If the random variables $X$ and $Y$ are normally distributed with mean $\mu = 0$ and standard deviation $\sigma = 1$, then define the random variable $Z = \min(X, Y )$. The problem is to prove that $Z^2$ follows a gamma distribution with parameters $(\alpha = 1/2, \lambda= 1/2)$ (i.e. a chi-squared distribution).

I have: $$P(Z^2 \le z) = P((\min\{X,Y\})^2 \le z) =P\left(-\sqrt{z}\le \min(X,Y)\le \sqrt{z}\right) \\ = P\left(-\sqrt{z}\le X\le \sqrt{z}, -\sqrt{z}\le Y\le \sqrt{z}\right) $$ But I am not sure where to go from here, since the random variables aren't assumed to be independent. I would appreciate any hints!

  • $\begingroup$ Your logic at the last step seems to go astray. If the minimum of (X,Y) is within $\pm \sqrt z$ (as you have in your second-last line), that doesn't make both of them in that interval, as you have in your last line. You only need one of the two of them in there. $\endgroup$
    – Glen_b
    Nov 28, 2016 at 3:57
  • $\begingroup$ @glen_b What then would be the interval on the other one? $\endgroup$
    – Mjt
    Nov 28, 2016 at 14:08
  • $\begingroup$ That's not a productive way to approach writing the event. But first you might want to ponder whether anything has been left out of your framing of the question (which is possibly why whuber is able to offer a counterexample). $\endgroup$
    – Glen_b
    Nov 28, 2016 at 15:55

2 Answers 2


The result is not true.

As a counterexample, let $(X,Y)$ have standard Normal margins with a Clayton copula, as illustrated at https://stats.stackexchange.com/a/30205. Generating 10,000 independent realizations of this bivariate distribution, as shown in the lower left of the figure, produces 10,000 realizations of $Z^2$ that clearly do not follow a $\Gamma(1/2,1/2)$ distribution (in a Chi-squared test of fit, $\chi^2=121, p \lt 10^{-16}$). The qq plots in the top of the figure confirm that the marginals look standard Normal while the qq plot at the bottom right indicates the upper tail of $Z^2$ is too short.


The result can be proven under the assumption that the distribution of $(X,Y)$ is centrally symmetric: that is, when it is invariant upon simultaneously negating both $X$ and $Y$. This includes all bivariate Normals (with mean $(0,0)$, of course).

The key idea is that for any $z \ge 0$, the event $Z^2 \le z^2$ is the difference of the events $X \ge -z\cup Y \ge -z$ and $X \ge z \cup Y \ge z$. (The first is where the minimum is no less than $-z$, while the second will rule out where the minimum exceeds $z$.) These events in turn can be broken down as follows:

$$\Pr(Z^2 \le z^2) = \Pr(X\ge -z) - \Pr(Y \le -z) + \Pr(X,Y\lt -z) - \Pr(X,Y\gt z).$$

The central symmetry assumption assures the last two probabilities cancel. The first two probabilities are given by the standard Normal CDF $\Phi$, yielding

$$\Pr(Z^2 \le z^2) =1 - 2\Phi(-z).$$

That exhibits $Z$ as a half-normal distribution, whence its square will have the same distribution as the square of a standard Normal, which by definition is a $\chi^2(1)$ distribution.

This demonstration can be reversed to show $Z^2$ has a $\chi^2(1)$ distribution if and only if $\Pr(X,Y\le -z) = \Pr(X,Y\ge z)$ for all $z\ge 0$.

Here is the R code that produced the figures.


n <- 1e4
xy <- qnorm(rCopula(n, claytonCopula(1)))
colnames(xy) <- c("X", "Y")
z2 <- pmin(xy[,1], xy[,2])^2

cutpoints <- c(0:10, Inf)
z2.obs <- table(cut(z2, cutpoints))
z2.exp <- diff(pgamma(cutpoints, 1/2, 1/2))
rbind(Observed=z2.obs, Expected=z2.exp * length(z2))
chisq.test(z2.obs, p=z2.exp)

qqnorm(xy[,1], ylab="X"); abline(c(0,1), col="Red", lwd=2)
qqnorm(xy[,2], ylab="Y"); abline(c(0,1), col="Red", lwd=2)

plot(xy, pch=19, cex=0.75, col="#00000003", main="Data")

qqplot(qgamma(seq(0, 1, length.out=length(z)), 1/2, 1/2), z2,
       xlab="Theoretical Quantiles", ylab="Z2",
       main="Gamma(1/2,1/2) Q-Q Plot")
abline(c(0,1), col="Red", lwd=2)

Let $M=min(X,Y)^2$, $$ P(M<m) = P(M<m,X<Y) + P(M<m,X>Y) ~~~~~~~~~~~\\ = P(M<m|X<Y)P(X<Y) + P(M<m|X>Y)P(X>Y) \\ = P(X^2<m)P(X<Y) + P(Y^2<m)P(X>Y) ~~~~~~~~~~~~~~~~~~~~~~\\ = \frac{1}{2}P(X^2<m) + \frac{1}{2}P(Y^2<m) \quad \quad \quad\quad\quad\quad\quad\quad\quad\quad~ \\ =P(X^2<m) \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ~~~ $$


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