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Got stuck...need help on the following question

Given: $Y \sim f_Y(y)=\frac{1}{\theta }e^{-\frac{y}{\theta }}$, $y>0$.

$R\sim g(r)$, $r>0$

$Y$ and $R$ are independent random variables.

$Z=YR$

Q: Find the joint density of $(Z,R)$

It is quite weird to me...

for the independency: $$f(Z)=f(Y)g(R)=\frac{1}{\theta }e^{-\frac{y}{\theta }}g(r) =\int_0^\infty f(z,r) \, dr \tag{$*$} $$

How to get

$$f(z,r)=\frac{1}{\theta }e^{-\frac{z}{r\theta }}g(r)\frac{1}{r},\qquad z>0,r>0\tag{$**$}$$

for that the fundamental theorem of Calculus cannot apply on the integration in (*), even if it could, it doesn't look like I could get $(**)$.

Any hint? Thanks for your time.

Michael Hardy
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Liz Sugar
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  • for the independency: f(Z)=f(Y)g(R) $\qquad$ This is incorrect. The density of the product $Z=YR$ is not the product of the densities $f(Y)$ and $g(R)$.

     – Dilip Sarwate Dec 11 '14 at 01:05

1 Answers1

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Hint:

Use the transformation

$$Z = YR, \,\, V = R,$$

where $Y > 0$ and $R > 0$.

The inverse transformation is

$$Y = \frac{Z}{V}, \,\, R = V,$$

with Jacobian

$$J(z,v)=\frac{\partial(Y,R)}{\partial(Z,V)}=\left| \begin{array}{ccc} 1/v & -z/v^2 \\ 0 & 1 \\ \end{array} \right|= \frac{1}{v}.$$

The joint density of $Z$ and $V$ is

$$f_{ZV}(z,v)= f_{YR}[(Y(z,v),R(z,v)]|J(z,v)|.$$

Now use what you know about the joint density $f_{YR}$.

RRL
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