Jtdylktuy

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.

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 $sumlimits_{k=0}^infty e^{-lambda} frac{lambda^{k+1}}{k!} = lambda not=1$, 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...) $$dfrac{2cdot 1}{0cdot 3+1cdot 2+2cdot1}=dfrac{2cdottfrac 16}{0cdot tfrac 36+1cdot tfrac 26+2cdottfrac 16}$$

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

Or $mathsf E(Scdotmathbf 1_{(S=j+1)})/mathsf E(S)$ And that is ...$$dfrac{(j+1)cdotdfrac{lambda^{j+1}je^{-lambda}}{(j+1)!}}{lambda}=dfrac{lambda^je^{-lambda}}{j!}qquadBig[jin{0,1,2,ldots}Big]$$

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 $dfrac{mathsf P(S=j+1)}{mathsf E(Smid S>0)}$ and as a reality check $sum_{j=0}^inftydfrac{mathsf P(S=j+1)}{mathsf E(Smid S>0)}=dfrac{1-e^{-lambda}}{lambda+e^{-lambda}-1}$.

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