Stochastic Calculus problem with three processes? (Itô calculus)

Question

Can someone help me solve this following Itô Calculus problem?

Let $Z(t):= [B(t)*X(t)]/S(t)$

We have the following dynamics of B(t), X(t) and S(t):

$dS(t)=\alpha S(t)dt+\sigma S(t)dW(t)$

$dB(t)=rB(t)dt$

$dX(t)=\alpha_X X(t)dt + \sigma_X X(t) dW(t)$

where W is a brownian motion.

I want to determine $dZ(t)$.

According to the answer it should be,

$$ dZ(t) = Z(t)(\sigma_X - \sigma)\left(\frac{\sigma^2 - \alpha + r_f + \alpha_X -\sigma\sigma_X}{\sigma_X-\sigma} dt + dW(t)\right) $$

but I do not know how to even begin to approach this problem.

Would appreciate a thorough solution. Thank you! :)

Edit: With the help of siou0107 and this lovely community I solved the process. Solution below

I write all this so that I learn too.

The money account $B_f(t)=r_fB(t)dt$ is defined as foreign money account and I want to determina the exchange rate dynamics $X(t)$ under the EMM with the stock as numeraire. The stock is in domestic economy. In my post and question I just wanted to solve the $Z(t)$ process defined as $Z(t):=\frac{B_f(t)X(t)}{S(t)}$. Dynamics of S and X are defined above.

With Itô's lemma and the help of this kind community I get,

$dZ(t)=-\frac{X(t)B_f(t)}{S^2(t)}dS(t)+\frac{X(t)}{S(t)}dB_f(t)+\frac{B_f(t)}{S(t)}dX(t)+[\frac{X(t)B(t)}{S^3(t)}\sigma^2 S^2(t)-\frac{B_f(t)}{S^2(t)}\sigma S(t)\sigma_X X(t)]dt$

$dZ(t)=-Z(t)(\frac{dS(t)}{S(t)}+r_fdt)+Z(t)(\alpha_X dt+\sigma_X dW(t))+[Z(t)(\sigma^2 + \sigma \sigma_X)dt]$

Gathering all the terms finally gives us,

$dZ(t)=Z(t)(\sigma_X-\sigma)dW(t)+Z(t)(\sigma_X+\sigma^2-\alpha-r_f-\sigma \sigma_X)dt$

Which is basically the answer stated in the picture.

Answer 1

You have $$dZ_t = df\left(S_t, B_t, X_t\right) = \frac{\partial f}{\partial s}dS_t + \frac{\partial f}{\partial b}dB_t + \frac{\partial f}{\partial x}dX_t + \frac{1}{2}\left[\frac{\partial^2 f}{\partial s^2} d\langle S\rangle_t + 2\frac{\partial^2 f}{\partial s\partial x} d\langle S, X\rangle_t \right]$$ since $d\langle B\rangle_t$, $d\langle B, S\rangle_t$, $d\langle B, X\rangle_t$ and $ \partial_{xx}^2 f(s, b, x)$ are null.

$$dZ_t = -\frac{X_tB_t}{S_t^2}dS_t + \frac{X_t}{S_t}dB_t + \frac{B_t}{S_t}dX_t + \frac{1}{2}\left[2\frac{X_tB_t}{S_t^3} \sigma^2S_t^2 - 2\frac{B_t}{S_t^2} \sigma S_t\sigma_X X_t \right]dt$$

Factor by $\frac{X_tB_t}{S_t}$ and then it is straightforward.

Answer 2

If you are happy to try the brute force approach, then here are the relevant formulae:

In ordinary calculus, you have the product rule for the differential of two variables:

$$d \left( x_1 x_2\right)=x_1 dx_2+x_2 dx_1$$

The general version of this for differential of products of n variables is:

$$d \prod_{i=1}^{n}{x_i}=\sum_{i=1}^{n}{ \prod_{j=1 j \ne i}^{n}{x_j}dx_i}$$

The stochastic equivalent of the product rule for 2 variables is:

$$d \left( X_1 X_2\right)=X_1 dX_2+X_2 dX_1+dX_1dX_2$$

And the general version of this for a product of n variables is:

$$d \prod_{i=1}^{n}{X_i}=\sum_{i=1}^{n}{ \prod_{j=1, j \ne i}^{n}{X_j}dX_i} +\sum_{i=1,k=1,k>i}^{n}{ \left( \prod_{j=1, j \ne i,k}^{n}{X_j} \right) dX_idX_k}$$

For n=3 this expands as follows:

$$d \left( X_1 X_2 X_3\right)=X_2 X_3 dX_1+X_1 X_3 dX_2+X_1 X_2 dX_3+ X_3 dX_1dX_2+X_2 dX_1 dX_3+ X_1 dX_2dX_3$$

Looks formidable, but would simplify for your system of three equation because the second equation (dB) is deterministic, so it won't contribute to the cross terms.