1
$\begingroup$

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)?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .