# Square of Random Variable with t-Distribution has F-Distribution

## Theorem

Let $k$ be a strictly positive integer.

Let $X \sim t_k$ where $t_k$ is the $t$-distribution with $k$ degrees of freedom.

Then:

$X^2 \sim F_{1, k}$

where $F_{1, k}$ is the $F$-distribution with $\tuple {1, k}$ degrees of freedom.

## Proof

Let $Y \sim F_{1, k}$.

We aim to show that:

$\map \Pr {Y < x^2} = \map \Pr {|X| < x}$

for all $x \ge 0$.

That is:

$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$

for all $x \ge 0$.

We have:

 $\ds \map \Pr {Y < x^2}$ $=$ $\ds \int_0^{x^2} \frac {k^{k/2} 1^{1/2} u^{\paren {1/2} - 1} } {\paren {k + u}^{\paren {1 + k}/2} \map \Beta {1/2, k/2} } \rd u$ Definition of F-Distribution $\ds$ $=$ $\ds \frac {k^{k/2} } {k^{k/2} \sqrt k \map \Beta {1/2, k/2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u$ extracting a factor of $k^{-\paren {1 + k}/2}$, rewriting $\ds$ $=$ $\ds \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt k \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u$ Definition of Beta Function $\ds$ $=$ $\ds \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u$ Gamma Function of One Half $\ds$ $=$ $\ds \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x \frac {2 v} {\sqrt {v^2} } \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v$ substituting $u = v^2$ $\ds$ $=$ $\ds \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x 2 \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v$ $\ds$ $=$ $\ds \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-x}^x \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v$ Definite Integral of Even Function $\ds$ $=$ $\ds \map \Pr {-x < X < x}$ Definition of Student's t-Distribution

$\blacksquare$