Distribution of Chi-Square divided by its degrees of freedom?
I have the following:
$$\frac{2n}{\chi^2_{\nu=2n}}$$
Does this simplify to be a $\chi^2_{\nu=1}$ distribution by any chance? Or is there a rule to get rid of the $2n$? Any help would be appreciated. I have not done too much with transformations of random variables.
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$\begingroup$Dividing a chi-square-distributed random variable by its degrees of freedom is merely rescaling; it doesn't change the shape parameter in the gamma distribution. The expected value does become the same as that of a $\chi^2_1$ distribution, but the shape of the density function is quite different. If you find the probability that that random variable is $<1/2$, you'll get a far bigger number with a $\chi^2_1$ than with $\chi^2_{2n}/(2n)$.
Dividing the degrees of freedom by the chi-square random variable results in a distribution of quite a different shape, not merely a rescaled chi-square distribution.
This comes up when one thinks about the F-distribution (The "F" stands for "Fisher", as in Ronald Aylmer Fisher, one of the most famous 20th-century scientists). The F-distribution is one of the great work-horses of applied statistics. The F-distribution with $\nu$ and $\xi$ degrees of freedom is $F=(\chi^2_\nu/\nu)/(\chi^2_\xi/\xi)$, where the two chi-square random variables are independent. One can show fairly easily that a simple rational function of a random variable with an F-distribution actually has a Beta distribution.
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