COURSE NOTES Financial Mathematics MTH3251 Modelling in Finance and Insurance ETC 3510. Lecturers: Andrea Collevecchio and Fima Klebaner School of Mathematical Sciences Monash University Semester 1, 2016
89
Embed
Financial Economics pt 2 Modelling in Insurance and Finance
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
1/89
Modelling in Finance and Insurance ETC 3510.
Lecturers: Andrea Collevecchio and Fima Klebaner School of
Mathematical Sciences
Monash University
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
2/89
1.2 Application in Finance . . . . . . . . . . . . . . . . . . . .
. . . . . 2 1.3 Application in Insurance . . . . . . . . . . . . .
. . . . . . . . . . . 2
2 Review of probability 4 2.1 Distribution of Random Variables.
General. . . . . . . . . . . . . . 4 2.2 Expected value or mean . .
. . . . . . . . . . . . . . . . . . . . . . 5 2.3 Variance Var, and
SD . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.4
General Properties of Expectation . . . . . . . . . . . . . . . . .
. 7 2.5 Exponential moments of Normal distribution . . . . . . . .
. . . . . 8 2.6 LogNormal distribution . . . . . . . . . . . . . .
. . . . . . . . . . . 9
3 Independence. 10 3.1 Joint and marginal densities . . . . . . . .
. . . . . . . . . . . . . . 10 3.2 Multivariate Normal
distributions . . . . . . . . . . . . . . . . . . . 11 3.3 A linear
combination of a multivariate normal . . . . . . . . . . . . 12 3.4
Independence . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . 13 3.5 Covariance . . . . . . . . . . . . . . . . . . . . . .
. . . . . . . . . . 15 3.6 Properties of Covariance and Variance .
. . . . . . . . . . . . . . . 15 3.7 Covariance function . . . . .
. . . . . . . . . . . . . . . . . . . . . . 16
4 Conditional Expectation 16
4.1 Conditional Distribution and its mean . . . . . . . . . . . . .
. . . 16 4.2 Properties of Conditional Expectation . . . . . . . .
. . . . . . . . 17 4.3 Expectation as best predictor . . . . . . .
. . . . . . . . . . . . . . 18 4.4 Conditional Expectation as Best
Predictor . . . . . . . . . . . . . . 18 4.5 Conditional
expectation with many predictors . . . . . . . . . . . . 20
5 Random Walk and Martingales 22 5.1 Simple Random Walk . . . . . .
. . . . . . . . . . . . . . . . . . . . 22 5.2 Martingales . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 23 5.3
Martingales in Random Walks . . . . . . . . . . . . . . . . . . . .
. 23
5.4 Exponential martingale in Simple Random Walk (
q
p)
. . . . . . . 25
6 Optional Stopping Theorem and Applications 26 6.1 Stopping Times
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 6.2
Optional Stopping Theorem . . . . . . . . . . . . . . . . . . . . .
. 27 6.3 Hitting probabilities in a simple Random Walk . . . . . .
. . . . . . 28 6.4 Expected duration of a game . . . . . . . . . .
. . . . . . . . . . . . 29 6.5 Discrete time Risk Model . . . . . .
. . . . . . . . . . . . . . . . . . 29 6.6 Ruin Probability . . . .
. . . . . . . . . . . . . . . . . . . . . . . . 30
1
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
3/89
7 Applications in Insurance 30 7.1 The bound for the ruin
probability. Constant R. . . . . . . . . . . . 32 7.2 R in the
Normal model . . . . . . . . . . . . . . . . . . . . . . . . . 32
7.3 Simulations . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . . . 34 7.4 The Acceptance- Rejection method . . . . . . . . .
. . . . . . . . . 35
8 Brownian Motion 36 8.1 Definition of Brownian Motion . . . . . .
. . . . . . . . . . . . . . . 36 8.2 Independence of Increments . .
. . . . . . . . . . . . . . . . . . . . 37
9 Brownian Motion is a Gaussian Process 38 9.1 Proof of Gaussian
property of Brownian Motion . . . . . . . . . . . 38 9.2 Processes
obtained from Brownian motion . . . . . . . . . . . . . . 41 9.3
Conditional expectation with many predictors . . . . . . . . . . .
. 42
9.4 Martingales of Brownian Motion . . . . . . . . . . . . . . . .
. . . . 44
10 Stochastic Calculus 46 10.1 Non-differentiability of Brownian
motion . . . . . . . . . . . . . . . 46 10.2 Ito Integral. . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . 46 10.3
Distribution of Ito integral of simple deterministic processes . .
. . 47 10.4 Simple stochastic processes and their Ito integral . .
. . . . . . . . 48 10.5 Ito integral for general processes . . . .
