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Let $Y(T)=\int_T B(t) dt$ an integral.

We wold like to evaluate $E(Y(T)^2)$

Now, $Y(T)$ may be Riemann integrated because dt have finite absolute variation and $B(t)$ is continuous. Then we can take that integral as the limit N going to infinite in the usual Riemman sum. Then the expected value of the variable $Y(T)^2$ may be written as a double sum of the expected value of $B(t)B(s)$, that is $\min(t,s)$. Performing the sum one can get $T^3/3$.

In contrat, one may take te Ito calculus as follows. One may represent the variable $Y(T)$ by using the integration by parts formula giving

$d(B(T) T)=t dB(t) + B(t) dt$

It seems that

$Y(T)=B(T) T - \int_T t dB(t)$

correct?

Take the square and use the Ito isometry, then

$E(Y(T)^2)=4/3 T^3$

Where is the point here?

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    *Take the square and use the Ito isometry:* You might want to expand this step of your proof.2012-05-19
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    The expected value of the squared may be written as E(B(T)^2 T^2) +Ito isometry of (int^T t dB(t))^2. The expected value of the cross term in the square is zero. The first term is T^3. The second is T^3/3. They add to 4/3 T^3. Where is the point? @Didier2012-05-19
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    *The expected value of the cross term in the square is zero*: Nope.2012-05-19
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    *Where is the point?* Irrespectively of the specifics of the question at hand, *the point* is to show what you tried.2012-05-19

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