Need help understanding relationship between the inverse cdfs of a unit normal and unit normal squared aka $\chi_1^2$

Question

I'm having problems understanding this comment I saw in a lecture.

With a little thought, you can see that because the graph is “folded over”, the $95$th percentile of the $\chi_1^2$ distribution is the square of the $97.5$th percentile of the standard normal distribution.

This seems to be true since

• $F_{\chi_1^2}^{-1}(.95) = 3.841459$

• $(\Phi^{-1}(.975))^2 = (1.959964)^2 = 3.841459$

Initially I thought maybe $(\Phi^{-1}(a))^2 = F_{\chi_1^2}^{-1}(a^2)$ however $.975^2 \neq .95$ exactly and also it doesn't work well for percentiles not close to $1$.

My question: What is the intuition behind the lecture comment? It is intuitive that $F_{\chi_1^2}^{-1}(a) > \Phi^{-1}(a)$ but I'm wondering what other relationship there is that makes the lecture note work. Perhaps there is a well known property. Thanks.

x <- c()
y <- c()

for (i in seq(0,1,.005))
{
  print(c(i, qnorm(i), qchisq(i,1) , sqrt(qchisq(i,1)), pnorm(sqrt(qchisq(i,1)))  ) )
  x <- c(x,i)
  y <- c(y,pnorm(sqrt(qchisq(i,1))))
}

plot(x~y)
abline(lm(x~y))

I ran this in R to try to see the relationship which seems to be close to linear maybe... I don't know.

Answer 1

With $X \sim \mathcal N(0,1)$ so $X^2 \sim \chi^2_1$, for any $a \gt 0$ you have $$\mathbb P(X^2 \gt a^2) = \mathbb P(X \gt a) + \mathbb P(X \lt -a) = 2 \mathbb P(X \gt a)$$ and $$\mathbb P(X^2 \le a^2) = \mathbb P(-a\le X \le a) = 2 \mathbb P(X \le a) -1$$

So if this latter expression is $0.95$, i.e. if $a^2$ is the $95$th percentile of a $\chi_1^2$ distribution,

then $\mathbb P(X \le a) = \frac{0.95 +1 }{2} = 0.975$, i.e. $a$ is the $97.5$th percentile of a standard normal distribution

In R:

pchisq(qnorm(0.975)^2, df=1)
# 0.95

pnorm(sqrt(qchisq(0.95, df=1)))
# 0.975

or with more examples:

pchisq(qnorm(seq(0.5, 1, 0.05))^2, df=1)
# 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
pnorm(sqrt(qchisq(seq(0, 1, 0.1), df=1)))
#0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

demonstrating the $\mathbb P(X^2 \le a^2) = 2 \mathbb P(X \le a) -1$ relationship