If $Y$ has a t-distribution, what is the distribution of $Y^2$?

Question

If $Y$ has a t-distribution with $n$ degrees of freedom, what is the distribution of that squared? I know that it should have an F-distribution $F(1,n)$, but how can I show that? I know the probability density function of Y, so do I just square the pdf of $Y$ and compare that pdf to the pdf of a distribution that follows $F(1,n)$? (If that is the case would I do something similar when $Y \sim F(m,n)$ and I want to find the distribution of $\frac{1}{Y}$?)

Answer 1

You certainly would not square the density function. Note that if $f$ is a probability density function then $$ \int_{-\infty}^\infty f(x)\,dx = 1, \text{ so } \int_{-\infty}^\infty (f(x))^2 \, dx \text{ typically} \ne 1, $$ and thus $f^2$ is not a probability density function.

If $Y \sim F(m,n)$ then $\dfrac 1 Y \sim F(n,m).$ One quick way to see that is that the distribution of $Y$ is that of $$ \frac{\chi^2_m/m}{\chi^2_n/n} \tag 1 $$ where the two chi-square random variables are independent of each other, so taking the reciprocal only has the effect of interchanging $m$ and $n.$

Now suppose $Y\sim t_n.$ Then the distribution of $Y$ is that of $$ \frac{Z}{\sqrt{\chi^2_n/n}} \tag 2 $$ where $Z\sim N(0,1)$ and the numerator and denominator are indpendent of each other. Now square the expression in $(2)$ and recall that $Z^2 \sim \chi^2_1$ and apply line $(1)$ above and you've got that the square of line $(2)$ is distributed as $F(1,n).$

If you want to derive the density of the $F(1,n)$ distribution from that of the $t_n$ distribution, then let $f$ be the density of the $t_1$ distribution and let $g$ be the density of the $F(1,n)$ distribution. Then \begin{align} g(x) & = \frac d {dx} \Pr( F(1,n) \le x) = \frac d {dx} \Pr( -\sqrt x \le T \le \sqrt x) = \frac d {dx} \int_{-\sqrt x}^{\sqrt x} f(u) \,du \\[10pt] & = 2 \frac d {dx} \int_0^{\sqrt x} f(u) \, du = 2 f(\sqrt x) \cdot \frac d {dx} \sqrt x = \cdots. \end{align}