Calculating the expected value of the square of an integral of Brownian Motion

[JohanL]

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

$$E\bigg[\Big(\frac{1}{T}\int\limits_0^TW_t\, dt\Big)^2\bigg]$$I can't figure out how the middle term simplifies.

$$
\mathsf E\left(\int_0^T W_t\mathrm dt\right)^2 = \mathsf E\left[T^2W_T^2\right] - 2T\mathsf E\left[W_T\int_0^T t\mathrm dW_t\right]+\mathsf E\left(\int_0^T t\mathrm dW_t\right)^2
$$
$$
= T^3- 2T\int_0^Tt\mathrm dt+\int_0^Tt^2\mathrm dt
$$

i.e. why is

$$
\mathsf 2T\mathsf E\left[W_T\int_0^T t\mathrm dW_t\right]
= 2T\int_0^Tt\mathrm dt
$$
?

 

[StoneTemplePython]

will you tell us what $T$ is? The standard usage of a capital letter $T$ would be for a stopping time (i.e. a random variable), but this seems to contradict other equations... you also haven't consistently used brackets for the expectations operator, which makes this not so easy to read.

 

[JohanL]

will you tell us what $T$ is? The standard usage of a capital letter $T$ would be for a stopping time (i.e. a random variable), but this seems to contradict other equations.

I found the exercise and solution online. They don't say anything about T.

Im guessing its just the upper limit of integration and not a stopping time if you say it contradicts the other equations.

you also haven't consistently used brackets for the expectations operator, which makes this not so easy to read.

Sorry about that. Seems like i can't edit now?

 

[BWV]

Is this what you are looking for?

http://tullo.ch/articles/the-ito-isometry/

 

[JohanL]

Is this what you are looking for?

http://tullo.ch/articles/the-ito-isometry/

Maybe. But i don't know how to apply it in this case.

edit: solved it. ty!