sigma_2^2+t_1t_2\sigma_{1,2}}$$

Note that

$$\left.\frac{\partial \left(e^{x+ty}\right)}{\partial t}\right|_{t=0}=ye^x$$

thus,

\begin{align} \mathrm{E}\left(e^{I(t,T)}I(t,U)\right)&=\mathrm{E}\left(e^{x_1}x_2\right)\\ &=\left.\frac{\partial \varphi(t_1=1,t_2)}{\partial t_2}\right|_{t_2=0}\\ &=e^{\frac{1}{2}\sigma_1^2}\sigma_{1,2}\\ &=\frac{1}{3}\rho(U-t)^3e^{\frac{1}{6}(T-t)^3} \end{align}

## 5. Putting all together

Thus, for Brownian motions $$W^1_t, W_2^t$$ with $$dW_1dW_2=\rho dt$$

\begin{align} \mathrm{Cov}\left(e^{\int\limits_{s=t}^TW^1_s\mathrm{d}s}\int\limits_{x=t}^UW^2_x\mathrm{d}x\right)&=\mathrm{E}\left(e^{x_1}x_2\right)-\mathrm{E}\left(e^{x_1}\right)\mathrm{E}(x_2)\\ &=\mathrm{E}\left(e^{x_1}x_2\right)\\ &=1/3\rho(U-t)^3e^{\frac{1}{6}(T-t)^3} \end{align}

• I believe there is a slight inaccuracy in the computation of Expectation of I(t,T)I(t,U), but thanks for the detailed answer and the ideas that you have explained. Thumbs up from me. Aug 26, 2021 at 15:47
• Hi, you are welcome to point that one out! My Ito calculus is a bit rusty Aug 26, 2021 at 15:51

Hint (too long for a comment). The integrals $$X_1:=\int_t^{T_1}W^1_s\,ds\,,\quad\quad X_2:=\int_t^{T_2}W^2_s\,ds$$ are two normals with expectation zero, variances $$\sigma_i^2=\int_t^{T_i}\int_t^{T_i}\min(u,s)\,du\,ds\quad\quad\text{(can be solved) }\,,\quad\quad i=1,2$$ and covariance $$\gamma=\rho\int_t^{T_1}\int_t^{T_2}\min(u,s)\,du\,ds\quad\quad\text{(can be solved). }$$ It should be straightforward to calculate $${\rm Cov}(e^{X_1},X_2)\,.$$

• Hi, I think you beat me to a comment yesterday (the exp expansion9, and you beat me to an answer today. :-D. Aug 26, 2021 at 9:37
• @Kermittfrog : what comment was that ? Aug 26, 2021 at 10:00
• Maybe I mistook you for somebody else- sorry, then. Aug 26, 2021 at 10:34