Question

Problem 3: You have access to two investment opportunities. Mutual Fund A, which promises 20% expected return with a variance of 0.36, and Mutual Fund B, which promises 15% expected return with a variance 0f 0.12. The correlation between the two is 0.084.

2. In addition to the funds A and B in the previous question, now you decide to include fund C to your portfolio. Its expected return is 10%, its variance 0.0625, its correlation with A is 0.1050 and its correlation with B is 0.07. You want to achieve an expected return of 16% on your portfolio, with the minimum possible risk (measured by the standard deviation). Derive analytically (that is, without the help of solver, but through calculus) the weights of such desired portfolio, and its standard deviation.

Answer 1

Given:

	Returns	Variance	Std	Weights
A	20%	0.3600	0.6	wA
B	15%	0.1200	0.34641	wB
C	10%	0.0625	0.25	wC

Correlations	A&B	B&C	A&C
	0.084	0.07	0.105

Let wA , wB and wC be the weights of each Mutual Fund in the portfolio

Portfolio variance V is given by

V = w_A²σ_A² + w_B²σ_B² + w_C²σ_C²+ 2w_Aw_BCovariance(A,B)+ 2w_Bw_C Covariance(B,C)+ 2w_Aw_C Covariance(A,C)

Covariance(A,B) = σ_A* σ_B * Correlation of A &B    = 0.6*0.346*0.084 = 0.0174

Covariance(A,C) = σ_A* σ_C * Correlation of A & C    = 0.6*0.25*0.105 = 0.0156

Covariance(B,C) = σ_B* σ_C * Correlation of B & C    = 0.346*0.25*0.07 = 0.0061

Substituting in V we get

V = w_A²*0.36 + w_B²*0.12 + w_C²*0.0625+ 2w_Aw_B*0.0174+ 2w_Bw_C*0.0061+ 2w_Aw_C *0.0156

For simplicity, let W_A = A, W_B = B, W_C = C. Then

V =0.36A² + 0.12B² + 0.0625C² + 0.0348AB + 0.0122BC + 0.0312AC

We need to minimise V such that 20A + 15B + 10C = 16 (Expected return)

10C = 16 - 20A - 15B

C = 1.6 -2A -1.5B,    Eq 1

Also A+B+C = 1, (sum of weights is 1)

C = 1 – A – B, Eq 2

From Eq 1 and 2 we get

1.6 – 2A – 1.5B = 1 – A – B

A+0.5B = 0.6

0.5B = 0.6 -A

B = 1.2 – 2A,

Sub in Eq 2, we get

C = 1 – A – (1.2 – 2A)

C = A – 0.2

Substituting the values of B and C in the V equation, we get

V = 0.36A² + 0.12(1.2-2A)² + 0.0625(A-0.2)² + 0.0348A(1.2-2A) + 0.0122(1.2-2A)(A-0.2) + 0.0312A*(A-0.2)

V = 0.36A² + 0.12(1.44+4A²-4.8A)+ 0.0625(A² +0.04 -0.4A)+ 0.04176A-0.0696A² + 0.0122(1.2A-0.24-2A²+0.4A)+ 0.0312A²-0.00624A

V=0.36A² + 0.1723+0.48A²-0.576A+ 0.0625A² -0.025A+0.0025+ 0.04176A-0.0696A² + 0.01464A-0.0244A² – 0.0029 +0.0049A + 0.0312A²-0.00624A

V = 0.8397A² -0.546A +0.1719

To minimise this equation, we should take the derivative

dV/A = 0.8397*2A -0.546=0

1.6794A = 0.546

Therefore A = 0.276 , C=A-0.2= 0.058 , B=1.2-2A = 1.2-2*0.276 = 0.648

A = 0.276 , B = 0.648 , C = 0.058