According to Wikipedia and some other articles I’ve seen the mean of the CIR model : $dr_t=a(b-r_t)dt+σ\sqrt r_t dW_t$
Is the following : $E(r_t) =r0e^{-at}+b(1-e^{-at})$
And the variance : $ Var(r_t)=r0\frac{σ^2}{a}(e^{-at}-e^{-2at})+\frac{bσ^2}{2a}(1-e^{-at})^2$
I have tried to derive it with Ito’s lemma by taking a function $r_te^{at}$ and solving this function with the Ito lemma formula however my mean and variance results were very different. I’m preparing for an exam and my professor only wants us to solve these stochastic differentials with Ito's lemma
I’m super frustrated I can’t come to the correct answer. Could anyone more experienced with stochastic calculus enlighten me ? Thanks :)
- I have seen a similar question on here however it wasn’t solved by itto’s lemma
My approach was:
- to take the function $x=e^{at}r_t$
And use the Itos lemma formula:
$e^{at}dr_t +ae^{at}dt$
Now substituting the original CIR model $r_t$ into drt $abe^{at}dt+ae^{at}r_tdt+e^{at}σ\sqrt r_t dW_t+ae^{at}r_t dt$
$ \int_{0}^{t} abe^{at}\,ds + \int_{0}^{t} ae^{at}r_t\,ds + \int_{0}^{t} e^{at}σ\sqrt r_t dW_t\,ds $
E(X) = $x0+2e_te^{ab}+be^{at}-ab-2ar_t+ \int_{0}^{t} e^{at}\sqrt r_t dW_t\,dt$
V(X) = $[x0+2e_te^{ab}+be^{at}-ab-2ar_t+ \int_{0}^{t} e^{at}\sqrt r_t dW_t\,dt]^2$
My second approach was taking the function $e^{bt}r_t$ I followed the same procedure , after taking ittos lemma , $be^{bt}r_t dt + e^{bt}dr_t$ And substituting drt with the CIR equation
$be^{bt}r_t dt+ ab e^{bt}dt+ ar_te^{bt}dt+e^{bt}σ\sqrt r_t dW_t$
$\int_{0}^{t} be^{bt}r_t \,ds + \int_{0}^{t} ab e^{bt} \,ds + \int_{0}^{t} ar_te^{bt} \,ds + \int_{0}^{t} e^{bt}σ\sqrt r_t dW_t \,ds$
$E(x) = x0 +(e^{bt}-1)rt+ ae^{bt}-a + \frac{ar_te^{bt}}{b}-ar_t + \int_{0}^{t} e^{bt}σ\sqrt r_t dW_t \,ds $
$V(x) = [x0 +(e^{bt}-1)rt+ ae^{bt}-a + \frac{ar_te^{bt}}{b}-ar_t + \int_{0}^{t} e^{bt}σ\sqrt r_t dW_t \,ds]^2 $