Expected number of siblings, given the expectation of children in a family

expected valueprobabilityprobability distributionsprobability theoryvariance

The number of children in a family is a random variable $X$, where
$ \mathbb E(X) = 1.8 $ and $Var(X) = 0.36 $. If we randomly choose a child, which is the expected number of its siblings?

(it is given that the answer $> 0.8$)

Let the number of a child's siblings be a random variable $Y$, so we are looking for $ \mathbb E(Y) $.

If found that,

$$
\mathbb E(Y) = \sum_{k=0}^{\infty}k \; \mathbb P(Y=k)
= \sum_{k=0}^{\infty}k \; \mathbb P(X=k+1) =
$$
$$
= \sum_{k=0}^{\infty}k \; \mathbb P(X=k)-\sum_{k=0}^{\infty}\mathbb P(X=k)
+\mathbb P(X=0)= \mathbb E(X)-1+\mathbb P(X=0)
$$
which is indeed greater than $0.8$, if $\; \mathbb P(X=0)>0$.

But, I don't know how to make use of the variance to find the final answer.

Thank you in advance

Best Answer

  • If there are $n$ families then there are $1.8 n$ children
  • If there are $n_k$ families with $k$ children then $\sum_k n_k =n$ and $\sum_k k \,n_k =1.8n$ and $\sum_k k^2 \,n_k =0.36 n +1.8^2n = 3.6 n$
  • A child in a size $k$-child family has $k-1$ siblings so there are $\sum_k k(k-1) \,n_k = 3.6n-1.8n = 1.8n$ siblings to count (including multiple cases when there are $3$ or more children in a family)
  • So the average number of siblings per child is $\frac{1.8n}{1.8n}=1$

In general the average number of siblings per child is $\mu-1+\frac{\sigma^2}{\mu}$. If the children were Poisson distributed, this would be $\mu$.

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