Sex ratio in the long term

If the gender of a child is equally distributed among boys and girls, then it's equally distributed among boys and girls, regardless of whether or how parents decide to shape their family size.

You could make your life difficult by summing up the numbers of boys and girls by partitioning the population according to family size. Or you could just count the number of boys and number of girls leaving the delivery room. And we're told as a condition that these numbers will be equal.

If the first child born is a boy:

Probability of having the first child as a boy would be P(1st as a Boy) = 1/2

Ratio of a girl to a boy in this case would be 0/1.

Similarly,

Probability of having the second child as a boy would be P(2nd as a Boy) = 1/2 * 1/2

Ratio of a girl to a boy in this case would be 1/1.

So, expected value for the ratio of a girl to a boy in the long term can be written as a sum of infinite series:

$S=\left( \frac {1} {2}\right)\times{0} +\left( \frac {1} {2}\right) ^{2}\times 1+\left( \frac {1} {2}\right) ^{3}\times 2+\left( \frac {1} {2}\right) ^{4}\times 3+\ldots$

For the simplicity I am writing 1/2 as ${x}$

=> $S=0\cdot x+ x^{2}+2x^{3}+3x^{4}+\ldots +\left( n-1\right) x^{n}$

This looks like an arithmetic-geometric progression which can be solved by multiplying by ${x}$ both sides:

=> $S\cdot x = 0\cdot x^{2}+x^{3}+2x^{4}+\ldots +\left( n-2\right) x^{n}+\left( n-1\right) x^{n+1}$

=> $S\left( 1-x\right) =x^{2}+x^{3}+x^{4}+... +x^{n}-\left( n-1\right) x^{n+1}$

Since ${x}$ < 1

=> $S\left( 1-x\right) =\frac {x^{2}} {1-x}-\left( n-1\right) x^{n+1}$

and $n\rightarrow \infty \Rightarrow x^{n+1}=0$

=> $S\left( \frac {1} {2}\right) =\frac {\left( \frac {1} {2}\right) ^{2}} {1-\frac {1} {2}}-0$

=> $S=1$

The ratio of a girl to a boy is 1, so the sex ratio will be 1000.