3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Did you try applying Ito's lemma to $Z(t) = f(S(t), B(t), X(t))$ ? $\endgroup$ – byouness Dec 8 '19 at 19:51
  • $\begingroup$ Hi byuoness, To be honest I have so far only been applying Itô's lemma on product such as X*Y. This is on a whole new level. How do I start? I just began with the stochastic calculus course. Regards, $\endgroup$ – Dreason94 Dec 8 '19 at 20:15
  • $\begingroup$ It is the same logic, if you know how to compute the partial derivative of $(x,y) \mapsto x \times y$ then you know how to do it for $(x,y,é) \mapsto x \times y \times z$. Please check this for example: math.stackexchange.com/questions/1351527/… $\endgroup$ – byouness Dec 9 '19 at 8:57
  • $\begingroup$ Are there any possibilities for you to solve the problem for me? Thank you! $\endgroup$ – Dreason94 Dec 9 '19 at 13:02
  • 1
    $\begingroup$ The idea is not to solve the problem for you, but rather to help you solve it yourself, by answering your questions and clarifying anything that's not clear for you. Otherwise, you'll be clueless next time you face the same kind of problem :/ $\endgroup$ – byouness Dec 9 '19 at 13:25
2
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Please sir I would appreciate if you could see my edited post and guide me further how to solve he problem. I have come a bit further but do not how to complete it. $\endgroup$ – Dreason94 Dec 11 '19 at 22:15
  • 1
    $\begingroup$ You forgot the quadratic (co-)variations of $X$ and $S$. If your second derivatives $\partial_{xx}^2 f$, $\partial_{xs}^2 f$ and/or $\partial_{ss}^2 f$ are not null, they come into play because of the order 2 variation of the Wiener process which is $O\left(dt\right)$. Since $B$ is deterministic, it has 0 quadratic variation and second derivatives with respect to $b$ are not required here. $\endgroup$ – siou0107 Dec 11 '19 at 23:01
  • $\begingroup$ Thank you. I have taken the cross-variation into account. I do not know if you can follow my currently edited calculations, but my ambition in my edit is that I am splitting up the derivation in two parts. Firstly, instead of directly applying Itô's lemma on $Z(t)= \frac{B(t)X(t)}{S(t)}$, I solve $d(B(t)X(t))$. Secondly, say that I define $Y(t)=d(B(t)X(t))$, then I calculate the stochastic quotient $dZ(t)=d(\frac{Y(t)}{S(t)})=d(\frac{d(B(t)X(t))}{S(t)})$. Am i allowed to solve it this way? If you do not understand my explanation I can re-edit. $\endgroup$ – Dreason94 Dec 12 '19 at 14:52
  • 1
    $\begingroup$ I don't see the $d\langle\tilde{B}, S\rangle_t$ and $d\langle S\rangle_t$ terms in your edited post. You just take the first order $d\tilde{B}_t$ and $dS_t$ variations. $\endgroup$ – siou0107 Dec 12 '19 at 16:19
  • $\begingroup$ I tried the method I have been learning past days (the one that I explained to you about Itô's product rule) and it did not work well. Thus I went on using the "formula" you presented. Thank you very much. I re-edited. $\endgroup$ – Dreason94 Dec 12 '19 at 19:23
4
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. But what do I do with the quotient S(t). Surely I can write it as S(t)^-1, but I have never handled these kind of problems. Would appreciate a thorough solution. $\endgroup$ – Dreason94 Dec 9 '19 at 13:07
  • 2
    $\begingroup$ You need to apply Ito to $f(S)$ where $f: x \mapsto \frac{1}{x}$ to get the SDE of $Z = \frac{1}{S}$ $\endgroup$ – byouness Dec 9 '19 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.