### PROBABILITY GENERATING FUNCTIONS:

From Wikipedia:

In probability theory, the probability generating function of a discrete random variable is a power series representation (the generating function) of the probability mass function of the random variable. Probability generating functions are often employed for their succinct description of the sequence of probabilities $$\Pr(X = i)$$ in the probability mass function for a random variable X, and to make available the well-developed theory of power series with non-negative coefficients.

If $$X$$ is a discrete random variable taking values in the non-negative integers $$\{0,1, \dots\}$$, then the probability generating function of $$X$$ is defined as:

$\color{blue}{\displaystyle G(z)=\mathbb{E} \left(z^{X}\right)=\sum_{x=0}^{\infty }p(x)\;z^{x}}$

where $$p$$ is the probability mass function of $$X$$. Note that the subscripted notations $$G_X$$ and $$p_X$$ are often used to emphasize that these pertain to a particular random variable $$X$$, and to its distribution.

CHARACTERISTICS:

From this video:

1. IT GIVES YOU PROBABILITIES by differentiating:

$\color{blue}{\large p_i = \left. \frac{1}{i!}\quad\frac{d^i \, G(z)}{dx^i} \right|_{z=0}=\frac{1}{i!} \;G^{(i)}\;(0)}$

1. $$G\,(1)=1$$ because $\displaystyle\sum_{i=0}^\infty p_i \; 1^i=1$

2. First differential

$G^{(1)}(z) =\frac{d}{dz}\mathbb E\left[z^X\right]=\mathbb E\left[X\,z^{X-1}\right]$

1. The first differential evaluated at $$1$$ gives you the mean: $G^{(1)}(1) =\left.\mathbb E\left[X\,z^{X-1}\right]\right|_{z=1}=\mathbb E\left[X\quad1^{X-1}\right]= \mathbb E[X].$

2. The second derivative evaluated at $$1$$ is the factorial momment, and is NOT the variance, because the second term is not squared.

\begin{align}G^{(2)}\;(1) &=\frac{d^2}{dz^2}\; \left.\mathbb E\left[z^X\right]\right|_{z=1}\\[2ex]&=\mathbb E\left[X\;(X-1)\;z^{X-2}\right]\\[2ex]&=\mathbb E\left[X\;(X-1)\right]\\[2ex]&=\mathbb E\left [X^2-X\right ]\\[2ex]&=\mathbb E\left[X^2\right] - \mathbb E\left[X\right]\end{align}

1. Generalizing, then, the $$i$$-th derivative evaluated at $$1$$ is the $$i$$-th factorial moment:

$G^{(i)}\;(1)= \mathbb E\left[X\;(X-1)\;\cdots\;(X-i+1)\right]$

1. To get the variance,

\begin{align}\sigma^2 &= \mathbb E\left[X^2\right]-\mathbb E\left[X\right]^2 \\[2ex] &=G^{(2)}\;(1)+G^{(1)}\;(1)-\left[G^{(1)}\;(1)\right]^2 \end{align}

1. We can get raw moments by differentiating the pgf and multipling it by $$z$$:

$\mathbb E\left[X^i\right]= \left. \left( z\;\left(\frac{d}{dz}\right)^i \; G(z)\right)\right|_{z=1}$

#### PGF FOR BERNOULLI DISTRIBUTION

$f(i,p)= p^i\;(1-p)^{(1-i)} \text{ for }i\in \{0,1\}$

The PGF is:

\begin{align}G(z)&=\sum_{i=0}^\infty p_i\;z^i\\[2ex]&= p_0\;z^0+p_1\;z^1\\[2ex] &= p^0\;(1-p)^{(1-0)}\times z^0 + p^1\;(1-p)^{(1-1)}\times z^1\\[2ex]&= (1-p)+p^1\times z^1\\[2ex]&= (1-p)+pz\\[2ex]&=\color{red}{q + pz}\end{align}

Differentiating and evaluating at zero it gives probabilities:

$p_i = \left.\frac{1}{i!}\frac{d^iG(z)}{dz^i}\right|_{z=0}=\frac{1}{i!}G^{(i)}(0)$

so

$\color{red}{p_1= \frac{1}{1!}G^{(1)}(0)=\left.\frac{1}{1!}\frac{d}{dz}(q+pz)\right|_{z=0}=\;p}$

and

$\color{red}{p_0= \frac{1}{0!}G^{(0)}(0)=\left.\frac{1}{0!}(q+pz)\right|_{z=0}=\;q}$

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