Problem Set I - Solutions, Ejercicios de Matemática Financiera

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Matemàtiques II

Curs 2011 - 2012

PS1 Solutions

1. By definition of the Euclidean norm for

x ∈ R n , with

x = (x 1 , ...., xn) let

x || =

x 2 1 +^ ...^ +^ x

2 n ∈^ [ 0,^ +∞).^ such that

(a) ||

v || = ||(1, 2)|| =

12 + 2^2 =

(b) ||

v || = ||(3, 4)|| =

32 + 4^2 =

(c) ||

v || = ||(− 1 , 2) =

(−1)^2 + 2^2 =

(d) ||

v || = ||(− 1 , 0 , 3)|| =

(−1)^2 + 0^2 + 3^2 =

(e) ||

v || = ||(− 2 , − 1 , −3)|| =

(−2)^2 + (−1)^2 + (−3)^2 =

(f) ||

v || = ||(0, 0 , −3)|| =

02 + 0^2 + (−3)^2 =

(g) ||

v || = ||(1, 2 , − 1 , 4)|| =

12 + 2^2 + (− 12 ) + 4^2 =

(h) ||

v || = ||(3, 0 , − 2 , 8)||=

32 + 0^2 + (−2)^2 + 8^2 =

(i) ||

v || = ||(− 1 , 6 , − 2 , 4)|| =

(−1)^2 + 6^2 + (−2)^2 + 4^2 =

2. By definition of the Euclidean distance between P, Q ∈ R n , :

d(P, Q) =

(P 1 − Q 1 )^2 + .... + (Pn − Qn)^2 =‖

P − Q ‖∈ [ 0, +∞),

we obtain:

(a) d(3, 1) =

(3 − 1)^2 =

(b) d((1, 2), (1, −2)) =

(1 − 1)^2 + (2 − (−2))^2 =

(c) d((1, 2), (− 2 , 3)) =

(1 − (−2))^2 + (2 − 3)^2 =

32 + 1^2 =

(d) d((0, 1), (− 1 , −2)) =

(0 − (−1))^2 + (1 − (−2))^2 =

(e) d((3, 2), (− 3 , −2)) =

(3 − (−3))^2 + (2 − (−2))^2 =

62 + 4^2 =

(f) d((1, 1), (6, 3)) =

(1 − 6)^2 + (1 − 3)^2 =

(g) d((1, 0 , 3), (− 1 , − 1 , 2)) =

(1 − (−1))^2 + (0 − (−1))^2 + (3 − 2)^2 =

22 + 1^2 + 1^2 =

(h) d((1, 1 , −1), (1, 0 , 3)) =

(1 − 1)^2 + (1 − 0)^2 + (− 1 − 3)^2 =

02 + 1^2 + (− 42 ) =

(i) d((1, − 1 , 0 , 6), (1, 2 , 1 , 0)) =

(1 − 1)^2 + (− 1 − 2)^2 + (0 − 1)^2 + (6 − 0)^2 =

02 + (−3)^2 + (−1)^2 + 6^2 =

5. a) The interval [0, 1] ⊂ R.

Set is closed since it contains its boundary points (∂A ⊂ A). Also, ist

complement A c is open. It is bounded since the ball B(0, 2) which is the

open interval (− 2 , 2) contains, [0, 1] ⊂ (− 2 , 2). Hence the set is compact.

b) The coordinate axes R^2.

A =

(x, y) ∈ R^2 | x = 0

(x, y) ∈ R^2 | y = 0

, the set is closed since

(∂A = A) but it is not bounded (not compact).

c) A = {(x, y) ∈ R | x + y ≤ 1 }.

Closed but not bounded.

d) A = {(x, y) ∈ R^2 | x − y ≤ 0 } ∪ {(x, y) ∈ R^2 | x ≤ 0 }.

Closed but not bounded

g) A = {(x, y) ∈ R | y ≤ x + 1 , x ≤ 1 , y ≥ −x − 1 }.

Is compact.

h) A = {(x, y) ∈ R^2 | − 1 ≤ y ≤ 1 , − 1 ≤ x ≤ 1 }.

Is compact.

i) A = {(x, y) ∈ R | x

• y = 4}.

Is compact.

j) A = {(x, y) ∈ R^2 | x^2 + y^2 > 1 }.

Open set, unbounded.

m) A = {(x, y) ∈ R | y − x

0 , x < 1 , y < 1 }.

Open, unbounded.

n) A = {(x, y) ∈ R^2 | y − ex^ ≥ 0 , x ≥ 0 , y + x^2 − 2 ≤ 0 }.

Compact.

o) A = {(x, y) ∈ R | x ≤ 2 , y + 2 − x ≥ 0 , y − ln x ≤ 0 }.

Compact

6. A set C ⊂ R n is convex if for all points in x, y ∈ C the line segment xy :=

{z ∈ R n |z = λx + (1 − λ)y, 0 ≤ λ ≤ 1 }, belongs to C.

For any two points a, b in

A = (x, y, z) ∈ R 3 |x + y + z = 1

with a = (x a , y a , z a )

and b = (x b , y b , z b ) call the combination c with c = λa + (1 − λ)b = (λx a

(1 − λ)x b , λy a

• (1 − λ)y b , λz a
• (1 − λ)z b ). Clearly, the sum of the three

components of c is λ1 + (1 − λ)1 = 1, so c must be part of A for any 0 ≤ λ ≤ 1.

Hence A is indeed convex.

c) A = {(x, y) ∈ R 2 | y ≥ e 2 x }. Convex set.

d) A = {(x, y) ∈ R | x − y ≤ 2 }. Convex set.

e) A = {(x, y) ∈ R^2 | y ≥ x^4 + 2x^2 − x − 3 }. Convex set.

f) A = {(x, y) ∈ R^2 | y ≤ sin(x)}. Not convex.

i) A = {(x, y) ∈ R | (y

• 1)x ≥ 1 , y ≥ x }. Convex.

j) A = {(x, y) ∈ R 2 | x ≤ −y 6 − 3 x 2

• 12, y ≥ 4 }. The conditions cannot

be satisfied simultaneously, hence A is the empty set (∅), and thus convex.

8. If A and B are subsets of R n , the sum is defined as: A + B = {z ∈ R n | z =

a + b, a ∈ A, b ∈ B}, hence if A and B are convex, then also A + B.

Proof

A + B convex⇔ for all z, w ∈ A + B ⇒ zw ⊂ A + B,

z ∈ A + B ⇐⇒ z = m + n, m ∈ A, n ∈ B

w ∈ A + B ⇐⇒ w = a + b, a ∈ A, b ∈ B

zw = λz + (1 − λ)w = λ(m + n) + (1 − λ)(a + b)

λ(m + n) + (1 − λ)(a + b)

? ∈ A + B

−→ zw = λm + (1 − λ)a ︸ ︷︷ ︸ A

• λn + (1 − λ)b ︸ ︷︷ ︸ B

→ zw ⊂ A + B