## expectation of a stochastic variable (Ito's formula)

This question is from Mathematics for Finance, a chapter on Ito's Formula. I could apply Ito's to calculate the required variable, but I could not follow through the second part, to find its expectation. Perhaps I don't understand the Brownian motion (most likely that I don't). Unfortunately I don't have the solution to this question.

PS $\displaystyle e^{rt}$ is a multiplication factor to find the value of a deposit/sum of money after t periods, given that the deposit pays interest rate r, and assuming continous compounding.

Question.

The stochastic differential equation for the rate of inflation I is given by

$\displaystyle dI=\mu{I}dt+\sigma{I}dz_t$

Find the equation satisfied by the real interest rate R defined to be $\displaystyle R=\frac{B}{I}$, where $\displaystyle B=e^{rt}$.

[Hint: Consider $\displaystyle f(x,t)=\frac{e^{rt}}{x}$.]

Show that

$\displaystyle E(\frac{\Delta{R}}{R})=r-\mu+\sigma^2$.

I will apply Ito's formula to find dR.

Let $\displaystyle R=\frac{e^{rt}}{I}, -> R=f(I,t)=\frac{e^{rt}}{I}$

$\displaystyle \frac{\partial{f}}{\partial{t}}=\frac{1}{I}re^{rt}$

$\displaystyle \frac{\partial{f}}{\partial{I}}=-e^{rt}\frac{1}{I^2}$

$\displaystyle \frac{{\partial}^2{f}}{\partial{t}^2}=2e^{rt}\frac {1}{I^3}$

therefore, we calculate df(I,t) from the equation of dI by applying Ito's formula:

$\displaystyle dR_t=[\frac{\partial{f}}{\partial{t}}(I,t)+\mu{I}\frac{\ partial{f}}{\partial{I}}+\frac{1}{2}\frac{{\partia l}^2{f}}{\partial{t}^2}\sigma^2I^2]dt+\frac{\partial{f}}{\partial{I}}\sigma{I}dz_t=(\ frac{re^{rt}}{I}-\frac{e^{rt}\mu}{I}+\frac{e^{rt}\sigma^2}{I})dt-\frac{e^{rt}\sigma}{I}dz_t$

by dividing this by $\displaystyle R=\frac{e^{rt}}{I}$, we get

$\displaystyle \frac{dR}{R}=(r-\mu+\sigma^2)dt-\sigma{dz_t}$

Here where it gets tricky,

$\displaystyle E(\frac{dR}{R})=E[(r-\mu+\sigma^2)dt]-E[\sigma{dz_t}]= ??? -0$

I know that E(dz)=0 as it is a Wiener process with mean=0

But what do I do with the remaining expectation on dt???

Somehow I need to show that $\displaystyle E[(r-\mu+\sigma^2)dt]=r-\mu+\sigma^2$ and I can only see that if I have

$\displaystyle \int_0^{T=1}(r-\mu+\sigma^2)dt=r-\mu+\sigma^2$ BUT why T=1???

Is there an implicit assumption that values of $\displaystyle \sigma, \mu, r$ are given on a per annum basis, therefore T=1?

thanks