A and B are two competing companies. An investor decides whether to buy (a) 100 shares of A, or (b) 100 shares of B, or (c) 50 shares of A and 50 shares of B. A profit made on 1 share of A is a random variable X with the distribution P(X = 2) = P(X = −2) = 0.5. A profit made on 1 share of B is a random variable Y with distribution P(Y=4)=0.2 and P(Y=-1)=0.8
If X and Y are independent, compute the expected value and variance of the total profit for strategies (a,) (b,) and (c.)

a) 100 shares of A

E(100X) = 0

Var(100X) = 40,000

b) 100 shares of B

E(100Y) = 0

Var(100Y) = 40,000

c) 50 shares of A and 50 shares of B

E(50X + 50Y) = 0

Var(50X + 50Y) = 20,000

They all give the same expected profit of 0, but option 3 shows the lowest variation in expected profit with variance of 20,000, so, it's the most advisable investment option.

Step-by-step explanation:

Expected value is given by

E(X) = Σ xᵢpᵢ

where xᵢ = each possible variable/sample space

pᵢ = probability of each possible variable/sample space happening.

Variance = Var(X) = Σxᵢ²pᵢ − μ²

where μ = E(X)

A profit made on 1 share of A is a random variable X with the distribution P(X = 2) = P(X = −2) = 0.5.

A profit made on 1 share of B is a random variable Y with distribution P(Y=4)=0.2 and P(Y=-1)=0.8

Expected profit on one share of A = E(X) = (2×0.5) + (-2×0.5) = 0

Var(X) = (2² × 0.5) + [(-2)² × 0.5] - 0² = 2 + 2 = 4

Expected profit on one share of B = E(Y) = (4×0.2) + (-1×0.8) = 0

Var(Y) = (4² × 0.2) + [(-1)² × 0.8] - 0² = 3.2 + 0.8 = 4

a) 100 shares of A

Expected value of profit on 100 shares of A = E(100X)

E(100X) = 100 E(X) = 100 × 0 = 0

Var(100X) = 100² Var(X) = 100² × 4 = 40,000

b) 100 shares of B

Expected value of profit on 100 shares of B = E(100Y)

E(100Y) = 100 E(Y) = 100 × 0 = 0

Var(100Y) = 100² Var(Y) = 100² × 4 = 40,000

c) 50 shares of A and 50 shares of B

Since the two events are described to be independent,

E(50X + 50Y) = E(50X) + E(50Y) = 50E(X) + 50E(Y) = (50×0) + (50×0) = 0

Var(50X + 50Y) = Var(50X) + Var(50Y) = 50² Var(X) + 50² Var(Y) = 50² (4 + 4) = 20,000

Hope this Helps!!!

answer: no, but if it does it's a obvious pattern.

step-by-step explanation:

98%

step-by-step explanation: