Assume $X_t$ satisfies the SDE:
$$dX_t = X_tdt + dW_t$$, where $W_t$ is standard normal.
If we know that $X_0 = 1$ and there exists some pdf of $X_t$, $p(x,0)$, how can I find the the pdf of $X_t$ at time 1, $p(x,1)?$
I tried to use Fokker–Planck equation $$ \frac{\partial}{\partial t}p(x,t)= -\frac{\partial}{\partial x}[\mu(x,t)p(x,t)]+\frac12\frac{\partial^2}{\partial x^2}[\sigma^2(x,t)p(x,t)] $$, but I am not sure how I can utilize this equation without explicitly knowing $p(x,0)$.
Could you guys give me an advice on this problem? :D
The initial condition is given by $X_0 = 1$; in other words, the stochastic process $X_t$ is not random at the beginning. The distribution associated to a deterministic quantity is the Dirac delta, that is why one has $p(x,0) = \delta(x-1)$.
As stated by KurtG. in the comment section, the Fokker-Planck equation turns out to be some rather heavy machinery for the present purpose. In fact, $X_t$ is an Ornstein-Uhlenbeck process, whose closed form is well-known, namely $X_t = X_0e^t + \int_0^t e^{t-s} \,\mathrm{d}W_s$ $-$ you can apply Itô's lemma on the transformation $Y_t := X_te^{-t}$ in order to rederive this result.
Since $\mathrm{d}W_t \sim \mathcal{N}(0,\mathrm{d}t)$ is gaussian and normal law is stable, i.e. linear combinations of normal laws are also normal, we understand that $X_t$ is gaussian too $-$ because the integral is nothing else than a sum, $X_0$ a constant quantity and the exponentials some scalar prefactors.
Let's determine the mean and variance of $X_t$. Firstly, one has by linearity $$ \mathbb{E}[X_t] = \mathbb{E}\left[X_0e^t + \int_0^t e^{t-s} \,\mathrm{d}W_s\right] = \mathbb{E}[X_0]e^t + \int_0^t e^{t-s} \,\mathbb{E}[\mathrm{d}W_s] = e^t $$ since $\mathbb{E}[X_0] = 1$ and $\mathbb{E}[\mathrm{d}W_t] = 0$. Then, Itô's isometry permits to compute its variance : $$ \begin{array}{rcl} \mathrm{Var}[X_t] &=& \displaystyle \mathbb{E}[(X_t-\mathbb{E}[X_t])^2] \\ &=&\displaystyle \mathbb{E}\left[\left(\int_0^t e^{t-s} \,\mathrm{d}W_s\right)^2\right] \\ &=&\displaystyle \mathbb{E}\left[\int_0^t e^{2(t-s)} \,\mathrm{d}s\right] \\ &=&\displaystyle \int_0^t e^{2(t-s)} \,\mathrm{d}s \\ &=&\displaystyle \left.\frac{e^{2(t-s)}}{-2}\right|_0^t \\ &=&\displaystyle \frac{e^{2t}-1}{2} \end{array} $$ In conclusion, you have $X_t \sim \mathcal{N}\left(e^t,\frac{e^{2t}-1}{2}\right)$, hence $X_1 \sim \mathcal{N}\left(e,\frac{e^2-1}{2}\right)$ and $p(x,1) = \frac{1}{\sqrt{\pi(e^2-1)}} \exp\left(-\frac{(x-e)^2}{e^2-1}\right)$.