Question

(5.) You run a store that sells Christmas trees for Sd each. Suppose the number of customers that enter your store on a given day is described by a Poisson random variable with parameter λ. Further, suppose that each customer buys a tree with probability p, and that all customers are independent. (a) What is the expected number of customers that enter your store on a given day? (b) What is your expected revenue for a single day (revenue is the amount of money that your store takes in from sales). Justify your answer. (c) We might expect that the mean number of customers that enter your store on a given day will increase if you decrease the cost of a tree. Likewise, we might also expect that the probability of a customer purchasing a tree once they are in your store p increases if you decrease the cost of a tree. Suppose the mean number of customers that enter your store on a day is given by e-ad and that the probability of purchasing a tree is given by p-e-M (with a 0 and b 0). What selling price d maximizes your revenue on a given day? (Justify your answer)

Answer 1

(a)

The mean of Poisson random variable is $\lambda$

Expected number of customers entering the store on a single day = $\lambda$

(b)

Probability of buying a tree by a customer = p

Then, number of customer buying a tree ~ Binomial(p, $\lambda$)

Expected number of customer buying a tree = p$\lambda$

Expected revenue on a single day = p$\lambda$ * $d = $ p$\lambda$d

(c)

Expected number of customers entering the store on a single day for given cost d = $e^{-ad}$

Probability of buying a tree by a customer for given cost d = $e^{-bd}$

Expected number of customer buying a tree for a given d = $e^{-ad}$ * $e^{-bd}$ = $e^{-(a+b)d}$

Expected revenue on a single day = $e^{-(a+b)d}$ * d = $de^{-(a+b)d}$

Let E = $de^{-(a+b)d}$

For maximum E,

$\frac{\partial E}{\partial d} = e^{-(a+b)d} - d(a+b)e^{-(a+b)d} = 0$

$\Rightarrow 1 - d(a+b) = 0$

$\Rightarrow d(a+b) = 1$

$\Rightarrow d = 1/(a + b)$

Now,

$\frac{\partial^2 E}{\partial d^2} = -(a+b)e^{-(a+b)d} - (a+b)e^{-(a+b)d} -d(a+b)^2e^{-(a+b)d}$

As d > 0, a > 0, b > 0,

$\frac{\partial^2 E}{\partial d^2} < 0$

and thus, E is maximized at d = 1/(a+b)

Thus, the selling price d = 1/(a+b) maximizes the revenue on a given day.