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I have a problem with this exercise on Stochastic differential equation. I almost arrived to the result but I guess I am missing an argument.

I have a matrix $A=\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $b=\begin{bmatrix} 0 \\ b_2 \end{bmatrix}$. I have to prove that $$ E[||X_t||^2]=E[||X_0||^2]+b_2^2 t$$ where $X_t=\begin{pmatrix}X_t^{(1)} \\ X_t^{(2)}\end{pmatrix}$ is the solution of $dX_t = AX_t~dt + b~dB_t$ with $B_t$ Brownian motion.

I use the Ito's formula as follow $$d||X_t||^2 = 2X_t^{(2)} b_2~dB_t + b_2^2 ~dt$$ and then I consider $$||X_t||^2=||X_0||^2+2\int_0^tX_s^{(2)}b_2~dB_s+b_2^2 ~t$$ But now I can't manage to say that $E[\int_0^tX_s^{(2)}b_2~dB_s]=0$. I try to say that $E[\int_0^t(X_s^{(2)})^2~ds]<\infty$ in order to say that it is a martingale.

If anyone could help me, I would be very grateful.

Luca
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  • Recalling that the increments $\mathrm{d}B_t$ are independent of the integrand, one gets $\mathbb{E}\left[\int_0^t X_s^{(2)}b_2 ,\mathrm{d}B_s\right] = \int_0^t \mathbb{E}[X_s^{(2)}b_2] ,\mathbb{E}[\mathrm{d}B_s] = 0$, since $\mathbb{E}[\mathrm{d}B_s] = 0$.

    You could have also applied this argument directly to the SDE for $|X_t|^2$, so that $\mathrm{d}\mathbb{E}[|X_t|^2] = b_2^2,\mathrm{d}t$ (assuming that $b_2$ is not stochastic) before integrating.

     – Abezhiko Jul 18 '23 at 12:16

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