# Density Function of $T = (\bar{x})^2$ where $\bar{x}$ $\sim$ $N(\mu, 1/n)$ [duplicate]

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?

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