. . . . . . . . . . . . . . . . 49 10.6 Properties of Ito Integral
. . . . . . . . . . . . . . . . . . . . . . . . 49 10.7 Rules of
Stochastic Calculus . . . . . . . . . . . . . . . . . . . . . . 50
10.8 Chain Rule: Ito’s formula for f (Bt). . . . . .
. . . . . . . . . . . . . 51
10.9 Martingale property of Ito integral . . . . . . . . . . . . .
. . . . . 52
11 Stochastic Differential Equations 54 11.1 Ordinary Differential
equation for growth . . . . . . . . . . . . . . . 54 11.2
Black-Scholes stochastic differential equation for stocks . . . . .
. . 54 11.3 Solving SDEs by Ito’s formula. Black-Scholes equation.
. . . . . . . 55 11.4 Ito’s formula for functions of two variables
. . . . . . . . . . . . . . 56 11.5 Stochastic Product Rule or
Integration by parts . . . . . . . . . . . 57 11.6
Ornstein-Uhlenbeck process. . . . . . . . . . . . . . . . . . . . .
. . 57 11.7 Vasicek’s model for interest rates . . . . . . . . . .
. . . . . . . . . 58
11.8 Solution to the Vasicek’s SDE . . . . . . . . . . . . . . . .
. . . . . 59 11.9 Stochastic calculus for processes driven by two
or more Brownian
motions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . 59 11.10Summary of stochastic calculus . . . . . . . . . . .
. . . . . . . . . 60
12 Options 61 12.1 Financial Concepts . . . . . . . . . . . . . . .
. . . . . . . . . . . . 61 12.2 Functions x+ and x−. . .
. . . . . . . . . . . . . . . . . . . . . . . . 63 12.3 The problem
of Option price . . . . . . . . . . . . . . . . . . . . . . 63 12.4
One-step Binomial Model . . . . . . . . . . . . . . . . . . . . . .
. 64 12.5 One-period Binomial Pricing Model. . . . . . . . . . . .
. . . . . . 65
2
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
4/89
12.6 Replicating Portfolio . . . . . . . . . . . . . . . . . . . .
. . . . . . 65 12.7 Option Price as expected payoff . . . . . . . .
. . . . . . . . . . . . 66 12.8 Martingale property of the stock
under p . . . . . . . . . . . . . . . 67 12.9 Binomial
Model for Option pricing. . . . . . . . . . . . . . . . . . . 68
12.10Black-Scholes formula . . . . . . . . . . . . . . . . . . . .
. . . . . . 69
13 Options pricing in the Black-Scholes Model 71 13.1
Self-financing Portfolios . . . . . . . . . . . . . . . . . . . . .
. . . . 71 13.2 Replication of Option by self-financing portfolio .
. . . . . . . . . . 72 13.3 Replication in Black-Scholes model . .
. . . . . . . . . . . . . . . . 72 13.4 Black-Scholes Partial
Differential Equation . . . . . . . . . . . . . . 73 13.5 Option
Price as discounted expected payoff . . . . . . . . . . . . . . 74
13.6 Stock price S T under EMM Q
. . . . . . . . . . . . . . . . . . . . . 74
14 Fundamental Theorems of Asset Pricing 7614.1 I ntroduction . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 14.2
Arbitrage . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
. . . 76 14.3 Fundamental theorems of Mathematical Finance . . . .
. . . . . . . 77 14.4 Completeness of Black-Scholes and Binomial
models . . . . . . . . . 78 14.5 A general formula for option price
. . . . . . . . . . . . . . . . . . . 78 14.6 Summary . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . 80
15 Models for Interest Rates 81 15.1 Term Structure of Interest
Rates . . . . . . . . . . . . . . . . . . . 81 15.2 Bonds and the
Yield Curve . . . . . . . . . . . . . . . . . . . . . . . 81
15.3 General bond pricing formula . . . . . . . . . . . . . . . . .
. . . . 81 15.4 Models for the spot rate . . . . . . . . . . . . .
. . . . . . . . . . . 82 15.5 Forward rates . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . 83 15.6 Bonds in Vasicek’s
model . . . . . . . . . . . . . . . . . . . . . . . . 83 15.7 Bonds
in Cox-Ingersoll-Ross (CIR) model . . . . . . . . . . . . . . . 84
15.8 Options on bonds . . . . . . . . . . . . . . . . . . . . . . .
. . . . . 84 15.9 Caplet as a Put Option on Bond . . . . . . . . .
. . . . . . . . . . 85
0
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
5/89
1 Introduction.Intro
In order to study Finance and Insurance, we need mathematical
tools. We start with a review of probability theory: random
variables, their expected values, vari- ance, independence etc. We
introduce Random Walks, Martingales and Brownian motion, stochastic
differential equations. These are sophisticated mathematical tools.
We compromise: we are going to learn how to use these tools, which
are useful in other areas, such as Engineering and Biology.
1.1 Example of models
Let xt the amount of money in a savings account.
Suppose the interest rate is r, and
x0 > 0.
The evolution of xt is described by the
differential equation
dxt
dt = rxt.
We solve this equation as follows. Divide by xt to
get x′t/xt = r. We know that the derivative of ln
x
t equals x
t = rt + C ,
where C is a constant. Finally, xt =
eC ert.
