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I've just learned about stochastic integral and only know how to evaluate $\int\limits^{t}_{0} W(s)\mathrm{d}W(s)$. Could anyone give me some instruction on how to evaluate the following integrals? $$\mathbb{Var}\left(\int_{1}^2 (W_t)^2\mathrm{d}W_t \right)$$

$$\mathbb{Var}\left(\int_{0}^T (W_t)^2\mathrm{d}t \right)$$

Thanks a lot!!

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    Can you please format this question using TeX?2012-01-19
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    @Alex Becker Sorry I don't know who to type TeX. I really hope I can do that. Really sorry.2012-01-19
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    Anything that you want formatted as TeX should be included in dollar signs, as in \$ code \$. As for what specific code to use, you can learn a lot by poking around this site, right-clicking on mathematical expressions and selecting "show source" to see the TeX code for them.2012-01-19
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    @Sasha Thanks for editing. I am learning to do it myself.2012-01-19

1 Answers 1

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Let $X = \int\limits_1^2 W_t^2 \mathrm{d} W_t$.

Since $\mathbb{E}\left(X\right) = 0$, the variance $\mathbb{Var}(X) = \mathbb{E}(X^2)$. Here we use Ito isometry: $$ \mathbb{E}\left( \left(\int\limits_1^2 W_t^2 \mathrm{d} W_t \right)^2 \right) = \int_{1}^{2} \mathbb{E}(W_t^4) \mathrm{d} t = \int_1^2 3 t^2 \mathrm{d} t = 2^3 - 1^3 = 7 $$

Let $Y = \int\limits_0^t W_s^2 \mathrm{d} s$. Expectation is $\mathbb{E}(W_s^2) = s$, thus $\mathbb{E}(Y) = \int\limits_0^t s \mathrm{d} s = \frac{t^2}{2}$.

$$\begin{eqnarray} \mathbb{E}(Y^2) &=& \int_0^t \int_0^t \mathbb{E}(W_s^2 W_u^2) \mathrm{d} s \mathrm{d} u = 2 \int_0^t \int_0^s \mathbb{E}(W_s^2 W_u^2) \mathrm{d} u \mathrm{d} s \\ &=& 2 \int_0^t \int_0^s \mathbb{E}(W_u^2 (W_u + W_{s-u})^2) \mathrm{d} u \mathrm{d} s \\ &=& 2 \int_0^t \int_0^s \left( \mathbb{E}(W_u^4) + \mathbb{E}(W_u^2) \mathbb{E}(W_{s-u}^2) \right) \mathrm{d} u \mathrm{d} s \\ &=& 2 \int_0^t \int_0^s \left( 3 u^2 + u(s-u) \right) \mathrm{d} u \mathrm{d} s = 2 \int_0^t \left( s^3 + \frac{s^3}{6} \right) \mathrm{d} s = \frac{7}{12} t^4 \end{eqnarray} $$ Thus $\mathbb{Var}(Y) = \mathbb{E}(Y^2) - (\mathbb{E}(Y))^2 = \frac{7}{12} t^4 - \left(\frac{t^2}{2} \right)^2 = \frac{t^4}{3}$.

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    So I could always interchange the integral sign with expectation sign when the expectation is of form $E[\int_{0}^{T}f(W_t,t)dt]$?2012-01-19
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    This is called [Fubini's theorem](http://en.wikipedia.org/wiki/Fubini%27s_theorem).2012-01-19
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    @Sasha : In 5th line, you wrote 0s by writing $W_{u-s}$. Am I wrong?2012-02-16
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    @neticin Thank you, you are correct, of course. I messed up, and that unfortunately changes the answer. See my updated post.2012-02-16
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    $E[(w_t)^4]=3t^2 $ is this formula has a name ? can you give a reference (a book for example )2015-04-17
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    here is what you asked, https://en.wikipedia.org/wiki/Normal_distribution#Moments =)2015-12-10
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    Just to let you know since the OP did not see fit to inform you, your answer is discussed [there](https://math.stackexchange.com/q/2392027/).2017-08-13
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    Hi Sasha, I used your answer as a basis for a question [here](https://math.stackexchange.com/questions/2392027/expectation-of-int-0t-w-s2-dw-s-integral-of-square-of-brownian-wrt-to-bro) about the integration of a stochastic integral. (question answered, I just wanted to let you know).2017-08-16