Poisson distribution and price reduction

Question

A store owner has certain existence of an item and he decides to use the following scheme to sell it:

The item has a price of \$100. The owner will reduce the price in half for every customer that buys the item on a given day. That way, the first customer will pay \$50, the next one will pay $25, and so on...

Suppose that the number of customers who buy the item during a day follow a Poisson distribution, with $\mu = 2$ Find the expected price of the item at the end of a day.

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I defined a "final price" function like this: $$g(x) = \frac{100}{2^x}$$ where $x$ is the random variable of the Poisson distribution: $$f(x)=\frac{\lambda^xe^{-x}}{x!}$$

Since $\mu = \lambda = 2$, I can rewrite $f(x)$ like this: $$f(x) = \frac{2^x}{e^2x!}$$

And there's where I'm stuck. I see two possible roads:

A: Is solving for $x$ the way to go? If so, how can I deal with $x!$ ?

B: Should I go with $E(g(x))$?

Answer 1

Assuming the final price is the price reduction after the last customer buys it. So for example if only 1 customer buys the item, then the final price is 50, which gives formula $P(X)=100/2^X$. Then since $X$ is Poisson with parameter $\mu=2$,

$$E[P(X)]=\sum_{x\geq 0} \frac{100}{2^x}\frac{1}{e^\mu}\frac{\mu^x}{x!}=\frac{100}{e^2}\sum_{x\geq 0} \frac{1}{x!}=100/e.$$

Answer 2

I'll use capital $X$ for the random variable and lower-case $x$ for the argument to the probability mass function (thus if I write $\Pr(X=x)$ with both capital $X$ and lower-case $x$ you'll know what I mean). You have \begin{align} & \operatorname{E}\left( \frac {100}{2^X}\right) = \sum_{x=0}^\infty \frac{100}{2^x} \Pr(X=x) = \sum_{x=0}^\infty \frac{100}{2^x}\cdot \frac{2^x}{x!e^2} \\[10pt] = {} & \frac{100}{e^2}\sum_{x=0}^\infty \frac 1 {x!} = \frac{100}{e^2}\cdot e = \frac{100} e \approx 36.78794. \end{align}