Answer To: Sydney Company has made a portfolio of these three securities: SecurityCostE(R)s(R)Treasury...
David answered on Dec 20 2021
3.1. Sydney Company has made a portfolio of these three securities:
Security
Cost
E(R)
σ(R)
Treasury bond
$70,000
5%
0
Gold Coast Corporation
$60,000
15%
30%
Tweed Company
$30,000
17%
35%
The correlation coefficient between Gold Coast and Tweed is 0.5. If the returns are normally distributed, find the probability that the return of the portfolio is more than 17%.
35.03% ♥
Solution:
Total value of three securities = $70,000 + $60,000 + $30,000 = $160,000
Weight of treasury bond = $70,000/$160,000 = 0.4375
Weight of Gold Coast corporation = $60,000/$160,000 = 0.375
Weight of Tweed Company = $30,000/$160,000 = 0.1875
Expected return on the portfolio (Rp) is
Rp = ∑wiRi = 0.4375 x 5% + 0.375 x 15% + 0.1875 x 17% = 11%
Standard deviation on the portfolio σp = √ (0.4375)² (30)² + (0.1875)² (35)² + 2 x 0.5 x 0.4375 x 30 x 0.1875 x 35
= √245.4536
= 15.603
Z = (X – Rp)/ σp = (17 – 11)/15.603 = 0.3845
Using Z-tables, the probability is
P [Z > 0.3845] = 1 – 0.6497 = 0.3503 or 35.03%
3.2. Suppose you have $20,000 that you want to invest in two companies, Melbourne Oil Company and Brisbane Aluminum. Melbourne has a return of 12% and standard deviation 25%, while Brisbane has return of 16% with a standard deviation of 32%. The correlation coefficient between them is 40%. Your portfolio should have a return of 13%.
(A) Find the amount invested in each of the two companies.
Melbourne $15,000, Brisbane $5,000 ♥
(B) Find the standard deviation of this portfolio's returns.
23.14% ♥
Solution:
A) Let the proportion of amount invested in Melbourne Oil company be X
And the proportion of amount invested in Brisbane Aluminum be 1 – X
Rp = X (12%) + (1 – X) (16%)
13 = 12X + 16 – 16X
4X = 3
X = ¾ or 0.75
1 – X = 1 – 0.75 = 0.25
Amount invested in Melbourne Oil company = $20,000 x 0.75 = $15,000
Amount invested in Brisbane Aluminum = $20,000 x 0.25 = $5,000
b) Standard deviation on the portfolio (σp)
σp = √X² σ²M + (1 – X)² σ²B + 2 x r x X x (1 – X) x σ²B x σ²M
= √ (0.75)² (25)² + (0.25)² (32)²...