# Exponentials of stochastic processes and Brownian motions

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.

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

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 $\mathrm dV_t=$ $______$.

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