0

I am trying to understand this integrand $ \int_0^T W^2 ds $

I tried breaking this into a Riemann Sum and then re-arranging and this is what I got at the end, not sure if it's correct or not $\sum_{k=0}^{M-1} [(W(t(k-1)))^2 - (W(t(k)))^2][t(k)-T] $

Now what do I make of this? It's not Normally distributed as far as I can tell. I can try to find out the Expectation and Variance of this. But what will be the best way to describe this integration?

P.S. :- W is Brownian Motion

saz
  • 120,083
vicky113
  • 133
  • 3
  • $W=(W_s)_{s\geqslant 0}$ is a standard Brownian motion and then it is path-wise continuous, i.e., for each $\omega \in \Omega$, we have a continuous function $s \mapsto W_s(\omega)$. Then apply the classical Riemann integration for a continuous function $W_s$. – ntt Apr 02 '17 at 16:15
  • Yes I tried to do the Riemann way only and that is what I got, which I have shared in the original post. Now how do I go about it? – vicky113 Apr 02 '17 at 16:17
  • You could write the Riemann sum formula for a function $f(s)$ on $[0,T]$ and then replace $f(s)$ by $W(s)$, right? – ntt Apr 02 '17 at 16:21
  • $W_s$ has the same distribution as $s*W_1$ . It could help you guess the distribution of the integral and deduce the parameters after all... – RiezFrechetKolmogorov Apr 02 '17 at 19:57
  • 1
    @RiezFrechetKolmogorov Note that the process $(W_s)_{s \leq T}$ does not have the same distribution as $(\sqrt{s} W_1)$. For this reason we cannot simply use the scaling property here. – saz Apr 04 '17 at 12:47

1 Answers1

4

There is no need to approximate the integral by Riemann sums in order to determine the expectation and variance of the integral $X:=\int_0^T W_s^2 \, ds$. Note that this integral is a "standard" Riemann integral, i.e. for each point $\omega \in \Omega$ the integral

$$X(\omega) = \int_0^T W_s(\omega)^2 \, ds$$

is defined as a Riemann integral. Everything is well-defined since $s \mapsto W_s^2(\omega)$ is continuous, hence Riemann-integrable.

An application of Fubini's theorem shows

$$\mathbb{E}(X) = \mathbb{E} \left( \int_0^T W_s^2 \, ds \right) = \int_0^T \mathbb{E}(W_s^2) \, ds.$$

Since $W_s$ is Gaussian with mean $0$ and variance $s$, we have $\mathbb{E}(W_s^2) = s$. Consequently, $$\mathbb{E}(X)=T^2/2$$

Calculating the variance is slightly more complicated. Recall that

$$\text{var}(X) = \mathbb{E}(X^2) - (\mathbb{E}(X))^2 \tag{1}$$

and, since we already know $\mathbb{E}(X)$, it suffices to calculate $\mathbb{E}(X^2)$. To this end, note that

$$X^2 = \left( \int_0^T W_s^2 \, ds \right) \left( \int_0^T W_t^2 \, dt \right) = \int_0^T \int_0^T W_s^2 W_t^2 \, ds \, dt.$$

Applying Fubini's theorem another time, we get

$$\mathbb{E}(X^2) = \int_0^T \int_0^T \mathbb{E}(W_s^2 W_t^2) \, ds \, dt = 2 \int_0^T \int_0^t \mathbb{E}(W_s^2 W_t^2) \, ds \, dt. \tag{2}$$

Using that $W_t^2-t$ is a martingale, it is not difficult to see that

$$\mathbb{E}(W_s^2 W_t^2) = st +2s^2 \qquad \text{for all} \, \, s \leq t.$$

Plugging this into $(2)$ yields $\mathbb{E}(X^2)$, and hence, by $(1)$, the variance of $X$.

Did
  • 279,727
saz
  • 120,083
  • @Did Thanks for correcting the stupid mistake. – saz Apr 04 '17 at 16:34
  • You are welcome (that was easy). (+1) – Did Apr 04 '17 at 19:53
  • @saz - Thanks for the nice solution. But is there a way to solve this integral? Is a close form solution feasible for such integrals? – vicky113 Apr 07 '17 at 05:08
  • @DebdiptaMajumdar No, there isn't a closed form for this integral. – saz Apr 07 '17 at 05:15
  • @saz - Can you share some docs or links on Fubini's theorem on Stochastic Integrals, something not very complicated if possible – vicky113 Apr 07 '17 at 16:50
  • @DebdiptaMajumdar what exactly do you mean by Fubini for stochastic integrals...? In my answer I used the standard version of Fubini's theorem ... you can find it in any (reasonable) book on measure theory. – saz Apr 07 '17 at 20:28