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"Suppose X and Y are independent random variables, each exponentially distributed with parameter $\lambda$. Determine the probability density function for $Z=\frac{X}{Y}$."

Here is what I have so far: I recognise $f_X(x)=\lambda e^{-\lambda x}$, $f_Y(y)=\lambda e^{-\lambda y}$, $f_{XY}(x,y)=\lambda e^{-\lambda {(x+y)}}$. In addition to $Z=\frac{X}{Y}$, it was suggested I use a second equation V=X+Y to solve for X and Y in terms of V and Z. I know the algebra simplifies to $X=\frac{ZV}{Z+1}$ and $Y=\frac{V}{Z+1}$ (thankfully I remembered algebraic manipulation from high school physics). I believe I am now trying to find $f_{ZY}(z,y)=f_{XY}(x,y)|J[x(z,v),y(z,v)]|$, where |J[x(z,v),y(z,v)]| is the Jacobian.

I've determined the Jacobian to be $\frac{v}{(z+1)^2}$. Which makes $f_{ZY}(z,y)=\lambda e^{-\lambda {(x+y)}}\frac{v}{(z+1)^2}=\frac{v\lambda^2e^{-\lambda v}}{(z+1)^2}$ when the appropriate X and Y substitutions are made. I'm informed by the answer key this simplifies to $\frac{1}{(z+1)^2}\lambda^2e^{-\lambda v}$=$f_Z(z)f_V(v)$ where $f_Z(z)=\frac{1}{(z+1)^2}$.

I disagree, saying the probability density function should be $f_Z(z)=\frac{v}{(z+1)^2}$ based on the work I provided. Can someone please clarify this for me? Is it a misprint where $f_V(v)=v\lambda^2e^{-\lambda v}$ instead?

  • It is a misprint. Please mention support of the distributions while writing densities, without which they make no sense. Joint density of $(Z,V)$ is $f_{Z,V}(z,v)=\frac1{(1+z)^2}1_{z>0}\cdot\lambda^2 ve^{-\lambda v}1_{v>0}$, which implies that pdf of $Z$ is indeed $f_Z(z)=\frac1{(1+z)^2}1_{z>0}$. You can see this also from here. – StubbornAtom May 16 '20 at 07:18
  • Thank you for your comment. Do you have resources where I can read more about support of a function? I'm taking this stochastic modeling independent study before grad prob and stat I and II due to program time constraints. I have a few statistics tests through which I've been working (mainly Wackerly's Mathematical Statistics). –  May 16 '20 at 16:23
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    By support, I mean the domain of the density function. So for example, one should write $f_Z(z)=\frac1{(1+z)^2}$ for $z\ge 0$ and $f_Z(z)=0$ if $z<0$. – StubbornAtom May 16 '20 at 18:08
  • Thank you for the clarification. –  May 17 '20 at 02:42
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    If your problem is solved, you can post an answer and accept it so that this no longer remains in the unanswered queue. – StubbornAtom May 18 '20 at 06:34
  • Thanks, I did not realise I can do that. Doing it now. –  May 18 '20 at 16:42

1 Answers1

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If the Jacobian to be $\frac{v}{(z+1)^2}$, $f_{ZY}(z,y)=\lambda e^{-\lambda {(x+y)}}\frac{v}{(z+1)^2}=\frac{v\lambda^2e^{-\lambda v}}{(z+1)^2}$ when the appropriate X and Y substitutions are made. This simplifies to $\frac{1}{(z+1)^2}v\lambda^2e^{-\lambda v}$=$f_Z(z)f_V(v)$ where $f_Z(z)=\frac{1}{(z+1)^2}$.

It is verified there was a misprint where $f_V(v)=v\lambda^2e^{-\lambda v}$ should be this instead. Hence, $f_Z(z)=\frac{1}{(z+1)^2}$.