Expectation of time integral I've 2 questions:
let be $W_s$ a standard Brownian motion:


*

*using Ito's formula show that $\left( W_t,\int_0^t W_sds \right)$ has a normal distribution;

*and calculate $ E\left[e^{W_t}e^{\int_0^t W_sds}   \right] .$


For the first part, i know that $W_t$ and $\int_0^t W_sds$ 
have normal distribution with mean and variance respectively $(0,t)$ and $(0, t^3/3)$, but i need help with Ito's formula. 
For the second part i've tried to solve $E\left[e^{W_t}e^{\int_0^t W_sds}   \right]= \iint e^{W_t}e^{\int_0^t W_sds} \;\phi \left( W_t,\int_0^t W_sds \right)\: dW_t \int_0^t W_sds$... 
Is these the only way?
P.S. sorry for my poor english
 A: Let me expand on mpiktas answer. As you noted, $W_t$ and $\int_0^t {W_s ds}$ are zero mean normal with variance $t$ and $t^3/3$, respectively (the first is trivial). Now, if $(W_t,\int_0^t {W_s ds})$ is bivariate normal, then any linear combination $aW_t + b\int_0^t {W_s ds}$, $a,b \in \mathbb{R}$, is univariate normal. In particular, the sum $W_t + \int_0^t {W_s ds}$ is (zero mean) normal. Since you are given the variance of $W_t$ and $\int_0^t {W_s ds}$, the variance of the sum follows straight from the covariance ${\rm Cov}(W_t,\int_0^t {W_s ds})$, that is from ${\rm E}[W_t \int_0^t {W_s ds}]$. Now, assuming you found that variance, calculating ${\rm E}[\exp (W_t  + \int_0^t {W_s ds} )]$ amounts to calculating the moment-generating function at $1$ of a ${\rm N}(0,\sigma^2)$ variable. 
EDIT: Regarding the first question, let's explain why we should expect $(W_t,\int_0^t {W_s ds})$ to be (bivariate) normal. Since $W$ is continuous on the interval $[0,t]$, $\int_0^t {W_s ds}$ is a standard Riemann integral, and 
$\int_0^t {W_s ds}  \approx \frac{t}{n}\sum\nolimits_{k = 0}^{n - 1} {W_{kt/n} }$. Denote the right-hand side by $Y_t^n$. Then $(W_t,Y_t^n)$ is normal. Indeed, $W$ is a Gaussian process; hence, by definition, $(W_{t_1},\ldots,W_{t_m})$ is normal for any choice of times $t_1,\ldots,t_m$. But $(W_{t_1},\ldots,W_{t_m})$ is normal if and only if any linear combination $\sum\nolimits_{i = 1}^m {a_i W_{t_i } }$, $a_i \in \mathbb{R}$, is (univariate) normal, and the conclusion follows. Finally, since $(W_t,Y_t^n)$ is normal, we should expect that in the limit as $n \to \infty$, $(W_t,\int_0^t {W_s ds})$ is normal.
A: If you have already shown that $W_t$ and $\int_0^tW_sds$ are normal, find their covariance and use moment generating function of bivariate normal distribution to solve the second part of your problem.
A: If i've understood what u say, i've to calculate $E[W_t \int_0^t W_s ds]$. 
Tell me if i'm right doing this:
as $\int_0^t W_sds=tW_t-\int_0^t s dW_s$, i've to calculate $E[W_t (tW_t-\int_0^t s dW_s)]$, or better $E[t W^2_t]-E[W_t\int_0^t s dW_s)]$, and the first expectation is $t^2$.
as $W_t=\int_0^t dW_s$ the second expectation become $ E[\int_0^t dW_s\int_0^t s dW_s)]$ that is for Ito-isometry (hope that this is the correct translation) = $ E[\int_0^t s ds)]$, and this expectation is $t^2/2$.
So covariance is  $t-t^2/2=t^2/2$...  
