# How do you get hyperbola by squaring a bell?

This is a Gaussian bell (aka normal distribution).

Its square, I belive looks the same. Yet, I see that chi-square distribution, which is a sum of k such bell squares, looks like

Take a look at yellow chi-squared, with k=1. It should be just single bell squared. But, it looks like a hyperbola instead of Gaussian. Why?

PS I read that normally-distributed signal generates a uniformly distributed power. Is it the same kind of magic?

• This is a psychological question: "Why is my intuition contradicted by mathematical facts?" It's better to do the algebra first and then adjust your graphical intuition accordingly. May 26 '13 at 15:15
• Here is a simple algebra. Let's normal distribution is 1,2,1 -- a pretty nice Gaussian curve. Its square is 1,4,1 -- exactly the same Gaussian curve. Ok?
– Val
May 26 '13 at 15:17
• The normal is already very concentrated in a neighbourhood of $0$. Squaring magnifies that concentration a great deal. May 26 '13 at 15:17
• @Val - the square of the standard normal distribution (mean = 0, standard deviation = 1) has a $\chi^2_1$ distribution. If you square an arbitrary Gaussian distribution, you'll get something more complicated. May 26 '13 at 15:22
• @Val. Read what I wrote. "Gaussian" is the same as "normal". But "Standard normal" is a special case. Did's correct comment has nothing to do with this and is possibly the clue to your confusion. May 26 '13 at 15:28

Consider a random variable $X$ with density $f$, say $X$ gaussian, and $Y=X^2$. Then $Y\geqslant0$ almost surely and, for every $y\geqslant0$, $[Y\leqslant y]=[-\sqrt{y}\leqslant X\leqslant\sqrt{y}]$ hence $$P[Y\leqslant y]=\int_{-\sqrt{y}}^{\sqrt{y}}f(x)\mathrm dx.$$ Differentiating this yields the density of $Y$, namely, for every $y\geqslant0$, $$g(y)=\frac{f(\sqrt{y})+f(-\sqrt{y})}{2\sqrt{y}}.$$ When $X$ is normal (centered or not), $f$ is continuous at $0$ hence the behaviour of $g$ at $0$ is $$g(y)\sim\frac{f(0)}{\sqrt{y}}.$$ Note finally that $f^2$ is rarely a density function.

Edit: Recall that an integral depending on a parameter, say $$J(y)=\int_{u(y)}^{v(y)}w(x)\mathrm dx,$$ has derivative $$J'(y)=w(v(y))v'(y)-w(u(y))u'(y).$$

• I am not ready to consume this. But, thanks anyway. I see that you based your proof on the standard normal whereas enemy in comments suggested that discrepancy of X^2 from from normal distribution stems from non-standard Gaussians.
– Val
May 26 '13 at 16:19
• The centering or decentering of gaussians has little to do with the $c/\sqrt{y}$ behaviour, when $y\to0$, that my answer exhibits.
– Did
May 26 '13 at 16:21
• Thanks confirming that. But, unfortunately, I am too stupid to comprehend your answer. The $\sqrt y$ appears out of blue. I look at it like illusion, when a thing materializes out of nothing in the hands of magician. I guess that if we had power of 1 instead of two, i.e. y=x^1, you'd get g(y) = f(x)/y.
– Val
May 26 '13 at 17:42
• Re the appearance of $\sqrt{y}$, see my Edit. To sum up, $y\mapsto\pm\sqrt{y}$ is the inverse of $x\mapsto x^2$ one started with and the final $c/\sqrt{y}$ is more or less its derivative.
– Did
May 26 '13 at 17:48
• @Val - I am upvoting your last comment because there is some truth in it. May 26 '13 at 20:15

Let $Z$ be standard normal, or more generally a normal with mean $0$. But in fact we will use almost no properties of the normal.

Let $Y=Z^2$. We want to show that the density function $f_Y(y)$ of $Y$ is very large for $y$ near $0$. This density function is approximately $\frac{F_Y(y)}{y}$.

We have $F_Y(y)=2\Pr(Z\le \sqrt{y})$. For $y$ close to $0$, the density function of $Z$ is close to a non-zero constant $c$.

Thus $F_Y(y)$ is approximately $2c\sqrt{y}$. It follows that near $0$, $\frac{F_Y(y)}{y}$ is very large.