In order to find the value of eC we need to know x0. In
fact, by plugging t = 0, we have x0 =
eC . Hence, we get
xt = x0ert.
What is it for? It allows to predict xt at a future
time t. Or it allows to find rate r if both xt
and x0 are known.
What if we introduce a random perturbation?
dX t = rX tdt + dξ t,
where ξ t is a random process. This is a strong
generalization. We will introduce and study how to solve some cases
of this class of equations. They are called Stochastic Differential
Equations.
1
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
6/89
0
Observed prices of stocks as functions of time
Plot price at time t, S t of the
y-axis and time t on the x-axis.
A model for such functions. Notice that simulated functions of time
that look like stock prices.Simulations These are random functions,
continuous but not smooth, (not differentiable). Using models we
solve the problem of Options Pricing in Finance. Option is
a financial contract that allows to buy assets in the future for
the agreed price at present.
This is modern approach to risk management in markets used by Banks
and other large Financial Companies.
1.3 Application in Insurance
Consider a sequence of independent games, and suppose that your
payoff at the end of each game is X i, which is a random
variable. We assume that X i are identically
distributed. The Random Walk is simply
∑n i=1 X i for n ∈ N. This is
the discrete
counterpart of Brownian motion. Using Random Walk to model
Insurance surplus we can calculate the ruin
probability.
2
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
7/89
0
.
4
0
.
8
1
.
2
0
.
6
1
.
0
1
.
4
1
2
3
4
5
6
0
5
Figure 2: Computer simulations
The equation for surplus at the end of year n is
U n = U 0 + cn − n
k=1
X k,
where U 0 the initial funds, c is the
premium collected in each year and X k is the
amount of claims paid out in year k. The insurance company
wants to compute the probability of ruin, i.e. the probability that
soon or later the process (U n, n ≥ 1) to hit zero
or become negative.
This model allows to find sufficient initial funds to control the
probability of ruin.
3
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
8/89
2.1 Distribution of Random Variables. General.
A random variable refers to a quantity that takes different values
with some prob- abilities. A random variable is completely defined
by its cumulative probability distribution function.
Cumulative probability distribution function
F (x) = Pr(X ≤ x), x ∈ IR. cdf
The probability of observing an outcome in an interval A
= (a, b] is
Pr(X ∈ A) = F (b) − F (a).
Sometimes it is more convenient to describe the distribution by the
probability density function.
Probability density function for continuous random variables
f (x) = ddxF (x) pdf
Using the relation between the integral and the derivative we can
calculate probabilities of outcomes by using the pdf.
The probability of observing an outcome in the range (a,b] (or (a,
b)) is
Pr(a < X ≤ b) = F (b) − F (a) =
b a
f (x)dx.
Any probability density is a non-negative function,
f (x) ≥ 0 that integrates to 1
f (x)dx = 1.
Uniform(0,1) have density
0 otherwise.
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
9/89
F (x) =
f (x) =
0 otherwise.
F (x) = 0 if x
≤ 0
f (x) = 1√
x2
2
General Normal Distribution¸ involves two numbers (parameters)
µ and σ The density of normal N (µ,
σ2) distribution is given by
f (x) = 1√ 2πσ
e− (x−µ)2 2σ2
The cumulative probability function of Standard Normal is denoted
by Φ(x).
Φ(x) =
x −∞
f (u)du.
It cannot be expressed in terms of other elementary functions. It
is available in Excel and Tables.
2.2 Expected value or mean
The expected value or the mean is defined as
E (X ) =
xf (x)dx.
Interpretation, if f (x) is the mass density then
EX is the centre of gravity.
5
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
10/89
V ar(X ) = E (X − EX ) 2
.
SD = σ = √
E (X − EX )2.
SD shows how far on average the values are away from the
mean.
It turns out that for the N (µ, σ2) distribution the
mean is µ and variance σ2.
Theorem 1 If X
has N (µ, σ2) distribution then
E (X ) = µ, V ar(X ) = σ2, SD(X )
= σ.
Proof is an exercise in Calculus.
Normal Distribution N (µ, σ2) is obtained from the
standard Normal by a linearear transform transformation.
linearNorm Theorem 2 If Z has
standard Normal distribution N (0, 1), then random
variable
X = µ + σZ
has N (µ, σ2) distribution. Conversely,
if X has N (µ, σ2)
distribution, then
Z = X − µ
has standard Normal distribution N (0, 1).
Proof. Write P (X ≤ x) and
differentiate.
Exercise 1. Find distribution
of X = µ + N (0, 1). 2.
Find distribution of X = −N (0,
1).
6
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
11/89
This result allows to calculate probabilities for any Normal
distribution by using tables of Standard Normal.
It also allows to generate any Normal by using a standard Normal.
Example: X ∼ N (1, 2). Find P (X
> 0).
