# Probability of a boy is born to the $i^{th}$

Consider a population of $n$ couples where a boy is born to the $i^{th}$ couple with probability $p_i$ and $c_i$ is the expected number of children born to this couple. If we suppose that $p_i$ is constant with time for all couple and that sexes of successive children born to a particular couple are independent r.v's and no multiple births are allowed. The sex ratio is defined to be $$S= \frac{expected \ number \ of \ boys \ horn \ in \ the \ population \ of \ n \ couples}{expected \ number \ of \ children \ horn \ in \ the \ population \ of \ n \ couples}$$

Suppose $c_i=c$, $i=1,\ldots,n$ then find $S$ I find this $S$ as : $$S \equiv S_0= \frac{\sum_{i=1}^{n}\frac{1}{p_i}}{n}$$

and in the second part of the problem asks that:

If the parents of all couples decide to have children until a boy is born and then have no further children, Then show that $$S=S_1 = \frac{n}{\sum_{i=1}^{n}\frac{1}{p_i}} \leq S_0$$

I´m really stuck with this problem, I tried to model the birth of the children as a geometric distribution but I think this is wrong cause the births start in $1$ and the geometric distribution starts in $0$.

How can I compute the expected value of the above events?

Could someone help me to show this inequality pls? Thank for your time and help.

• Not sure I follow. Since $0<p_i≤1$, presumably, we have $\frac 1{p_i}≥1$. Thus $\sum \frac 1{p_i}≥n$ which implies $\left( \sum \frac 1{p_i} \right)^2≥n^2$. Is that all you wanted? – lulu Feb 19 '18 at 14:37
• why $\sum \frac{1}{p_i} \geq n$ ? – Rachel Feb 19 '18 at 14:40
• If you add $n$ numbers each of which is at least $1$ together you get at least $n$. Note: your sums incorrectly start at $i=0$. I believe you meant $i=1$. – lulu Feb 19 '18 at 14:41
• I´m just stuck trying to prove the inequality. I tried to compute the expected value of the ratio $S$ – Rachel Feb 19 '18 at 14:41
• It starts from $1$, from the book Samuel Karlin an introduction to SP. – Rachel Feb 19 '18 at 14:43

Wikipedia lists two definitions of the geometric distribution, but the first one starts at $1$ and seems like the correct way to go about solving this problem. It counts the number of trials until the first success is reached, and in order to have a success you must have at least one trial. In your case "success" is family $i$ having a boy, and the probability of success is $p_i$.