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Finding a marginal pmf

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose that X1 and X2 have the joint pmf
    [tex]f(x_{1},x_{2})=p^{2}q^{x_{2}},x_{1}=0,1,2,...,x_{2},x_{2}=0,1,2,...[/tex]
    with
    [tex]0<p<1,q=1-p[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I'm confused because the expression doesn't have x_1 in it. So usually, if I want to find f1x(1), say, I would add up all the x_2 probabilities where x_1=1 to get my marginal (summing over all values of x_2). But without having an x_1 in my pmf, I don't know how to do this. I'm tempted to say the marginal is just the original pmf, but how would I differentiate between p(x_1=1,x_2=2) from p(x_1=0,x_2=2)?
     
  2. jcsd
  3. Oct 22, 2012 #2

    jbunniii

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    There must be a typo somewhere. If you sum [itex]f(x_1, x_2)[/itex] over [itex]x_1[/itex] and [itex]x_2[/itex], the result will be infinity, not 1. Therefore this is not a valid probability mass function
     
  4. Oct 22, 2012 #3

    Ray Vickson

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    No, that is not the case: x1 is summed from 0 to x2, and x2 is summed from 0 to ∞.

    RGV
     
    Last edited: Oct 22, 2012
  5. Oct 22, 2012 #4

    jbunniii

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    Oops, you're right. I misread the range for [itex]x_1[/itex].
     
  6. Oct 22, 2012 #5
    okay, so my first thought is that the joint then should be
    [tex]\sum_{x_{2}=0}^{\infty}\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}[/tex]

    when I let p=.5, this term does indeed sum to 1, so that's looking alright.
    oh, I think I maybe figured this out. So I should have (summing over values in the domain
    [tex]f_{1}(x_{1})=\sum_{x_{2}=0}^{\infty}p^{2}q^{x_{2}}[/tex]
    [tex]f_{2}(x_{2})=\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}[/tex]

    No, I'm still confused. The f1 sums out to .5 (when I let p=.5). So then I should sum .5 from... 0 to x2 to get 1? But then I'll have x2 in my term, and not 1.
    When I integrate f2 from 0 to infinity, I do get one, which I think is good, but I'm still confused by not having an x1 term anywhere.
     
  7. Oct 22, 2012 #6

    jbunniii

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    Yes, that's the sum of the joint pmf.

    No, you aren't summing over the right range of indices. Try drawing a picture of the region where the joint pmf is nonzero. Then it will be easier to see what the upper and lower limits for [itex]x_2[/itex] should be in the sum.
    Yes, this is correct. The summand does not depend on [itex]i[/itex], so you are simply summing [itex]x_2 + 1[/itex] copies of the same thing. What is the result?
     
  8. Oct 22, 2012 #7
    Okay, so looking at a picture, I know that P(x1,x2)=0 whenever x1>x2 (and x1,x2>0). So I have kind of a triangular set up. So looking at these values, then, I'm thinking that my marginal for x should be
    [tex]f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{x_{1}}[/tex]
     
  9. Oct 23, 2012 #8

    jbunniii

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    Almost, except the exponent for [itex]q[/itex] is wrong: it should be [itex]q^i[/itex].

    You should be able to find a closed form expression for each pmf. For [itex]f_{1}(x_{1})[/itex], observe that you can pull [itex]p^2[/itex] outside the sum, and you're left with [itex]p^{2}\sum_{i=x_{1}}^{\infty}q^i[/itex]. How does this sum relate to a geometric series?
     
  10. Oct 23, 2012 #9
    okay, so finally, we have that
    [tex]f_{1}(x_{1})=\sum_{i=x_{1}}^{\infty}p^{2}q^{i}=p^{2}\sum_{i=x_{1}}^{\infty}q^{i}[/tex]
    so [tex]\sum_{i=0}^{\infty}q^{i}=\dfrac{1}{1-(1-p)}=\dfrac{1}{p}. \sum_{i=0}^{x_{1}}q^{i}=\dfrac{1-q^{x_{1}-1}}{1-q}=\dfrac{1-q^{x_{1}-1}}{p}[/tex], so taking the difference, we have
    [tex]f_{1}(x)=p^{2}\left(\dfrac{q^{x_{1}-1}}{p}\right)=pq^{x_{1}-1}[/tex].

    for f_2(x_2), we just have
    [tex]f_{2}(x_{2})=\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}=(x_{2}+1)p^{2}q^{x_{2}}[/tex]

    is this right? thanks for your help.
     
    Last edited: Oct 23, 2012
  11. Oct 23, 2012 #10

    jbunniii

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    Yes.
    The upper limit on your sum should be [itex]x_1 - 1[/itex], not [itex]x_1[/itex]. Otherwise you will lose the [itex]x_1[/itex] term when you take the difference.
    Yes, but you need to specify the range of values of [itex]x_2[/itex] for which this formula is valid. For example, if [itex]x_2[/itex] is negative, this formula isn't right.
     
  12. Oct 23, 2012 #11

    Ray Vickson

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    What you have written makes no sense: your summation gives the wrong answer
    [tex]\sum_{i=0}^{x_{2}}p^{2}q^{x_{2}}= (x_2+1) p^2 q^{x_2};[/tex] do you see why?
    Maybe you intended to say
    [tex]f_2(x_2) = \sum_{i=0}^{x_2}p^2 q^i,[/tex]
    in which case that is what you should have written.

    BTW: it will save you a lot of time and trouble if you use only the needed brackets in LaTeX. When the subscript or superscript is a single symbol, you do not need curly brackets, so it is OK to write f_2(x_2) and q^i, for example, but you DO need brackets when you have things like \sum_{i=0}^{x_2}, because the sub- or super-scripts have more than one symbol. However, you do NOT need to write ^{x_{2}}.

    RGV
     
  13. Oct 23, 2012 #12

    jbunniii

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    Assuming the joint pmf is correctly stated in the original post, this is in fact a valid expression for the pmf of [itex]x_2[/itex]. What makes you say that [itex](x_2+1) p^2 q^{x_2}[/itex] (for [itex]x_2 \geq 0[/itex], otherwise 0) is the wrong answer? It's certainly a valid pmf: it takes on only nonnegative values and it sums to 1.
     
  14. Oct 23, 2012 #13

    Ray Vickson

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    OK, right! I should not post anything before my first cup of coffee!

    RGV
     
  15. Oct 23, 2012 #14
    Thanks. As far as the latex goes, I usually type it in an editor, and then paste the code in here.
     
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