Density function of $ Z = \frac{ \ln(X+1)}{\ln[(X+1)(Y+1)]}$.

Question

I’ve got X, Y, which are independent random variables with density function given as: $$ f(x) = \frac{4}{(1+x)^{5}}, x >0$$ And 0 everywhete else. I’m looking for density function of $$ Z = \frac{ \ln(X+1)}{\ln[(X+1)(Y+1)]}.$$ It looks similar to random variable $C=\frac{X}{X+Y}$. I know how to find C( we will have a straight line through $(X,Y)$ plane, passing through $(0,0)$, then we calculate proper double integral. I will be realy grateful for any hint

Answer 1

One way to do this is the usual transformation involving jacobians. These are couple of steps for you to follow:

• Transform $X\to U$ such that $U=\ln (X+1)$, so that $U$ would turn out to be an exponential variable with mean $1/4$. Same thing happens with $V=\ln (Y+1)$. As $(X,Y)$ are i.i.d, so are $(U,V)$.

• Now change variables $(U,V)\to(Z,W)$ where $Z=\frac{U}{U+V}$ and $W=U+V$. From the joint density of $(Z,W)$ it would be easy to see that $Z$ and $W$ are also independent with $Z\sim\mathcal U(0,1)$, the uniform distribution over $(0,1)$.

Clearly, the $Z$ as defined by me is the same as your $Z$.