Z distribution is symmetric. Chi square distribution is not symmetric. Why?

Question

Z distribution is symmetric. Chi square distribution is not symmetric. Why?

Answer 1

A continuous distribution is symmetric if his density function verifies:

$$ \forall x \in R \quad f(-x) = f(x) $$

The support of the chi-square distribution is $ [0,\infty) $: this distribution can not be symmetric.

We can speak about symmetry about a point, $a \in R$ which is not the origin. In that case, the distribution is symmetric if verifies:

$$ \forall x \in R \quad f(a-x) = f(a+x) $$

Answer 2

There is a very simple reason which, although alluded to in comments and answers, deserves to be brought forward. To expose the key underlying concept, I will answer a generalization.

Let $Z$ be any random variable with unbounded support. Then $Z^2$ cannot have a symmetric distribution.

"Unbounded support" means that for any number $N$, $\Pr(|Z|\gt N) \gt 0:$ that is, no matter how large a possible bound $N$ might be, $Z$ still has some chance of exceeding $N$ in size. For brevity, I will refer to such variables as "unbounded."

Proof. Suppose, on the contrary, that $Z^2$ does have a symmetric distribution. This means (by definition) that there is some number $\mu$ for which $Z^2$ and $2\mu - Z^2$ have the same distribution. But then

$$\Pr(Z^2 \gt 2\mu) = \Pr(2\mu - Z^2\gt 2\mu) = \Pr(Z^2\lt 0) = 0$$

shows $Z^2$ must be bounded. This contradiction implies our initial assumption (that $Z^2$ is symmetric) is false, QED.

The idea of this proof is illustrated in the figures appearing at the end of this post.

Here is an easy consequence:

All unbounded symmetric variables are unbounded both positively and negatively.

To prove this, let $X$ be an unbounded symmetric random variable. Because it is symmetric, there is a number $\mu$ for which $2\mu-X$ has the same distribution as $X.$ If $X$ had a lower bound $M$ for which $\Pr(X \lt M) = 0,$ then $2\mu-M$ would be an upper bound of $X$ because by symmetry

$$\Pr(X \gt 2\mu-M) = \Pr(2\mu-X \lt M) = 0.$$

In the same way, any upper bound of $X$ would give a symmetric lower bound. Thus, $X$ has neither an upper nor a lower bound, QED.

---

Answer to the question:

All chi-squared variables are unbounded because when $X$ has a $\chi^2(\nu)$ distribution, the chance that $X$ exceeds any number $N$ is

$$\Pr(X \gt N) \propto \int_N^\infty t^{\nu/2-1}\,e^{-t/2}\,\mathrm{d}t = \int_N^\infty e^{(\nu/2-1)\log(t)\ -\ t/2}\,\mathrm{d}t \gt 0$$

because on the interval $(\max(N,0), \infty)$ the integrand (an exponential function of finite arguments) is strictly positive. Because (by definition) $X$ is bounded below by $0,$ $X$ cannot have a symmetric distribution, QED.

---

Interesting followup.

Nevertheless, we know that when $X$ has a $\chi^2(\nu)$ distribution, the limiting distribution of $Z = (X - \nu)/\sqrt{2\nu}$ as $\nu$ grows large is standard Normal, which is symmetric. Thus, for large $\nu,$ $X = \sqrt{2\nu}Z + \nu$ is close to symmetric.

The dotted red line is the density of $2\nu - X.$ Because it is very nearly the same as that of $X,$ we see that $X$ is nearly symmetric. However, it cannot be perfectly symmetric because to the left of $0$ $X$ has no probability even though it has positive probability arbitrarily far to the right. We can see this by plotting the log density: