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

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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

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

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

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    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Nov 22, 2018 at 10:50

2 Answers 2

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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

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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.

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