Expected value of average of Brownian motion

Question

For a standard one-dimensional Brownian motion $W(t)$, calculate:

$$E\bigg[\Big(\frac{1}{T}\int\limits_0^TW_t\, dt\Big)^2\bigg]$$

Note: I am not able to figure out how to approach this problem. All i can think of is that the term $\frac{1}{T}\int\limits_0^TW_t\,dt$ is like 'average'. But not sure how to proceed ahead. I'm relatively new to Brownian motion. I tried searching the forum for some hints..but could not find one. I will really appreciate if you could please guide me in the right direction. Thanks!

Answer 1

If you recall that $\mathrm d(t W_t) = W_t\mathrm dt + t\mathrm dW_t$ you can write your integral in the other form $$ \int_0^T W_t\mathrm dt = TW_T - \int_0^Tt\mathrm dW_t. $$ If we forget about the factor $\frac1T$ as it does not affect the derivation much, we obtain $$ \mathsf E\left(\int_0^T W_t\mathrm dt\right)^2 = \mathsf E\left[T^2W_T^2\right] - 2T\mathsf E\left[W_T\int_0^T t\mathrm dW_t\right]+\mathsf E\left(\int_0^T t\mathrm dW_t\right)^2 $$ $$ = T^3- 2T\int_0^Tt\mathrm dt+\int_0^Tt^2\mathrm dt $$ where we applied the Ito isometry a couple of times. Hopefully, the most hard part now is done an you can finish the derivations.

Answer 2

Another approach would be to show that the random variable

$$\omega \mapsto \int_0^T W_t(\omega) \, dt$$

is a centered normal random variable with variance $\int_0^T (s-T)^2 \, ds=\frac{T^3}{3}$. You can find a proof here. (The proof is a bit lenghthy, but you don't need Itô-Calculus to prove it.)

Answer 3

Expand the square as $$ \left(\int_0^TW_t\, \mathrm dt\right)^2=\int_0^T\int_0^t2W_tW_s\,\mathrm ds\mathrm dt, $$ and use Fubini theorem and the identity $\mathbb E(W_tW_s)=s$ for every $s\leqslant t$ to conclude that the expectation you want to compute is $$ \frac1{T^2}\int_0^T\int_0^t2s\,\mathrm ds\mathrm dt=\frac{T}3. $$