Covariance of the product of log normal process and normal procces

Question

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 $u<t$, 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 ?

Thanks in advance.

Answer 1

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 $dW_1dW_2=\rho dt$

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

Answer 2

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 $$ {\rm Cov}(e^{X_1},X_2)\,. $$