# Marginal PMG of of 2 random variables with Joint PMF

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1. Oct 25, 2015

### whitejac

1. The problem statement, all variables and given/known data
Consider two random variables X and Y with joint PMF given by:
PXY(k,L) = 1/(2k+l), for k,l = 1,2,3,....

A) Show that X and Y are independent and find the marginal PMFs of X and Y
B) Find P(X2 + Y2 ≤ 10)

2. Relevant equations
P(A)∩P(B)/P(B) = P(A|B)
P(A|B) = P(A) if independent

3. The attempt at a solution
Choosing two arbitrary numbers to show P(A|B) = P(A)

P(x<2) ∩ P(y≤1) / P(y≤1)
=P(1,1) + P(1,2) + P(1,3)... + P(1, L) ∩ P(1,1) + P(1,2) + P(1,3)...+ P(k,1) / P(y≤1)
=P(1,1) / P(1,1) + P(1,2) + P(1,3)... + P(k,1)
Note: Geometric series
=(1/4) / (1/2)(1/2^k)
=(1/4) / (1/4)(1 / (1 - (1/2)))
=(1/4) / 2
= 1/2

P(A) = P(x<2) = P(1,L) =
=1/4 (1/2L)
=1/4 (1 / (1 - (1/2)))
=1/4 (2)
= 1/2

X and Y are independent.

How does one find the Marginal PMF of this equation then? The ones I've seen before in discrete sections were pre-made and finite... meaning that they were a table of results for X =1,2,3... and Y = 1,2,3... for the range of each. Should I be finding a summation for each value of X + Y?

I haven't considered part B yet.

2. Oct 25, 2015

### Ray Vickson

The marginal pmf of $X$ is $P_X(k) = P(X = k) = P(X = k \: \& \; Y= \text{anything})$. Note that for $l = 1,2,3 \ldots$ the different events $\{ X = k, Y = l \}$ are disjoint (mutually exclusive), so the probability of $\{ X = k \: \& \; Y \in \{ 1,2,3, \ldots \} \}$ is just the sum of their probabilities for different $l$.

3. Oct 25, 2015

### whitejac

How does this differ from PXY, because that's what it sounds like. If the marginal pmf is P(X = k and Y equals anything), it sounds like I'm looking for PXY(k,l)

4. Oct 26, 2015

### whitejac

Sorry, it differs by definition. PXY(K,L) = PX(K)PY(L) which, in this case, equals:
1 / (2K) * 1 / (2L).

5. Oct 26, 2015

### Ray Vickson

Right: and $P_X(k) = \sum_{l=1}^{\infty} P_{XY}(k,l)$, etc.

By the way, the fact that $P_{XY}(k,l) = P_X(k) P_Y(l)$ is what makes $X$ and $Y$ independent. You did not convincingly show independence in your previous post, because you only showed a few examples of $P(X \in A \; \& \; Y \in B) = P(X \in A) \, P(Y \in B)$ for one or two special cases of $A$ and $B$. To prove independence, you need to establish this for all possible choices of $A$ and $B$.