# plugging binomial moment function into poisson moment function

I have no idea how the author got from $$E[(pe^s+(1-p)e^t)^x]\rightarrow e^{\lambda(pe^s+(1-p)e^t-1)}$$

The author writes $a=e^t$, and then sets $a=pe^s+(1-p)e^t$ and plugs it in where $e^t$ normally goes in a Poisson moment function.

Why does this make sense? I don't get how $e^t=pe^s+(1-p)e^t$

• I cannot find in the proof what you describe. Which of the equalities mentioned in the proof is unclear to you? – drhab May 22 '18 at 11:36
• The part where they plug into the poisson moment function – Frank May 22 '18 at 17:13

If $X$ has Poisson distribution with parameter $\lambda$ then: $$\mathbb Ee^{tX}=e^{\lambda(e^t-1)}\text{ for every }t\in\mathbb R\tag1$$

This can also be written as $\mathbb E(e^t)^X=e^{\lambda(e^t-1)}$ and substitution $a=e^t$ then tells us that $(1)$ is the same statement as: $$\mathbb Ea^X=e^{\lambda(a-1)}\text{ for every }a>0\tag2$$

This means that we can substitute for $a$ every expression that results in a positive real number and we choose for the expression $pe^s+(1-p)e^t$.

This gives us the result:$$\mathbb E(pe^s+(1-p)e^t)^X=e^{\lambda(pe^s+(1-p)e^t-1)}\tag3$$

It has been shown earlier in the proof that the LHS of this equality equals $\mathbb E(e^{sX_c+t(X-X_c)})$.

Further the RHS of the expression equals $e^{\lambda p(e^s-1)}e^{\lambda (1-p)(e^s-1)}$.

So our final result is:$$\mathbb E(e^{sX_c+t(X-X_c)})=e^{\lambda p(e^s-1)}e^{\lambda (1-p)(e^s-1)}\tag4$$