1. (12 marks) Consider the properties of quasiconcavity and quasiconvexity.a. Define quasiconcavity and quasiconvexity as they relate to f(xy, ..., xp).b. Prove that convexity implies quasiconvexity...

1 answer below »

View more »
Answered 1 days AfterDec 03, 2022

Answer To: 1. (12 marks) Consider the properties of quasiconcavity and quasiconvexity.a. Define quasiconcavity...

Banasree answered on Dec 05 2022
45 Votes
1.a,
A function f is}for any pair of distinct points u and v in the (convex-set) domain of f, and for 0<θ<1,
f(v)>=f(u) → f[θu+(1-θ)v){ }
to adapt this definition to strict quasiconcavity and quasiconvexity, the two weak inequalities on the right should be change into strict inequalities
Now, if c
onsider the given condition then, f(x), where x is vector variables x =(x1….xn)
Which follows, three theorems
1. If f(x) is quasiconcave(strictly quasiconcave), then -f(x) is quasiconvex (strictly quasiconvex).
2. (concavity versus quasiconcavity) Any concave (convex) function is quasiconcave (quasiconvex), but the converse is not true. Similarly, any strictly concave
(strictly convex) function is strictly quasiconcave (strictly quasiconvex),but the converse is not true.
3. (linear function) If f(x) is a linear function, then it is quasi concave as well as quasiconvex.
1.b) Prove
Theorem I follow from the fact that multiplying an inequality by -1 reverses the sense of inequality.
Let f(x) be quasiconcave, with f(v) >= f(u).
Then, f[θu + (1 -θ)v] >= f(u).
So, function -f(x) is concerned,
(after multiplying the two inequalities through by -1) -f(u) >= -f(v)
and
-f[θu +(1 -θ)v]<= -f(u).
Interpreting -f(u) the height of point N, and -f(v) as the height of M ,
the function -f(x) satisfies the condition for quasi convexity . This prove Theorem I
Theorem 2
f(x) be concave,
then,
f[θu+(1-θ)v]>=θf(u)+(1-θ)f(v)
Let assume,
f(v)>=f(u), then any weighted average of f(u) and f(v) cannot possibly be less than f(u),
θf(u)+(1-θ)f(v)>=f(u)
equating these two, we know that, by transitivity,
f[θu+(1-θ)v]>=f(u)
for f(v)>=f(u)
this satisfies the definition of quasiconcavity.
Once theorem 2 proved, 3 follows immediately.
We know that a linear function is both concave and convex. Not strictly. Therefore, theorem 2 a linear function must also be both quasiconcave and quasiconvex. Not strictly.
1.c)
A differentiable function f(x1…..xn) is { }
u = (u1…….un)
v =(v1….vn)
f(v)>=f(u) [ ] >=0
fj = df/dxi
for strict quasiconcavity and quasiconvexity, the weak inequality on the right should be change to the inequality>0
for function f(x,y)
applying bordered Hessian
|H| = |B1| + |B2|
Where,
|B1| = | | and |B2| = | |
To characterize the configuration of that function
f(x,y) = log x + log y
Here two conditions, one is necessary and the other is suffiecient.
Both relate to quasiconcavity on domain domain consisting only of the nonnegative orthant.
Where,
|B1|<= 0
|B2|>=0
Also
|B1|<0 and |B2|>0
These automatically satisfy the theorem 1
1.d)
Assue,
F(v)>=f(u)
Or
v>=u
(v,u >=0
So partial derivatives
f1 = x1 and f2(x2)
f1(u)(v1-u1)+f2(u)(v2-u2) = u2(v1-u1)+u1(v2-u2)>=0
rearrangement,
u2(v1-v2)>=u1(u2-v2)
where,
1. u1=u2=0
2. u1=0, u2>0
then,
u2v1>=0, which satisfied u2 and v1 both non negative
3. u1>0 and u2=0
then 0>=-u1v2
so,
v2(vl -u) >= u1(u2 -v2)
three possibilities,
1. u2=v2 the v1>=u1
2. u2>v2 the v1>u1
then,
u2(v1 -u1) >- u2/v2 *u1| (u2...
SOLUTION.PDF

Answer To This Question Is Available To Download

Related Questions & Answers

More Questions »

Submit New Assignment

Copy and Paste Your Assignment Here
January
January
February
March
April
May
June
July
August
September
October
November
December
2025
2025
2026
2027
SunMonTueWedThuFriSat
29
30
31
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
1
00:00
00:30
01:00
01:30
02:00
02:30
03:00
03:30
04:00
04:30
05:00
05:30
06:00
06:30
07:00
07:30
08:00
08:30
09:00
09:30
10:00
10:30
11:00
11:30
12:00
12:30
13:00
13:30
14:00
14:30
15:00
15:30
16:00
16:30
17:00
17:30
18:00
18:30
19:00
19:30
20:00
20:30
21:00
21:30
22:00
22:30
23:00
23:30