I need confirmation on the following problem: Take a SDE of the form: \begin{equation} dX_t=a(X_t,t)dt+b(X_t,t)dW_t \end{equation}
where all the conditions, such that the solution $X_t$ is defined and is a diffusion, are satisfied.
The process $X_t$ has a distribution at each time $t$ with a density $p_t(x)$. This tells me about the probability that the process (which can be viewed at $t$ as the random variable $X$) takes different values $x$.
Imagine I am at some time $s<t$ and I want to compute $p_t(x)$. I can use Fokker-Plank equation to find the transition density $p(x,t;y,s)$. Now, by multiplying the transition probability density by the density function of the process $p_s(y)$ at time $s$ I obtain my density at time $t$. \begin{equation} p_t(x) =p(x,t;y,s)p_s(y) \end{equation}
If $s$ is the initial time, that is $s=t_0$ I can say that $p_s(y)$ is the Dirac delta function $\delta(x-x_0)$ and I can solve the FPE with the initial condition $\delta(x-x_0)$ to get $p(x,t;x_0,t_0)$.
My question is: Is the density $p_t(x)$ at time $t$ equal to the transition density $p(x,t;x_0,t_0)$? If so, what is the role of $\delta(x-x_0)$ in this story?
I agree with the previous answer that $p_t(x)$ is actually
$$ p_t(x)=\int p(x,t;y,s)p_s(y)\mathrm dy. $$
So if you substitute
$$ p_s(y) = \delta(y-x_0) $$
where $s=t_0$, then the delta function does the integration so as you suggest,
$$ p_t(x)=\int p(x,t;y,t_0)\delta(y-x_0)\mathrm dy = p(x,t;x_0,t_0). $$
The 'role' of the delta function is to do the integration.
Incidentally, here's a tip that I find useful. If I ever get confused about this kind of stuff, I just revert to the simplest possible case, which in this case would be Brownian motion, i.e.
$$ dX_t=dW_t $$
Doing that, all the distributions become Gaussian etc which is always easy to understand.