# Expected Value of Products of Processes

Suppose I have two processes.

$A_t = A_0 \exp((a-\frac{1}{2}\sigma_A^2)t+\sigma_A W_t^A$

$B_t = B_0 \exp((b-\frac{1}{2}\sigma_B^2)t+\sigma_B W_t^B$

I would like to calculate $E[A_s B_t]$ where s < t.

Attempt:

I can rewrite $B_t$ as $B_t = B_s \exp((b-\frac{1}{2}\sigma_B^2)(t-s) + \sigma_B W_{t-s}^B)$. Then, $A_sB_t = A_sB_s\exp((b-\frac{1}{2}\sigma_B^2)(t-s) + \sigma_B W_{t-s}^B)$.

Thus, $E(A_s B_t) = E(A_s B_s)E(\exp((b-\frac{1}{2}\sigma_B^2)(t-s) + \sigma_B W_{t-s}^B))$

$= A_0 B_0 \exp((a+b+\rho\sigma_A \sigma_B)s)+\sigma(b(t-s))$.

Could anyone confirm whether this approach is correct. How would I then be able do compute $E(A_s B_t B_r)$ where $s < t < r$. Would I simply do the same thing but use

$B_r = B_s\exp((b-\frac{1}{2}\sigma_B^2)(r-t)+\sigma_BW_{r-t}^B)\exp((b-\frac{1}{2}\sigma_B^2)(t-s)+\sigma_BW_{t-s}^B)\exp((b-\frac{1}{2}\sigma_B^2)s+\sigma_BW_{s}^B)$

My second part of this question relates to the forward. Assume $s<t<r$ and $F_{t,r}^A = E(A_r|A_t)$. How would I be able to compute $E(B_r F_{t,r}^A )$? Am I allowed to do $E(B_r)E( F_{t,r}^A )$ even though the time periods overlap because I am working with a forward?

• You need to use Cholesky decomposition for your Brownian motions, and then the computations becomes straightforward. – Gordon May 5 '16 at 14:49
• Would anyone be able to provide an example – Jim May 5 '16 at 21:39

Based on Cholesky decomposition, \begin{align*} W_t^A &= W_t^1,\\ W_t^B &= \rho W_t^1 + \sqrt{1-\rho^2}W_t^2, \end{align*} where $(W_t^1, t \ge 0)$ and $(W_t^2, t \ge 0)$ are two independent standard Brownian motions. Then \begin{align*} A_t &= A_0\exp\Big(\big(a-\frac{1}{2}\sigma_A^2\big)t + \sigma_A W_t^1 \Big),\\ B_t &= B_0\exp\Big(\big(b-\frac{1}{2}\sigma_B^2\big)t + \sigma_B\big( \rho W_t^1 + \sqrt{1-\rho^2}W_t^2\big) \Big). \end{align*} The remaining calculation are based on this decomposition. For example, assuming that $t<r$, note that \begin{align*} A_r &= A_0\exp\Big(\big(a-\frac{1}{2}\sigma_A^2\big)r + \sigma_A W_r^1 \Big)\\ &=A_0\exp\Big(\big(a-\frac{1}{2}\sigma_A^2\big)t + \sigma_A W_t^1 + \big(a-\frac{1}{2}\sigma_A^2\big)(r-t) + \sigma_A (W_r^1-W_t^1)\Big)\\ &= A_t \exp\Big( \big(a-\frac{1}{2}\sigma_A^2\big)(r-t) + \sigma_A (W_r^1-W_t^1)\Big). \end{align*} Similarly, \begin{align*} B_r &= B_t \exp\Big( \big(b-\frac{1}{2}\sigma_B^2\big)(r-t) + \sigma_B\big[\rho (W_r^1-W_t^1) + \sqrt{1-\rho^2}(W_r^2-W_t^2)\big]\Big). \end{align*} Then, \begin{align*} F_{t, r}^A &= E(A_r\mid A_t)\\ &= A_t\exp\big(a(r-t)\big). \end{align*} Consequently, \begin{align*} E(B_r F_{t, r}^A) &=E(B_r E(A_r\mid A_t))\\ &=\exp\big(a(r-t)\big) E(B_r A_t)\\ &=\exp\big(a(r-t)\big) E\big(E(B_r A_t \mid \mathcal{F}_t)\big)\\ &=\exp\big((a+b)(r-t)\big) E(A_t B_t)\\ &=A_0B_0\exp\big((a+b)(r-t)\big) \\ &\quad E\left(\exp\Big(\big(a-\frac{1}{2}\sigma_A^2\big)t + \sigma_A W_t^1 + \big(b-\frac{1}{2}\sigma_B^2\big)t + \sigma_B\big( \rho W_t^1 + \sqrt{1-\rho^2}W_t^2\big)\Big) \right)\\ &=A_0B_0\exp\big((a+b)\,r + \rho\, \sigma_A\sigma_B\, t\big). \end{align*}
As for $E(A_sB_tB_r)$, where $s<t<r$, you can use the tower law: first consitional on $\mathcal{F}_t$ , and then conditional on $\mathcal{F}_s$.
• Your notations are messy. Note that $W_{r-t}$ and $W_{t-s}$ are not independent, though $W_r-W_t$ and $W_t-W_s$ are independent. – Gordon May 6 '16 at 15:11