Suppose we have $\bar{x}$ $\sim$ $N(\mu, 1/n)$, its pdf then is $f(\bar{x}) = \sqrt{\frac{n}{2\pi}}e^{-n(\bar{x}-\mu)^2/2}$

Let $T = (\bar{x})^2$, we want to calculate its pdf.

My first thought is:

$P(T \le t) = P(\bar{x}^2 \le t) = P(-\sqrt{t} \le \bar{x} \le \sqrt{t}) = \Phi(\sqrt{t}) - \Phi(-\sqrt{t}) = 2\Phi(\sqrt{t}) -1 $

where $t \ge 0$

Therefore, its pdf is:

$f(t) = \frac{1}{\sqrt{t}}\sqrt{\frac{n}{2\pi}}e^{-n(\sqrt{t}-\mu)^2/2}$

However, if I use calculator, the integral is not equal to 1.

The correct pdf should be:

$f(t) = \frac{1}{2\sqrt{t}}\sqrt{\frac{n}{2\pi}}(e^{-n(\sqrt{t}-\mu)^2/2} + e^{-n(-\sqrt{t}-\mu)^2/2})$

This integral is 1.

Any thoughts how to get the correct pdf?

  • $\begingroup$ It is not $\Phi(\sqrt t)$ but $\Phi\left(\frac{\sqrt t-\mu}{\sqrt{1/n}}\right)$. $\endgroup$ Feb 2 '20 at 6:41
  • 1
    $\begingroup$ Expanding on that, $\Phi$ refers specifically to the CDF of the STANDARD normal. $\endgroup$
    – Dave
    Feb 2 '20 at 6:59