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Joint PMF of p x,y

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi, if you could offer help to any of these questions, it would be great, i was unable to attend my lectures this week and I had no idea how to do these.

    The joint pmf of x and y is p x,y = c(x^2 + y^2) x,y = 1,2,3 Find:

    a. c and the marginal pmfs of x and y
    b. e(x)
    c. the pmf of 3x-2y
    d. e(3x-2y) in two ways


    2. Relevant equations



    3. The attempt at a solution
    From first investigation, I think that I have to create a table with
    1,1 1,2 1,3
    2,1 2,2, 2,3
    3,1 3,2 3,2
    and x^2 + y^2 = 1 and the p = 1/9 since theres only 9 options.

    i have no idea how to continue.
     
  2. jcsd
  3. Aug 15, 2011 #2

    HallsofIvy

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    Science Advisor

    " p = 1/9 since theres only 9 options." I have no idea what you mean by this. "p" is a function, not a number. Did you mean "c= 1/9"?

    No, [itex]x^2+ y^2[/itex] is NOT 1. [itex]x^2+ y^2= 1+ 1= 2[/itex] when x= y= 1, [itex]x^2+ y^2= 4+ 1= 5[/itex] when x= 2, y= 1, etc.

    Start by actually filling in your table: What is p(1, 1), p(1, 2), etc.? Those will have a c in them. Then find c by using the fact that the sum of all the probabilities must be 1.
     
  4. Aug 15, 2011 #3
    Thanks for your reply,

    I'll give it a shot now.


    x,y 1 2 3
    1 2c 5c 10c
    2 5c 8c 13c
    3 10c 13c 18c
    is this right so far?
    if so, then is it just 83c = 1? c=1/84


    i hope it is...
     
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