Find the joint probability density given the support set

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.

• Yes everything you tried is correct. Were you able to calculate those integrals? – Gregory Grant Feb 14 '16 at 1:42
• @GregoryGrant: No I couldn't figure out the bounds of the integrals. – user3727610 Feb 14 '16 at 1:44
• $\int_0^\infty\int_0^{e^{-x/3}} 1dydx$ – Gregory Grant Feb 14 '16 at 1:46
• So $f_{X}(x)=\int_{0}^{e^{-x/3}}dy$ and $f_{Y}(y)=\int_{0}^{-3ln(y)}dx$ – user3727610 Feb 14 '16 at 1:51
• Shucks looks like going out to a Saturday night movie cost me getting the credit. – Gregory Grant Feb 14 '16 at 5:13

$(X,Y)$ 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, $$f_X(x) = \int_{0}^{e^{-x/3}}\frac{1}{3}dy = \frac{1}{3}e^{-x/3}\to X \sim\mathcal{E}xp(1/3).$$
• So the joint pdf $f_{X,Y}(x,y) = \frac{1}{3}$? – user3727610 Feb 15 '16 at 2:46