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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.

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 $E(g(x))$?

paranoidhominid
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  • Is there a reason why you're using two Greek letters $\mu$ and $\lambda$? Are they supposed to be two different things? ${}\qquad{}$ – Michael Hardy Jul 28 '15 at 23:37
  • Yes, $\mu$ is the mean and $\lambda$ is the Poisson parameter (Schaum). – paranoidhominid Jul 28 '15 at 23:39
  • $\mu$ is the mean of what? Of the Poisson-distributed random variable? Of the price? Of something else? If it's of the Poisson distributed random variable, then it's the "Poisson paragemeter", so there's no occasion for using two distinct letters rather than just one. ${}\qquad{}$ – Michael Hardy Jul 28 '15 at 23:48

2 Answers2

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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}

Michael Hardy
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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 $\mu=2$,

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

Alex R.
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