The CIR process follows

$$dr_t = (\alpha - \beta r_t)dt + \sigma \sqrt{r_t}dW_t.$$

It can be proved that there exists a unique solution for this SDE but it's not possible to get an expression for that solution. However, according to the Shreve (page 152) we can obtain the expected value of the solution. Applying Ito's lemma with the function $f(t,x)=e^{\beta t}x$ we obtain $$e^{\beta t}r_t = r_0 + \frac{\alpha}{\beta}(e^{\beta t} -1) + \sigma \int_0^t (e^{\beta t}-1)+ \sigma \int_0^t e^{\beta u} \sqrt{r_u}dW_u.$$ So he obtains $$e^{\beta t}E [r_t] = r_0 + \frac{\alpha}{\beta}(e^{\beta t} -1)$$ because he says that $$E \left[\sigma \int_0^t e^{\beta u} \sqrt{r_u}dW_u \right]=0.$$ He says that the expectation of an Ito integral is zero, but this is not always true. If $$E\left[ \int_0^T |\sigma e^{\beta u} \sqrt{r_u}|^2du \right]< \infty \tag*{($\star$)}$$ then the process $\{I_t\}:=\left\{\sigma \int_0^t e^{\beta u} \sqrt{r_u}dW_u ; 0 \leq t \leq T \right\}$ is a martingale and the expectation is zero. But if ($\star$) is not true, then the process $\{I_t\}$ is just a local martingale, not necessarily a (true) martingale and the expectation above may not be zero. Since we don't have the expression for the solution, I don't think the condition ($\star$) can be verified. Is there any other argument that allow us to conclude that the expectation is zero? Do you know of any papers/books where they treat this issue?


1 Answer 1


Please have a look at this question:
You have there two different proofs for your question.
But in general case, you can try to prove that your lebesgues integral is finite by using the fubini theorem that allows you to move the expectation inside the integral as the stuff inside the integral is positive.

  • $\begingroup$ Thank you very much! $\endgroup$
    – user39119
    Commented Feb 4, 2020 at 1:23
  • $\begingroup$ ah ok. delete it and note that i tried to help you for FREE. $\endgroup$ Commented Feb 11, 2020 at 11:58

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