Chapter 6 Homework Problem Solutions - ISyE Homeshackman/isye4803A_F07/Ch6Solns.pdf · Chapter 6...

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Chapter 6 Homework Problem Solutions ISyE 4803 Problem 3 The covariance matrix is Σ= 0.0225 0.0045 0.0045 0.0900 ! , and its inverse is Σ -1 = 1 0.0020048 0.0900 -0.0045 -0.0045 0.0225 ! . The minimum variance portfolio x * is proportional to Σ -1 e, which means that it is proportional to the vector (0.0855, 0.0180). After normalization, the minimum variance portfolio equals (0.826, 0.174). The variance corresponding to the minimum variance portfolio equals 1 e T Σ -1 e = 0.0020048 0.09 - 2(0.0045) + 0.0225 =0.0193696, and thus the standard deviation σ =0.1392 or 13.92%. The expected return of the portfolio x * is 11.392%. Problem 5 The expected rate of return if Gavin buys u units of insurance is 0.50(3000000) + 0.50u 1000, 000 + 0.50u . If Gavin purchases 3000000 units of insurance, he will guarantee a cash flow of 3000000 next year. Since his initial investment will be 2500000 his risk-free return will be 20%. Problem 7 The minimum variance portfolio is proportional to 0.7500 -0.5000 0.2500 -0.5000 1.0000 -0.5000 0.2500 -0.5000 0.7500 1 1 1 = 0.50 0 0.50 . (1) The vector on the right-hand side of (1) is already normalized, and so it is the minimum vari- ance portfolio. (The solution can also be found by directly solving the system of equations given by Σx = e.) Now ˆ r = r - 0.2e = (0.2, 0.6, 0.6). The tangent portfolio is proportional to 1

Transcript of Chapter 6 Homework Problem Solutions - ISyE Homeshackman/isye4803A_F07/Ch6Solns.pdf · Chapter 6...

Chapter 6 Homework Problem SolutionsISyE 4803

Problem 3

The covariance matrix is

Σ =

(0.0225 0.00450.0045 0.0900

),

and its inverse is

Σ−1 =1

0.0020048

(0.0900 −0.0045−0.0045 0.0225

).

The minimum variance portfolio x∗ is proportional to Σ−1e, which means that it is proportionalto the vector (0.0855, 0.0180). After normalization, the minimum variance portfolio equals(0.826, 0.174). The variance corresponding to the minimum variance portfolio equals

1eT Σ−1e

=0.0020048

0.09− 2(0.0045) + 0.0225= 0.0193696,

and thus the standard deviation σ = 0.1392 or 13.92%. The expected return of the portfoliox∗ is 11.392%.

Problem 5

The expected rate of return if Gavin buys u units of insurance is

0.50(3000000) + 0.50u1000, 000 + 0.50u

.

If Gavin purchases 3000000 units of insurance, he will guarantee a cash flow of 3000000 nextyear. Since his initial investment will be 2500000 his risk-free return will be 20%.

Problem 7

The minimum variance portfolio is proportional to 0.7500 −0.5000 0.2500−0.5000 1.0000 −0.5000

0.2500 −0.5000 0.7500

1

11

=

0.500

0.50

. (1)

The vector on the right-hand side of (1) is already normalized, and so it is the minimum vari-ance portfolio. (The solution can also be found by directly solving the system of equationsgiven by Σx = e.) Now r̂ = r − 0.2e = (0.2, 0.6, 0.6). The tangent portfolio is proportional to

1

Σ−1r̂ = (0, 0.2, 0.2), which after normalization gives the tangent portfolio as (0, 0.5, 0.5).

Problem 9

The payoff equals 1 regardless of the outcome of the wheel. Since the investment equalsΣn

i=11/Ai and the payoff is certain the risk-free rate of return is

1Σn

i=11/Ai− 1. (2)

For example, if the numbers on the wheel are 2, 5 and 10, then the risk-free rate is 25%.

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