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I'm checking over solutions for probability homework and the question reads:

A store owner has overstocked a certain item and decides to use the following promotion to decrease the supply. The item has a marked price of 100. For each customer purchasing the item during a particular day, the owner will reduce the price by a factor of one-half. Thus, the first customer will pay 50 for the item, the second will pay 25, and so on. Suppose that the number of customers who purchase the item during the day has a Poisson distribution with mean 2. Find the expected cost of the item at the end of the day. Hint: The cost at the end of the day is $100(1/2)^Y,$ where $Y$ is the number of customers who have purchased the item.

The answer key solution does not make much sense to me:

Since $Y$ is Poisson with $\lambda=2,$ $$E(C)=E\left[100\left(\frac12\right)^{\!\!Y\,}\right]=\sum_{y=0}^{\infty}\frac{100(\frac{2}{2})^y e^{-1}}{y!}=100e^{-1}\sum_{y=1}^{\infty}\frac{1^ye^{-1}}{y!}=100e^{-1}.$$

Why is it that I can't just say the expected number of people to buy the item is 2 so the expected cost at the end of the day is $100(1/2)^2 = 25?$

Also in the solution how did the fraction become $2/2$ in the first summation?

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  • $\begingroup$ Is it possible that the solution manual is wrong? I don't understand why the answer isnt 25 $\endgroup$
    – VicVer09
    May 11, 2018 at 16:47
  • $\begingroup$ What is the answer given in your manual? $\endgroup$ May 11, 2018 at 17:21

2 Answers 2

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If $f$ is a function and $X$ is a random variable, then in general $$\mathbb{E}[f(X)] \neq f(\mathbb{E}[X]).$$ (You can make quantitative statements about the relationship between these two quantities when $f$ has certain structure, e.g. equality when $f$ is linear or Jensen's inequality when $f$ is convex.)

By attempting to plug $2$, the expected value of $Y$, into $f(x) = 100 (1/2)^{x}$, you are making this error. Instead you should compute $\mathbb{E}[f(Y)]$ directly, as in your solution.

The fraction became $2/2$ by inserting the pmf of a Poisson random variable with rate $2$ into the definition of the expected value. That is, $$\mathbb{P}[Y = y] = \frac{2^y e^{-2}}{y!}.$$

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  • $\begingroup$ Perfect! makes sense $\endgroup$
    – VicVer09
    May 11, 2018 at 16:56
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Let $X$ represent the price of the product at any given time of purchase. We have to find $E[X]$.

We notice that $x$ (the value associated with $X$) is a function of the number of people (say $n$, with its random variable denoted as $N$) who bought the product prior, and if the original price of the product is F, then $x=g(n)=\frac{F}{2^n}$.

We are given $E[N]=2=\lambda$ which follows a Poisson distribution.

Now, $$E[X]=E[g(N)]=\sum_{k=0}^{\infty} g(n=k)\frac{\lambda^k e^{-\lambda}}{k!}$$ $$ \Rightarrow E[X]= \sum_{k=0}^{\infty} \frac{F}{2^k} . \frac{2^k e^{-\lambda}}{k!}$$ $$ \Rightarrow E[X]= Fe^{-\lambda} \sum_{k=0}^\infty \frac{1}{k!}$$ $$ \Rightarrow E[X]= Fe^{-\lambda} (e)$$

Putting values, we get

$$ \Rightarrow E[X]= 100 e^{-2} (e)= 100 \frac{1}{e}$$ $$ \Rightarrow E[X]= 100*0.3678=36.78$$

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