I am trying to understand, intuitively, the shape of the chi-square distribution with 1 degree of freedom.
Let $X$ be a random variable whose distribution is given by the standard normal distribution. Let the possible values of $X$ be the set of reals $\mathbb{R}$.
Now, the distribution of the random variable $Z=X^2$ is given by the chi-square distribution $\chi^2$ with 1 degree of freedom, df=1. It is the same as a gamma distribution parameterised with the values ($1/2$, $1/2$).
This distribution happens to be asymptotic as $Z$ approaches 0.
From the formula of the gamma distribution with ($\alpha = 1/2, \beta = 1/2$), I can see why it is asymptotic as $Z$ approaches zero. The equivalence between the $\chi^2$ and gamma($1/2, 1/2$) distributions I'm not too concerned about.
However, what I'm having trouble understanding intuitively and processing visually is how to map the distribution of a normally distributed random variable $X$ onto the y-axis asymptotic distribution we see for the distribution of $X^2$.
What this seems to be saying is that, is spite of the one-to-one correspondence between $X$ and Z = $X^2$, as $Z \rightarrow 0$, the frequency of each smaller value increases at a very, very large rate. Why?
I presume it wouldn't be asymptotic as $Z \rightarrow 0$ for a transformation such as $Z= |X|$?
Best Answer
Yes, there is a vertical asymptote in the PDF for the $\chi^2$ distribution at $0$. (One wouldn't normally call this "asymptotic", but I can see how you might use that word.)
The $\chi^2$ distribution with $k$ degrees of freedom is the sum of the squares of $k$ independent samples from a standard normal distribution (mean = 0, s.d. = 1). Let's first talk about the "easy" part of the change of variable that goes from the PDF of the normal distribution to the PDF of the square of a single sample. If a normal sample lies in $[-1,1]$ (which happens with probability about 68%), its square lies in $[0,1]$. If the normal sample lies in $[-2,-1] \cup [1,2]$ (which happens with probability about 27%), its square lies in $[1,4]$, so if we were to graph these two PDFs, the area under the normal PDF over the two intervals with total length $2$ is spread over an interval with total length $3$ under the $\chi^2$ distribution. If the normal sample lies in $[-3,-2] \cup [2,3]$ (about 4.2% probability), its square lies in $[4,9]$, so the area is stretched horizontally by about 2.5-times and we expect the height to scale down by about 2.5-times (nonuniformly). So, for the "tails" the probability is stretched over larger intervals and we expect to see the values of the PDF of the $\chi^2$ distribution reduced (when we compare the PDF for a value in the normal distribution with its corresponding value in the $\chi^2$ distribution).
Now what's happening near zero? 68% of the probability mass is over $[-1,1]$ in the normal distribution, so 68% of the probability mass is over $[0,1]$ in the $\chi^2$ distribution. Now about 38% starts over the interval $[-1/2,1/2]$ and the squares of these values are in the interval $[0,1/4]$. So more than half the probability mass is sent to one-fourth the interval -- we expect the $\chi^2$ PDF to be double the corresponding normal PDF. But, about 19% starts in the interval $[-1/4,1/4]$, which squares to the interval $[0,1/16]$, so about half the probability mass from $[-1/2,1/2]$ is sent to an interval only one-fourth as big, so we expect to double the PDF of the $\chi^2$ distribution again over that interval.
We can continue this coarse argument. Since the PDF of the normal distribution is stationary at 0 (meaning, it has a derivative of zero), the "about half as much probability mass" in the above becomes more and more accurate. And each time we halve the starting interval about half the mass is sent to one quarter of the previous interval and about half to three-quarters of the previous interval. That is, as we zoom in on small intervals around zero, they are sent to much smaller intervals around zero, forcing the PDF of the $\chi^2$ distribution to diverge to infinity at zero.
We could be more rigorous, but you are seeking an intuitive understanding. I don't believe increasing the rigour is intuitively helpful.