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Let the probability $p_n$ that a family has exactly $n$ children be $\alpha p^n$ when $n\geq1$, and $$p_0=1-\alpha p(1+p+p^2+\cdots).$$ Suppose that all the sex distributions have the same probability. Show that for $k\geq1$ the probability that a family has exactly $k$ boys is $2\alpha p^k/(2-p)^{k+1}$.

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    $\begingroup$ Please don't write your questions as if you are giving us a homework assignment! Show us what you already tried and where you got stuck. $\endgroup$
    – sxd
    Nov 26, 2011 at 7:45
  • $\begingroup$ Assuming the probability of a boy being born=probability of a girl being born=1/2, I am getting Required probability$=\Sigma\alpha p^n nCk*(1/2^n)$.,summation over n extending from 1 to infinity.Is it the correct interpretation? $\endgroup$ Nov 26, 2011 at 8:00
  • $\begingroup$ Shouldn’t $p_0$ be $1-\alpha(p+p^2+\dots)$? $\endgroup$ Nov 26, 2011 at 8:16
  • $\begingroup$ Summation over n should be from k to infinity in my last comment. $\endgroup$ Nov 26, 2011 at 8:32
  • $\begingroup$ I have made the correction as suggested by B. M scott $\endgroup$ Nov 26, 2011 at 11:16

2 Answers 2

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Extended hint: We sketch an argument that uses only basic notions. Note that the probability $b_k$ of $k$ boys is, by a conditional probability argument, given by $$b_k=\sum_{n=1}^\infty \alpha p^n \binom{n}{k} \left(\frac{1}{2}\right)^{k}\left(\frac{1}{2}\right)^{n-k}.$$ This simplifies to $$b_k=\sum_{n=1}^\infty \alpha \binom{n}{k}\left(\frac{p}{2}\right)^n.\qquad\qquad(\ast)$$ (We define $\binom{n}{k}$ to be $0$ if $n<k$.) There are many tools for evaluating $(\ast)$. We do it using not much machinery. Recall the combinatorial identity $$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}.$$ Substitute for $\binom{n}{k}$ in $(\ast)$. We obtain $$b_k=\sum_{n=1}^\infty \alpha \binom{n-1}{k}\left(\frac{p}{2}\right)^n +\sum_{n=1}^\infty \alpha \binom{n-1}{k-1}\left(\frac{p}{2}\right)^n.\qquad\qquad(\ast\ast)$$ The first term in $(\ast\ast)$ is just $\dfrac{p}{2}b_k$. The second term is $\dfrac{p}{2}b_{k-1}$. So we have derived the recurrence $$b_k=\frac{p}{2}b_k+\frac{p}{2}b_{k-1}$$ or equivalently $$b_k=\frac{p}{2-p}b_{k-1}.$$ This almost settles things: each time we increment $k$ by $1$, the probability gets multiplied by $\dfrac{p}{2-p}$. To get the process started, we need $b_1$. We have $$b_1=\alpha\sum_{n=1}^\infty n \left(\frac{p}{2}\right)^n.$$ There is a trick for finding $\sum_{n=1}^\infty n x^{n}$. Using the fact that for $|x|<1$, $$1+x+x^2+x^3+ x^4+ \cdots=\frac{1}{1-x},$$ we find, by differentiating, that $$\frac{1}{(1-x)^2}=1+2x+3x^2+ 4x^3+\cdots.$$ It follows that $$x+2x^2+3x^3+4x^4+\cdots =\frac{x}{(1-x)^2}.$$

Comment: Alternately, one could evaluate the sum $(\ast)$ by a repeated differentiation argument that generalizes the method we used for $b_1$.

The recurrence $b_k=\dfrac{p}{2}b_k+\dfrac{p}{2}b_{k-1}$ can also be obtained directly, bypassing the series manipulation.

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  • $\begingroup$ There is a minor typo:It should be $b_1=\alpha\Sigma_1^{\infty}np^n$. The coefficient $\alpha$ has been left out. This does not reduce the elegance of the proof $\endgroup$ Dec 3, 2011 at 12:51
  • $\begingroup$ There is a minor typo:It should be $b_1=\alpha\Sigma_1^{\infty}n(p/2)^n$. The coefficient $\alpha$ has been left out. This does not reduce the elegance of the proof $\endgroup$ Dec 3, 2011 at 12:58
  • $\begingroup$ @Anamitra Palit: Thank you for noticing the typo, and telling me about it. Fixed. $\endgroup$ Dec 3, 2011 at 15:11
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Required probability=$\alpha \Sigma [(p/2)^n *nCk]$

[Summation on “n” running from k to infinity]

The above expression=$\alpha \Sigma [(k+r) Ck *(p/2)^{k+r} ]$

[Summation on r running from zero to infinity]

Expression =$\alpha (p/2)^k \Sigma (p/2)^r (k+r)Ck$ -------(1)

Now,

$\Sigma (p/2)^r (k+r)Ck$

=The coefficient of $x^k$ in the sum given below:

$(p/2+x)^0+(p/2+x)^1+(p/2+1)^2 ……..$ ------- ---- (2)

[We have an infinite number of terms in the above series].

We choose x such that:

$p<p/2+x<1$ --------[inequality A]

$=>0<p/(2-p)<2x/(2-p)<1$ --------[Inequality B]

The series given by (2) evaluates to

$1/(1-x-p/2) = 2/(2-2x-p)$

$=2(2-2x-p)^{-1}$

$=[2/(2-p)] [1-2x/(2-p)]^{-1}$

Inequality A ensures the convergence of the series expressed by (2).

Coefficient of $x^k$ in the above expression is:

$2/(2-p) 2^k/(2-p)^k$ ----(3)

Inequality B ensures the workability of the binomial expansion to obtain (3)

Using expression (3) in (1) we obtain the probability=

$\alpha (p/2)^k*2/(2-p) 2^k/(2-p)^k$

$=2\alpha p^k/(2-p)^{k+1}$

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    $\begingroup$ Your writing style and presentation leave much to be desired. $\endgroup$ Nov 26, 2011 at 22:06
  • $\begingroup$ I have put out my calculations over here just to assert that I have been working on the problem $\endgroup$ Dec 3, 2011 at 12:55
  • $\begingroup$ @Anamitra Palit: Nice work. The convergence issues you spent some time on turn out not to matter, since yours is essentially a generating function argument. $\endgroup$ Dec 3, 2011 at 15:16
  • $\begingroup$ @AnamitraPalit Why is sumation running from $k$ to $\infty$ ? $\endgroup$ Nov 22, 2017 at 17:31
  • $\begingroup$ @Mathematics :we are considering the probability of exactly 'k' boys out of 'n' children[n>=k].Probability of 'n' children=alpha*p^n. The stated probability for n childeren tends to zero for n tending to infinity, p being a fraction.This is apparent from the series defining p_0: the series can converge only if p is a fraction.Now 'n' could be any integral value greater than or equal to k. So we have considered the summation from n=k to n=infinity $\endgroup$ Nov 24, 2017 at 4:42

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