# Stochastic Integral

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? –  Alex Becker Jan 19 '12 at 4:17
@Alex Becker Sorry I don't know who to type TeX. I really hope I can do that. Really sorry. –  BVFanZ Jan 19 '12 at 4:19
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. –  Alex Becker Jan 19 '12 at 4:23
@Sasha Thanks for editing. I am learning to do it myself. –  BVFanZ Jan 19 '12 at 4:35

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 $E[\int_{0}^{T}f(W_t,t)dt]$? –  BVFanZ Jan 19 '12 at 5:14
This is called Fubini's theorem. –  Did Jan 19 '12 at 6:10
@Sasha : In 5th line, you wrote 0<u<s. However in the next line you assume that u>s by writing $W_{u-s}$. Am I wrong? –  neticin Feb 16 '12 at 15:30
@neticin Thank you, you are correct, of course. I messed up, and that unfortunately changes the answer. See my updated post. –  Sasha Feb 16 '12 at 15:44