# Determining PDF and CDF of multivariable function

If $$X$$ and $$Y$$ are independent and identically distributed RVs with pdfs $$f_X(x)=e^{-x}U(x)$$ and $$f_Y(y)=e^{-y}U(y)$$, find the PDF of $$Z$$ if:

a. $$Z=X-Y$$

b. $$Z=\min(X,Y)/\max(X,Y)$$

$$U$$ represents the unit step function.

For the first part, I tried looking at the range of values that $$y$$ could be, and I got $$y\geq x-z$$ since $$x-y \leq z$$ for $$F_Z(z)$$. I set up my CDF integral to be

$$\int_0^\infty\int_0^{z+y}e^{-x-y} dx dy$$

and got $$F_Z(z)= 1-e^{-z}/2$$. I differentiated this to get my PDF, which is $$e^{-z}/2$$. This answer is incorrect, and I'm not sure what I did wrong.

For the second part, I again tried looking at the range of values that $$y$$ could be, and I got $$y \leq zx$$ where $$0 \leq z \leq 1$$. I set up my CDF integral to be $$\int_0^\infty \int_0^{y/z}e^{-x-y} dx dy$$

and got $$F_Z(z)= 1-\frac {z} {z+1}$$. I differentiated this to get $$\frac {-1} {(z+1)^2}$$. This answer is also incorrect. I'm not sure what to do on either of these two problems at this point.

For the first case you should consider when, for any given $$z\in \Bbb R$$, when $$Y > X-z$$. When $$z>0$$ you have the situation depicted below.

Thus, for $$z>0$$, $$\begin{eqnarray} F_Z(z) &=& \int_0^{+\infty}\int_0^{z+y} e^{-x-y} dx dy=\\ &=&1-\frac12 e^{-z}. \end{eqnarray}$$

For $$z<0$$ you can look at the figure below.

Then you can compute, in such interval, e.g. as $$\begin{eqnarray} F_Z(z) &=& \int_{0}^{+\infty}\int_{x-z}^{+\infty} e^{-x-y} dy dx=\\ &=&\frac12 e^{z}. \end{eqnarray}$$

Differentiation yields $$\boxed{f_Z(z) = \frac12 e^{-|z|}}.$$

As for the second question, consider that, if $$X>Y$$ you need to consider the event $$Y; whereas, if $$X, your event becomes $$Y>\frac1{z}X$$.

It is thus immediate to conclude that, for $$z>1$$, $$F_Z(z) = 1.$$

For $$0, consider then the shaded area in the figure below.

Therefore, for $$0, $$\begin{eqnarray} F_Z(z) &=&1- \int_0^{+\infty} \int_{zx}^{\frac{x}{z}} e^{-x-y}dy dx=\\ &=&\frac{2z}{1+z}. \end{eqnarray}$$

And, finally, by differentiation $$\boxed{f_Z(z) = \frac2{(1+z)^2}}, \ \ 0

• Ah, so I was just was missing half of the integral on both steps. These answers make sense. Thanks! Commented Oct 7, 2019 at 12:49

a) Your CDF integral is almost correct.

It should be:$$P\left(X-Y\leq z\right)=\int_{0}^{\infty}\int_{0}^{\infty}\left[x\leq y+z\right]e^{-x-y}dxdy=\int_{0}^{\infty}\int_{0}^{\max\left(0,y+z\right)}e^{-x-y}dxdy$$For calculation you must discern the cases $$z\geq0$$ and $$z<0$$.

I haven't looked at b) but there is a good chance that you made a similar mistake there.

• I understand that I need to discern the two cases, but I'm not sure I understand the new integral bound min(0,y+z). If the min was 0 on the upper bound, wouldn't you just be integrating 0? Commented Oct 7, 2019 at 12:29
• @SD'Anc You are wright. There was a mistake in my answer (sorry). $\min$ must be $\max$. I repaired. Commented Oct 7, 2019 at 13:42