Forward Kolmogorov Equation and Normal Distribution I read through my textbook and find one interesting statement. The normal pdf function $\frac{1}{\sqrt[]{2\pi\sigma^2}\ t}\exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$ satisfies the forward Kolmogorov equation. I find the forward Kolmogorov equation and try to prove it. But I'm confused about how to deal with the derivatives, especially the $\frac{\partial P}{\partial t}$ since I never see the t in other normal pdf. Could someone give me some ideas about how to go forward? Thank you very much.
$$\frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial x^2} - \mu\frac{\partial P}{\partial x}=\frac{\partial P}{\partial t}$$
 A: Ok, the problem isn't hard if you know what those partials mean, it's just a huge mess haha  To take the partial with respect to $t$ (or any variable) you just treat the other variables like constants.  For example, if $P = xt^2$ then $\partial P/\partial t = 2xt$.  So to compute the partial $\partial P / \partial t$ first use the product rule and take the derivative of the left summand:
$$\partial P / \partial t = \frac{\partial}{\partial t}\left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right) =$$
$$= \left( \frac{-1}{t^2 \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right)$$
Now to take the derivative on the right, we use the chain rule for $(e^{f(t)})' = f'(t)e^{f(t)}$:
$$\frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right) = \frac{\partial}{\partial t} \left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right) =$$
$$ = - \left( \frac{-4\mu \sigma^2 t(x-\mu t) - 2\sigma^2(x-\mu t)^2}{4 \sigma^4 t^2}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right) =$$
$$ = \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t^2}\cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right)$$
Where the first equality is just the well-known quotient rule for derivatives, and the second equality is just distributing the minus sign and reducing.  I'd double-check these calculations if I were you, I just woke up.  Putting these together and writing $A = \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$, we get the atrocious:
$$\partial P/\partial t = \left( \frac{-1}{t^2 \cdot \sqrt{2\pi\sigma^2}}\right) \cdot A + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t^2} \cdot A =$$
$$ =\frac{A}{t^2 \cdot \sqrt{2 \pi \sigma^2}} \cdot \left( \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t} - 1\right) =$$
$$= \frac{A}{t^2 \cdot \sqrt{2 \pi \sigma^2}} \cdot \left( \frac{x^2 - \mu^2t^2}{2 \sigma^2 t} - 1\right)$$
Now you need to check that the functional equation is satisfied, taking the partials wrt $x$.  Hopefully it just falls right out - which it basically always does when something like that is true.  But without some heavy theorems I see no way to prove this without doing the monster calculations.  The partial wrt $x$ is a lot simpler, thankfully.
