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Marginal PMG of of 2 random variables with Joint PMF

  1. Oct 25, 2015 #1
    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. jcsd
  3. Oct 25, 2015 #2

    Ray Vickson

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    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##.
     
  4. Oct 25, 2015 #3
    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)
     
  5. Oct 26, 2015 #4
    Sorry, it differs by definition. PXY(K,L) = PX(K)PY(L) which, in this case, equals:
    1 / (2K) * 1 / (2L).
     
  6. Oct 26, 2015 #5

    Ray Vickson

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    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##.
     
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