Expectation of $\sqrt{r_t}$ in $\textrm{d}r_t=\kappa(\theta-r_t)\,\textrm{d}t+\sigma\sqrt{r_t}\,\textrm{d}B_t$ I'm currently looking at the CIR model and trying to calculate the expectation of $r_t$. I know that by Fubini's theorem, $\mathbb{E}\left(\int_0^t\kappa(\theta-r_s)\,\textrm{d}s\right)=\kappa\theta t-\int_0^t\mathbb{E}(r_s)\,\textrm{d}s$ and we can create an ODE from it, but as for the second term I know I can't use Fubini's theorem, so what can I do? I know one way is to multiply the SDE by $\textrm{e}^{\kappa t}$ but I haven't seen how it helps with the square root term, and Shreve claims it is an Ito integral but I don't see how $\sqrt{r_t}\in\mathcal{L}^2_{\mathcal{F}}$. All help is appreciated, thanks!
 A: We need the following Lemma.
Lemma.
Consider the SDE
$$dX_t = f(X_t)dt + g(X_t)dB_t, \quad X_0 = x_0 \in \mathbb R^{+}. \tag{1}$$
with solution $\{x_t; 0 \leq t \leq T\}.$ Assume that the linear growth condition holds, i.e.
There exists a constant $K \in \mathbb R$ such that for all $x \in \mathbb R$
$$|f(x)|^2 \bigvee |g(x)|^2 \leq K(1+x^2).$$
Then there exists a constant $C$ such that
$$E[\sup_{0 \leq t \leq T} X_t^2] \leq C.$$
(For a proof of this lemma see for example lemma 3.2 - page 51 in Xuerong Mao's book)
Now going to your problem: your SDE is of the SDE (1) type with $f(x)= \kappa(\theta -x)$ and $g(x) = \sigma \sqrt{x}.$ We can see that clearly the linear growth condition holds, so the lemma applies and there exists $C>0$ such that $E[\sup_{0 \leq t \leq T} r_t^2] \leq C.$ This implies that $E\int_0^T (\sigma \sqrt{r_t})^2dt < \infty.$ Thus the process $I_t:= \int_0^t (\sigma \sqrt{r_t})dB_t, 0 \leq t \leq T$ is a martingale and has constant expectation. Therefore, $E[I_T]=E[I_0]=0.$
