PORTFOLIO THEORY - EXERCISES 4 Dr. Andrea Rigamonti
EXERCISE 1
Given the following series of returns
/?! = 0.05, R2 = -0.02, R3 = -0.01, R4 = 0.1
and the following series of risk-free rates
Rfl = 0.005, Rfi2 = 0.005, Rfi3 = 0, RfA = 0
compute the Sharpe ratio
The Sharpe ratio is given by:
E[Rp — Rf] mean excess return
SR =
SD(Rp — Rf)    standard deviation of excess returns First, we compute the excess returns:
Rexl = 0.05 - 0.005 = 0.045 Rex,2 = -0-02 - 0.005 = -0.025 Rex,3 = -o.oi - 0 = -0.01
RexA = 0.1 - 0 = 0.1 Then we compute the mean excess return and the standard deviation of excess returns:
_ 0.045 - 0.025 -0.01 + 0.1
Rex =-^-= 0.0275
T
°2 — T _ 1 ^{RexX — Rex)
t=l
(0.045 - 0.0275)2 + (-0.025 - 0.0275)2 + (-0.01 - 0.0275)2 + (0.1 - 0.0275):
3
0.0003 + 0.0028 + 0.0014 + 0.0053
* 3
a = V0.0032 « 0.0566
So, the Sharpe ratio is:
0.0275 SR = ——— « 0.48 0.0566
0.0032
EXERCISE 2
Given the following series of returns, compute the downside deviation with benchmark B = 0.01 Rt = 0.04, R2 = -0.02, R3 = -0.01, R4 = 0.1, R5 = 0.005
The semivariance is given by:
T
°* =^[Min(Kt-S,0)]2
t=i
which means that every return above the benchmark must be replaced with 0 in the computations. Therefore, we subtract the benchmark from each return:
R1-B = 0.04-0.01 = 0.03 R2-B = -0.02 - 0.01 = -0.03 R3-B = -0.01 - 0.01 = -0.02
R4-B = 0.1-0.01 = 0.09 R5-B = 0.005 - 0.01 = -0.005
Now we apply the formula:
_    0 + (-0.03)2 + (-0.02)2 + 0 + (-0.005)2 <Tg =---= 0.000265
From this we get the downside deviation, which is simply the square root of the semivariance:
oB = V0.000265 « 0.0163
EXERCISE 3
Consider the set of weights
	0.2		0.2
	0.4	wt =	0.2
	.0.1.		.0.1.
Compute the turnover taking into account the effect of the realized returns Rt^ =
r—o.i"
0.05
. 0.2 .
We have to use the formula
N
wit - w,
i,t-i\
1=1
where
First, we compute the second term in the numerator, and the term in the denominator:
N
^ Wi,t-1 =
N
0.2 + 0.4 + 0.1 = 0.7
i=l
N
^(w£,t-i + wt,t-i x Ri,t-i) = [°-2 + 0.2 x (-0.1)] + [0.4 + 0.4 x 0.05] + [0.1 + 0.1 x 0.2] = 0.72
Now we compute the weights in the previous periods accounting for the realized returns:
0.2 + 0.2 x (-0.1) 0.18 w^i =--- X 0.7 =        X 0.7 = 0.175
0.72 0.72
0.4 + 0.4 x 0.05 0.42 wtt-i =-^rzz-X 0.7 =        X 0.7 « 0.408
0.72 0.72 0.1 + 0.1x0.2 0.12
3>-i = -—-X °"7 = X °"7 * °-117
0.72 0.72 We can finally apply the formula for the turnover:
N
TO,
i=i
This means we need to trade 25% of our wealth in order to update the weights.
)t = ^Jwut - w^.jI = |0.2 - 0.175| + |0.2 - 0.408| + |0.1 - 0.117| = 0.25
EXERCISE 4
Consider an equally weighted portfolio of two assets, A and B, which experience the following monthly returns over three periods:
RA1 = 0.1, RA2 = -0.05, RA3 = 0.15
RBil = 0, RBi2 = 0.05, RBi3 = 0.1
There are proportional transaction costs equal to 10 basis points.
If we invested 10000 euro in such portfolio (i.e., 5000 in A and 5000 in B) at t = 0, how much money would we have at period t = 3, net of transaction costs (ignore the initial transaction costs required to start investing at time t = 0)?
To compute the portfolio returns net of transaction costs we first need the turnover at each period:
N
where the formula for        simplifies to
N
TO,
i=l
,+
w,
+
Wift_! + Wix-t X
i,t-l y"W
liLiCwj,^! + Wift_! X R^)
because the weights always sum to 1.
At time t = 1 the turnover is zero. Therefore, we start from time t = 2:
0.5 + 0.5x0.1 0.55
wt-i =-=-« 0.524
A1    (0.5 + 0.5 x 0.1) + (0.5 + 0.5 x 0)    0.55 + 0.5
0.5 + 0.5x0 0.5
w„+, =-=-« 0.476
8,1    (0.5 + 0.5 x 0.1) + (0.5 + 0.5 x 0)    0.55 + 0.5
So, the turnover at time t = 2 is
2
T02
= ^K,2 -       = |0.5 - 0.524| + |0.5 - 0.476| = 0.048
and the transaction costs at time t = 2 are
TC2 = 0.048 X 0.001 = 0.000048
We do the same computations for t = 3:
+ _ 0.5 + 0.5 x (-0.05) _°-475 _
Wa2 ~ [0.5 + 0.5 x (-0.05)] + (0.5 + 0.5 x 0.05) ~    1   ~ °-475
+ _ 0.5 + 0.5x0.05 _ 0.525 _
Wfi'2 ~ [0.5 + 0.5 x (-0.05)] + (0.5 + 0.5 x 0.05) ~ ~T~ ~ °-525
So, the turnover at time t = 3 is
3
T03
= ^K,3 - w^2| = |0.5 - 0.475| + |0.5 - 0.525| = 0.05
and the transaction costs at time t = 3 are
TC3 = 0.05 x 0.001 = 0.00005 We can now compute the returns net of transaction costs in all three periods:
Ri.net = 0.5 x 0.1 + 0.5 x 0 - 0 = 0.05 Ri.net = 0.5 x (-0.05) + 0.5 x 0.05 - 0.000048 = -0.000048 R3.net = 0.5 x 0.15 + 0.5 x 0.1 - 0.00005 = 0.12495
Finally, we compute the value of the investment at t = 3:
3
v3 = v0 + v0 1
t=i
= 10000 + 10000[(1 + 0.05)(1 - 0.000048)(1 + 0.12495) - 1] = 11811.41