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?
