Question

1. The World Hockey League (WHL) is looking to increase their terrible viewership, to increase their also terrible revenues. They hire a statistician, who estimates that if they spend at least $100,000 on marketing, then their additional revenue will be somewhere between $0 and $1,000,000. Fur- thermore, the statistician gives the WHL an estimate for the distribution of their revenues: Let R be the additional revenue they bring in, and let c be the amount they spend on marketing. Then the probability density function of their additional revenue is given by: 3,000,000,000,000 – 3R2 – 3c2 + 6cR p(R) 2,000,000,000,000,000,000+ 3,000,000,000,000c – 3,000,000c2 a) Find the smallest c for which the probability that their additional revenue exceeds $100,000 is at least 80%. b) One way to look at whether this is a worthwhile expense is to co er their additional mean profit. Their additional mean profit is their additional mean revenue (i.e., the mean revenue gained by running their marketing campaign) minus the associated costs of running the cam- paign. For example, if c = $100,000, then their mean revenue is $411, 894.27, and so their mean profit in this case is $411, 894.27 – $100,000 = $311, 894.27. What level of spending on marketing will maximize their mean profit?

Answer 1

a) What we need here is to find the smallest c such that:

$Prob(R \ge 100000) \ge 0.8$

$\implies \int_{10^5}^{10^6}p(R)dR \ge 0.8$

$\implies \int_{10^5}^{10^6} (3\cdot 10^{12} - 3R^2 - 3c^2 + 6cR)dR \ge 1.6 \cdot 10^{18} + 2.4 \cdot 10^{12} c - 2.4 \cdot 10^6 c^2 \\ \implies 0.3c^2 - 0.57 \cdot 10^6 c - 0.101\cdot 10^{12} \le 0 \\ \implies (c-2063178)(c+163178) \le 0 \\$

Therefore the smallest value of c that satisfies the above inequality is c = 0. And indeed the $Prob(R \ge 100000) = 0.8505$ when c = 0.

b) In part (b) we have to find c such that it maximizes the following expression:

$\int_{0}^{10^6} (R-c)p(R) dR$

$= \frac{0.75 \cdot 10^{18} - 1.5 \cdot 10^6 c^2 + 2c\cdot 10^{18}}{2\cdot 10^{24} + 3\cdot 10^{12}c - 3 \cdot 10^{6}c^2} - c$

Now this function is decreasing in c. Hence it attains a maximum at c = 0. The maximum mean profit comes out to be $375,000.