# Distribution of no. of siblings of a random child if the no. of children of a family is Poisson distributed

Consider a large population of families, and suppose that the number of children in the different families are independent Poisson random variables with mean $$\lambda$$. Show that the number of siblings of a randomly chosen child is also Poisson distributed with mean $$\lambda$$.

My approach:

For any random child, if it has $$k$$ siblings, it implies that its parent had $$k+1$$ children. Hence, if $$S =$$ no. of siblings and if $$C =$$ no. of children

I'm not sure how to proceed after this. I tried evaluating the mean of S, by computing

I'm not sure where I'm going wrong and how to proceed.

EDIT: Found an answer in one of the solution manuals. Basically,

The probability of choosing a child that has $$j$$ siblings is the fraction of total children that have $$j$$ siblings.

Now, if $$Z$$ is the total number of families, hence the total no. of children would be $$\lambda Z$$. Also, if $$P(j+1)$$ is the probability that a family has $$j+1$$ children, then the no. of families with $$j+1$$ children is $$Z \cdot P(j+1)$$. Also, each of this family has $$(j+1)$$ children, each of whom have $$j$$ siblings. Hence, there are in total

children each having $$j$$ siblings. This divided by total number of children gives the fraction of children with $$j$$ siblings, i.e.

Clearly my answer is wrong in the first step itself. However I'm not able to articulate why the initial step is incorrect.

• Please read this tutorial on how to typeset mathematics on this site. Nov 22, 2018 at 10:50

As you say, the problem is in the first step

The likelihood that a randomly chosen family has $$m=k+1$$ children is proportional to $$e^{-\lambda} \frac{\lambda^m}{m!}$$

but the likelihood that a randomly chosen child is in a family with $$m$$ children is proportional to $$m e^{-\lambda} \frac{\lambda^m}{m!}$$ since there are more children in larger families than in smaller famalies, and in particular you cannot choose a child from families with $$0$$ children

so the likelihood that a randomly chosen child has $$k=m-1$$ siblings is proportional to $$(k+1) e^{-\lambda} \frac{\lambda^{k+1}}{(k+1)!} = e^{-\lambda} \frac{\lambda^{k+1}}{k!}$$, which is not what you have as you have $$(k+1)!$$ in the denominator

This is not the exact probability unless $$\lambda=1$$, as taking the sum $$\sum\limits_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k+1}}{k!} = \lambda \not=1$$, so we need to divide the expression by $$\lambda$$ to give the probability that a randomly chosen child is in a family with $$m$$ children as $$e^{-\lambda} \frac{\lambda^{k}}{k!}$$, as expected

If I have three boxes with no apples, two boxes with one apple, and a box with two apples, then the probability that a randomly selected apple comes from the later box is: the count for all apples in boxes containing two apples divided by the total count for apples. (Note: not just the count for apples per box of...) $$\dfrac{2\cdot 1}{0\cdot 3+1\cdot 2+2\cdot1}=\dfrac{2\cdot\tfrac 16}{0\cdot \tfrac 36+1\cdot \tfrac 26+2\cdot\tfrac 16}$$

Likewise the probability that a random child has $$j$$ siblings (ie from a family with $$j+1$$ children) is: $$\dfrac{(j+1)~\mathsf P(S=j+1)}{\mathsf E(S)}\qquad\Big[j\in\{0,1,2,\ldots\}\Big]$$

Or $$\mathsf E(S\cdot\mathbf 1_{(S=j+1)})/\mathsf E(S)$$ And that is ...$$\dfrac{(j+1)\cdot\dfrac{\lambda^{j+1}je^{-\lambda}}{(j+1)!}}{\lambda}=\dfrac{\lambda^je^{-\lambda}}{j!}\qquad\Big[j\in\{0,1,2,\ldots\}\Big]$$

Clearly my answer is wrong in the first step itself. However I'm not able to articulate why the initial step is incorrect.

You tried to evaluate $$\dfrac{\mathsf P(S=j+1)}{\mathsf E(S\mid S>0)}$$ and as a reality check $$\sum_{j=0}^\infty\dfrac{\mathsf P(S=j+1)}{\mathsf E(S\mid S>0)}=\dfrac{1-e^{-\lambda}}{\lambda+e^{-\lambda}-1}$$.

You did not account for the size of the families , and you needlessly eliminated 0 size families.