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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?

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  • $\begingroup$ The Forward Kolmogorov is a differential equation and to solve for it you need an initial condition. The transitional density by itself is not defined at $ t=0 $; hence, how can you evolve the density if you do not have a starting point. This is why DIrac Delta is used which is a distribution infinitely tall with zero width, and the area under it is equal to 1. $\endgroup$
    – Edv Beq
    Nov 21, 2015 at 22:40

2 Answers 2

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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.

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  • $\begingroup$ Thank you. The link between FPE and Chapman-Kolmogorov eq. was a bit fuzzy to me, but I guess now is clear. $\endgroup$
    – KAT
    Apr 28, 2013 at 12:34
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...what is the role of $\delta(x-x_0)$ in this story?

None, and this part of your post is rather confusing. Instead, use the identity $$ p_t(x)=\int p(x,t;y,s)p_s(y)\mathrm dy, $$ for every $x$ and any $s\lt t$.

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