Chi square distribution when $n=1$

Question

The PDF of $X \sim \chi^2(n)$ distribution is

$$ P(\chi^2) = \frac{\Gamma(n/2)}{2^{n/2}} \, \exp(-x/2) x^{n/2-1}. $$

In this PDF, for $n=2$ :-

\begin{align*} P(\chi^2) &= \frac{\Gamma(2/2)}{2^{2/2}} \, \exp(-x/2) x^{2/2-1} \\ &= \frac{\Gamma(1)}{2}\, \exp(-x/2) \\ &= \frac{1}{2} \, \exp(-x/2), \end{align*}

which is the PDF of exponential with mean $2$.

And my problem is :-

Normal distribution is a particular case of $\chi^2$-distribution when $n=1$, since for $n=1$,

$$ P(\chi^2) = \frac{\Gamma(1/2)}{2^{1/2}} \, \exp(-x/2) x^{1/2-1} = \frac{1}{\sqrt{2\pi}} \, \exp(-x/2) x^{-1/2}. $$

But extra $1/\sqrt{x}$ term is there in the equation and it is not a standard normal distribution PDF.

How to get it with $n=1$ from chi-square distribution PDF.

If there anything wrong in my calculations then correct me.

Answer 1

The normal distribution is not a special case of the chi-squared distribution. What is true is that if $Z$ is standard normal, then $Z^2$ is chi-squared with $n=1$.

Answer 2

Your calculation is correct and confirms the fact that ChiSquared(m) $\equiv$ Gamma(m/2, 1/2). For this case, ChiSquared(1)=Gamma(1/2, 1/2), which has pdf

$$\frac{\left(\frac 12\right)^{1/2}}{\Gamma(1/2)}e^{-1/2 x}x^{1/2-1}= \frac{1}{\sqrt {2\pi}}e^{-\frac 12 x}x^{-1/2}$$

Note that it is not just the extra $x^{-1/2}$ term but the $x$ in the exponential is not squared as in the normal pdf.