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Compute $$\mathbb{E} \left( \int_0^1 \left| B_s \right|^{\frac{1}{2}} dB_s \right)^2.$$

Let $f(x) = x^2$, and for reference, we note that $f'(x) = 2x$ and $f''(x) = 2$. Furthermore, let $X_t$ be the process defined by $$X_t : = \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s.$$ Ito's formula gives us that $$f(X_t) - f(0) = \int_0^t f'(X_s) dX_s + \frac{1}{2} \int_0^t f''(X_s) ds.$$ Therefore, we see that \begin{eqnarray*} \left( \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s \right)^2 &=& \int_0^t 2 \left( \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s \right) dX_s + \frac{1}{2} \int_0^t 2 ds \\ &=& 2\left( \int_0^t \left |B_s \right|^{\frac{1}{2}} dB_s \right) \left( X_t - X_0 \right) + t \\ &=& 2 \left( \int_0^t \left| B_s \right|^{\frac{1}{2}} dB_s \right)^2 + t \end{eqnarray*}

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I believe that the authors of this problem meant that you should write the Itô formula to $f(X_t)$, where $f(x) = x^2$, $X_t = \int_0^t |B_s|^{1/2}ds$: $$ \left(\int_0^1 |B_s|^{1/2}dB_s\right)^2 = f(X_1) = f(X_0) + \int_0^1 f'(X_s) dX_s + \frac12 \int_0^1 f''(X_s)(dX_s)^2 \\ = 2\int_0^1 X_s |B_s|^{1/2} dB_s + \int_0^1 |B_s|ds. $$ Can you compute the expectation of the rhs?

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  • $\begingroup$ I am still confused, I have edited my working. I do not see how $\frac{1}{2} \int_0^1 f''(X_s) (dX_s)^2 = \int_0^1 \left| B_s \right| ds$. And I still don't know how to compute the expectation of the RHS. Thank you for your time thus far. $\endgroup$
    – user466491
    Sep 27, 2017 at 0:52
  • $\begingroup$ @YinPeiTo, 1) you're trying to apply the usual change-of-variable formula in what you wrote. This is not possible. 2) The rule for multiplying differentials: $(dt)^2 = dt\cdot dB_t = 0$, $(dB_t)^2 = dt$. Therefore, $(dX_t)^2 = |B_t|dt$. 3) To compute the expectation, use that $E[\int\dots dB_t] = 0$ and $E[\int\dots dt] = \int E[\dots]dt$. $\endgroup$
    – zhoraster
    Sep 27, 2017 at 11:24

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