# How to compute the variance of this stochastic integral?

I'm new to stochastic calculus and I did an exercise but I don't know if it is correct, so I need somebody with more experience to check if it is true.

I am trying to compute the variance of the following random variable:

$$Z=\int _0^T e^{W_t} dW_t$$

So we have:

$\text{Var}(Z)=\text{Var}\left(\int _0^T e^{W_t} dW_t\right)$

By Itō's isometry we have:

$$\mathbb{E}\left[\int _0^T e^{2W_t} dt\right]$$

we can then bring inside the expectation to get:

$$\int _0^T \mathbb{E}\left[e^{2W_t}\right] dt = \int_0^T e^{2t} dt = \frac{e^{2T}}{2}-\frac{1}{2}$$

Moreover, if the above result is correct, what should I get instead of the problem asked me to compute

$$\text{Var}\left(\int _0^T e^{W_t} dt \right)$$

It should simply be the variance of a lognormal distributed random variable, computed in the extrema of the interval, or not?

• $Var(Z) = E(Z^2) - [E(Z)]^2$. – Gordon May 31 '15 at 0:48
• I know this basic stuff, please tell me if what I did is correct. I've just applied the basic relationships of stochastic calculus – james42 May 31 '15 at 0:51
• That is correct. – Gordon May 31 '15 at 1:25
• @Gordon What about the second question? – BCLC Dec 29 '16 at 0:51
• What does "if the above result is correct, what should I get instead of the problem asked me to compute" mean? – SRKX Dec 29 '16 at 10:02

To compute the variance $$\text{Var}\left(\int _0^T e^{W_t} dt \right),$$ we need to compute \begin{align*} E\left( \left(\int _0^T e^{W_t} dt \right)^2 \right) &= \int_0^T\!\!\!\!\int_0^T E\left(e^{W_s+W_t} \right) ds\,dt. \end{align*} Note that, for $0 \le s, t \le T$, \begin{align*} W_s+W_t = \begin{cases} W_t -W_s + 2 W_s, & \text{ if } s \le t,\\ W_s -W_t + 2 W_t, & \text{ else}. \end{cases} \end{align*} That is, as a sum of two independent normal random variables, $W_s+W_t$ is normal, with mean $0$ and variance \begin{align*} \text{Var}(W_s+W_t) = \begin{cases} t+3s, & \text{ if } s \le t,\\ s+3t, & \text{ else}. \end{cases} \end{align*} Then \begin{align*} E\left( \left(\int _0^T e^{W_t} dt \right)^2 \right) &= \int_0^T\!\!\!\!\int_0^T E\left(e^{W_s+W_t} \right) ds\,dt\\ &=\int_0^T\left[\int_0^t E\left(e^{W_s+W_t} \right) ds+\int_t^T E\left(e^{W_s+W_t} \right) ds\right]dt\\ &=\int_0^T\left[\int_0^t e^{\frac{1}{2}t + \frac{3}{2}s} ds+\int_t^T e^{\frac{1}{2}s + \frac{3}{2}t} ds\right]dt. \end{align*} The remaining is simple calculus.

$$Var(\int _0^T e^{W_t} dt)$$

$$= E[(\int _0^T e^{W_t} dt)^2] - (E[\int _0^T e^{W_t} dt])^2$$

Now

$$E[\int _0^T e^{W_t} dt] = \int _0^T E[e^{W_t}] dt$$

Recall that $W_t$ is normal. use mgf

As for

$$E[(\int _0^T e^{W_t} dt)^2]$$

I'll try following this:

$$E[(\int_0^T e^{W_t} dt)^2]$$

$$= E[(\int_0^T e^{W_t} dt)(\int _0^T e^{W_s} ds)]$$

$$= E[\int_0^T \int_0^T e^{W_t} e^{W_s} dt ds]$$

$$= \int_0^T \int_0^T E[e^{W_t} e^{W_s}] dt ds$$

Without loss of generality, suppose $s < t$. Then by considering the exponential martingale, we have

$$E[e^{W_t} e^{W_s}] = E[E[e^{W_t} e^{W_s}|\mathscr F_s]]$$

$$= E[e^{W_s}E[e^{W_t} |\mathscr F_s]]$$

$$= E[e^{W_s}e^{\frac{t}{2}}E[e^{\frac{-t}{2}}e^{W_t} |\mathscr F_s]]$$

$$= E[e^{W_s}e^{\frac{t}{2}}e^{\frac{-s}{2}}e^{W_s}]$$

$$= e^{\frac{t}{2}}e^{\frac{-s}{2}}E[e^{W_s}e^{W_s}]$$

$$= e^{\frac{t}{2}}e^{\frac{-s}{2}}E[e^{2W_s}]$$

Note that $2W_s$ is normal too. Use mgf again

Thus

$$\int_0^T \int_0^T E[e^{W_t} e^{W_s}] dt ds$$

$$= \int_0^T e^{\frac{t}{2}} dt \int_0^T e^{\frac{-s}{2}}E[e^{2W_s}] ds$$

• $W_t$ is not log-normal. $e^{W_t}$ is. – SRKX Dec 29 '16 at 9:55
• Also, the conclusion is not clear. You should keep the left part of the equation at least in the first line. – SRKX Dec 29 '16 at 10:04