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Let $X=(X_t)_{t\geq0}$ be a non negative stochastic process solving $$dX_t=2dt+2\sqrt{X_t}dB_t \text{ with} \space X_0=0$$

I am trying to work out $E[X_t]$, $E[X_t^2]$ and $E[X^3_t]$. I believe I have solved $E[X_t]$, by applying Ito's formula to $f(x)=x$, which gives

$$X_t=\int^t_0 dXt$$

$$=\int^t_02dt + \int^t_02\sqrt{X_t}dB_t$$

Applying expectation gives: $$E[X_t]=2t+\int^t_02E[\sqrt{X_t}]dB_t$$ This is where I get stuck, this is similar to where I get stuck in trying to work out $E[X_t^2]$, so my question is can someone offer advice on how to proceed for this and for the further parts?

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    $\mathbb{E} \left( \int_0^t 2 \sqrt{X_s} \, dB_s \right) \neq \int_0^t 2 \mathbb{E}(\sqrt{X_s}) \,d B_s$ – 2017-01-14
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    How do I go about solving $E[\int^t_0 2 \sqrt{X_s}dB_s]$ – 2017-01-14

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The stochastic integral $M_t:=\int_0^t\sqrt{X_s}\,dB_s$ is a local martingale; if we knew that $\int_0^t X_s\,ds$ had finite expectation, then by the Ito isometry we would have $\Bbb E\left[\left(\int_0^t\sqrt{X_s}\,dB_s\right)^2\right]=\Bbb E\int_0^t X_s\,ds<\infty$ and $M$ would be a true martingale, with $\Bbb E[M_t]=0$. This in turn would yield $\Bbb E[X_t]=2t$. To break out of the vicious circle here (to compute $\Bbb E[X_t]$ we must first know that it is finite...) define $T_n:=\inf\{t: X_t>n\}$ and let $X^{(n)}$ be $X$ stopped at time $T_n$. Then $$ X^{(n)}_t=2(T_n\wedge t)+2\int_0^{T_n\wedge t}\sqrt{X_s}\,dB_s. $$ Notice that the integrand on the right in this display is bounded by $\sqrt{n}$, so the stochastic integral on the right has mean zero, yielding $\Bbb E[X^{(n)}_t]=2\Bbb E[T_n\wedge t]\le 2t$. The sequence of random variables $(X^{(n)}_t)_{n\in\Bbb N}$ converges pointwise to $X_t$ and is uniformly integrable because $$ \Bbb E[(X^{(n)}_t)^2]\le 4t^2+4\Bbb E\int_0^{T_n\wedge t}X_s\,ds\le 4t^2+4\int_0^t 2s\,ds=8t^2<\infty. $$ It follows that $\Bbb E[X_t]=\lim_n\Bbb E[X^{(n)}_t]=2t$.

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    Do I apply this same methodology to work out $E[X^2_t]$ and $E[X^3_t]$? @JohnDawkins – 2017-01-14
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    Yes, similar reasoning will work there. – 2017-01-15
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    Does the $T_n$ work for any non-negative function? Also i'm not sure on why it is bounded by $\sqrt{n}$ – 2017-01-19
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    This $T_n$ should work for any continuous function. Because $X_s\le n$ for $0\le s\le T_n$, you have $\sqrt{X_s}\le\sqrt{n}$ for $0\le s\le T_n\wedge t$. – 2017-01-19
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    But what if $t>T_n$? or are we holding $t$ fixed and only n is increasing? – 2017-01-19
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    Time $t>0$ is fixed. – 2017-01-19