Generating function of $X+\sum\limits_{i=1}^{X}D_i$ with $X$ and every $D_i$ Poisson

Question

The following model can be used to describe the number of women (mothers and daughters) in a given area. The number of mothers is a random variable $X \in Po (\lambda)$. Independently of the others, every mother gives birth to a $Po(\mu)$- distributed number of daughters. Let $Y$ be the total number of daughters and hence $Z=X+Y$ be the total number of women in the area.

Find the generating function of $Z$.

The answer, according to the book (An intermediate course in probability, Allan Gut) is : $$g_{Z}(t)=\exp(\lambda (t\cdot e^{\mu (t-1)}-1))$$ But this is not what i get, so i would like to know if its me or the book who made a mistake.

My approach was the following:

$Z=X+Y= X +\sum_{i=1}^{X}D_i $, where $D_i \sim Po (\mu)$.

$\begin{align}g_{Z}(t) & =g_{X+Y}(t) \\ & =g_{X}(t)\cdot g_{X}(g_{D_1}(t))\\ & =\exp(\lambda (t-1))\cdot \exp(\lambda (e^{\mu (t-1)}-1))\\ &=\exp(\lambda (e^{\mu (t-1)} + t -2))\end{align}$

I fail to see where I would have done any mistakes, so I would like to believe that the book has done a mistake. Any ideas? Thanks!!

Answer 1

Let $Y_k$ be the count of daughters for mother $k$ so $Y=\sum_{k=1}^X Y_k$ and $(Y_k)\overset{\text{iid}}\sim\mathcal{Po}(\mu)$

The generating function of a sum of random variables is equal to the product of their generating functions only when the random variables are independent.

$Y=\sum_{k=1}^X Y_k$ means that $Y$ is rather dependent on $X$, so the step $\mathsf G_{X+Y}(t)=\mathsf G_X(t)\,\mathsf G_Y(t)$ is invalid.

Instead, return to the definition of probability generation, and use the identical distribution of all $Y_k$, and the total independence of all $Y_k$ and $X$.

$\begin{align}\mathsf G_Z(t) &= \mathsf E(t^Z) \\ &=\mathsf E(\mathsf E(t^{X+\sum_{k=1}^X Y_k}\mid X)) \\ & =\mathsf E(\phantom{t^{X(1+Y_1)}}) \\ & =\mathsf G_X(\phantom{\mathsf G_{1+Y_1}(t)}) \\ & =\mathsf G_X(t\,\mathsf G_{Y_1}(t)) \\ & = \exp\Big(\lambda \big(t\exp(\mu (t-1))-1\big)\Big)\end{align}$

Answer 2

Let $D_j$ be the number of daughters for the $j$th mother. Note that $G_{D_j}(x)$ is independent of $j$ since the $D_j$ all have the same distribution. We'll write this as $G_D(x)$.

Define:

$$h(x,y)=\sum_{i,j=0}^{\infty} a_{ij}x^iy^j$$

where $a_{ij}=P(X=i,D_1+D_2+\cdots+D_i=j)$ is the probability that there are $i$ mothers and the $i$ mothers have a total of $j$ daughters.

For fixed $i$, you have:

$$\begin{align}\sum_{j=0}^{\infty}a_{ij}y^j&=P(X=i)\sum_{j=1}^{\infty}P(D_1+D_2+\cdots+ D_i=j)y^j\\ &=P(X=i)G_{D_1+\cdots+D_i}(y) \\&=P(X=i)(G_{D}(y))^i\end{align}$$

because the $D_i$ are independent.

Then we get: $$h(x,y)=\sum_{i=0}^{\infty} P(X=i)x^i(G_{D}(y))^i = G_{X}(xG_{D}(y))$$

Now just show that the value you are seeking is:

$$G_{X+Y}(x)=h(x,x)=\sum_{i,j} a_{ij}x^ix^j = G_{X}(xG_{D}(x))$$

In your case, $G_X(x)=e^{\lambda(x-1)}$ and $G_D(x)=e^{\mu(x-1)}$, and thus your PGF is:

$$G_{X+Y}=\exp\left(\lambda\left(xe^{\mu(x-1)}-1\right)\right)$$

Note that:

$$G_Y(x)=h(1,x)=G_X(G_D(x))=\exp\left(\lambda\left(G_D(x)-1\right)\right)$$

If $X$ is Poisson.

So we get: $$\begin{align}\frac{G_{X+Y}(x)}{G_Y(x)}&=\frac{\exp\left(\lambda \left(xG_D(x)-1\right)\right)}{\exp\left(\lambda\left(G_D(x)-1\right)\right)}\\ &=\exp\left(\lambda\left((x-1)G_D(x)\right)\right)\\ &=G_{X}(x)^{G_D(x)} \end{align}$$

This holds whenever $X$ is Poisson, whatever $D$ is.