提供高质量的essay代写,Paper代写,留学作业代写-天才代写

首頁 > > 詳細

代寫MATH11158代寫Optimization Methods in Finance

MATH11158 May 2018 MAT-MSc-OMF
Optimization Methods in Finance
1. Risk Neutral Probability Measure and Pricing. Let S0, S1, . . . , Sn
be securities and ? = {ω1, ω2, . . . , ωm} be a state space describing the possi-
ble “states of the world” at time t = 1 (each ωj has a positive probability of
occurring). By Si1(ωj) we denote the price/value of security S
i at time t = 1
in state ωj . By S
i
0 we denote the price of security S
i at time t = 0 (present).
Security S0 is the risk-free asset paying interest rate r ≥ 0 between time 0 and
1. Finally, assume that S00 = 1 and S
0
1(ωj) = 1 + r = R for all j.
(a) A risk-neutral probability measure on ? is a vector p = (p1, . . . , pm)
?
satisfying certain properties. List the properties. [5 marks]
(b) Define Type A and Type B arbitrage using the notation established above.
[5 marks]
(c) State the First Fundamental Theorem of Asset Pricing. [4 marks]
(d) Assume n = 2 (three securities, including the risk-free asset), m = 2 (two
states), S10 = 9, S
1
1(ω1) = 9, S
1
1(ω2) = 12, S
2
1(ω1) = 22, S
2
1(ω2) = 11 and
r = 0.1 (i.e. R = 1 + r = 1.1).
Either by using a no-arbitrage/replication argument or by calculating the
risk neutral probability measure directly, find S20 (that is, the price of the
2nd security at time t = 0). What is the risk neutral probabiliy measure?
Is there arbitrage?
[11 marks]
Comment: Material covered in lecture and a numerical problem similar to
one solved in class.
1
Solution: Solution:
(a) (i) pj > 0 for all j = 1, 2, . . . ,m
(ii)

j pj = 1
(iii) Si0 =
1
R
∑m
j=1 pjS
i
1(ωj), for all i = 0, 1, . . . ,m
(b) Type A arbitrage: A trading strategy (a portfolio y0, . . . , yn) having posi-
tive initial cashflow (

i S
i
0yi < 0) and no risk of loss later on (
∑n
i=1 S
i
1(ωj)yi ≥
0 for all j = 1, 2, . . . ,m).
Type B arbitrage: A trading strategy (a portfolio y0, . . . , yn) having non-
negative initial cashflow (

