- Portfolio Management 103
and (5.8) take the form
μV=(1−s)μ 1 +sμ 2 , (5.11)
σV^2 =(1−s)^2 σ 12 +s^2 σ 22 +2s(1−s)ρ 12 σ 1 σ 2. (5.12)Obviously,μVas a function ofsis a straight line andσ^2 Vis a quadratic function
ofswith a positive coefficient ats^2 (namelyσ^21 +σ^22 − 2 ρ 12 σ 1 σ 2 >σ 12 +σ^22 −
2 σ 1 σ 2 =(σ 1 −σ 2 )^2 ≥0). The problem of minimising the varianceσ^2 V (or,
equivalently, the standard deviationσV) of a portfolio is solved in the next
theorem. First we find the minimum without any restrictions on short sales.
If short sales are not allowed, we shall have to take into account the bounds
0 ≤s≤1 on the parameter.
Theorem 5.5
For− 1 <ρ 12 <1 the portfolio with minimum variance is attained at
s 0 =σ^21 −ρ 12 σ 1 σ 2
σ 12 +σ^22 − 2 ρ 12 σ 1 σ 2. (5.13)
If short sales are not allowed, then the smallest variance is attained at
smin=
0ifs 0 < 0 ,
s 0 if 0≤s 0 ≤ 1 ,
1if1<s 0.Proof
We compute the derivative ofσV^2 with respect tosand equate it to 0:
−2(1−s)σ 12 +2sσ^22 +2(1−s)ρ 12 σ 1 σ 2 − 2 sρ 12 σ 1 σ 2 =0.Solving forsgives the aboves 0. The second derivative is positive,
2 σ^21 +2σ 22 − 4 ρ 12 σ 1 σ 2 > 2 σ 12 +2σ^22 − 4 σ 1 σ 2 =2(σ 1 −σ 2 )^2 ≥ 0 ,which shows that there is a minimum ats 0. It is a global minimum because
σV^2 is a quadratic function ofs.
If short sales are not allowed, then we need to find the minimum for 0≤
s≤1. If 0≤s 0 ≤1, then the minimum is ats 0 .Ifs 0 <0, then the minimum
is at 0, and ifs 0 >1, then it is at 1, sinceσ^2 Vis a quadratic function ofswith
a positive coefficient ats^2. This is illustrated in Figure 5.2. The bold parts of
the curve correspond to portfolios with no short selling.