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Suppose that the support set of $(X,Y)$ is $$S_{X,Y}=\{(x,y)\in\mathbb{R}^2: x \geq 0 \text{ and } 0 \leq y \leq e^{-x/3}\}$$

$(X,Y)$ is uniformly distributed on $S_{X,Y}$.

a) Find the joint probability density function for $(X,Y)$.

b) Find the marginal PDFs for X and Y.

c) Are X and Y independent? Explain.

What I have tried

a) Is the joint PDF $\int \int e^{-x/3}dxdy$?. If so, what are the bounds?

b) Fix X, integrate over all Y and vice versa.

c) Check if the joint PDF is the product of the marginals.

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  • $\begingroup$ Yes everything you tried is correct. Were you able to calculate those integrals? $\endgroup$ – Gregory Grant Feb 14 '16 at 1:42
  • $\begingroup$ @GregoryGrant: No I couldn't figure out the bounds of the integrals. $\endgroup$ – user3727610 Feb 14 '16 at 1:44
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    $\begingroup$ $\int_0^\infty\int_0^{e^{-x/3}} 1dydx$ $\endgroup$ – Gregory Grant Feb 14 '16 at 1:46
  • $\begingroup$ So $f_{X}(x)=\int_{0}^{e^{-x/3}}dy$ and $f_{Y}(y)=\int_{0}^{-3ln(y)}dx$ $\endgroup$ – user3727610 Feb 14 '16 at 1:51
  • $\begingroup$ Shucks looks like going out to a Saturday night movie cost me getting the credit. $\endgroup$ – Gregory Grant Feb 14 '16 at 5:13
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$(X,Y)$ are uniformly over $S$, so $$ \int\int_S Cdydx = C\int_{x=0}^{\infty}\int_{y=0}^{e^{-x/3}} dydx = 3C= 1 \to C=1/3. $$ Therefore, the marginal distributions are given by $$ f_Y(y) = \int_{0}^{-3ln(y)}\frac{1}{3}dx = - ln(y), $$ similarly, $$ f_X(x) = \int_{0}^{e^{-x/3}}\frac{1}{3}dy = \frac{1}{3}e^{-x/3}\to X \sim\mathcal{E}xp(1/3). $$

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    $\begingroup$ So the joint pdf $f_{X,Y}(x,y) = \frac{1}{3}$? $\endgroup$ – user3727610 Feb 15 '16 at 2:46
  • $\begingroup$ @user3886450 yep $\endgroup$ – V. Vancak Feb 15 '16 at 8:20

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