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    Ratio of chi squared distribution.




    U_1 and U_2 are both \chi^2 distribution with 1 and 2 degrees of freedom respectively. Both U_1 and U_2 are independent. Denote \frac{U_1}{U_2} by X. Then 2X \sim F(1,2) or X \sim \frac{1}{2}F(1,2). Does this mean that pdf and cdf of X are half of F(1,2)? And is expectation and variance of X both infinity?
    Last edited by geniusacamel; 11-08-2016 at 11:24 AM.

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    Re: Ratio of chi squared distribution.


    Yes, the pdf would just be one-half the height (without being scaled by one-half) of the pdf for the F-distribution (1,2). It is straight-forward - because the degrees of freedom are known - that the pdf would be (when scaled by 1/2):

    f\left ( x \right )=\frac{1}{4\sqrt{2}\left ( 1+\frac{x}{2} \right )^{\frac{3}{2}}\sqrt{x}}.

    The mean and variance do not exist because you need greater than 2 (and 4) degrees of freedom (in the denominator) for the mean (and variance) to exist. Try integrating the function and you will see that the pdf will not converge to get the mean and variance.
    Last edited by Dragan; 11-09-2016 at 11:00 AM.

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