Evaluate the PDF (Forward Kolmogorov Equation)
Solve the forward kolmogorov equation
\[ \frac{\partial p(y, t)}{\partial t} = c^2\frac{\partial^2p(y, t)}{\partial y^2} \]
for the probability density function \( p(y, t) \).
We can use similarity reduction method to solve this equation. Let us say the solution for the problem is of the form
\[ p(y, t) = t^af(\frac{y}{t^b}) \]
where we need to evaluate the constants \( a \& b \) and the function \( f \) using the boundary conditions. In this case the properties of probability density functions act as boundary conditions.
Setting \( \xi = \frac{y}{t^b} \), we have
\[ p(\xi, t) = t^af(\xi) \]
\[ \frac{\partial \xi}{\partial t} = -ybt^{-b-1} = \frac{-b\xi}{t} \& \frac{\partial \xi}{\partial y} = t^{-b} \]
\( \frac{\partial p}{\partial t} = at^{a-1}f(\xi) + t^{a}\frac{\partial f}{\partial t} = at^{a-1}f(\xi) + t^{a}\frac{df}{d\xi}\frac{\partial \xi}{\partial t} = at^{a-1}f(\xi) - b\xi t^{a-1}\frac{df}{d\xi} \)
\( \frac{\partial p}{\partial y} = t^{a}\frac{\partial f}{\partial y} = t^{a}\frac{dy}{d\xi}\frac{\partial \xi}{\partial y} = t^{a-b}\frac{df}{d\xi} \)
\( \frac{\partial^2p}{\partial y^2} = \frac{\partial}{\partial y}(\frac{\partial p}{\partial y}) = \frac{\partial }{\partial y}(t^{a-b}\frac{df}{d\xi}) = \frac{\partial }{\partial \xi}(t^{a-b}\frac{df}{d\xi})\frac{\partial \xi}{\partial y} = t^{a-2b}\frac{d^2f}{d\xi^2} \)
The change of variable helped us to convert the partial differentials into ordinary differentials in the equation.
Substituting in the equation:
\( \frac{\partial p}{\partial t} = c^2\frac{\partial^2p}{\partial y^2} \)
\( \implies at^{a-1}f(\xi) - b\xi t^{a-1}\frac{df}{d\xi} = c^2t^{a-2b}\frac{d^2f}{d\xi^2} \)
\( \implies af(\xi) - b\xi\frac{df}{d\xi} = c^2t^{-2b+1}\frac{d^2f}{d\xi^2} \)
The LHS above is a function of only \( \xi \) and the RHS is a function of both \( t \) and \( \xi \). This equality between these two expressions implies that the RHS should also be independent of \( t \) making
\[ t^{-2b+1} = 1 \implies b = \frac{1}{2} \]
Our equation and the expression for probability now reduce to
\[ af(\xi) - \frac{1}{2}\xi\frac{df}{d\xi} = c^2\frac{d^2f}{d\xi^2} \]
\[ p = t^{a}f(\xi) = t^{a}f(\frac{y}{\sqrt{t}}) \]
\( \int_{-\infty}^{\infty}p(y, t)dy = 1 \implies t^a\int_{-\infty}^{\infty}f(\frac{y}{\sqrt{t}})dy = 1 \)
\( y = x\sqrt{t} \implies dy = \sqrt{t}dx \)
\( t^{a+\frac{1}{2}}\int_{-\infty}^{\infty}f(x)dx = 1 \)
This change of variable helped make the expression inside the integral independent of \( t \). For a given function \( f(x) \), the LHS above has to evaluate to 1 for all \( t \). We can only have that when the LHS is independent of \( t \)
\[ t^{a+\frac{1}{2}} = 1 \implies a = -\frac{1}{2} \]
Our equation and expression of probability now reduce to
\[ -\frac{1}{2}f(\xi) - \frac{1}{2}\xi\frac{df}{d\xi} = c^2\frac{d^2f}{d\xi^2} \implies -\frac{1}{2}\frac{d(\xi f(\xi))}{d\xi} = c^2\frac{d^2f}{d\xi^2} \]
\[ p = t^{-\frac{1}{2}}f(\xi) = t^{-\frac{1}{2}}f(\frac{y}{\sqrt{t}}) \]
\( -\frac{1}{2}\frac{d(\xi f(\xi))}{d\xi} = c^2\frac{d^2f}{d\xi^2} \)
This can be integrated directly to get
\( -\frac{1}{2}\xi f(\xi) = c^2 \frac{df(\xi)}{d\xi} \)
The equation now becomes a variable separable equation and we can write it as
\( \frac{df(\xi)}{f(\xi)} = -\frac{1}{2c^2}\xi d\xi \)
\( \implies d(ln(f(\xi))) = -\frac{1}{4c^2}d(\xi^2) \)
\( \implies ln(f(\xi)) = -\frac{\xi^2}{4c^2} + K \)
\( \implies f(\xi) = Ae^{-\frac{\xi^2}{4c^2}} \)
\( p = \frac{A}{\sqrt{t}}e^{-\frac{\xi^2}{4c^2}} = \frac{A}{\sqrt{t}}e^{-\frac{y^2}{4c^2t}} \)
\( \int_{-\infty}^{\infty}p(y, t) dy = 1 \implies \frac{A}{\sqrt{t}}\int_{-\infty}^{\infty}e^{-\frac{y^2}{4c^2t}}dy = 1 \)
\( x = \frac{y}{2c\sqrt{t}} \implies dy = 2c\sqrt{t}dx \)
\( \implies \frac{A}{\sqrt{t}}2c\sqrt{t}\int_{-\infty}^{\infty}e^{-x^2}dx = 1 \implies 2cA\sqrt{\pi} = 1 \implies A = \frac{1}{2c\sqrt{\pi}} \)
So our probability density function \( p \) is
\[ p(y, t) = \frac{1}{2c\sqrt{\pi t}}e^{-\frac{y^2}{4c^2t}} \]
