I am trying to prove that the following integral is almost surely finite.
Assume $X$ is a stochastic process such that $$ dX_t = \kappa (\theta - X_t)dt + \sigma\sqrt{X_t}dW_t . $$ I would like to prove that for $ n\in\mathbb{N} $ and $T>0$ we have that $$ \int_0^T |X_t|^ndt < \infty \quad\mathbb{P}-a.s. $$ I do not know whether it is true or not! Thank you in advance!
As mentioned here Does finite expectation imply bounded random variable?, it suffices to show that
$$ \int_0^T E[|X_t|^n]dt < \infty. $$
We will simply use Application of the Burkholder Davis Gundy inequality
Let $X$ be a weak solution of $$dX_t=b(t,X_t)dt+\sigma(t,X_t)dW_t$$ With $b$ and $\sigma$ continuous and satisfying the linear growth condition: $$|b(t,x)|+|\sigma(t,x)|\leq K(1+|x|)$$ Then for any finite time $T$ and $p\geq 2$ there is a constant $C$ such that $$\mathbb{E}\sup_{t\leq T}|X_t|^p\leq Ce^{CT}(1+\mathbb{E}|X_0|^p)$$
This is the case here since $\sqrt{x}\leq |x|$ for all large enough $x$. So for $n\geq 2$, we bound
$$ \int_0^T E[|X_t|^n]dt \leq TCe^{CT}(1+\mathbb{E}|X_0|^n). $$
For $n=1$, we simply bound $E|X|\leq \sqrt{E|X|^{2}}$.
