Question

2. The standard deviation of a probability distrubution measures how bunched together the data is. The larger the standard deviation, the further away from the mean the data tends to be. Consider the two density functions below: 1 P1(c) 2-12,000 2 (3,204 e 1 2-22,000 e 10) 3, 204721 P2(x) 9) 12,060727 Suppose that these give the density functions for the return on investment for two different invest- ment options. Say we have two investment firms with different investing philosophies: Firm 1: This firm invests conservatively. They are interested in taking safe investments with lower risk, even if that means taking investments with lower expected yields. They are interested in avoiding taking any losses. Firm 2: This firm is are willing to take risks. For this investment, they want to maximize the proba- bility that they make at least $25,000 back. Explain which investment each firm chooses, and why. How can you explain these choices in terms of the standard deviations of each distribution?

Answer 1

The p.d.f. of X ~N($\small \mu, \sigma$) = $\small (1/\sigma \sqrt{2\pi})exp[(-1/2)((x - \mu)/\sigma)^2]$

It is easy to see that p1(X) and p2(x) are normal distributions.

The mean and standard deviation for each are:

For p1(x), $\small \mu1 = 12000, \sigma1 = 3204$

For p2(x), $\small \mu2 = 22000, \sigma2 = 12060$

For p1(x), expectation E(x) = $\small = \mu1 = 12000$

For p2(x), expectation E(x) = $\small = \mu2 = 22000$

Hence, Firm1 chooses p1(x) as it has lower expected yield(E(x)) with lower risk($\small \sigma$).

The probability of making atleast $25,000 is:

For p1(x), P(x > 25000) = P(z >= (25000 - 12000)/3204) = P(z > 4.05742821473) = 1 - P(z < 4.05742821473) = 0 (approximately)

For p2(x), P(x > 25000) = P(z >= (25000 - 22000)/12060) = P(z > 0.2487562189) = 1 - P(z < 0.2487562189) = 0.4018

Hence, Firm2 chooses p2(x) as the probability of making atleast $25,000 is higher.