When f is a probability density function

Question

How to determine $C$ if $f$ is a probability density function? $$f(x,t)=\dfrac {1}{C\sqrt{t}}e^{-\dfrac{{x}^2}{4t}}$$

Should I integrate the integral $$\int_{-\infty}^{\infty}f(x,t) \ \ dx$$

Answer 1

This answer is not really structural but can be enough if you have allready some knowledge about PDF-s.

Do you know that $$\frac{1}{\sigma\sqrt{2\pi}}e^{-\dfrac{x^{2}}{2\sigma^{2}}}$$ is a PDF for $\sigma>0$ linked with distribution $Norm(0,\sigma^{2})$?

Compare it with your $f(x,t)=\frac{1}{C\sqrt{t}}e^{-\dfrac{x^{2}}{4t}}$.

By setting $t=\frac{1}{2}\sigma^{2}$ we come to $\frac{2}{C\sigma\sqrt{2}}e^{-\dfrac{x^{2}}{2\sigma^{2}}}$ which can only be a PDF if $\frac{2}{C\sigma\sqrt{2}}=\frac{1}{\sigma\sqrt{2\pi}}$ or equivalently $C=2\sqrt{\pi}$.

Answer 2

By definition, a probability density function integral is $1$. That is:

$$\int_{-\infty}^{+\infty}f(x,t)\text{d}x = 1$$

In your case, the integral depends on the constant $C$.

So, you have to solve the following: $$\int_{-\infty}^{+\infty}f(x,t)\text{d}x = g(C) = 1$$

In your case, $g(C) = 2\frac{\sqrt{\pi}}{C}$, and hence $C = 2\sqrt {\pi}$