Function with variable of Poisson distribution

Question

Let be a function $f= 100e^{-0.25X}.$

Assume $X$ follows Poisson distribution with $λ =0.5.$

What is the mean or expected value of the function $f$?

Since we have a Poisson distribution then mean will be equal to $λ$ and $0.5$ consequently. Can we just plug $0.5$ into equation? This seems really confusing and too easy.

Answer 1

You have a function $f$ which associated to any value $x$ the value $f(x) = 100 \cdot e^{-0.25 \cdot x}$. Now you wish to know what the expected value of $f(N)$ is with $N$ Poisson$(\lambda)$. For this you have the following definition:

$$ \mathbb{E}[f(N)] = \sum_{n=0}^{\infty} f(n) \mathbb{P}\{N = n\}, $$ with $\mathbb{P}\{N = n\}$ the probability that the poisson distribution $N$ is equal to $n$. For further reference see:

https://imai.princeton.edu/teaching/files/Expectation.pdf

and

https://en.wikipedia.org/wiki/Poisson_distribution

Answer 2

\begin{align} & \operatorname{E} \left(100e^{-0.25X} \right) = \sum_{x=0}^\infty 100 e^{-0.25 x} \Pr(X=x) = \sum_{x=0}^\infty 100 e^{-0.25 x} \frac{\lambda^x e^{-\lambda}}{x!} \\[10pt] = {} & 100e^{-\lambda} \sum_{x=0}^\infty \frac{\left( e^{-0.25}\lambda \right)^x}{x!}= 100e^{-\lambda} e^{-0.25\lambda} = 100 e^{-1.25\lambda}. \\& \quad \uparrow \quad \text{This can be done because $100e^{-\lambda}$ does not change as $x$ changes.} \end{align}