According to Wikipedia, $Y\sim F_{1,v}$ has an F-distribution if $Y=X^2$ and $X\sim t(v)$ has a Student's t distribution. The proof of this makes sense to me. But I don't understand the relationship between their quantiles. Why, for instance, is it that $$F_{0.80;\ 1,30}=(t_{0.90;\ 30})^2=1.31^2=1.72$$ Specifically, the difference between $0.8$ for the $F$ and $0.9$ for the $t$ distribution is what I don't understand.
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1$\begingroup$ Why would you expect it to change? $\endgroup$– RepmatJul 8, 2015 at 21:00
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$\begingroup$ Edited the question for specificity. Why is it .8 for F and .9 for t? $\endgroup$– HatshepsutJul 9, 2015 at 0:23
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1$\begingroup$ When you square the $t$, 10% from the lower tail and 10% from the upper tail will both move above the square of its 90th percentile, while the 80% in between will remain below that, making the square of the 90th percentile of the $t$ the 80th percentile of the distribution of the square. $\endgroup$– Glen_bJul 9, 2015 at 1:58
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1 Answer
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It's because quantiles are only preserved under monotone transformations, and the square function fails to be monotone when we're dealing with positive and negative numbers (a $t$ random variable can be both).
If we look at the $0.8$ quantile of the $F$ distribution $q_{0.8}$ then we know $80\%$ of the probability mass lies between this point and zero. But that means $80\%$ of the probability mass of the corresponding $t$ distribution lies between $-\sqrt{q_{0.8}}$ and $\sqrt{q_{0.8}}$, and so $\sqrt{q_{0.8}}$ is not the $0.8$ quantile of the $t$ distribution. This value corresponds to a larger quantile since we are not including the probability below $-\sqrt{q_{0.8}}$. Because the $t$ distribution is symmetric about zero the extra probability we would be adding is $(1 - 0.8) / 2 = 0.1$, which explains the $0.9$.
