# Joint Probability mass function

I have been searching for some information on a problem I'm working on. Unfortunately i cannot find anything that is similar. All of the questions I find on PMF are related to coin flips, or dice rolls. And I cannot figure out how to apply the same logic to my question:

Discrete RV $X$ and $Y$ are positive integers that have a JPMF: $Px,y(x,y) = 2^{-x-y}$

1. Determine the Marginal probability mass functions $px(x), py(y)$

2. Are $x$ and $y$ Independent?

3. Find $E[x], E[y]$ and $E[xy]$

I know that $P(x,y) = P(X=x, Y=y)$ and $P(X=x) = \sum p(x,y)$.

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Do you have some idea of what one means when one says that two random variables are independent? –  Did Feb 25 '13 at 17:14
yes, it means that they do not effect each other. –  Special--k Feb 25 '13 at 17:16
No. This is mathematics, not word plays, hence you are supposed to have a precise definition at your disposal (and to base your solution on it). –  Did Feb 25 '13 at 17:19
X and Y are independent if P(XY) = P(X)$*$P(y) –  Special--k Feb 25 '13 at 17:24
?? What is P(XY)? What are P(X) and P(y)? Numbers? Functions? –  Did Feb 25 '13 at 17:26

The independence can be asserted by obtaining the marginal distributions from the joint distribution and checking for product form as follows.

$$P_{XY}(x,y)=2^{-x-y}\\ P_X(x)=\sum_y 2^{-x}2^{-y}=2^{-x}\\ P_Y(y)=\sum_x 2^{-x}2^{-y}=2^{-y}\\ P_{XY}(x,y)=P_X(x)P_Y(y)$$

Thus the marginal pdf $P_X(x)=2^{-x},\ x\ge1$ and $X$ and $Y$ are iid.

$$E(X)=E(Y)=\sum_{x=1}^{\infty}x2^{-x}=\frac{0.5}{(1-0.5)^2}=2\\ E(XY)=E(X)E(Y)=4$$

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PDF and PMF are not the same thing? –  Special--k Feb 25 '13 at 17:36
Edited... Just a name - pdf for cont. and pmf for discrete. –  Bravo Feb 25 '13 at 17:39
If $P_{X,Y}(x,y) = g(x)h(y)$, why does it necessarily follow that $P_X(x) = g(x)$ and $P_Y(y) = h(y)$? As a variant of this problem, consider $P_{X,Y}(x,y)= 2^{-x+1}3^{-y}$. Are the pmfs of $X$ and $Y$ given by $2^{-x+1}$ and $3^{-y}$? –  Dilip Sarwate Feb 25 '13 at 17:40
@DilipSarwate: Is your example a valid joint distribution function? We need $\sum_x \sum_y P(x,y)=1$. –  Bravo Feb 25 '13 at 17:48
$$2^{-x+1}3^{-y} = 2^{-x}\cdot\left(\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{y-1}\right)$$ is the product of the mass functions of independent geometric random variables of parameters $\frac{1}{2}$ and $\frac{2}{3}$ respectively. –  Dilip Sarwate Feb 25 '13 at 18:35