# Finding solution to a stochastic PDE

I am going to cite a an Example in Finance textbook:

$$dX_t = \sigma dW_t$$ where $$W_t$$ is a Wiener process. The Fokker-Planc equation is: $$\frac{\partial p (s,y;x,t)}{\partial t}=\frac{1}{2}\sigma^2\frac{\partial^2 p (s,y;x,t)}{\partial x^2} (1)$$ and it is easily checked that the solution is given by the Gaussian density $$p(s,y;x,t) = \frac{1}{\sigma\sqrt{2\pi(t-s)}}\exp \left[-\frac{1}{2} \frac{(x-y)^2}{\sigma^2(t-s)} \right] (2)$$

So the transition $$p$$ must satisfy the Fokker-Planc and this leads to the PDE in (1): This I do understand.

But how can I use the information about $$X_t$$ process to show that (2) is the solution to the PDE given by (1)?

## 1 Answer

So many ways to look at that problem , first one , we can write $$X_t=X_s+\sigma(W_t-W_s)$$ with $$s. Therefore , if $$X_s=y$$, we have

$$X_t=y+\sigma(W_t-W_s)$$ and the distribution of $$X_t$$ is normal centered in $$y$$ and variance $$\sigma^2(t-s)$$ by definition of the Brownian motion, therefore the density function is $$p(s,y;x,t) = \frac{1}{\sigma\sqrt{2\pi(t-s)}}\exp \left[-\frac{1}{2} \frac{(x-y)^2}{\sigma^2(t-s)} \right]$$

The second way is to look at the relationship between Ito-processes and PDE, through the Fokker-Planck, as you showed in your equation (1). You first connected equation(1) to $$X_t$$, now take equation(2) and differentiate it wrt $$t$$, then do it twice wrt $$x$$

$$\frac{\partial p (s,y;x,t)}{\partial t}=-\frac{1}{2\sigma\sqrt{2\pi}}\frac{1}{(t-s)^{\frac{3}{2}}}\exp \left[-\frac{1}{2} \frac{(x-y)^2}{\sigma^2(t-s)} \right] \\+\frac{1}{2\sigma^3\sqrt{2\pi}}\frac{(x-y)^2}{(t-s)^{\frac{3}{2}}}\exp \left[-\frac{1}{2} \frac{(x-y)^2}{\sigma^2(t-s)} \right]$$

$$\frac{\partial p (s,y;x,t)}{\partial x}=-\frac{1}{\sigma\sqrt{2\pi(t-s)}}\frac{(x-y)}{\sigma^2(t-s)}\exp \left[-\frac{1}{2} \frac{(x-y)^2}{\sigma^2(t-s)} \right]$$ $$\frac{\partial^2 p (s,y;x,t)}{\partial x^2}=-\frac{1}{\sigma^3\sqrt{2\pi}}\frac{1}{(t-s)^{\frac{3}{2}}}\exp \left[-\frac{1}{2} \frac{(x-y)^2}{\sigma^2(t-s)} \right]\\+\frac{1}{\sigma\sqrt{2\pi(t-s)}}\frac{(x-y)^2}{\sigma^4(t-s)^2}\exp \left[-\frac{1}{2} \frac{(x-y)^2}{\sigma^2(t-s)} \right]$$

Multiply $$\frac{\partial^2 p (s,y;x,t)}{\partial x^2}$$ by $$\frac{1}{2}\sigma^2$$ and you get the result wanted.
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