How to find the distribution of the following stochastic integral of a geometric Brownian motion? $K_{\phi,\lambda}(r)=\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}dB_s$
where $W$ and $B$ are independent standard Brownian motions, and $(\phi,\lambda) \in \mathbb{R} \times \mathbb{R}_+ $
The answer is $\mathcal{N}(0,\frac{e^{2cr}-1}{2c})$, where $c=\phi+\lambda^2$
 A: Partial answer: We will derive the unconditional variance. By the Ito isometry, we get that
$
\mathbb{E}(\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}dB_s)^2 = 
\mathbb{E}\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}^2ds
$
Noting that $\exp\{r\phi + \lambda W_r\}$ is a constant, we move that outside of the integral. 
$\mathbb{E}\int_{0}^{r}\exp\{(r-s)\phi+\lambda(W_r-W_s)\}^2ds = \mathbb{E}\exp\{2(r \phi + \lambda W_r)\}\mathbb{E}\int_{0}^{r}\exp\{2(-\phi s-\lambda W_s))\}ds$ 
We can split the expectation into two by independence. Now, since $W_r \sim \mathcal{N}(0, r)$, the exponential is log-normally distributed:
$ \mathbb{E}\exp \{2(r \phi + \lambda W_r)\} = \exp \{ 2r\phi + \frac{4\lambda^2}{2} \} = \exp \{2cr\}$.
The expectation operator is just an integral with respect to the $\mathbb{P}$-measure, so we can exchange the integrals by Fubini's theorem (under regularity assumptions): 
$\mathbb{E}\int_{0}^{r}\exp\{2(-\phi s-\lambda W_s)\}ds = \int_{0}^{r}\mathbb{E}\exp\{2(-\phi s-\lambda W_s)\}ds$,
and again using the distributional properties of the log-normal distribution, we get
$\int_{0}^{r}\mathbb{E}\exp\{2(-\phi s-\lambda W_s))\}ds = \int_0^r \exp \{-2(\phi s+ \lambda^2 s) \}ds = \int_0^r \exp \{-2(\phi + \lambda^2)s\} ds = \frac{1-\exp\{-2cr\}}{2c}$.
Multiplying both of these together, we get that the variance of $K_{\phi, \lambda}$ is
$\frac{\exp\{2cr\}-1}{2c}$
which is the desired result. 
