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Suppose I have the following SDE.

$dX_t=-k\cdot X_tdt+\sigma\sqrt{X_t}dB_t$

If I want to find a bound at any $t$ ofthe expectation of $X_t^2$, given $X_0=0$, is it legitimate to do the following? I use Dynkin's Theorem.

$\mathbb{E}^x\left(X_t^2 \right)=x+\mathbb{E}^x\int_0^t\left( -2k+\sigma^2\right)X_s^2+\sigma^2ds$

Then I take the expectation inside the integral by Fubini's Theorem and conclude that the integral is bounded for $-2k+\sigma^2<0$ .

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I think you missed to transfer the square root in the diffusion term. By Ito, $$ d(X_t^2)=2X_tdX_t + d\langle X_t\rangle = [-2kX_t^2\,dt + 2σX_t^{3/2}\,dB_t] + σ^2X_t\,dt\tag1 $$ so that under expectations $\newcommand{\E}{\mathbb E^x}$ $$ d\E(X_t^2)=-2k\E(X_t^2)\,dt + σ^2\E(X_t)\,dt.\tag2 $$ We need to compute the expectation formula of $X_t$ first, $$ d\E(X_t)=-k\E(X_t)\,dt\implies \E(X_t)=xe^{-kt}.\tag3 $$ By using the integrating factor $e^{2kt}$ in the first order ODE (2) we get to $$ e^{2kt}\E(X_t^2)=\frac{σ^2x}{k}(e^{kt}-1)\implies \E(X_t^2)=\frac{σ^2x}{k}e^{-kt}(1-e^{-kt}) $$ All this under the assumption that $X_t>0$ for all $t>0$.

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  • $\begingroup$ Thank you, much easier way to do it. $\endgroup$ – thaumoctopus Dec 17 '18 at 15:31

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