P (X > 0) = P ( X − 1√
2 >
) = P (Z > − 1√ 2
) = 1 − Φ(−0.707) = 0.76
Example Consider the process X t =
√ tZ , where Z is N (0, 1).
Give the distri- bution of X t. Give the
distribution of the increments of X t.
2.4 General Properties of Expectation Expectation
1. Expectation is linear E (aX +
bY ) = aE (X )
+ bE (Y )
2. If X
3. If X = c then
E (X ) = E (c) = c.
4. Expectation of a function of a rv.
Eh(X ) =
h(x)f X (x)dx.
5. Expectation of an indicator I A(X )
(I A(X ) = 1 if X ∈ A and 0
if X /∈ A)
EI A(X ) = P (X ∈ A)
These properties are established from the definition of
expectation. Remark that if h(x) = xn then
Eh(X ) = E (X n) is called the n-th
moment of
rv. X .
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
12/89
2.5 Exponential moments of Normal distribution
Exponential moment of a random variable X
is EeuX for a number u. It is also known as the
moment generating function of X when
considered as a
function of the argument u.
mgfNormal Theorem 3 Exponential moment
of N (µ, σ2) distribution is given
by eµu+ σ2
2 u 2 .
Proof: By the Property 4 of the expectation (Expectation of
a function of a rv. Eh(X ) =
∫ h(x)f X (x)dx) with h(x)
= eux
EeuX =
eux 1√
2πσ e−
(x−µ)2
2σ2 dx.
The rest is an exercise in integration. By putting exponential
terms together and
completing the square,mgf of Normal
EeuX =
2σ2 dx
2σ2 dx
2σ2 dx
2σ2 dx.
By taking the term which does not include x outside, we
have
= eµu+(σ2u2)/2
2σ2 dx.
Recognising that the function under the integral is probability
density of N (µ + σ2u, σ2) distribution, and
that its integral equals to 1,
= eµu+(σ2u2)/2 × 1 = eµu+(σ2u2)/2.
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
13/89
2.6 LogNormal distribution LogNormal
X is Lognormal LN (µ, σ2) if log
X is Normal N (µ, σ2). In other words,
X = e Y
,
where Y is Normal N (µ, σ2). Since
ex > 0 for any x, the lognormal variable is
always positive. Lognormal density is given by the formula: for
x > 0
f (x) = 1√
2σ2
Exercise Derive this formula by using the definition and the
normal density.
Example: X ∼ LN (1, 2). Find P (X
> 1).X = eY where
Y ∼ N (1, 2). Then
P (X > 1) = P (eY > 1)
= P (Y > 0) = 0.76,
where the last value by using the previous example.
Theorem 4 If X
has LN (µ, σ2) distribution then its
mean mean of LN
EX = eµ+σ2
2 (eσ 2
2 .
Proof: This is just mgf of N (µ, σ2)
evaluated at u = 1.
9
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
14/89
3 Independence. Concept of independence of random variables
involves their joint distributions.
When we model many Random Variables together we can look at them as
a a vector X = (X 1, X 2,...,X n).
It takes values x = (x1, x2,...,xn)
according to some probability distribution, called the
joint distribution. Its proba- bility density is a
function of n variables f (x) which is
non-negative and integrates to 1.
Similar to the one dimension, probabilities, by definition
of f (x), are given by the multiple integral. For a
set B in Rn
Pr(X ∈ B) =
f (x)dx1dx2...dxn
Note that this formula is only sometimes used for calculations. The
probability density functions for each X i are
called marginal density func-
tions .
Consider case n = 2.
Theorem 5 If X
and Y have a joint
density f (x, y) then marginal densities
are given by integrating out the other variable.
f X (x) = ∫ ∞ −∞ f (x, y)dy,
and f Y (y) =
∫ ∞ −∞ f (x, y)dx.
y −∞
x −∞
f (u, v)dudv.
Differentiating with respect to x gives the formula for
marginal density of X .
10
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
15/89
3.2 Multivariate Normal distributions
Multivariate Normal distribution is a collection of a number of
Normal distribu- tions, which are correlated with each other.
Definitionultivariate N Multivariate Normal distribution is
determined by its mean vector and its co-
variance matrix: X = (X 1,X 2,....,X d)
is N (µ, Σ), where
µ = (EX 1, EX 2, . . . , E X d),
Σ =
Cov(X i, X j) i,j=1,...,d
if its probability density function is given by
f X(x) = 1
(X−µ)T .
det(Σ) is the determinant of the square matrix Σ and
Σ−1 is the inverse, ΣΣ−1 = I .
Example A bivariate normal.
µ = 0 and Σ =
1 ρ ρ 1
Standard Multivariate Normal Z is N (0,
I )
It is easy to see that Z is N (0,
I ) is a vector of independent standard Normals
Z i. |I | = 1, I −1 = I ,
and
f Z(z) = 1
(2π) d 2
11
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
16/89
Exercise If the random
variables (X 1,X 2,....,X d) are
jointly Normal, then they are independent if and only if they
are uncorrelated.
