Maximum Density in Normal Probability Density where x has been replaced with $\mu+y^{(1/2)}$

Question

Thanks for reading.

Let's say you have the normal probability density:

$P(x|\mu,\sigma)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-(x-\mu)^2}{2\sigma^2}}$

Now, if you set $ y = (x-\mu)^2$, what's the probability density of y, and where is y greatest?

Here's where i have got to:

$x =\mu +/- y^{\frac{1}{2}} $

so plugging back into the equation we get:

$P(y|\mu,\sigma)=p(y|\sigma)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-y}{2\sigma^2}}$

But as far as I can see the derivative of this function is never equal to zero, and so has no maximum density.

I think I've got something wrong but can't figure it out - could I, for example, replace $\sigma^2$ with $\int{y}$?

Would really appreciate some help! Thanks in advance!

Jonny

Answer 1

First, $X$ is $N(\mu,\sigma)$. So $X-\mu$ is $N(0,\sigma)$. By definition of standard normal, $X-\mu=\sigma Z$. Therefore, $(X-\mu)^2=\sigma^2Z^2$. This distribution is Chi-squared distribution of degree $1$ freedom.

Using the the density function of chi-squared distribution, you can find the maximum at 1.