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Suppose we have the following SDE

$$dS(t) = S(t)(\mu(t)dt + \sigma(t)dW(t))=:S(t)dX(t)$$

where $W$ is a Brownian Motion and the processes $\mu,\sigma$ are well defined, such that the expression makes sense. Then I know that

$$S(t)=S(0)\mathcal{E}(X)(t)$$

where $\mathcal{E}$ denotes the stochastic exponential. Additionally we have

$$dB(t) = B(t)r(t)dt$$

hence

$$B(t) = \exp\left(\int_0^tr(s)ds\right)$$

Now I want to compute $d(\frac{S(t)}{B(t)})$ using the product formula:

$$d(U(t)Y(t))=U(t)dY(t)+Y(t)dU(t)+\langle U,Y\rangle (t)$$

here I use $U(t)=S(t)$ and $Y(t)=\frac{1}{B(t)}$. Hence

$$d(\frac{S(t)}{B(t)})=S(t)d\frac{1}{B(t)}+\frac{1}{B(t)}dS(t)+\langle S,\frac{1}{B}\rangle(t)$$

this should be equal

$$\frac{S(t)}{B(t)}(\mu(t)-r(t))dt+\frac{S(t)}{B(t)}\sigma(t)dW(t).$$

How do I get this equality?

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First of all, note that $\langle \cdot,\frac1B\rangle = 0$ no matter which process you place instead of $\cdot$. Thus $$ \mathrm d \frac{S}{B} = S\,\mathrm d\frac1B+\frac1B\,\mathrm dS = \frac1B\left(-S\frac{\mathrm d B}{B} + \mathrm dS\right) = \frac SB\left(-r\mathrm dt+\mu\mathrm dt+\sigma\mathrm dw_t\right) $$ as needed.

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Thanks for your answer. I just have two small questions: Why is it true that $\langle \cdot, \frac{1}{B}\rangle = 0$? Is $\frac{1}{B}$ of finite variation (if so, why)? And the second question is, why is $\frac{-SdB}{B^2}=Sd\frac{1}{B}$? –  user20869 Jan 3 '13 at 12:39
    
@hulik: the process $B$ is of finite variation as you can see it from the explicit form for $B(t)$ you gave. Second, for the Ito differential you have $$ \mathrm df(B_t) = f'(B_t)\mathrm dt $$ yet again since $B_t$ is of finite variation. Just use $f(x) = \frac1x$ –  Ilya Jan 3 '13 at 12:42
    
Ah, thanks a lot! I didn't think about to use $Itô$ once more. –  user20869 Jan 3 '13 at 12:48
    
Sorry, but looking again at your argument why $1/B$ should be of finite variation, I struggled. Note $r$ is not deterministic. Why should it be increasing without knowing that $r$ is positive? –  user20869 Jan 29 '13 at 13:00
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