Show that the marginal density function of $f_V(v)$ if $V=X-Y $ is $$f_{V}(v)= \frac{1}{(1+|v|)^2}$$ for $ -\infty < v < \infty $. When the bivariate density function $f_{X,Y}(x,y)$ is given as: $$f_{X,Y}(x,y) = \begin{cases}\frac{2}{(1+x+y)^3} \ &\text{ for } x,y>0\\0\ &\text{ otherwise } \end{cases}.$$


My solution strategy is quite clear as the joint density function for $f_{U,V}(u,v)$ can be obtained from the transformation formula for continuous variables as stated below: $$f_{U,V}(u,v)=f_{X,Y}(x(u,v),y(u,v))\text{ }|\text{ }J \text{ }|$$ where $U$ is an auxiliary variable and $J$ the Jacobian. From this joint distribution the marginal density function $f_{V}(v)$ is derived by integration over the domain of the auxiliary variable $u$. $$f_{V}(v)=\int_{\in u} f_{U,V}(u,v) \text{ }du $$

I am currently struggling with the choice of auxiliary variable $U$ and the integral that follows. So far my attempts include $U=Y$ and $U=X+Y$ with the corresponding density functions: $$f_{U,V}(u,v)=\begin{cases} \frac{2}{(1+v+2y)} \ & u = y\\ \frac{1}{(1+u)^3} \ & u=x+y\end{cases}$$

I am solving this as an exercise in my probability course. Any help is greatly appreciated!

  • 1
    $\begingroup$ Heard something about the convolution fromula? $\endgroup$ – Qwerty Jun 28 '16 at 11:55

$$ f_{X,Y}(x,y) = \begin{cases}\frac{2}{(1+x+y)^3} \ &\text{ for } x,y>0\\0\ &\text{ otherwise } \end{cases}. $$ What you want to compute is $$ f_V(v)=\iint_0^\infty dx dy \frac{2}{(1+x+y)^3} \delta(v-(x-y))\ . $$ Resolving the Dirac delta $$ f_V(v)=\int_0^\infty dy \frac{2}{(1+(v+y)+y)^3}\Theta(v+y)\ , $$ where $\Theta(z)=1$ if $z>0$ and $0$ otherwise.

Therefore $$ f_V(v)=\int_b^\infty dy \frac{2}{(1+(v+y)+y)^3}=\frac{1}{2 (2 b+v+1)^2}\ , $$ where $b=\max(0,-v)$.

So, for $v>0$ $$ f_V(v)=\frac{1}{2 (v+1)^2} $$ and for $v\leq 0$ $$ f_V(v)=\frac{1}{2 (-v+1)^2}\ . $$ We can check normalization $$ \int_{-\infty}^\infty dv f_V(v)=\int_{-\infty}^0 dv\frac{1}{2 (-v+1)^2}+\int_0^\infty dv \frac{1}{2 (v+1)^2}=1\ , $$ as it should. [The pdf you wrote in your question for $V$ cannot be correct, as it is not normalized for $-\infty<v<\infty$]

  • $\begingroup$ Thanks! And nicely spotted with that pdf which is indeed incorrect. $\endgroup$ – Camram Aug 10 '16 at 10:33

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