A mean square derivative I'm doing an exercise where I have to check some properties about these two stochastical processes:
$X(t)=At+B\;\;$  and  $\;\;Y(t)=\frac{1}{t}\displaystyle\int_{0}^{t}X(\tau)\;d\tau$, $t>0$.
Where $A$ and $B$ are uncorrelated 2-random variables with $\mathbb{E}[A]=m_A$, $\mathbb{E}[B]=m_B$, $\mathbb{V}[A]=\sigma_A^2$ and $\mathbb{V}[B]=\sigma_B^2$.
I have checked that both verify that are continuous, differentiable and integrable, and I have their correlation functions $\Gamma_X$ and $\Gamma_Y$.
I have checked too using the definition that that $X'(t)=A$. 
Now, I need to check that $Y'(t)=A\frac{t}{2}+B$. I have tried using the definition ( that is, checking if $\mathbb{E}[(\frac{Y(t+\epsilon)-Y(t)}{\epsilon}-(A\frac{t}{2}+B))^2]$ goes to zero when $\epsilon$ approaches zero), but i don't know how to manage this...
Thanks for any help.
 A: I think there is something wrong here. Things are not complicated if A and B don't depend on t. If so there is nothing stochastic in the integrals, just come back to integral(at+b) with a and b "real numbers"
A: By definition and the linearity of the integral,
$$\begin{align*} Y(t)(\omega) &= \frac{1}{t} \int_0^t X(\tau)(\omega) \, d\tau = \frac{1}{t} \int_0^t A(\omega) \cdot \tau \, d\tau +\frac{1}{t}  \int_0^t B(\omega) \, d\tau \\ &= A(\omega) \cdot \frac{t}{2} + B(\omega) \end{align*}$$
i.e. $Y$ is of the same form as $X$. You already proved $X'(t)=A$; exactly the same calculation shows $Y'(t) = A/2$.

On the other hand, if we consider
$$Y(t) = \int_0^t X(\tau) \, d\tau$$
it's not difficult to show that
$$\frac{Y(t+\varepsilon)-Y(t)}{\varepsilon} = A \frac{2t +\varepsilon}{2} + B$$
using a similar reasoning as above. This implies
$$\mathbb{E} \left[ \left( \frac{Y(t+\varepsilon)-Y(t)}{\varepsilon}- (A \cdot t + B) \right)^2 \right] \stackrel{\varepsilon \to 0}{\to} 0$$
i.e. $Y'(t) = A \cdot t+B=X(t)$.

This shows that in both cases the mean-square derivative coincides with the pointwise derivative. Nevertheless, you can not simply apply the fundamental theorem of calculus; you have to check the $L^2$-convergence.
