Variance of F-Distribution

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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:

\(\ds \expect {\paren {\dfrac {Y / n} {Z / m} }^2}\) \(=\) \(\ds \int_0^\infty \int_0^\infty \frac {y^2 / n^2} {z^2 / m^2} \map {f_{Y, Z} } {y, z} \rd y \rd z\)
\(\ds \) \(=\) \(\ds \frac {m^2} {n^2} \int_0^\infty \int_0^\infty \frac {y^2} {z^2} \map {f_Y} y \map {f_Z} z \rd y \rd z\)
\(\ds \) \(=\) \(\ds \frac {m^2} {n^2} \paren {\int_0^\infty \frac {\map {f_Z} z} {z^2} \rd z} \paren {\int_0^\infty y^2 \map {f_Y} y \rd y}\) rewriting
\(\ds \) \(=\) \(\ds \frac {m^2} {n^2} \paren {\frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z} \paren {\frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2 + 1} e^{-y / 2} \rd z}\) Definition of Chi-Squared Distribution

Note that the integral:

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

converges if and only if:

$\dfrac m 2 - 3 > -1$

That is:

$m > 4$

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

\(\ds \frac 1 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty z^{m / 2 - 3} e^{-z / 2} \rd z\) \(=\) \(\ds \frac 2 {2^{m / 2} \map \Gamma {\frac m 2} } \int_0^\infty \paren {2 u}^{m / 2 - 3} e^{-u} \rd u\) substituting $z = 2 u$
\(\ds \) \(=\) \(\ds \frac {2^{m / 2 - 3} } {2^{m / 2 - 1} \map \Gamma {\frac m 2} } \int_0^\infty u^{m / 2 - 3} e^{-u} \rd u\)
\(\ds \) \(=\) \(\ds \frac 1 4 \times \frac {\map \Gamma {\frac m 2 - 2} } {\map \Gamma {\frac m 2} }\) Definition of Gamma Function
\(\ds \) \(=\) \(\ds \frac 1 4 \times \frac {\map \Gamma {\frac m 2 - 2} } {\paren {\frac m 2 - 1} \paren {\frac m 2 - 2} \map \Gamma {\frac m 2 - 2} }\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds \frac 1 {\paren {m - 2} \paren {m - 4} }\)

Note that the integral:

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

converges if and only if:

$\dfrac n 2 + 1 > -1$

That is:

$n > -4$

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

With that, we have:

\(\ds \frac 1 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty y^{n / 2 + 1} e^{-y / 2} \rd z\) \(=\) \(\ds \frac 2 {2^{n / 2} \map \Gamma {\frac n 2} } \int_0^\infty \paren {2 v}^{n / 2 + 1} e^{-v} \rd v\) substituting $y = 2 v$
\(\ds \) \(=\) \(\ds \frac {2^{n / 2 + 1} } {2^{n / 2 - 1} \map \Gamma {\frac n 2} } \int_0^\infty v^{n / 2 + 1} e^{-v} \rd v\)
\(\ds \) \(=\) \(\ds 4 \times \frac {\map \Gamma {\frac n 2 + 2} } {\map \Gamma {\frac n 2} }\) Definition of Gamma Function
\(\ds \) \(=\) \(\ds 4 \times \frac n 2 \paren {\frac n 2 + 1} \frac {\map \Gamma {\frac n 2} } {\map \Gamma {\frac n 2} }\) Gamma Difference Equation
\(\ds \) \(=\) \(\ds n \paren {n + 2}\)

We therefore have:

\(\ds \expect {X^2}\) \(=\) \(\ds \frac {m^2 n \paren {n + 2} } {n^2 \paren {m - 2} \paren {m - 4} }\)

We therefore have:

\(\ds \var X\) \(=\) \(\ds \expect {X^2} - \paren {\expect X}^2\) Variance as Expectation of Square minus Square of Expectation
\(\ds \) \(=\) \(\ds \frac {m^2 \paren {n + 2} } {n \paren {m - 2} \paren {m - 4} } - \frac {m^2} {\paren {m - 2}^2}\)
\(\ds \) \(=\) \(\ds \frac {m^2 \paren {n + 2} \paren {m - 2} } {n \paren {m - 2}^2 \paren {m - 4} } - \frac {m^2 n \paren {m - 4} } {n \paren {m - 2}^2 \paren {m - 4} }\) aiming to write as a single fraction
\(\ds \) \(=\) \(\ds \frac {m^2 \paren {\paren {n + 2} \paren {m - 2} - n \paren {m - 4} } } {n \paren {m - 2}^2 \paren {m - 4} }\) factoring $m^2$
\(\ds \) \(=\) \(\ds \frac {m^2 \paren {n m + 2 m - 2 n - 4 - n m + 4 n} } {n \paren {m - 2}^2 \paren {m - 4} }\) expanding brackets
\(\ds \) \(=\) \(\ds \frac {m^2 \paren {2 m + 2 n - 4} } {n \paren {m - 2}^2 \paren {m - 4} }\) simplifying
\(\ds \) \(=\) \(\ds \frac {2 m^2 \paren {m + n - 2} } {n \paren {m - 2}^2 \paren {m - 4} }\) factoring $2$

$\blacksquare$

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