How to determine the PDF of the chi distribution using the Jacobian formula?

Question

The Question is:

Assume $X$ has the $\chi^2$ distribution ("chi-squared") with $n$ degrees of freedom. What is the distribution of $X^{1/2}$?

I know that I can use the Jacobian formula since $\sqrt{\cdot}: (0,\infty) \rightarrow (0,\infty)$ is bijective and thus invertible and so on... Such that it satisfies The conditions of the Theorem. Then I get the probability density function $$f_{X^{1/2}}(u) = K u^{n-1} e^{-u^2/2}$$ with $K \in \mathbb{R}$, but only for $u \in (0,\infty)$. How do I then determine the probability density function in the Interval $(-\infty,0]$ ? Intuitively I would say that it is zero there, since the probability density function $f_X$ of $X$ is $0$ since $X$ is $\chi^2$ distributed. But how do I know for sure? How can I show this?

Thanks a lot in advance!

Answer 1

If $X \sim \mathsf{Chisq}(df=n)$ then its PDF is $0$ on $(-\infty,0).$ The square root transformation maps $(0,\infty)$ to itself. (Notice that if $X$ had negative values, you couldn't use the square root transformation for those negative values.) So $Y = X^{1/2}$ also has support $(0,\infty)$ and its PDF is also $0$ on $(-\infty,0).$

Addendum. Here is a demo for 5 df using R statistical software: Generate 100,000 values of $X,$ square them to get $Y$'s, make a histogram of the $Y$'s, plot the PDF of $Y$ through the histogram to see if it fits. (With the correct $K$ for the PDF of $Y,$ it does.)

m = 10^5;  x = rchisq(m, 5); y = sqrt(x)
hist(y, prob=T, br=30, col="skyblue2", ylim=c(0,.6))
k = 1/(2^(5/2-1)*gamma(5/2))
yy = seq(0, 6, by=.001); pdfy = k*yy^4*exp(-yy^2/2)
lines(yy, pdfy, type="l", lwd=2, col="red")