Variance of Student's t-Distribution/Proof 2

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 $k \gt 1$:

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:

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

Therefore:
 * $\dfrac {k - 2} 2 \gt 0 \leadsto k \gt 2$