Consider a population of $n$ couples where a boy is born to the $i^\text{th}$ 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 $$S= \frac{\text{expected number of boys born in the population of $n$ couples}}{\text{expected number of children horn in the population of $n$ couples}}$$
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.
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 $p_i$.