Stat 134 Fall 2011: Distributions of numbers of children

Stat 134 Fall 2011: Distributions of numbers of children
Michael Lugo
November 16, 2011
We considered the following model in class today: in a certain town, the probability
that a random family has t children is P (T = t) = (1 − p) t p. All children have probability
1/2 of being a boy; the gender of each child is independent of all other children and of the
number of children the family has. So if B is the number of boys that a family has, then
P (B = b|T = t) = 2 −t( t
b)
. We want to know the distribution of the number of boys.
By the definition of conditional probability we can say
P (B = b) = ∑ t
= ∑ t≥b
P (B = b|T = t)P (T = t)
( ) t
2 −t (1 − p) t p
b
but this sum is tricky! To be able to do it let’s make it even harder; we’ll find ∑ b≥0 P (B =
b)z b , the generating function of B. This is given by the double sum
∑
P (B = b)z b = ∑ ( ∑
( ) )
t
2 −t q t p z b
b
b≥0 n≥0 t≥b
where q = 1 − p. Often with these sums it turns out to be useful to interchange the order of
summation. If we do this we get
∑ t∑
( ) t
2 −t q t pz n
n
t≥0 n=0
and pulling otu constants gives
∑ ( q t
t∑
( t p z
2)
n)
n .
t≥0
n=0
Now the inner sum (over n) can be done by the binomial theorem; it’s (1 + z) t . So we’re left
with
∑ ( q
) t
p(1 + z) t .
2
t≥0
1

This is just a geometric series, which is a little more obvious if it’s written in the form
( ) t q(1 + z)
.
p ∑ t≥0
2
Its sum is therefore
and after some rewriting this is
p
1 − q (1 + z)
2
(
1 −
2p
1+p
1−p
1+p
)
z.
Of course this is just a geometric series; we can expand it to get
2p
1 + p + 2p
1 + p
1 − p
1 + p z + 2p
1 + p
and the coefficient of z b is P (B = b); this is
P (B = b) =
2p
1 + p
( ) 2 1 − p
z 2 + · · ·
1 + p
( ) b 1 − p
.
1 + p
In particular, the number of boys that a family has is geometric, with parameter 2p/(1 + p).
Also, if you recall that the mean of a geometric(p) is just (1 − p)/p, then you find that
E(T ) = (1 − p)/p and
E(B) = 1 − 2p
1+p
2p
1+p
= 1 + p − 2p
2p
= 1 − p
2p .
So E(B) = E(T )/2, which is of course what you’d expect – the expected number of boys is
exactly half the expected number of children. What’s surprising is that B turns out to be
geometric. We can work out P (T = t|B = b) as well:
P (T = t)P (B = b|T = t)
P (T = t|B = b) =
P (B = b)
= (1 − p)t p2 −t( )
t
b
(
=
=
2p
1+p
1−p
1+p
) b
( ) t ( ) ( ) b 1 − p t 1 + p 1 + p
.
2 b 2 1 − p
( ) ( ) b+1 ( ) t−b (t + 1) − 1 1 + p 1 − p
.
(b + 1) − 1 2 2
2

The reason for writing it int his form is for comparison with the negative binomial. The
number of trials T r until the rth success in Bernoulli(π) trials has distribution
( ) t − 1
P (T r = t) = π r (1 − π) t−r
r − 1
and so comparing these, we see that, given B = b, the distribution of T +1 is the distribution
of the waiting time until the (b + 1)st success in Bernoulli((1 + p)/2) trials.
In particular that means that
and so we get the formula
E(T + 1|B = b) = b + 1
1+p
2
E(T |B = b) = b + 1
1+p
2
− 1 = 2b − p + 1 .
p + 1
In particular, as p → 0 (which corresponds to large families) we get
2b − p + 1
lim
p→0 p + 1
= 2b + 1
and so if we learn a family has b boys, we expect them to have 2b + 1 children; the 2 here is
probably not surprising but the 1 is. In the limit as p → 1 (small families) we have
2b − p + 1
lim
p→1 p + 1
If people just don’t have children at all, then if we learn a family has, say, three boys, it’s
very likely that that’s all of their children.
= b.
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