Ratio of chi squared distribution.

#1
[math]U_1[/math] and [math]U_2[/math] are both [math]\chi^2[/math] distribution with 1 and 2 degrees of freedom respectively. Both [math]U_1[/math] and [math]U_2[/math] are independent. Denote [math]\frac{U_1}{U_2}[/math] by [math]X[/math]. Then [math]2X \sim F(1,2)[/math] or [math]X \sim \frac{1}{2}F(1,2)[/math]. Does this mean that pdf and cdf of [math]X[/math] are half of F(1,2)? And is expectation and variance of [math]X[/math] both infinity?
 
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Dragan

Super Moderator
#2
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):

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

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.
 
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