# Finding probability density function using CDF method

I'm working on the following problem, but I'm not really sure how to approach it - it's different from anything I've seen before! The problem is as follows: Consider the probability density function

$f_{X,Y}(x,y) = \left\{\frac{8+xy^3}{64}\right\}$ if $-1<x<1, -2<y<2$, with probabilility $0$ otherwise.

What I'm trying to do is find the PDF of $W=2X+Y$, which is causing me some trouble - in fact I hardly know where to start! So I know the support of X is $-1<x<1$ and the support of Y is $-2<y<2$, since the region is a square. I think this means that the support of W is $-3<w<3$, since $W=2X+Y$.

This is where I start to get confused. I believe in order to find the PDF, I first want to find the CDF of W, and then take the derivative of that. In order to find the CDF, I want to evaluate a double integral in terms of X and Y with the given PDF. However, I don't know what to set the bounds of these integrals to! In fact, I'm not really sure how to even begin; I feel like it might involve solving for X and Y in terms of W $(y=2x-w)$ and $(x=\frac{y-w}{2})$ but I don't know exactly what (if anything) to do with these!

Thank you so much for your help - I really appreciate it!

Sarah

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Why go from the PDF of (X,Y) to the PDF of W through the CDF of W (with its morass of cases) instead of proceeding directly from PDF to PDF? Are you asked to proceed thus? – Did Nov 28 '12 at 8:27
Unfortunately, I'm not yet familiar with the techniques used to go from one pdf to another (so I suppose I am, in a sense, asked to proceed this way). – SarahJ Nov 28 '12 at 22:49
These are illogical and unfortunate pedagogical choices. – Did Nov 29 '12 at 6:00

Begin by drawing a sketch of the $x$-$y$ plane and marking on it the region where $f_{X,Y}$ is nonzero. (Hint: it is not a square).

STOP. Do not proceed if you have ignored the previous statement and have not made a sketch. Reading beyond this point is useless.

You are not correct when you say that $W \in (-3,3)$. $W$ can take on values in $(-4,4)$ since $2\times 1 + 2 = 4$, , So you know already that $F_W(w)=0$ for $w-<4$ and $1$ for $W\geq 4$. Now, pick a fixed number $w \in(-4,4)$, and draw the line $2x+y=w$. Then, $W=2X+Y\leq w$ if the point $(X,Y)$ lies on or below this line. Can you integrate to find $P(W\leq w)=F_W(w)$ for your chosen value of $w$? (Hint: you may need to break up the integral into two or more parts to carry out the calculation). Lather, rinse, repeat for different choices of $w$ which will give you different lines and different integrals to compute. You may be able to see a trend developing and ultimately be able to write a complete expression for $F_W(w)$ like $$F_W(w) = \begin{cases}0, & w < -4,\\ \cdots,& -4 < w < ?,\\ \cdots,& ? < w <??\\ \cdots & \cdots\\ 1, & w \geq 4,\end{cases}$$ after just a few tries with different values of $w$.

Then, differentiate to find the pdf $f_W(w)$ and you are done.

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Thank so much! I just started reading your comment and am in the process of drawing the sketch - from what I can tell $f_{x,y}$ takes on a zero value only when x=1 and y=-2 and x=-1 and y=2; am I missing something else here? – SarahJ Nov 28 '12 at 2:48
@SarahJ $f(x,y)=0$ if $|x|\geq 1$ or $|y|\geq 2$. That's the meaning of "with probability $0$ otherwise" (Does your homework really say it the way you state it, or did you try to put your own interpretation on what the problem actually says?) – Dilip Sarwate Nov 28 '12 at 3:03
My apologies, I think I was unclear in my question - I was simply looking at the region bounded by -1<x<1 and -2<y<2; am I wrong in believing it comes out to be a rectangle with width 2 and height 4? (I believe we can ignore the two points on the upper right and lower left edges where $f_[x,y]$=0? – SarahJ Nov 28 '12 at 3:09
$f(x,y)$ is nonzero on the interior of the rectangle you describe, and so you can ignore all the sides, not just the corner points. But keep in mind that a single point or even a curve will contribute $0$ to the value of the integrals that you have to compute, and you will get the same answers whether you include or exclude the sides of the rectangles in your integrals. – Dilip Sarwate Nov 28 '12 at 3:15
Sorry for continuing to bother you, I (hopefully!) just have one more question! I've drawn the sketch and highlighted the non-zero points, and now I'm working on drawing the lines for various W; it looks like the line w=0 (so 0=2x+y) is where the 'split' in the CDF you mentioned above will occur; does it make sense to do something like $\int_{-2}^2\int_{-1}^{y/2}[givenPDF]dxdy$ for the area below that line (-4<w<0) and $\int_{-2}^2\int_{y/2}^{1}[givenPDF]dxdy$ for the area above (0<w<4)? Thanks again! – SarahJ Nov 28 '12 at 4:02