Variance of F-Distribution

Theorem
Let $n, m$ be strictly positive integers.

Let $X \sim F_{n, m}$ where $F_{n, m}$ is the F-distribution with $\tuple {n, m}$ degrees of freedom.

Then the variance of $X$ is given by:


 * $\var X = \dfrac {2 m^2 \paren {m + n - 2} } {n \paren {m - 4} \paren {m - 2}^2}$

for $m > 4$, and does not exist otherwise.

Proof
Since $m > 4 > 2$, we have by Expectation of F-Distribution:


 * $\expect X = \dfrac m {m - 2}$

We now aim to compute $\expect {X^2}$ with a view to apply Variance as Expectation of Square minus Square of Expectation.

Let $Y$ and $Z$ be independent random variables.

Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.

Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.

Then:


 * $\dfrac {Y / n} {Z / m} \sim F_{n, m}$

Therefore:


 * $\expect {X^2} = \expect {\paren {\dfrac {Y / n} {Z / m} }^2}$

Let $f_Y$ and $f_Z$ be the probability density functions of $Y$ and $Z$ respectively.

Let $f_{Y, Z}$ be the joint probability density function of $Y$ and $Z$.

From Condition for Independence from Joint Probability Density Function, we have for each $y, z \in \R_{\ge 0}$:


 * $\map {f_{Y, Z} } {y, z} = \map {f_Y} y \map {f_Z} z$

We therefore have:

Note that the integral:


 * $\ds \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z$

converges :


 * $\dfrac m 2 - 3 > -1$

That is:


 * $m > 4$

With that, we have for $m > 4$:

Note that the integral:


 * $\ds \int_0^\infty y^{n / 2 + 1} e^{-y / 2} \rd z$

converges :


 * $\dfrac n 2 + 1 > -1$

That is:


 * $n > -4$

This is ensured by the fact that $n \in \N$.

With that, we have:

We therefore have:

We therefore have: