Prove that ${(1-p+pe^t)}^n = E(e^{t\sum Z_i}))$

Question

Let $Z_1,...,Z_n$ be independent random variables such that $ Z_i \in \vert0,1\vert$. Prove that ${(1-p+pe^t)}^n = E(e^{t\sum Z_i}))$ where $p=\sum{\frac{E(Z_i)}{n}}$

Not quite sure how to do this. Do I write out the formula (as a summation) for $E(e^{tZ_i})$ and then use Taylor series to get bounds on $e^{tZ_i}$ maybe? I just don't see it at the moment. AM-GM??

would like some help

Answer 1

Hint:

Independence of the $Z_i$ (hence also of the $e^{tZ_i}$) says: $$\mathbb E(e^{t\sum Z_i})=\mathbb E(e^{tZ_1})\times\cdots\times\mathbb E(e^{tZ_n})$$ If the $Z_i$ are identically distributed then this leads to:$$\mathbb E(e^{t\sum Z_i})=\mathbb E(e^{tZ_1})^n$$Also note that here: $$p=\mathbb E(Z_1)$$