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This is my first time looking at problems in stochastic calculus, so please bare with the simplicity of the question. As always, any help is greatly appreciated.

1) Given $X_t=\int_0^ur_sds$ for a given stochastic process $r$ and defining $R_t=e^{X_t}$ what is $dR/R$?

The only thing I have worked out so far is that $dX=rdt$, but I wouldn't consider myself 100% sure of that either. I have worked with exponentials of normal random variables, but exponentials of stochastic variables is brand new to me.

Is $R$ here the expectation? I know that for a standard Brownian motion, the drift is zero and the variance is equal to the time interval, so if I take the expectation of an exponential of a normal I get $e^{\mu + \sigma^2/2}$ but I don't know if that helps me.

2) Now given $dS/S=\mu dt +\theta dB$ where $B$ is a Brownian motion and both $\mu, \theta$ are stochastic. If I define a new process $Y_t=log S_t$ what is $dY$?

So, if I multiply through by $S$ I get

$dS=S\mu dt +S\theta dB$ and this implies:

$S=S_0+\int_o^u S\mu dt+\int_0^u S\theta dB$

So, then $Y=log S$ gives: $Y=log(S_0)+\int_o^u log(S)\mu dt+\int_0^u log(S)\theta dB$

and Ito tells me that:

$dY=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial B}dB+1/2\frac{\partial^2 f}{\partial B^2}dt$

Again, I am brand new at this and am still working to get an underlying understanding of Ito's formula and stochastic calculus in general.

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  • $\begingroup$ Both questions are direct consequences of Itö's formula. Do you know this statement? $\endgroup$
    – Did
    Oct 13, 2013 at 15:12
  • $\begingroup$ I guess not well enough.For the first question, I have a result that $dR/R=rdt$ but I don't understand the derivation. $\endgroup$
    – Justin
    Oct 13, 2013 at 15:28
  • $\begingroup$ Also, for part 2, I will add some thoughts: $\endgroup$
    – Justin
    Oct 13, 2013 at 15:29
  • $\begingroup$ First thing to do (if you ask me): state Itö's formula. $\endgroup$
    – Did
    Oct 13, 2013 at 15:33
  • $\begingroup$ I added some more thoughts/work above. $\endgroup$
    – Justin
    Oct 13, 2013 at 15:38

1 Answer 1

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Itô's formula states that, under suitable regularity conditions, for every semi-martingale $U$, $V=G(U)$ is again a semi-martingale,such that $$ \mathrm dV_t=\vec\nabla G(U_t)\cdot\mathrm dU_t+\tfrac12\vec\nabla^2 G(U_t)\cdot\mathrm d\langle U,U\rangle_t. $$ A point worth noting here is that this works for multi-dimensional processes, for example, if $G:\mathbb R^n\to\mathbb R$, then $$ \mathrm dV_t=\sum_{k=1}^n\partial_kG(U_t)\mathrm dU^{(k)}_t+\tfrac12\sum_{k=1}^n\sum_{\ell=1}^n\partial^2_{k,\ell}G(U_t)\mathrm d\langle U^{(k)},U^{(\ell)}\rangle_t. $$ For example, if $B$ is a one-dimensional Brownian motion, then $U_t=(B_t,t)$ defines a two-dimensional semi-martingale hence, for every regular function $G:\mathbb R^2\to\mathbb R$, $$ \mathrm dG(B_t,t)=\partial_1G(B_t,t)\mathrm dB_t+\partial_2G(B_t,t)\mathrm dt+\tfrac12\partial^2_{11}G(B_t,t)\mathrm dt. $$ When $G:\mathbb R\to\mathbb R$, the formula reduces to $$ \mathrm dV_t=G'(U_t)\cdot\mathrm dU_t+\tfrac12G''(U_t)\cdot\mathrm d\langle U,U\rangle_t. $$

In your first case, $G(v)=\mathrm e^v$ and $\mathrm dU_t=r_t\mathrm dt$ hence $\mathrm d\langle U,U\rangle_t=0$. This yields $\mathrm dV_t=$ $______$ and $\mathrm dV_t/V_t=$ $______$.

In your second case, $G(v)=\log v$ and $\mathrm dU_t=\theta U_t\mathrm dB_t+\mu U_t\mathrm dt$ hence $\mathrm d\langle U,U\rangle_t=\theta^2 U_t^2\mathrm dt$. This yields $\mathrm dV_t=$ $______$.

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  • $\begingroup$ so the first two blanks are just $e^v rdt$ and $rdt$? $\endgroup$
    – Justin
    Oct 13, 2013 at 18:29

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