0
$\begingroup$

Question: Let X and Y be two continuous random variables with joint probability density function

$$f(x,y)=\begin{cases}\frac{1}{2} & \text{if} \ \lvert x \rvert + \lvert y \rvert \le 1 & \\ 0 :\ \ \ \ \ \ \ \ \ \text{otherwise}\end{cases}$$

Find the joint probability density function of $U=X+Y$ and $V=X-Y$

My attempt: I know the general method with these types of questions is very similar however I'm not sure how to calculate any of it in this case.

$\endgroup$
  • $\begingroup$ Did you draw a picture to understand the (U,V)-domain? $\endgroup$ – Did Apr 12 '16 at 15:13
  • $\begingroup$ Yes I did, but I didn't help me to deal with this question $\endgroup$ – THISISIT453 Apr 12 '16 at 15:20
  • $\begingroup$ Which (U,V)-domain did you get? $\endgroup$ – Did Apr 12 '16 at 15:44
  • $\begingroup$ I don't understand what you mean by $(U,V)$ domain. I meant that I drew out the domain of $f(x,y)$ and got a rhombus. How would this relate to U and V? $\endgroup$ – THISISIT453 Apr 12 '16 at 15:46
  • $\begingroup$ Maybe a square? What kind of transformation is $u = x + y,\, v = x - y$? $\endgroup$ – BruceET Apr 13 '16 at 6:45
1
+50
$\begingroup$

Here is a somewhat related transformation. Does it give you any ideas?

Suppose the joint distribution of $U$ and $V$ is uniform on the square with vertices at $(0,0)$ and $(1,1).$ Then consider the joint distribution of $X = (U + V)/2$ and $Y = (U-V)/2.$

Here are plots of 20,000 such points:

  u = runif(20000);  v = runif(20000)
  x = (u + v)/2;  y = (u - v)/2
  par(mfrow=c(1,2), pty = "s")
  plot(u,v, pch="."); plot(x, y, pch=".")
  par(mfrow=c(1,1), pty = "m")

enter image description here


Addendum: Because of continuing interest in this Question, I am appending a demonstration of the exact transformation described (rather than a 'related' transformation as a hint.)

The joint distribution of $X$ and $Y$ has the density function $f_{X,Y}(x,y) = 1/2,$ for $x$ and $y$ in the square with vertices at $(1, 0),\, (0, 1),\, (-1, 0),$ and $(0, -1),$ and $0$ elsewhere. The program below uses an acceptance-rejection method to generate 20,000 points in the support of $(X, Y)$. The distribution is illustrated by these points in the left-hand panel of the figure below.

Then it uses the transformation $U = X + Y,\,$ $V = X - Y$ of these points to illustrate the jointly uniform distribution of $U$ and $V$ on the square with diagonal vertices at $(-1, -1)$ and $(1,1).$ The target joint PDF is $f_{U,V}(u,v) = 1/4,$ for $(x,y)$ in that square, and $0$ elsewhere. This distribution is illustrated in the right-hand panel of the figure below.

The transformation can be described as a reflection about the $x$-axis followed by a 45 degree clockwise rotation (so far orthonormal), and then a scaling to double the area of the support of $(U,V).$ A point and its image under transformation are plotted in the same color to help show the nature of the transformation. (Red components of the RGB color specification are proportional to $x$ and blue components proportional to $y$; the 'background' green component is constant.)

I guess, all that remains for the requested 'canonical' derivation of the PDF of $(U,V)$ is to show the Jacobian of the transformation and algebra to verify the support of $(U,V).$

enter image description here

The R code is shown below, in case it is of interest.

 x.cand = runif(50000, -1, 1); y.cand = runif(50000, -1, 1)
 accept = (abs(x.cand) + abs(y.cand) < 1)
 mean(accept)    
 ## 0.4978          # about half accepted 
 x = x.cand[accept];  y = y.cand[accept]
 x = x[1:20000]; y = y[1:20000]   # keep 20,000
 rr = round((x+1)*255/2)  # redscale magnitudes to color accepted points
 bb = round((y+1)*255/2)  # bluescale magnitudes, scaled btw 0 and 255

 par(mfrow=c(1,2), pty="s")  # two panels per figure, both square
 plot(x,y, pch=".", col=rgb(rr,255/2,bb,max=255))   # plot of X and Y
    abline(h=0, col="green4"); abline(v=0, col="green4")

 u = x + y;  v = x - y       # transformation
 plot(u,v, pch=".", , col=rgb(rr,255/2,bb,max=255)) # plot of U and V
     abline(h=0, col="green4"); abline(v=0, col="green4")
 par(mfrow=c(1,1), pty="m")  # return to default plotting
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.