A multivariate Normal is a linear transformation of the standard
multivariate Normal, just like in one dimension.
Theorem 6 If X =
(X 1,X 2,....,X d) is N (µ,
Σ) then
X = µ + AZ,
where Z is standard multivariate Normal,
and A is a matrix square root of Σ
satisfying Σ = AAT.
Matrix square root is not unique (cf √
4 = ±2). Proof: This is an exercise in multivariable
calculus. The probability density
3.3 A linear combination of a multivariate normal A linear
combination of components of a multivariate vector
is aX for a nonrandom a. It is is a scalar random
variable
aX =
a1X 1 + a2X 2 + ... + adX d
aX = N
linearcomb Theorem 7 If X is
multivariate Normal N (µ, Σ)
then aX is N (aµ,
aΣaT ).
This theorem can be proved by using transforms of distributions
given later. Note that in this theorem the joint distribution is
multivariate normal, and it is
not enough that the marginal distributions (ie distribution of each
X i) is normal.
A counterexample: Z is standard normal and let
X 1 = Z and X 2 =
−Z . Then both X 1 and X 2
are standard normal also, but
X 1 + X 2 = 0.
Example. Find the distribution of (X 1 + X 2),
and specify its variance, where X 1, X 2 are
correlated normals.
X = (X 1,X 2) is N (µ, Σ),
Σ =
σ2
ρσ1σ2 σ2 2
Note that the sum can be written as a scalar product
X 1 + X 2 = aX, where
a = (1, 1).
12
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
17/89
1 + 2ρσ1σ2 + σ2 2.
as it should be, as can be verified directly that V
ar(X 1 + X 2) = σ2
1 + 2ρσ1σ2 + σ2 2
Example. The average of Normals, even if they are correlated, is
again Normal, but not so for LogNormals (eN (µ,σ2)).
If X is N (µ, Σ) find distribution
of
X = 1
eX i.
Remark Let X be multivariate Normal
and U = B X, for a non random matrix B.
Using Theorem 7 show that U is multivariate
Normal with mean BµX and covariance
matrix BΣX B
T .
3.4 Independence Independence
Events A1 and A2 are independent if the
probability that they occur together, is given by the product of
their probabilities,
P (A1 ∩ A2) = P (A1)P (A2).
Random variables X and Y are
independent, if the joint probability distribution
is a product of marginal probabilities, and in terms of densities
f X(x, y) = f X (x)f Y (y).
In general it is not enough to know the distribution of each
variable X and Y in order to know the
distribution of the random vector (X, Y ).
But if variables X and Y are
independent then their marginal distributions determine their joint
distribution (by the product formula).
An important corollary to independence is
13
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
18/89
expindepend Theorem 8 If random
variables X and Y are
independent then
E (XY ) = E (X )E (Y ).
Proof:
=
Independence can be formulated as a property of expectations.
Theorem 9 X and Y are
independent if and only if for any bounded functions h
and g
E (h(X )g(Y )) = Eh(X )Eg(Y ).
Independence for many variables
Events A1, A2, . . . , An are independent if for any
their subcollection the proba- bility that they occur together, is
given by the product of their probabilities.
Random variables are independent, if the joint probability
distribution is a product of marginal probabilities, and the the
joint density function is a product of marginal density
functions.
f (x) = f 1(x1)f 2(x2)....f n(xn).
In general it is not enough to know the distribution of each
variable X i in order to know the distribution of
them all, the random vector (X 1, X 2, . . . , X
n).
But if variables X 1, X 2, . . . , X n
are independent then their marginal distribu- tions determine their
joint distribution (by the product formula).
14
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
19/89
3.5 Covariance Covariance
Let X and Y be two random
variables with finite second moments E (X 2)
< ∞ and E (Y 2) <
∞ . Their covariance is defined as
Cov(X, Y ) = E (X − EX )(Y −
EY ).
Theorem 10 Cov(X, Y ) = E (XY ) −
E (X )E (Y )
Proof:
XY − Y EX − XEY
+ EXEY
Now use the property of expectation that constants can be taken out
E (aX ) = aEX
= E (XY ) −
Now Theorem 8 has the following
Corollary If X
and Y are independent then they are
uncorrelated.
3.6 Properties of Covariance and Variance Covariance
1. Cov(X, Y ) = E (XY ) −
E (X )E (Y ).
2. Covariance is bilinear (as multiplying polynomials)
Cov(aX +bY,cU +dV ) = acCov(X,
U )+adCov(X, V )+bsCov(Y, U )+bdCov(Y,
V )
3. V ar(X ) = C ov(X, X )
4. V ar(X ) = E (X 2) −
(E (X ))2 = E ((X − E (X ))2) It
is always nonnegative
5. V ar(X + Y ) = V ar(X ) +
2Cov(X, Y ) + V ar(Y )
6. If X and Y are
independent or uncorrelated, then
V ar(X + Y ) = V ar(X ) + V
ar(Y )
15
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
20/89
3.7 Covariance function
Definition The covariance function of a random
process X t is defined by
γ (s, t) = C ov(X t, X s
= E (X t − EX t(X s − EX s
= E
( X tX s
− EX tEX s
4 Conditional Expectation
Recall expectation or mean
xf X (x)dx.
