Hi,

I have the following problem:

There is a random Gaussian variable $\displaystyle L=a \int_0^T \varphi(s) dB_s + (1-a) \epsilon$ with $\displaystyle \varphi \in L^2(\mathbb{R}_+,ds)$ and $\displaystyle \epsilon$ is a standard nomally distributed variable with mean 0 and variance 1.

Under some filtration $\displaystyle \mathcal{F}_t$ generated by $\displaystyle B$, I have to find $\displaystyle Law(L|\mathcal{F}_t)$.

Therefore I calculated the mean (with use of martingale property):

$\displaystyle
E[L|\mathcal{F}_t]=E[a \int_0^T \varphi(s) dB_s|\mathcal{F}_t]+ E[(1-a)\epsilon|\mathcal{F}_t]=E[a \int_0^T \varphi(s) dB_s|\mathcal{F}_t]=a \int_0^t \varphi(s) dB_s
$

Now I have to calculate the variance (Var):

$\displaystyle
Var L = Var (a \int_0^T \varphi(s) dB_s)+ Var((1-a) \epsilon)+2 Cov(a \int_0^T \varphi(s) dB_s, (1-a)\epsilon) = a^2\int_t^T \varphi(s)^2 ds +(1-a)^2 + 2 Cov(a \int_0^T \varphi(s) dB_s, (1-a)\epsilon)
$

To determine the covariance I proceed as follow:
$\displaystyle
Cov(a \int_0^T \varphi(s) dB_s, (1-a)\epsilon) = E[(a \int_0^T \varphi(s) dB_s - E[a \int_0^T \varphi(s) dB_s])((1-a)\epsilon)-E[(1-a)\epsilon])] = E[(a \int_t^T \varphi(s) dB_s)((1-a)\epsilon)] = 0$ because of the independency ?

Is that correct?