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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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    *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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$\mathrm E\left(TB_T\int_0^Tt\,\mathrm dB_t\right)=T\int_0^Tt\,\mathrm E(B_T\,\mathrm dB_t)=T\int_0^Tt\,\mathrm dt=\frac{T^3}2\ne0 $