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$X$ and $Y$ are iid r.v.s where $f(x)=\lambda e^{-\lambda x}$, $x>0$. We are given $Z=X/Y$ and asked to find the CDF and density for $Z$.

I tried doing this using the multivariate transformation formula, where

$f_{W_1,W_2}(w_1,w_2)=f_{X_1,X_2}(x_1,x_2)|J|$, but I ended up getting that $f_Z(z)=\frac{1}{(1+z)^2}$, which results in a negative CDF. I was hoping to see a different method of doing this. Any help appreciated.

Edit: Here is my work for obtaining my answer:

$Z=X/Y$; we then let $W=Y$ and $X=ZW$. The Jacobian is then calculated to be $w$. That means we have:

\begin{align} f_Z(z) & =\int_0^\infty f_X(x)f_Y(y)w\,dw \\[10pt] & =\int_0^\infty \lambda e^{-\lambda x}\lambda e^{-\lambda y}w \, dw \\[10pt] & =\lambda^2 \int_0^\infty e^{-(z+1)\lambda w}w\,dw \\[10pt] & =\frac 1 {(z+1)^2} \end{align}

Michael Hardy
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Taylor
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  • For future reference, you should post all your work to be proofread. We can't guess what your errors were. – Em. Feb 03 '16 at 02:28
  • stats.stackexchange.com anyone? – JimLohse Feb 03 '16 at 02:44
  • I didn't know stats.stackexchange existed! Thanks for pointing that out; I'll use it in the future. And I have posted my work to be proofread, as recommended. Thanks for your input. – Taylor Feb 03 '16 at 03:25

2 Answers2

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Anyway, like I said, you should post your work to get proofread. This reduces redundancies and helps you get an answer faster.

The pdf seems fine. As for the cdf, \begin{align*} F_Z(z)&=\int_0^z \frac{1}{(1+t)^2}\,dt\\ &=-\left.\frac{1}{1+t}\right|_0^z\\ &=-\left[\frac{1}{1+z}-1\right]\\ &=\frac{1+z-1}{1-z}\\ &=\frac{z}{1+z} \end{align*} for $z\geq 0$.

I don't see where it is "negative".

Em.
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Since $X$ and $Y$ are independent, you can say that $1/Y$ and $X$ are independent as well. Now $W=1/Y$, then $f(w)=\lambda e^{-\lambda/w} (1/w^2), w>0$

$$Z=X/Y = XW$$ $$f(Z=z)= \int_{x>0}f(X=x)f(W=z/x) \frac{1}{|x|} dx$$ from product distribution formula.
which gives $$f(Z=z)= \int_{x>0}\lambda^2 e^{-\lambda (x+x/z)}(x/z^2) dx$$

Using formula for gamma function gives $f(Z=z)=\frac{1}{(1+z)^2} $

AnandJ
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