Variance of Student's t-Distribution/Proof 2

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 the variance of $X$ is given by:

$\var X = \dfrac k {k - 2}$

for $k > 2$, and does not exist otherwise.

Proof

From the definition of the Student's t-Distribution, $X$ has probability density function:

$\map {f_X} x = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {1 + \dfrac {x^2} k}^{-\frac {k + 1} 2}$

with $k$ degrees of freedom for some $k \in \R_{> 0}$.

From Variance as Expectation of Square minus Square of Expectation:

$\ds \var X = \int_{-\infty}^\infty x^2 \, \map {f_X} x \rd x - \paren {\expect X}^2$

So:

$\ds \var X = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-\infty}^\infty \dfrac {x^2} {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x - \mu^2$

Recall from Expectation of Student's t-Distribution that $k \gt 1$, so for $k \gt 1$:

\(\ds \var X\)

\(=\)

\(\ds \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-\infty}^\infty \dfrac {x^2} {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x\)

From Expectation of Student's t-Distribution, $\mu = 0$

\(\ds \)

\(=\)

\(\ds \dfrac {2 \map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {\int_0^\infty \dfrac {x^2} {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x }\)

Definite Integral of Even Function/Corollary

\(\ds \)

\(=\)

\(\ds \dfrac {2 \map \Gamma {\frac {k + 1} 2} } {\sqrt {k } \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\int_0^\infty \dfrac {x^2} {\paren {1 + \dfrac {x^2} k}^{\frac {k + 1} 2} } \rd x }\)

Gamma Function of One Half

\(\ds \)

\(=\)

\(\ds \dfrac {2 \map \Gamma {\frac {k + 1} 2} } {\sqrt {k } \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\int_0^\infty \paren {\dfrac x {\paren {1 + \dfrac {x^2} k}^{\frac 1 2} } } \paren {\dfrac 1 {\paren {1 + \dfrac {x^2} k}^{\frac {k - 4} 2} } } \paren {\dfrac x {\paren {1 + \dfrac {x^2} k}^2 } } \rd x }\)

Rewriting

Let:

$u = \dfrac {\dfrac {x^2} k} {\paren {1 + \dfrac {x^2} k} }$

Then by Quotient Rule for Derivatives, we have:

$\dfrac k 2 \rd u = \dfrac {x \rd x} {\paren {1 + \dfrac {x^2} k}^2 }$

And:

$\paren {1 - u } = \dfrac 1 {\paren {1 + \dfrac {x^2} k} }$

We can re-write $u$ as:

$u = \dfrac 1 {\paren {1 + \dfrac k {x^2}} }$

And in this form, we can see that as $x \to 0$, $u \to 0$ and as $x \to \infty$, $u \to 1$

Plugging these results back into our integral above, we have:

\(\ds \)

\(=\)

\(\ds \dfrac {2 \map \Gamma {\frac {k + 1} 2} } {\sqrt {k } \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\int_0^1 \paren {ku}^{\frac 1 2} \paren {1 - u}^{\frac {k - 4} 2} \paren {\dfrac k 2 \rd u } }\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 \map \Gamma {\frac {k + 1} 2} \frac {k \sqrt k} 2} {\sqrt {k } \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\int_0^1 u^{\frac 1 2} \paren {1 - u}^{\frac {k - 4} 2} \rd u }\)

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma {\frac {k + 1} 2} k} {\map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\int_0^1 u^{\frac 3 2 - 1} \paren {1 - u}^{\frac {k - 2} 2 - 1} \rd u }\)

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma {\frac {k + 1} 2} k} {\map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\map \Beta {\dfrac 3 2, \dfrac {k - 2} 2} }\)

Definition 1 of Beta Function

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma {\frac {k + 1} 2} k} {\map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\dfrac {\map \Gamma {\frac 3 2} \map \Gamma {\frac {k - 2} 2} } {\map \Gamma {\frac {k + 1 } 2} } }\)

Definition 3 of Beta Function

\(\ds \)

\(=\)

\(\ds \dfrac {\map \Gamma {\frac {k + 1} 2} k} {\map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \paren {\dfrac {\map \Gamma {\frac 1 2} \frac 1 2 \map \Gamma {\frac k 2} } {\map \Gamma {\frac {k + 1 } 2} \frac {k - 2} 2 } }\)

Definition of Gamma Function $\map \Gamma {\frac {k - 2} 2} = \dfrac {\map \Gamma {\frac k 2} } {\frac {k - 2} 2} $

\(\ds \)

\(=\)

\(\ds \dfrac k {k - 2}\)

Note: from the Definition of Beta Function, the Beta function is only defined for $\map \Re x, \map \Re y > 0$

Therefore:

$\dfrac {k - 2} 2 \gt 0 \leadsto k \gt 2$

$\blacksquare$