Similarly, the conditional expectation is the integral with respect
to the condi- tional distribution
E (X |Y = y) =
xf (x|y)dx.
The conditional distribution is defined as follows. Let
X , Y have joint density f (x, y)
and marginal densities f X (x), and
f Y (y). The conditional distribution of
X given Y = y is defined by
the density
f (x|y) = f (x, y)
f Y (y) ,
at any point y where
f Y (y) > 0. It is easy to see that so
defined f (x|y) is indeed a probability density, as it
is nonnegative and integrates to one.
The expectation of this distribution, when exists, is called the
conditional ex- pectation
of X given Y = y, and
is given by the above formula.
Example Let X and Y have
a standard bivariate normal distribution with pa- rameter ρ.
Then
1. The conditional distribution of X given
Y = y is normal N (ρy, (1 −
ρ2)). 2. E (X |Y = y) = ρy,
E (X |Y ) = ρY .
Proof: 1. The joint density f (x, y) = 1
2π √
,
√ 2π e−y2/2.
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
21/89
Hence the conditional distribution of X given
Y = y is f (x,
y)/f Y (y)
f X | y(x) =
1√ 2πe−y2/2
= 1√ 2π(1 − ρ2)
} .
Conditional expectation as a random variable
E (X |Y = y) is a function
of y. If g denotes this function,
that is, g(y) = E (X |Y = y),
then by replacing y by Y we obtain a
new random variable g(Y ), which is called the
conditional expectation of X given
Y ,
E (X |Y ) = g(Y ).
In the above example
E (X |Y ) = ρY.
Similarly to this example, in the case of the multivariate normal
vector the conditional expectation of E (X|Y) is a
linear function of Y, Theorem 23.
4.2 Properties of Conditional Expectation Prop. E(X|Y)
1. Conditional expectation is linear in X
E (aX 1 + bX 2|Y )
= aE (X 1|Y )
+ bE (X 2|Y ).
2. E (E (X |Y ))
= E (X ). The law of double expectation.
3. If X is a function
of Y (also said as “Y -measurable”),
then
E (X |Y ) = X.
4. If U is Y -measurable, then
“it is treated as a constant”
E (XU
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
22/89
E (X |Y ) = EX,
that is, if the information we know provides no clues about
X , then theconditional expectation
of X is simply its mean value.
4.3 Expectation as best predictor
Let X denote a random variable. If we predict the
outcome of X by a number c then the
difference between the actual and the predicted outcomes is
(X −c). This
represents the error in our prediction. If we want to predict an
outcome so that theerror irrespective of its size is smallest we
minimize mean-squared error E (X −
c)2.
Theorem 11 The best mean-square predictor
of X is its
mean X = E (X ).
Proof: The mean-squared error is a function
of c, E (X − c)2. Minimize
in c.
E (X − c)2 = E (X 2 − 2cX +
c2) = E (X 2) + c2 −
2cE (X ).
The number c that minimizes this is found by
differentiating and equating to zero.
d
dc
Equate to 0, to have c = E (X ).
4.4 Conditional Expectation as Best Predictor
We now look for the best possible predictor
of X based on Y , some function
of Y , h(Y ). We define the optimal predictor
or estimator X as the one that minimizes the
mean-squared error, i.e. for any random variable Z ,
which is a function of Y
E (X − X )2 ≤ E (X −
Z )2.
It turns out that the best predictor of X
based on Y is the conditional expec-
tation of X given Y ,
denoted E (X |Y ).
18
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
23/89
est Predictor Theorem 12 The best predictor (optimal
estimator) X based on Y
is given by X = E (X |Y ), in
other words for any random
variable Z , Y -measurable (a
function of Y )
E (X −
For the proof we need the following result.
Theorem 13 Any random variable Z which
is Y -measurable (a function of Y )
is uncorrelated with X −
X = X − E (X |Y ).
Proof:
Cov(Z, X − X ) = E (Z (X −
X )) − E (Z )E (X −
X )
The second term is zero because by the law of double
expectation
E (X − X ) = E (X ) −
E (E (X |Y )) = E (X ) −
E (X ) = 0.
Thus Cov(Z, X − X ) = E (Z (X −
X ))
= E (ZX ) − E (Z X ) = 0
by the law of double expectation
E (ZX ) = E
E (ZX |Y )
since Z is Y -measurable. Finally
Cov(Z, X − X ) = 0.
Example Consider the process X t =
√
tZ , where Z is N (0, 1). Give the
distri- bution of X t. Give the distribution of the
increments of X t.
In particular, we have a
Corollary X = E (X |Y )
and X − X = X −
E (X |Y ) are uncorrelated.
