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Consider the following problem and solution. (I am stuck at the modified problem.)

Problem

There are exactly two phone shops, $A$ and $B$, serving a town of 1000 people. Both shops sell an iPhoneX at the same price. Assuming that every consumer is equally likely to purchase an iPhoneX from either of the stores, what is the minimum number of iPhoneX's that each shop should stock up so that the probability that customers are turned away because there were not enough iPhoneXs for sale is less than 0.01?

Solution

Define variable $X_i$ as follows: $$ X_i = \begin{cases}1, &\text{the ith customer buys from shop $A$}\\ 0, & \text{the ith customer does not buy from shop $A$} \end{cases} $$

Then $X_1, X_2, \dots, X_{1000}$ are i.i.d. variables and $\sum_{i=1}^{1000} X_i$ gives the number of customers who buy from shop $A$. $X_i$ follows a Bernoulli distribution and has $E(X_i)=0.5, Var(X_i)=0.25$.

Applying the Lindeberg-Levy CLT, we have $$ \frac{\sum_{i=1}^{1000}X_i-1000\times 0.5}{\sqrt{1000\times 0.25}}\sim N(0,1) $$ approximately. Let the minimum number of iPhoneXs each shop should stock up be $b$, such that $$ P \left\{\sum_{i=1}^{1000}X_i>b \right\}<0.01 $$ Then we have $$ P \left\{\sum_{i=1}^{1000}X_i>b \right\}= P\left\{ \frac{\sum_{i=1}^{1000}X_i-1000\times 0.5}{\sqrt{1000\times 0.25}} \le \frac{b-1000\times 0.5}{\sqrt{1000\times 0.25}}\right\} >0.99 $$ which gives $b>536.7$. Therefore the minimum number of iPhoneXs each shop should stock up is $537$.

Modified Question

Suppose now that there $n$ stores serving the town of 1000 people, all selling an iPhoneX at the same price, and that each consumer is equally likely to purchase an iPhoneX from either of the stores. Obtain an expression for the minimum number of iPhoneXs each store should stock up so that the probability that customers are turned away because of a shortage of iPhoneXs is less than 0.01.

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$\texttt{Comment:}$ I suspect I need to implement the multinomial distribution here, but I do not know how to keep define the variables to keep track of. Do I also have to use the Multidimensional Central Limit Theorem?

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  • $\begingroup$ @owen88 :your answer helped, thanks alot! $\endgroup$ – NetUser5y62 Nov 12 '18 at 11:11
  • $\begingroup$ You're welcome! $\endgroup$ – owen88 Nov 12 '18 at 11:32
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There is no need to use a multidimensional central limit theorem. As you did in your original problem, we can focus on a single one of the stores (due to the symmetry/exchange-ability of the problem) which we will call store $A$.

As before we let $X_i = 1$ if customer $i$ buys the phone at store $A$, otherwise $X_i = 0$. Now $\mathbf P[X_i = 1] = \frac1n,$ and we have

$$\mu = \mathbf E[X_i] = \frac1n, \qquad \sigma^2 = \text{Var}(X_i) = \frac1n\left(1- \frac1n\right).$$

Denoting $S_A = \sum_{i=1}^{1000} X_i$ for the number of sales at store $A$, the CLT asserts

$$S_A \approx N\left( \sqrt{1000} \mu, 1000 \sigma^2\right)$$

And we can calculate the required of phones to achieve a 99% likelihood of having sufficient stock via:

$$1000 \mu + \sqrt{1000 \sigma^2} \Phi^{-1}(0.99)$$

where $\Phi^{-1}$ is the inverse CDF of a $N(0,1)$ variable.

Note that in the case $n=2$ (i.e. the original problem) this is:

$$\left( 1000 \times \frac12\right) + \sqrt{ 1000 \times \frac14} \Phi^{-1}(0.99) \sim 536.78$$

The plot below shows how the number of phones store $A$ has to stock reduces as the number of competing stores increases.

Plot of approximate number of phones required given n competing stores

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