# MATH11154 | Stochastic Analysis in Finance

Stochastic Analysis in Finance

MATH11154
1. All random variables in the following questions are defined on a probability space (⌦,F,P),

and G denotes a -algebra contained in F.

1. (a)  Let X and Y be random variables such that E|X| < 1. Define precisely the conditional expectations E(X|G) and E(X|Y ). [7 marks]
2. (b)  Let X be a random variable with finite second moment, and consider R = X E(X|G), the di↵erence between the “true value” of X and the “predicted value” of X based on the “information” G. Compute ER and E(R|G). Show that

where Z = E(X|G).

[7 marks]

E|X2| < 1. Set X = (X1 + X2)/2, and determine E(↵X1 + (1 ↵)X2|X)

ER2 = EX2 EZ2,

(c) Let X be a random variable with finite second moment, and let L2(G) denote the space of G- measurable random variables with finite second moment. Show that Z = E(X|G) minimises the mean square distance of X from L2(G), i.e.,

E(XZ)2 =min{E(XY)2 :Y 2L2(G)}.
(d) Let X1 and X2 be independent identically distributed random variables such that E|X1| =

̄

as a function of X ̄ for any constant ↵ 2 R. [Hint: You may use without proof that due to ̄ ̄

symmetry, E(X1|X) = E(X2|X).] [5 marks] Solution:

G 2 G. (ii) E(X|Y ) = E(X|(Y )) where (Y ) is the -algebra generated by Y . (i) ER=E(XE(X|G))=EXEE(X|G)=EXEX=0,

[7 marks]

[7 marks]

(ii) E(R|G) = E(X E(X|G|G) = E(X|G) E(X|G) = 0 2

(iii) Notice that E(XZ) = EE(XZ|G) = EZ . Hence 222222

ER =E(XZ) =EX 2E(XZ)+EZ =EX EZ .

2222

̄

May 2019

[6 marks]

(i) Z := E(X|G) is a G-measurable random variable such that E(Z1G) = E(X1G) for every

(a) (b)

(c)
E(XY) =E(XZ+ZY) =E(XZ) +2E{(XZ)(ZY)}+E(ZY) .

Notice that E{(X Z)(Z Y )} = EE ((X Z)(Z Y )|G) = E ((Z Y )E (X Z|G)) = 0. Consequently,

E(XY)2 =E(XZ+ZY)2 =E(XZ)2 +E(ZY)2 E(XZ)2,
which proves the statement since Z 2 L2(G). [6 marks]

1

MATH11154

May 2019

Stochastic Analysis in Finance

̄ ̄

E(Xi|X) =

Consequently,
E(↵X1 + (1 ↵)X2|X) = ↵E(X1|X) + (1 ↵)E(X2|X) = ↵X + (1 ↵)X = X.

[5 marks] 2. We want to model the evolution of the instantaneous interest rate by an Itˆo process r = (rt)t0

satisfying following conditions:

(i)r0 =4and3rt 5almostsurelyforallt0; (ii)Ert =4forallt0;
(iii) E(|rt 4|2)  1/200 for all t 0.

(d) By symmetry E(X1|X) = E(X2|X). Hence ̄1 ̄ ̄ ̄ ̄ ̄

2

{E(X1|X) + (X2|X)} = E(X|X) = X ̄ ̄ ̄ ̄ ̄ ̄

for i = 1, 2.

The solution r of the stochastic di↵erential equation
drt =100(4rt)dt+(|rt 5||rt 3|)1/2dWt,

r0 =4 is suggested as a suitable model, where W = (Wt)t0 is a Wiener process.

1. (a)  State precisely a comparison theorem for SDEs, and applying it to suitable SDEs, show that property (i) holds for the solution r. [12 marks]
2. (b)  Prove that r satisfies property (ii). [6 marks]
3. (c)  Setting Yt := rt 4 and using Itoˆ’s formula, write an expression for the stochastic di↵erential of e100tYt. Hence, estimating E(|e100tYt|2), or otherwise, deduce that r satisfies property (iii). [7 marks]

Solution: