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

Question

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

Answer 1

• 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 $\mu$.

Answer 2

The problem is ill-posed. You answered one possible interpretation of it, but apparently another interpretation is intended. To say that a child is chosen “randomly” tells us nothing without specifying the distribution according to which it is chosen. Typically, when no distribution is specified, the intention is that a uniform distribution is implied. This becomes problematic if there is more than one candidate uniform distribution.

The distribution that you seem to be assuming is one where first a family is chosen uniformly randomly among all families, and then a child in that family is chosen uniformly randomly. (I gather that you're assuming this distribution from your use of $X=Y+1$.)

The intended distribution (which to my mind seems closer to the actual wording) is likely the uniform distribution over all children.

Answer 3

Let $p_k=P(X=k)$. If all children have equal probability of being chosen, then the probability of being in a family with $k$ children is $$\frac{kp_k}{\sum ip_i}= \frac{kp_k}{1.8}$$ and so the expected number of siblings is$$\frac{\sum (k-1)kp_k}{1.8}=\frac{0.36+1.8^2-1.8}{1.8}=1.$$