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Conditional expectation and partitioning

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data
    I'm told that of n couples, each of whom have at least one child, with couples procreating independently and no limits on family size, births single and independent, and for the ith couple the probability of a boy is p_i and of a girl is q_i with p_i + q_i = 1.

    1. Show the expected family size if the ith couple stop when have had both sexes is 1/(p_iq_i) - 1.

    2. If all n couples stop when have children of both sexes, what is the expected number of girls.

    2. Relevant equations

    E(X) = Sum(i=1..n) E(X|A_i) P(A_i)
    E(X|A) = Sum(over x) xP(X=x|A)

    3. The attempt at a solution

    So for 1 I've got:
    Let X be the number of births until a girl and boy
    A1 = boy born 1st
    A2 = girl born 1st
    E(X) = E(X|A_1)P(A_1) + E(X|A_2)P(A_2) = (p_i/q_i) + (q_i/p_i) = 1/(p_iq_i) -2
    Do I add 1 as I'm considering 2 births not 1?

    For 2:
    I'm not too sure how to go about this at all, I can use the second formula with X being the number of girls and A being that both sexes are born, but how do I know P(X=x|A)?

    Earlier in the question I calculated the expected family size if the family stopped after a boy or a girl.

    Thanks.
     
  2. jcsd
  3. Jan 19, 2010 #2

    tiny-tim

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    Hi Kate2010! :wink:
    hmm … start again …

    (with these problems, it's just a question of how to count …)

    what's the probability that the first n-1 births are the same sex, and the nth birth is different? :smile:
     
  4. Jan 19, 2010 #3
    Would this be (p_i ^(n-1))q_i + (q_i ^(n-1))p_i ?
     
  5. Jan 19, 2010 #4

    tiny-tim

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    (have a sigma: ∑ and try using the X2 and X2 tags just above the Reply box :wink:)

    Yup! :smile:

    So the expected family size is … ?
     
  6. Jan 19, 2010 #5
    I still dont know where to go from here. Do I use E(X) = ∑ xP(X=x)?
     
  7. Jan 19, 2010 #6

    tiny-tim

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  8. Jan 19, 2010 #7
    π²³ ∞ ° → ~ µ ρ σ Ω √ ∫ ≤ ≥ ± ∃ … · θ φ ψ ω Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô

    E(X) = ∑ n (pin-1qi + qin-1pi)?

    If this is correct, which I'm really not sure about, how do I do it?
     
  9. Jan 19, 2010 #8

    tiny-tim

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    The left-hand part is qi times ∑ npin-1

    how can you calculate that ∑ ? :smile:
     
  10. Jan 19, 2010 #9
    Would I do it as a geometris series which then differentiates to 1/(1-pi)2?

    So it would be altogether qi/(1-pi)2 + pi/(1-qi)2 , but this equals 1?
     
  11. Jan 19, 2010 #10

    tiny-tim

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    (i'm going out for the evening now, so this is my last post for some hours)

    I haven't checked it properly, but did you take care to start your ∑ pn from the right point (ie does it start at n = 0, n = 1, or n = 2) ?
     
  12. Jan 20, 2010 #11
    I'm sorry I'm still struggling with this question.
    I have qi∑ npin-1 + pi∑ nqin-1
    These are geometric sums so can I use the formula ∑ (to infinity) xn = 1/(1-x) so ∑ (to infinity) nxn-1 = d/dx(1/(1-x)) = 1/(1-x)2 . However, I'm confused about what I'm summing from and 2. Am I summing from n=2 to infinity?
     
  13. Jan 20, 2010 #12

    tiny-tim

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    oh no, this is long and complicated and i don't want to have to write it all out myself …

    you write it all out, including the ∫s and the ∑n=?s and i'll check it :smile:
     
  14. Jan 20, 2010 #13
    I want qi∑[tex]^{n=infinity}_{2}[/tex] npin-1 + pi∑[tex]^{n=infinity}_{2}[/tex] nqin-1

    ∑[tex]^{infinity}[/tex] xn= 1/(1-x) so ∑[tex]^{infinity}[/tex] nxn-1[/SUP = 1/(1-x)2

    So qi∑[tex]^{infinity}_{n=1}[/tex] npin-1 + pi∑[tex]^{infinity}_{n=1}[/tex] nqin-1 = qi/(1-pi)2 + pi/(1-qi)2 = 1/qi + 1/pi = 1/piqi

    I now need to subtract the sum to 1 of each, i.e. qi∑[tex]^{n=1}_{1}[/tex] npin-1 + pi∑[tex]^{n=1}_{1}[/tex] nqin-1 = 1+1 = 2

    So I get 1/piqi -2, but I wanted 1/piqi -1.

    I've only done discrete random variable so far and have not used [tex]\int[/tex] in probability.
     
    Last edited: Jan 20, 2010
  15. Jan 20, 2010 #14
    The third line down the sums should be the other way around but I can't make it format properly.
     
  16. Jan 20, 2010 #15

    tiny-tim

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    (have an infinity: ∞ :wink:)
    No, pi0 = qi0 = 1, so it's qi + pi = 1. :wink:

    (and no need to say "the sum to 1", just say "the first term" !)
     
  17. Jan 20, 2010 #16
    Thank you so much!

    For question 2:
    If all n couples stop when have children of both sexes, what is the expected number of girls.

    Do I use the forumula E(X|A) = [tex]\sum[/tex][tex]_{x}[/tex] xP(X=x) where X = number of girls and A = both sexes?

    If so, I'm not sure how to calculate P(X=x), is it just 1/2? And would I sum from 2 to n (I'm considering n couples)?
     
  18. Jan 20, 2010 #17

    tiny-tim

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    Each family is independent, so just add the expectation for each family, i = 1 to n.

    Use a similar method as before … if the last birth is a girl, the number is 1, if it's a boy, the number is … ? :smile:
     
  19. Jan 20, 2010 #18
    Is P(X=x|A) always 1 as if there have been both sexes born there will always be a girl? Then it would just be [tex]\sum[/tex]x? But I don't know what to sum between.
     
  20. Jan 20, 2010 #19

    tiny-tim

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    I'm confused … where does A come into it? :confused:

    Just use E(X) = ∑x xP(X=x)
     
  21. Jan 20, 2010 #20
    I thought I should use conditional expectation.

    If I use E(X) = ∑x xP(X=x) with X being the being the number of girls, do I do ∑[tex]^{n}_{x=1}[/tex] xqi?

    This would be qin(n+1)/2?
     
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