Why is that if $X \sim \operatorname{Poisson}(1)$ then $E(3^X) = e^{2} $?

Question

Why is that if $X \sim \operatorname{Poisson}(1)$ then $E(3^X) = e^{2} $?

I'm trying to derive a general formula for $E(3^X) $, but without luck. Any help is appreciated!

Answer 1

Recall that for a Poisson distribution, $$\sum_{k = 0}^\infty e^{-\lambda} \frac{\lambda^k}{k!} = 1$$ for some $\lambda>0$.

Since $X$ is a discrete random variable, then using the formula $$E[g(X)] = \sum_{k = 0}^\infty g(k)P(X = k),$$ we have $$E[3^X] = \sum_{k = 0}^\infty 3^k\cdot e^{-1}\frac{1^k}{k!} = e^{-1}\cdot e^{3}\underbrace{\sum_{k=0}^\infty e^{-3}\frac{3^k}{k!}}_{1} = e^2.$$

Answer 2

I think we have to follow the definition $E(g(X)) = \sum_{x=0}^{\infty} g(x) \cdot f(x)$. But we have that $f(x) = e^{-\lambda} \cdot \frac{\lambda^x}{x!}.$

Thus, setting $g(x) =3^x$, we have: $$E(3^X) = \sum_{x=0}^{\infty} 3^x \cdot e^{-1} \cdot \frac{1^x}{x!}=\frac{1}{e}\sum_{x=0}^\infty \frac{3^x}{x!} =\frac{1}{e}\cdot e^{3}=e^2.$$