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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, 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 and the support of Y is $-2, since the region is a square. I think this means that the support of W is $-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?2012-11-28
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    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).2012-11-28
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    These are illogical and unfortunate pedagogical choices.2012-11-29

1 Answers 1

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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 $w$.

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?2012-11-28
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    @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?)2012-11-28
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    My apologies, I think I was unclear in my question - I was simply looking at the region bounded by -1$f_[x,y]$=0? – 2012-11-28
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    $f(x,y)$ 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.2012-11-28
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    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$\int_{-2}^2\int_{y/2}^{1}[givenPDF]dxdy$ for the area above (02012-11-28
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    @SarahJ You have the right idea that it makes sense to compute $P\{W\leq w\}=F_W(w)$ for $w\in(-4,0)$ and $P\{W>w\}=1-F_W(w)$ for $w\in(0,4)$ by integrating over triangular regions, but your integral limits are incorrect. For $w\in(-4,0)$, the triangle you are integrating over has vertices $(-1,-2), (1+w/2,-2)$ and $(-1,w+2)$ and so you have $$F_W(w) = P\{W \leq w\}=\int_{y=-2}^{w+2}\int_{x=-1}^{(w-y)/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy.$$2012-11-28
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    Thanks! Fully understand it now; I really appreciate your help!2012-11-29