Kolmogorov forward equations and PDEs Let $$u(t, x) = \mathbb{E}{(f(x e^{W_{t}^{2}- \frac{1}{2}t}))}$$
whence $f$ is a twice continuously differentiable function such all its derivatives up to order $2$ are bounded. 
I would like to find a partial differential equation such that $u := u(t, x)$ is the solution for the boundary value problem $$u(0, x) = f(x)$$. 
For example, one can consider a simplier case: let $$u(t, x) = \mathbb{E}{(f(x+W_{t}))}$$
Since the $W_{t} \sim N(0, t)$, then
$$ u(t, x) = \mathbb{E}{(f(x+W_{t}))} = \int_{\mathbb{R}}{f(x+y) \frac{1}{\sqrt{2 \pi}t} e^{-\frac{y^{2}}{2t}}dy}$$
Here we notice that the latter integral is the solution of the boundary value problem for the heat equation $$u_{t} = u_{xx}$$ with $$u(0, x) = f(x)$$.
My questions are: 
(1) In order to proceed in the similar fashion one should understand, how does the distribution of $$X_{t} = e^{W_{t}^{2} - \frac{1}{2}t}$$ look like for a fixed $t$. Are there any simple ways to derive the latter? (i'm familiar with the brownian bridge, but that does not seem to be helpful though)
(2) One can use the Fokker-Plank-Kolmogorov equation that, endowed with the Ito formula, gives the result, but how one can derive the answer without applying a sort of 'overkilling' techniques?
 A: So, finally, seems as if i have figured out the right way to answer my question.
Let $X_{t} = e^{W_{t}^{2}-\frac{1}{2}t}$
(1)  What can we say about the distribution of $X_{t}$?
Lemma  If $$X \sim N(\mu, \sigma^{2})$$ then $$aX+b \sim N(a\mu+v, a^{2} \sigma^{2})$$
(there are many ways to prove it, for example, using characteristic functions)
Since $W_{T} \sim N(0, T)$, then, applying the lemma, we obtain $$W_{T} - \frac{1}{2}T \sim N(-\frac{1}{2}T, T)$$
Finally, $e^{W_{T} - \frac{1}{2}T}$ is log-normally distributed with $\mu = -\frac{1}{2}T$ and $\sigma^{2} = T$
(2) What can we say about $u(x, t)$?
The "high-tech" approach
Lemma
Let $X_{t}$ be the solution of the SPDE
$$dX_{t} = \mu(X_{t}, t) dt + \sigma(X_{t}, t) dW_{t}$$
$$X_{0} = X$$
Then the function $u(t, x) = \mathbb{E}(f(X_{t}))$ is the solution of the following PDE
$$u_{t} = \mu(x, t) u_{x} + \frac{1}{2} \sigma^{2}(x, t) u_{xx}$$
$$u(x, 0) = f(x)$$
(this result based on a derivation of so-called Kolmogorov forward equation)
Thus, by Ito formula
$$dX_{t} = -\frac{1}{2}xe^{W_{t} - \frac{1}{2}t} dt + x e^{W_{t} - \frac{1}{2}t} dW_{t} + \frac{1}{2} x e^{W_{t}-\frac{1}{2}t} (dW_{t})^{2} = x e^{W_{t} - \frac{1}{2}t} dW_{t}$$
The latter means that $\mu = 0$ and $\sigma = x$, thus, $u := u(t, x)$ is the solution of the following PDE
$$ u_{t} = \frac{1}{2}x^{2}u_{xx}$$
$$ u(0, x) = f(x)$$
The low-tech approach
Note that $$u := u(x, t) = \mathbb{E}(f(X_{t})) = \frac{1}{\sqrt{2 \pi t}} \int_{\mathbb{R}}{f(x e^{y-\frac{1}{2}t}) e^{-\frac{y^{2}}{4t^{2}}}dy}$$
Under some mild conditions (we should be able to differentiate w.r.t the parameter under the integral sign) one can check that the function $u(x, t)$ indeed satisfy to the equation above.
