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 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.


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


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

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

Show that



I will apply Ito's formula to find dR.

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



\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:

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 R=\frac{e^{rt}}{I}, we get


Here where it gets tricky,

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 E[(r-\mu+\sigma^2)dt]=r-\mu+\sigma^2 and I can only see that if I have

\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 \sigma, \mu, r are given on a per annum basis, therefore T=1?