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How can I sketch the joint p.d.f. of X and Y?

$f(x,y)=(\frac{1}{3}$ if $1<x<3$ and $0<y<1$ ; $\frac{1}{6}$ if $6<x<8$ and $0<y<1$ ; $0$ otherwise)

After this is graphed, find the marginal p.d.f.'s of X and Y?

jerry2144
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    The joint pdf is two rectangular surfaces at heights $\frac 13$ and $\frac 16$ above the $x$-$y$ plane. So you could draw a three-dimensional sketch showing unsupported rectangles above the plane (sort of like flying carpets in Disney movies based on tales from 1001 Arabian nights), or rectangular prisms sitting on the plane etc. – Dilip Sarwate Oct 17 '14 at 04:20
  • OP: Did you try to follow the suggestion in @DilipSarwate's comment or did you decide to offer a bounty instead? – Did Oct 20 '14 at 18:36
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    It worked! Question with zero personal input + bounty = full answer ready to be handed back. – Did Oct 23 '14 at 22:18

1 Answers1

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For the PDF of the marginals, you just have to integrate:

$$ f_X(x) = \int f(x,y) dy= \begin{cases} \int_0^1 \frac 13 dy = \frac 13 &\text{ if }&1<x<3\\ \int_0^1 \frac 16 dy = \frac 16 &\text{ if }&6<x<8\\ 0&\text{ otherwise}\\ \end{cases} $$

$$ f_Y(y) = \int f(x,y) dx= \begin{cases} \int_1^3 \frac 13 dy + \int_6^8 \frac 16 dy = \frac23 + \frac 26 = 1 &\text{ if }&0<y<1\\ 0&\text{ otherwise}\\ \end{cases} $$

Bonus: you can check the relations $$ \int f_X(x)dx = \int_1^3 \frac 13 dx + \int_6^8 \frac 16 dx = 1 \\ \int f_Y(y)dy = \int_0^1 dy = 1 $$

mookid
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