Proof: of the Theorem that best predictor is
X = E (X |Y ). Take
any Z , which is a function of Y . We
need to show that
E (X − X )2 ≤ E (X −
Z )2.
E (X − Z )2 = E
X − X + X − Z 2
= E (X
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
24/89
= E (X − X )2
+ E ( X − Z )2, by the previous
result
≥ E (X − X )2.
Thus X = E (X
4.5 Conditional expectation with many predictors
Let X, Y 1, Y 2, . . . , Y n be random
variables. By definition, the optimal predictor
X minimizes the mean square error, i.e.
for any Z function of Y ’s
E (X − X )2 ≤ E (X −
Z )2.
Theorem 14 The best predictor X based
on Y 1, Y 2, . . . , Y n is given
by
X = E (X |Y 1, Y 2, . . . Y
n).
Conditional expectation given many random variables is defined
similarly as the mean of the conditional distribution. It is
denoted by
E (X |Y 1, Y 2, . . . Y n) Notation: If
we denote the information generated by Y 1, Y 2, .
. . , Y n by F n then
E (X |Y 1, Y 2, . . . Y n)
= E (X |F n).
Note that often it is hard to find a formula for the conditional
expectation. But in the multivariate Normal case it is known and is
established by direct calculations.
Cond.Exp.MvN Theorem 15 (Normal Correlation)
Suppose X and Y jointly form
a multi- variate normal distribution. Then the vector of
conditional expectations is given by the
following
E (X|Y) = E (X) + Cov(X, Y)Cov−1(Y, Y)(Y −
E (Y)).
Cov(X, Y) denotes the matrix with elements
Cov(X i, Y j), Cov−1(Y, Y)
denotes the inverse of the covariance matrix
of Y .
20
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
25/89
Example Best predictor of X based
on Y in Bivariate Normal best Pred
Direct application of the formula
E (X |Y ) = E X +
Cov(X, Y )
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
26/89
5 Random Walk and Martingales
5.1 Simple Random Walk RW
A model of pure chance is served by an ideal coin being tossed with
equal proba- bilities for the Heads and Tails to come up. Introduce
a random variable Y taking values +1 (Heads)
and −1 (Tails) with probability 1
2 . If the coin is tossed n times
then a sequence of random variables Y 1, Y 2, . . .
, Y n describes this experiment. All Y i
have exactly the same distribution as Y 1,
moreover they are all independent. Random walk is the process
X n, defined by
X n =
X 0 + Y 1 + Y 2 + .... + Y n.
X n gives the fortune of a player in a game of chance
after n plays, where a coin is tossed and one wins $1 if
Heads come up and loses $1 when Tails come up. Random walk is the
central model for stock prices, the standard assumption is that
returns on stocks follow a random walk
Random Walk
A more general Random Walk X n =
X 0 + Y 1 + Y 2 + .... + Y n,
where Y i’s are i.i.d. (not necessarily ±1). RW is
unbiased if EY i = 0 and biased
otherwise.
Mean and Variance of Random Walk
Since E (Y i) = 0, and V ar(Y i)
= E (Y 2) = 1, the mean and the variance of the
random walk are given by E (X n)
= X 0 + E (∑n
i=1 Y i) = ∑n i=1 E (Y i) = X 0,
V ar(X n) = V ar(X 0 + n
i=1
V ar(Y i) = nV ar(Y 1) = n.
Useful tools. Strong law of large numbers and central limit
theorems. In general if X 1, X 2, . . . , X
n, . . . are i.i.d. random variables with finite mean,
we have
lim n
22
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
27/89
Moreover if X i have finite variance
σ2, we have
lim n
−∞
5.2 Martingales Martingales
Definition A process (X n), n
= 0, 1, 2, . . . is called a martingale if for
all n E |X n| < ∞, and the martingale
property holds
E (X n+1|X 1, X 2, . . . , X n)
= X n.
Martingale property of Random Walk
Since
i=1
E |Y i| = nE |Y 1|,
X n is integrable provided E |Y 1| <
∞. For any time n given X n,
E (X n+1|X 1, X 2, . . . , X n)
= X n + E (Y n+1|X 1, X 2,
. . . , X n).
Since Y n+1 is independent of the past, and
X n is determined by the
first n variables, Y n+1 is independent
of X n. Therefore, E (Y n+1
| X 1, X 2, . . . , X n)
= E (Y n+1). It now
follows that if E (Y n+1) = 0, then
E (X n+1|X n)
= X n + E (Y n+1|X 1, X 2,
. . . , X n) = X n + 0 = X n.
Thus X n is a martingale.
5.3 Martingales in Random Walks
Some questions about Random Walks, such as ruin probabilities can
be answered with the help of martingales.
RWMG’s Theorem 16 Let X n, n
= 0, 1, 2, . . . be a Random Walk. Then the
following processes are martingales.
1. X n − µn, where µ =
E (Y 1). In particular, if the Random Walk is
unbiased ( µ = 0), then it is itself is a
martingale.
