16 November 2011

In which I declare four things which my probability class is not about

In class today, I said approximately this:

So people decide whether to have children by flipping a coin, and if it comes up tails they have a kid, and if it comes up heads they don't. They repeat this until it comes up heads. This is probably not a good model of how people decide whether or not to have children, but maybe it's good in the aggregate. And anyway this isn't a class about how people decide whether to have kids.

Then there are two kinds of children, girls and boys -- well, not always, but this isn't a class about that -- and each child is equally likely to be a boy or a girl -- well, wait, that's not exactly true, but it's not a horrible assumption about how reproduction works on a cellular level, but this isn't a class about that either.

And people's decisions to stop having kids is independent of the sex of the children they've had -- which says this isn't China, because people do interesting things under the one-child policy -- but this isn't a class about that.

(Then I actually did some math -- namely, assume that the number of children a random family has is geometrically distributed with some parameter p, and assume that all children are equally likely to be male or female and that their genders are independent of the gender of any other children or the number of children in the family. Pick a random family with no boys. What is the distribution of the number of children they have?)


Rafael said...

Let the number of homes with one child be 2x, then half would have no boys. If the family has two children, then the probability of having no boys would be (1/2)ˆ2; with three children, the probability would be (1/2)ˆ3 etc.

The distribution would be a function Y, being equal to 2x*(1/2)^n. And n could assume values from 1 to infinity. In the case of n=1, the number of families with no boy would be x.

Panos Ipeirotis said...

Assume t is the total number of children, and b is the number of boys.

Pr(t) = p^t*(1-p)

Since we have equal probability of boys and girls, if we know the total number of children t:

Pr(b|t) = binomial(t,b) * 0.5^b * 0.5^(t-b)

Now, we want the number of children, given the number of boys:

Pr(t|b) = Pr(b|t) * Pr(t) / Pr(b)

= binomial(t,b) * 0.5^t * p^t*(1-p) / sum)t (Pr(b|t) *P(t)

For b=0, we have (1-p/2)*p^t*0.5^t

I have no idea what is the name of this distribution though :-)

Panos Ipeirotis said...

And, of course, being an idiot, I could not see that this is a geometric with parameter 0.5*p :-)