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Stuck on this problem from my Econometrics homework, would appreciate some help!

Given the equation of the $F$ random variable $F = \frac{(X_1/k_1)}{(X_2/k_2)}$, show that:

$E(F) = E\left(\frac{1}{(X_2/k_2)}\right)$,

Can you conclude that $E(F) = 1$?

http://imgur.com/a/QuQLQ

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    Can you please write the definition of the $F$ random variable from equation B.43? What is $X_1, X_2, k_1, k_2$ etc.? You are at the risk of being down-voted for not supplying all the details and your own ideas/thoughts. – 2017-01-18
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    https://i.imgur.com/qEP3tMz.png – 2017-01-18
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    First use the independence; Second use the result from Chi-Square distribution to arrive the result of first part. For the next part apply Jensen's inequality. – 2017-01-19
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    @BGM: For me anyhow, without elaboration, this Comment does not seem helpful. – 2017-01-19

1 Answers 1

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I assume that $X_1$ and $X_2$ are independently chi-squared with degrees of freedom $k_1$ and $K_2,$ respectively. Then $F$ has Snedecor's F-distribution with $k_1$ and $k_2$ degrees of freedom. Wikipedia on 'F distribution' says that $E(F) = k_2/(k_2 - 2),$ for $k_2 > 2.$ So, "not exactly."

The F-statistic (or "variance ratio" statistic) $F = S_1^2/S_2^2,$ where $S_1^2$ and $S_2^2$ are variances of independent samples from two normal populations, is often used to test whether the population variances are equal ($H_0: \sigma_1^2 = \sigma_2^2$) or not ($H_a: \sigma_1^2 \ne \sigma_2^2$). In that context, one often says, roughly, that $H_0$ is true if $S_1^2/S_2^2 \approx 1.$ But that is different from saying that $E(F) = 1$ under $H_0.$

If you have something else in mind, please edit your Question.

Reality check: A brief simulation in R statistical software with a million realizations of $Q_1 \sim Chisq(10)$ and independently $Q_2 \sim Chisq(8):$

m = 10^6;  k1 = 10;  k2 = 8
q1 = rchisq(m, k1);  q2 = rchisq(m, k2)
f = (q1/k1)/(q2/k2)
mean(f)
## 1.334032  # aprx E(F) = 1.3333
k2/(k2-2)
## 1.333333

mean(1/(q2/k2))
## 1.333556

This illustrates the Wikipedia formula for $E(F)$. Just from the simulation, it does seem that $E[1/(Q_2/k_2)]$ is the same. But, under my interpretation of your question, I do not see a reasonable scenario in which these means would be exactly unity. (Yes, close to 1 for very large $k_2$, but not exactly 1.)