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Full Disclosure: I am a graduate student taking a class in probability. I am working through extra problems in the Hoel, Port and Stone book which has answers in the back of the textbook. I am not doing these problems for homework, merely as a supplement to the Ross Book which I find a bit confusing.

This is problem #7 in Chapter 6:

$$f(x,y) = \alpha^2 e^{-\alpha y}, \qquad0 \leq x \leq y \leq 1$$ Find the Joint Distribution of X and Y.

The answer is given as: $$F_{x,y}(x,y) = \begin{cases}1 - e^{-\alpha x} - \alpha x e^{-\alpha y},& 0 \leq x \leq y\\\\ 1 - e^{-\alpha y}(1 + \alpha y),& 0 \leq y \leq x\end{cases}$$

I have been struggling to find the proper limits of integration and would appreciate some help.

Thank you.

Jimmy R.
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Clan_Destine
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1 Answers1

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Hint: This works always, when you do not want to mess up with integration limits: Draw a picture, not of $f$ (so do not worry about how $f$ looks) but of the domain on which $f$ lives!

Jimmy R.
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  • Thank you for the hint, Stefanos. I remind myself all of the time to draw a picture as it really helps a lot.

    I am still unsure on the limits, however, as I am unsure how to add in the constraint. I have drawn the unit square as well as the diagonal to represent y = x. I have tried integrating with dy from 0 to y, dx from 0 to x as well as dy from 0 to y, dx from 0 to y and I just cannot get it to come out right.

     – Clan_Destine Nov 17 '14 at 20:20
  • Fix some x and fix some y. No, try to see the limits of the tetrahedron between the lines X=0,X=x,Y=y and y=x. Does it help. Of course the solution conveys that the result will be different depending on whether x<y or y<x. Does it help? – Jimmy R. Nov 17 '14 at 20:25
  • It is important to fix a x and y and point the area with X<x, Y<y. – Jimmy R. Nov 17 '14 at 20:25
  • The limits that you suggest, contained within that area, are the ones that I tried initially: x from 0 to x, y from x to y. Upon integration, I get a different result than the answer. When I do this integration, I get ax(exp(-ax) - exp(-ay)). Have I done something wrong or is this simply the same quantity in a different format? – Clan_Destine Nov 17 '14 at 20:42
  • No, X does not go from 0 to x. It goes from 0 to y for y<x and from 0 to x (indeed) for y>x. You have to determine the tetrahedron I mentioned above. Do not simplify things. If X goes from 0 to x and Y from x to y then you have a triangle.... – Jimmy R. Nov 17 '14 at 20:51
  • I have drawn this out a few times and the area that I can see is a triangle + a rectangle under the curve for y < x, as seen in the following image: link. Integrating this in WolframAlpha to be sure I am not missing something, I see this as my final answer: link – Clan_Destine Nov 17 '14 at 22:25
  • No, false, your function lives above the line x=y, never below. Now, fix x_1 and y_1<x_1 and then (second case) y_2>x_1 to see what happens. But you are above the line x=y, Nothing exists below the line... no, false, correct it. – Jimmy R. Nov 17 '14 at 22:28