First find $E(U), E(\frac{1}{V}), E(U^2),E(\frac{1}{V^2})$.

When I consider finding $E(U)$ I feel as though integrating over the pdf of the F distribution multiplied by $u$ will leave me with a spare $u$. Is there a better strategy to consider? I know that the F distribution is made up of two independent Chi-Square distributions, perhaps I should calculate the expected values separately?


$\newcommand{\E}{\operatorname{E}}$ $$ F = \frac{U/r_1}{V/r_2} $$ where $U\sim\chi^2_{r_1}$ and $V\sim\chi^2_{r_2}$ and $U,V$ are independent. \begin{align} & \E(F) = \overbrace{\E\left( \frac{U/r_1}{V/r_2} \right) = \E(U/r_1)\E\left( \frac 1{V/r_2} \right)}^{\text{because of independence}} \\[8pt] = {} & \frac {r_2}{r_1} \E(U)\E\left(\frac1V\right) = \frac{r_2}{r_1} r_1 \E\left(\frac 1 V\right) = r_2\E\left(\frac 1 V\right) \end{align} Here I have assumed you know that $\E(\chi^2_{r_1})=r_1$. To find $\E\left(\frac 1 V\right)$, evaluate $$ \int_0^\infty \frac 1 v f(v)~dv $$ where $f$ is the $\chi^2_{r_2}$ density, i.e. $$ \frac 1 {\Gamma(r_2/2)} \int_0^\infty \frac 1 v \left(\frac v 2\right)^{(r_2/2)-1} e^{-v/2}\frac{dv}2. $$ Letting $w=v/2$, this becomes \begin{align} & \frac 1 {\Gamma(r_2/2)} \int_0^\infty \frac 1 {2w} w^{(r_2/2)-1} e^{-w}~dw \\[8pt] = {} & \frac 1 {2\Gamma(r_2/2)} \int_0^\infty w^{(r_2/2)-2} e^{-w}~dw \\[8pt] = {} & \frac{\Gamma((r_2/2)-1)}{2\Gamma(r_2/2)} = \frac{1}{2\left( \frac{r_2}2-1 \right)} = \frac{1}{r_2-2}. \end{align}

In a similar way, one finds $\E(F^2)= \E\left(\frac{U^2}{V^2}\right)$. Finally, $\operatorname{var}(F) = \E(F^2)-(\E(F))^2$.

  • $\begingroup$ This is very helpful thank you! $\endgroup$
    – Chris
    Jan 1 '15 at 14:38
  • $\begingroup$ Neat move to set $w=\frac{v}{2}$, how did you think of it? Are there other moves you would consider to evaluate the integral? $\endgroup$
    – Chris
    Jan 1 '15 at 14:45
  • $\begingroup$ Because we had to given $e^{-\frac{v}{2}}$. Nice. $\endgroup$
    – Chris
    Jan 1 '15 at 16:03
  • $\begingroup$ @Chris : I think of the Gamma distribution as $\dfrac1{\Gamma(\alpha)} \left(\frac x \beta\right)^{\alpha-1} e^{-x/\beta}\dfrac{dx}\beta$ for $x>0$, putting one of the $\beta$s with $dx$ like that. I've never seen anyone else write it quite that way, but it's the natural thing to do with a scale parameter. Or alternatively $\dfrac1{\Gamma(\alpha)}(\lambda x)^{\alpha-1} e^{-\lambda x}\Big(\lambda\,dx\Big)$ for $x>0$. ${}\qquad{}$ $\endgroup$ Jan 1 '15 at 19:13
  • $\begingroup$ I believe this same form is used by Joe Blitzstein of Stats 110 at Harvard: projects.iq.harvard.edu/stat110/youtube $\endgroup$
    – Chris
    Jan 2 '15 at 22:17

In answer to your doubt:

When I consider finding E(U) I feel as though integrating over the pdf of the F distribution multiplied by u will leave me with a spare u.

Well here's the 'direct' way which I tried:

If $U\sim F\left(r_1,r_2\right)$, then, $$\begin{align} E(U)=&\int_0^\infty \frac{u\left(\frac{r_1}{r_2}\right)^{r_1/2}u^{\left(r_1/2\right)-1}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)\left(1+\frac{r_1}{r_2}u\right)^{\left(r_1+r_2\right)/2}}du\\ =&\frac{\left(\frac{r_1}{r_2}\right)^{r_1/2}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{u^{\left(r_1/2\right)}}{\left(1+\frac{r_1}{r_2}u\right)^{\left(r_1+r_2\right)/2}}du\\ \end{align}$$

Now, substitute $\frac{r_1}{r_2}u=t$. Then, we're getting:

$$\begin{align} E(U)=&\frac{\left(\frac{r_1}{r_2}\right)^{r_1/2}}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{\left(\frac{r_2}{r_1}t\right)^{r_1/2}}{\left(1+t\right)^{\left(r_1+r_2\right)/2}}\left(\frac{r_2}{r_1}\right)dt\\ =&\frac{\left(\frac{r_2}{r_1}\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \frac{t^{r_1/2}}{\left(1+t\right)^{\left(r_1+r_2\right)/2}}dt\\ =&\frac{\left(\frac{r_2}{r_1}\right)B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\int_0^\infty \underbrace{\frac{t^{\left(r_1/2\right)+1-1}}{B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)\left(1+t\right)^{\overline{\left(r_1/2\right)+1}+\overline{\left(r_2/2\right)-1}}}}_{\text{pdf of a Beta distribution}}dt\\ =&\frac{\left(\frac{r_2}{r_1}\right)B\left(\frac{r_1}{2}+1,\frac{r_2}{2}-1\right)}{B\left(\frac{r_1}{2},\frac{r_2}{2}\right)}\\\end{align}$$

Now, if I use the relationship of the Beta function with the Gamma function, we'll end up getting something like this:

$$\begin{align} E(U)=&\left(\frac{r_2}{r_1}\right)\frac{\Gamma\left(\frac{r_1}{2}+1\right)\Gamma\left(\frac{r_2}{2}-1\right)}{\Gamma\left(\frac{r_1+r_2}{2}\right)}\frac{\Gamma\left(\frac{r_1+r_2}{2}\right)}{\Gamma\left(\frac{r_1}{2}\right)\Gamma\left(\frac{r_2}{2}\right)}\\ =&\left(\frac{r_2}{r_1}\right)\frac{\Gamma\left(\frac{r_1}{2}+1\right)\Gamma\left(\frac{r_2}{2}-1\right)}{\Gamma\left(\frac{r_1}{2}\right)\Gamma\left(\frac{r_2}{2}\right)}\\ =&\left(\frac{r_2}{r_1}\right)\left(\frac{r_1}{2}\right)\frac{1}{\frac{r_2}{2}-1}\\ =&\frac{r_2}{r_2-2}\\ \end{align}$$

This gives us the expectation of $U$ if $U\sim F\left(r_1,r_2\right)$.

Similarly, you can obtain $E\left(U^2\right)$ and ultimately find the $Var(U)=E\left(U^2\right)-(E(U))^2$.

I would still recommend Michael Hardy's method as that is far more efficient than this one.

P.S. Pardon me for any formatting errors that might've crept while writing the answer.


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