Variance of a Brownian Stochastic Integral with an Absolute-Value Integrand

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


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












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.

## Answer by clarkmaio (score 7, accepted)

https://quant.stackexchange.com/a/37332

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

## Answer by user217285 (score 3)

https://quant.stackexchange.com/a/37396

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 $t \geq 0$.