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So the problem is to showthat if $X$ and $Y$ are independent exponential random variables with $λ$ = $1$, then $X/Y$ follows an F distribution. Also, identify the degrees of freedom.

Here is the answer on the book:

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I understand all the steps in the solution expect the degree of freedom. I thought $F$ distribution is $W = \frac{U/m}{V/n}$ with $m$ numerator degrees of freedom and $n$ denominator degrees of freedom. But I don't see any numerator and denominator here in this answer except $1$ even if I plug in $v=x/y$. I appreacite if someone could help!

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3 Answers 3

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You can look up the PDF of the F distribution and see that the PDF derived matches the one for $2,2$ degrees of freedom.

Or, as an alternative to your book's solution, you can use the fact that a chi-squared on two degrees of freedom is the same as an exponential with mean $2$ (to see this, just look up the PDF for chi-squared). Then $2X$ and $2Y$ are independent chi-squareds on two degrees of freedom and $$ \frac{X}{Y} = \frac{\frac{2X}{2}}{\frac{2Y}{2}} \sim F_{2,2}$$ by the formula/definition $$ F_{d_1,d_2} = \frac{U_1/d_1}{U_2/d_2}$$ where $U_1$ and $U_2$ are independent chi-squareds on $d_1$ and $d_2$ degrees of freedom, respectively.

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That comes from the formula for the density of the $F$-distribution.

$$f(x) = \frac{1}{B(d_1/2, d_2/2)}\left(\frac{d_1}{d_2}\right)^{d_1/2}x^{(d_1/2) -1} \left(1 + \frac{d_1}{d_2} x\right)^{-(d_1+d_2)/2}$$

wjere $d_1$ and $d_2$ are the degrees of freedom and $B()$ is the beta function. If you plug in $d_1=d_2=2$ then everything reduces down to $(1+x)^{-2}$ since it happens that $B(1,1)=1$.

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The Answer by @Flounderer (+1) nicely shows how your final PDF is $\mathsf{F}(2,2).$ I will try to explain how the ratio $V = X/Y$ follows your definition of an F-distributed random variable.

$X, Y \stackrel{iid}{\sim} \mathsf{Exp}(\lambda=1).$ Consider $2X \sim \mathsf{Exp}(\lambda=1/2) \equiv \mathsf{Chisq}(df=2).$ Then consider $V = X/Y = 2X/2Y$ as the ratio of two $\mathsf{Chisq}(df=2)$ random variables each divided by its degrees of freedom. (Just noticed that, while I was typing this, with interruptions, @spaceisdarkgreen's essentially equivalent Answer (+1) appeared.)

Below is a plot of the PDF of $\mathsf{Chisq}(2)$ (dashed blue) overlaid on the PDF of $\mathsf{Exp}(1/2)$ (red).

enter image description here

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