Variance and expectation of the stochastic intergal

Question

Compute the unconditional expected value and variance, and describe, as far as possible, the distribution of the random variable

$Y_{t} = \int^{t}_{0} W_{s} ds $

with the hint below

$\int^{t}_{0} \tau dW_{\tau} $ = $ tW_{t} - \int^{t}_{0} W_{\tau} d\tau$

$ \int^{t}_{0} \tau dW_{\tau} $ = $ tW_{t} - \int^{t}_{0} W_{\tau} d\tau $ With this function f($W_t$)= $tW_t$

( by Ito's integral use $df_t = f_t dt + f_w dW_t + f_wt dt $

$df_t = W_t dt + t dW_t + dW_t dt$

And the last variable is equal to zero . Then the intergal of the above is equal to the hint.) The expectation I get is 0 and variance of $ \frac{2}{3} t^3 $ but have been told it is something different by other people.

Answer 1

\begin{align}var \int_0^t W_t dt &= \int_0^t\int_0^t \min(r,s)dr ds\\ &=\frac 13 t^3\\ E \int_0^t W_t dt &= \int_0^t E W_t dt\\ &=0 \end{align}

Now let's use the hint for the variance

\begin{align}var \int_0^t W_t dt &= var \Big(t\int_0^tdW_{\tau}-\int_0^t \tau dW_{\tau}\Big)\\ &=var \Big(t\int_0^tdW_{\tau}\Big)+var \Big(\int_0^t \tau dW_{\tau}\Big)-2cov \Big(t\int_0^tdW_{\tau},\int_0^t \tau dW_{\tau}\Big)\\ &=t^2 \times t + \int_0^t \tau^2 d \tau -2 t \int_0^t \tau d\tau\\ &=\frac13 t^3 \end{align} Using the hint for the mean:

\begin{align}E \int_0^t W_t dt &= E \Big(t\int_0^tdW_{\tau}-\int_0^t \tau dW_{\tau}\Big)\\ &=E \Big(t\int_0^tdW_{\tau}\Big)-E\Big(\int_0^t \tau dW_{\tau}\Big)\\ &=0+0\\ \end{align}