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I'm new to stochastic calculus. Could someone please explain how I would calculate the variance of

$\int_{t=o}^{T}\sqrt{|B(t)|}$ $dB(t)%$

I'm aware that I would first have to calculate the expectation, but I'm not sure as to how to go about this.

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    $\begingroup$ I'm voting to close this question as off-topic because it has no direct link to quant finance and is thus more suitable for Math SE, where it has already been double-posted math.stackexchange.com/questions/2561237. $\endgroup$ – LocalVolatility Dec 12 '17 at 7:45
  • $\begingroup$ It seems to be removed from Math.SE now. $\endgroup$ – Bob Jansen Dec 13 '17 at 11:47
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Ciao, cool exercise.

The best thing you can do is to see the integral as a SDE and then use Ito's Lemma. In this particular case you can write: $$ Y_t = \int_0^t \sqrt{\left|B_s \right|} dB_s $$ so that: $$ dY_t = \sqrt{\left|B_t \right|} dB_t $$

The mean is easy to compute since this stochastic process has no drift (the $dt$ term) so that it has $0$ mean. This means that the variance is just $\mathbb{E}[Y_t^2]$.

At this point the best thing is to apply Ito's derivative to $Y_t^2$ and then take the expected value of the result... let me do the explicit computation!

$$ \begin{align} dY_t^2 & = 2Y_t dY_t + \frac{1}{2} 2d \langle Y_t \rangle \\ & = 2Y_t \sqrt{|B_t|}dB_t + |B_t| dt \end{align} $$

Since we are going to take expected value of the terms we can ignore the first one since we know it has $0$ mean:

$$ \mathbb{E}[Y_t^2] = \mathbb{E}\left[ \int_0^t |B_s| ds\right] = \int_0^t \mathbb{E}[|B_s| ] ds $$

At this point you can compute the expected value using the distribution of the Brownian motion: see this. The result is:

$$ \mathbb{E}[Y_t^2] = \sqrt{\frac{2}{\pi}}\int_0^t \sqrt{s} ds = \frac{2\sqrt{2}}{3\sqrt{\pi}} \sqrt{t^3}. $$

Ok, maybe the last part about the absolute valute of $B_t$ is not so usefull in general but is always usefull try to express what are you integrating is terms of stochastic process and use Ito's Lemma. In this way you can always decompose the integrand in the part with zero mean (i.e. $\dots dW_t$) and the part which gives a not zero contribution to the expected value (i.e. $\dots dt$).

Ciao Ciao! AM

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  • $\begingroup$ @MarissaB the truth is that you can reach the result $\mathbb{E}[Y_t^2] = \int_0^t \mathbb{E}[|B_s|] ds$ just by using Ito' s isometry... I forgot that but the method I've written is general and it works also when you do not have such results as Ito's one :) $\endgroup$ – clarkmaio Dec 15 '17 at 11:13
  • $\begingroup$ clarkmaio, what's the relevance, if any, with the following please? How to compute expectation of square of Riemann integral of a random variable? $\endgroup$ – BCLC Jan 7 '18 at 7:44
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    $\begingroup$ @BCLC I dont't know if I've well understood what you meen. However wrt the question you've linked of course you can obtain the same result by using the technique I've used in this post. It is enough to consider the stochastich process: $Z_t = Y_t^2$ where $Y_t = \int_0^t W_t dt$. You have then: $dY_t = W_t dt$ and $dZ_t = 2Y_t W_t dt $. You get the final result taking the expected value of $Z_t$ and keeping in mind that $\mathbb{E}[W_t W_s] = \min(t,s)$ $\endgroup$ – clarkmaio Jan 7 '18 at 10:01
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This question already has a marked accepted answer, but it is worth noting that if $B_t$ is a standard Brownian motion with respect to a filtration $\{\mathcal{F}_t\}$, $A_t$ is an adapted process with continuous paths, and $$Z_t = \int_0^t A_s \,\mathrm{d}B_s,$$ then $\mathbb{E}[Z_t] = 0$ and $$ \mathrm{Var}[Z_t] = \mathbb{E}[Z_t^2] = \int_0^t \mathbb{E}[A_s^2] \,\mathrm{d}s.$$ In your case, denoting the integral as $Z_t$, \begin{align*} \mathrm{Var}[Z_t] &= \int_0^t \mathbb{E}[|B_s|] \,\mathrm{d}s \\ &= \mathbb{E}[|B_1|] \cdot \int_0^t \sqrt{s} \,\mathrm{d}s \\ &= \mathbb{E}[|B_1|] \cdot \frac{2t^{3/2}}{3}. \end{align*} The law of $|B_1|$ is that of a half-normal distribution, which is easily checked to have expectation $\sqrt{\frac{2}{\pi}}$, so that the final answer is $$ \mathrm{Var}[Z_t] = \frac{(2t)^{3/2}}{3\sqrt{\pi}},$$ in agreement with the computation done by user clarkmiao. Incidentically, since this variance is finite almost surely for all $t$, the process is a martingale for all $t \geq 0$.

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