Conditional expectation and partitioning

[Kate2010]

Homework Statement

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.

Homework Equations

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

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.

 

[tiny-tim]

Hi Kate2010!

… 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.

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?

 

[Kate2010]

Would this be (p_i ^(n-1))q_i + (q_i ^(n-1))p_i ?

 

[tiny-tim]

Would this be (p_i ^(n-1))q_i + (q_i ^(n-1))p_i ?

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

Yup!

So the expected family size is … ?

 

[Kate2010]

I still don't know where to go from here. Do I use E(X) = ∑ xP(X=x)?

 

[tiny-tim]

Yes.

 

[Kate2010]

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

E(X) = ∑ n (p_i^n-1q_i + q_i^n-1p_i)?

If this is correct, which I'm really not sure about, how do I do it?

 

[tiny-tim]

The left-hand part is q_i times ∑ np_i^n-1 …

how can you calculate that ∑ ?

 

[Kate2010]

Would I do it as a geometris series which then differentiates to 1/(1-p_i)²?

So it would be altogether q_i/(1-p_i)² + p_i/(1-q_i)² , but this equals 1?

 

[tiny-tim]

(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) ?

 

[Kate2010]

I'm sorry I'm still struggling with this question.
I have q_i∑ np_i^n-1 + p_i∑ nq_i^n-1
These are geometric sums so can I use the formula ∑ (to infinity) xⁿ = 1/(1-x) so ∑ (to infinity) nx^n-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?

 

[tiny-tim]

… I'm confused about what I'm summing from and 2. Am I summing from n=2 to infinity?

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

 

[Kate2010]

I want q_i∑^{n=infinity}_{2} np_i^n-1 + p_i∑^{n=infinity}_{2} nq_i^n-1

∑^{infinity} xⁿ= 1/(1-x) so ∑^{infinity} nx<sup>n-1[/SUP = 1/(1-x)2

So q_i∑^{infinity}_{n=1} np_in-1 + p_i∑^{infinity}_{n=1} nq_in-1 = q_i/(1-p_i)2 + p_i/(1-q_i)2 = 1/q_i + 1/p_i = 1/p_iq_i

I now need to subtract the sum to 1 of each, i.e. q_i∑^{n=1}_{1} np_in-1 + p_i∑^{n=1}_{1} nq_in-1 = 1+1 = 2

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

I've only done discrete random variable so far and have not used \int in probability.</sup>

 

[Kate2010]

The third line down the sums should be the other way around but I can't make it format properly.

 

[tiny-tim]

(have an infinity: ∞ )

I now need to subtract the sum to 1 of each, i.e. q_i∑^{n=1}_{1} np_i^n-1 + p_i∑^{n=1}_{1} nq_i^n-1 = 1+1 = 2

No, p_i⁰ = q_i⁰ = 1, so it's q_i + p_i = 1.

(and no need to say "the sum to 1", just say "the first term" !)

 

[Kate2010]

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) = \sum_{x} 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)?

 

[tiny-tim]

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) = \sum_{x} xP(X=x) where X = number of girls and A = both sexes?

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 … ?

 

[Kate2010]

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 \sumx? But I don't know what to sum between.

 

[tiny-tim]

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 \sumx? But I don't know what to sum between.

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

Just use E(X) = ∑_x xP(X=x)

 

[Kate2010]

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 ∑^{n}_{x=1} xq_i?

This would be q_in(n+1)/2?