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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?

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

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  • $\begingroup$ Thank you. If you look at the last part of my question, could you explain why that isn't applicable. $\endgroup$ – Jim May 6 '16 at 15:08
  • $\begingroup$ 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. $\endgroup$ – Gordon May 6 '16 at 15:11
  • $\begingroup$ Sorry you are definately right on that - they just have the same distribution. But surely a forward price can always be factored out no matter the time period right? $\endgroup$ – Jim May 6 '16 at 15:14
  • $\begingroup$ That depends on your definition of forward price. Note, for such expectations, you need to pay particular attention to independence of the concerned terms. If you can follow all steps in my answer, you should be able to figure out all the remaining ones. $\endgroup$ – Gordon May 6 '16 at 15:16
  • $\begingroup$ Right so I cannot just blindly assume independence between a stock process and a forward price? $\endgroup$ – Jim May 6 '16 at 15:21

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