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The F-Distribution has a Probability Density Function that can be defined as:

F-Distribution PDF

with an expected value of:

F-Distribution Expected Value

What would be a logical explanation for why the expected value only depends on the degrees of freedom associated with the Chi-square random variable in the denominator, and not from the numerator (I .e., depends on 'm' and not on both 'm' and 'n')?

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Hint:

if $X\sim \Gamma (a,\lambda)$ so $E(X^{k})=\frac{\Gamma(a+k)}{\Gamma(a)}\lambda^k$

so $X\sim \chi^{(2)}_{(n)}=\Gamma (\frac{n}{2},2)$ so $E(X^{k})=\frac{\Gamma(\frac{n}{2}+k)}{\Gamma(\frac{n}{2})}2^k$

$$F_{(n-1,m-1)}=\frac{\frac{\chi^{(2)}_{(n-1)}}{n-1}}{\frac{\chi^{(2)}_{(m-1)}}{m-1}}$$

$$E(F_{(n-1,m-1)})=E\left( \frac{\frac{\chi^{(2)}_{(n-1)}}{n-1}}{\frac{\chi^{(2)}_{(m-1)}}{m-1}} \right)=E\left( \frac{\chi^{(2)}_{(n-1)}}{n-1} \right)E\left( \frac{1}{\frac{\chi^{(2)}_{(m-1)}}{m-1}} \right)$$

note that $$E\left( \frac{\chi^{(2)}_{(n-1)}}{n-1} \right)=1$$

so

$$E(F_{(n-1,m-1)})=E\left( \frac{1}{\frac{\chi^{(2)}_{(m-1)}}{m-1}} \right) =(m-1)E\left\{\left( \chi^{(2)}_{(m-1)}\right)^{-1}\right\}$$

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  • $\begingroup$ This is just what I was looking for, thank you! $\endgroup$ – Ryan Rothman Feb 18 at 22:11
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    $\begingroup$ You are welcome $\endgroup$ – Masoud Feb 18 at 22:24

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