i S
i
0yi ≤ 0) and positive probability of profit (=
positive cashflow) in the future (there exists j such that
∑n
i=1 S
i
1(ωj)yi >
0).
(c) There is no arbitrage (i.e., no type A and no type B) if and only if there
exists a Risk Neutral Probability Measure.
(d) We form a portfolio P of ? shares of S1 and B amount of S0 (cash). At
time 1, the value of the portfolio is given by{
P1(ω1) = S
1
1(ω1)? + S
0
1(ω1)B = 9?+ 1.1B,
P1(ω2) = S
1
1(ω2)? + S
0
1(ω2)B = 12?+ 1.1B.
Note that we know the value of the portfolio at time 0:
P0 = S
1
0?+B = 9?+B.
Let us now choose ? and B so as to replicate the price/value of asset S2
at time 1. That is, we require that the following holds:
P1(ω1) = S
2
1(ω1), P1(ω2) = S
2
1(ω2).
This leads to the following system of 2 equations in 2 unknowns (?, B):
9? + 1.1B = 22, 12?+ 1.1B = 11.
The solution is: ?∗ = − 113 = −3 23 and B∗ = 50. Now, the portfolio is
indistinguishable from S2 at time 1; hence, the price of the portfolio at
time 0 (something we know) should be the same as the price of S2 at time
0 (something we do not know). So, we conclude that
S20 = P0 = 9?
∗ +B∗ = −9× 113 + 50 = 17
The risk neutral probabilities p1, p2 need to satisfy conditions (a)(iii)
above, so for cash S0 and asset S1 we have
1 = 11.11.1(p1 + p2)
9 = 11.1 (9p1 + 12p2)
the first equation gives p1 + p2 = 1 ⇒ p2 = 1 − p1, substituting into
the second equation gives 9.9 = 9p1 + 12(1 − p1) = 12 − 3p1 and thus
p1 = 0.7, p2 = 0.3. As the price for the second asset S
2
0 we get
S20 =
1
1.1 (22p1 + 11p2) =
1
1.1 (15.4 + 3.3) = 17
as before. Since p1, p2 > 0 these form a risk-neutral probability measure
and hence there is no arbitrage.
2
2. Mean Absolute Deviation.
(a) Assume that X is a space of random variables (which we take as repre-
senting loss of an investment). What is the definition of a coherent risk
measure ρ(X) on X ? [4 marks]
(b) Show that ρ(X) = IE[X ] is a coherent risk measure. [2 marks]
(c) Show that ρ(X) = IE[X ]+MAD[X ] satisfies all criteria for a coherent risk
measure apart from monotonicity (Note: MAD[X ] := IE[|X − IE[X ]|] is
mean absolute deviation). [6 marks]
(d) Given possible investment options A,B with returns given by random
variables rA, rB (where we use the convention that £1 invested in asset
i will yield £ri at the end of the investment period). Let xi denote the
fraction of the total investment to be invested in option i, and X = rAxA+
rBxB be the random variable giving the return on one unit of investment.
Assume that the joint return (rA, rB) for the assets will take values (0.98, 0.94)
and (1.03, 1.09) with probabilities 0.4 and 0.6 respectively. Consider the
following variant of the Markowitz model that uses MAD[X ] as a risk
measure:
min
x
MAD[X ], s.t. IE[X ] ≥ 1.02, xA + xB = 1, xA ≥ 0, xB ≥ 0.
Express MAD[X ] in terms of the decision variables xA, xB and show how
this objective can be modelled as minimizing a linear objective subject to
linear constraints. [13 marks]
Solution:
(a) A coherent risk measure ρ(X) has to satisfy
• Monotonicity: If X,Y are random variables representing loss with
X(ξ) ≤ Y (ξ) then
ρ(X) ≤ ρ(Y )
• Sub-additivity: If X,Y are random variables, then
ρ(X + Y ) ≤ ρ(X) + ρ(Y )
• Translation Equivariance: If X is a random variable representing
loss, c ∈ IR then
ρ(X + c) = ρ(X) + c
• Positive homogenity: If X is a random variable, α ≥ 0 then
ρ(αX) = αρ(X)
[4 marks]
3
(b) Monotonicity, Translation Equivariance and Positive Homogenity follow
from elemental properties of the Expectation. [1 mark]
For subadditivity we even have
IE(X + Y ) = IE(X) + IE(Y )
so ρ(X + Y ) ≤ ρ(X) + ρ(Y ) is satisfied. [1 mark]
(c) Sub-additivity:
ρ(X + Y ) = IE[X + Y ] + IE[|X + Y − IE[X + Y ]|]
≤ IE[X ] + IE[Y ] + IE[|X − IE[X ]|+ |Y − IE[Y ]], using |a+ b| ≤ |a|+ |b|
= ρ(X) + ρ(Y )
[2 marks]
Translation Equivariance:
ρ(X + c) = IE[X + c] + IE[|X + c− IE[X + c]|] = IE[X ] + c+ IE[|X − IE[X ]|]
= ρ(X) + c
[2 marks]
Positive Homogenity:
ρ(αX) = IE[αX ] + IE[|αX − IE[αX ]|] = αIE[X ] + IE[α|X − IE[X ]|]
= αρ(X), for α ≥ 0.
[2 marks]
(d) Introduce the additional variables R1, R2 and ER to denote the return
in each possible future scenario and the expected return. These can be
defined by the additional linear constraints
R1 = 0.98xA + 0.94xB (1)
R2 = 1.03xA + 1.09xB (2)
ER = 0.4R1 + 0.6R2 (3)
and introduce variables AD1, AD2 that denote the absolute deviation of
the return in each scenario from the expected return:
ADi = |Ri − ER|.
[7 marks]
These can be modelled by the linear constraints
ADA ≥ RA − ER, ADA ≥ ER−RA (4)
ADB ≥ RB − ER, ADB ≥ ER−RA (5)
Thus the objective minMAD[X ] can be modelled as
min 0.4ADA + 0.6ADB, s.t. (1), (2), (3), (4), (5).
[4 marks]
Note that since ADA, ADB are minimized, at least one side of each of the
constraints (4),(5) must be active at the solution, ensuring that (4),(5) is
indeed equivalent to ADi = |ri − ER|. [2 marks]
4
3. Scenario Generation
(a) State three different ways of generating a set of discrete scenarios that
approximate a (known) continuous distribution. Give at least two advan-
tages and disadvantages for each. [6 marks]
(b) Assume you want to find 4 scenarios that match a bivariate normal dis-
tribution with the following mean and variance(
h1
h2
)
∼ N (µ,Σ), µ =
(
0
1
)
, Σ =
(
2 1
1 4
)
,
by Moment Matching. Assume you want to match the five quantities
IE[h1], IE[h2], IE[h
2
1], IE[h
2
2], IE[h1h2]. Write down the moment matching
optimization problem. State clearly what are the variables to be optimised
over. [8 marks]
(c) How would you express the following constraints
(i) Each scenario should have at least a 10% probability
(ii) There must be two scenarios with an h1 value below or equal to 0
and two with an h1 value greater than or equal to 0.
[3 marks]
(d) Assume you want to approximate the bivariate uniform distribution on
[−2, 2]×[−2, 2] by the 5 discrete scenarios (0, 0), (1, 1), (1,−1), (−1, 1), (−1,−1).
If you are looking for the best match using Wasserstein/Kantorovich
distances
(i) How would you work out the probabilities for each of these scenarios?
A diagram might be helpful. [4 marks]
(ii) Give the optimal values for the probabilities. It is sufficient to give
geometric arguments to work out the exact values. [4 marks]
Solution:
(a) Possibilities include: Moment Matching, Historical Data, Wasserstein/Kantorovich/Clustering,
Construction by hand.
(b) The moment matching problem is

聯系我們
  • QQ:1067665373
  • 郵箱:1067665373@qq.com
  • 工作時間:8:00-23:00
  • 微信:Essay_Cheery
熱點文章
程序代寫更多圖片

聯系我們 - QQ: 1067665373 微信:Essay_Cheery
? 2021 uk-essays.net
程序代寫網!

在線客服

售前咨詢
售后咨詢
微信號
Essay_Cheery
微信
全优代写 - 北美Essay代写,Report代写,留学生论文代写作业代写 北美顶级代写|加拿大美国论文作业代写服务-最靠谱价格低-CoursePass 论文代写等留学生作业代做服务,北美网课代修领导者AssignmentBack 北美最专业的线上写作专家:网课代修,网课代做,CS代写,程序代写 代码代写,CS编程代写,java代写北美最好的一站式学术代写服务机构 美国essay代写,作业代写,✔美国网课代上-最靠谱最低价 美国代写服务,作业代写,CS编程代写,java代写,python代写,c++/c代写 代写essay,作业代写,金融代写,business代写-留学生代写平台 北美代写,美国作业代写,网课代修,Assignment代写-100%原创 北美作业代写,【essay代写】,作业【assignment代写】,网课代上代考