2. (X n − µn)2 − σ2n, where σ2
= E (Y 1 − µ)2 = V ar(Y 1).
23
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
28/89
3. For any u, euX n−nh(u),
where h(u) = ln E (euY 1) (exponential
martingales). Using the moment generating function
notation m(u) = E (euY 1), this mar-
tingale becomes (m(u))−neuX n.
Proof 1. Since, by the triangle
inequality |a + b| ≤ |a| + |b|,
E |X n−nµ| = E |X 0+ n
i=1
i=1
E |Y i|+n|µ|
= E |X 0|+n(E |Y 1|+|µ|),
X n − nµ is integrable provided
E |Y 1| < ∞, and E |X 0|
< ∞. To establish the martingale property consider
for any n
E (X n+1|X n)
= X n + E (Y n+1|X n).
Since Y n+1 is independent of the past, and
X n is determined by the
first n variables, Y n+1 is independent
of X n. Therefore,
E (Y n+1|X n) = E (Y n+1). It
now follows that
E (X n+1|X n)
= X n + E (Y n+1|X n)
= X n + µ,
and subtracting (n + 1)µ from both sides of the equation, the
martingale property is obtained,
E (X n+1 − (n + 1)µ|X n) = X n −
nµ.
2. This is left as an exercise. 3. Put M n =
euX n−nh(u). Since M n ≥ 0,
E |M n| = E (M n), which is given
by
E (M n) = EeuX n−nh(u)
= e−nh(u)EeuX n = e−nh(u)Eeu(X 0+ ∑n
i=1 Y i)
= euX 0e−nh(u)E n
i=1
= euX 0e−nh(u) n
The martingale property is shown by using the fact that
X n+1 = X n + Y n+1,
(1)
with Y n+1 independent of X n
and of all previous Y i’s i ≤ n, or
independent of F n. Using the properties of
conditional expectation, we have
E (euX n+1|F n) =
E (euX n+uY n+1|F n)
= euX nE (euY n+1|F n)
= euX nE (euY n+1)
= euX n+h(u).
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
29/89
Multiplying both sides of the above equation by e−(n+1)h(u),
the martingale property is obtained,
E (M n+1|F n) = M n.
5.4 Exponential martingale in Simple Random Walk (
q
p )X n
exp MG RW In the special case when P (Y i =
1 ) = p, P (Y i = −1) =
q = 1 − p choosing u = ln(q/p)
in the previous martingale, we have euY 1 =
(q/p)Y 1 and E (euY 1) = 1. Thus h(u) =
ln E (euY 1) = 0, and euX n−nh(u) =
(q/p)X n . Alternatively in this case, the martingale property
of (q/p)X n is easy to verify directly and is left as
exercise.
25
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
30/89
6.1 Stopping Times topping times
Let X 1, X 2, . . . , X n, . . . be a
sequence of random variables. A random time τ is
called a stopping time if for any n, one can decide whether
the event {τ ≤ n} (and hence the complementary
event {τ > n}) has occurred by observing the first n
variables X 1, X 2, . . . , X n.
Another way of expressing the fact that τ is a
stopping time is that for any n you can tell if
{ τ = n
} holds by looking at X 1, X 2, . . . , X
n. Alternatively, i.e.
equivalent definition, you can tell if {τ ≤
n} holds by looking at X 1, X 2, . . . , X
n
Let’s see first an example of a random variable that is not a
stopping time. Flip a fair coin 10 times. Denote by
τ the last time you observe Head. Is this a
stopping time? No. Looking at X 1 and
X 2 is not enough to tell
if τ = 2. In fact if X 1
= 0 (here 0 means Tail), X 2 = 1,
X i = 0 for all i ∈ {3, 4, . . . , 10}
then τ = 2. On the other hand
if X 1 = 0, X 2 = 1,
X i = 1 for all i ∈ {3, 4, . . . , 10}
then τ = 10. Hence the first two
observations X 1 and X 2 are not
enough to tell if τ = 2 holds.
Time of ruin is a stopping time.
τ = min{n : X n = 0}.
{τ > n} = {X 1 = 0, X 2 = 0, . . . , X
n = 0}.
If we can tell if τ > n we can also tell
if {τ ≤ n}. So by observing the capital
at times 1, 2, . . . , n, we can decide if the ruin by time n
has occurred or not, e.g. If X 1 = 0,
X 2 = 0, X 3 = 0 then τ
> 3.
The time when something happens for the first time is a stopping
time. E.g. first time Random Walk hits value 1 (or 100). Say you
gamble from 8pm to 11 pm, τ is first time you win $
100. By observing your winnings you can decide whether
τ has or has not occurred.
A stopping time is allowed to take value +∞ with a positive
probability. For example if τ is the first
time when a RW with a positive drift hits 0, then
P (τ =
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
31/89
One way to see that a random variable is finite is to establish
that it has finite mean.
If E (τ ) < ∞ then
P (τ < ∞) = 1.
If τ 1 and τ 2 are stopping
times then their minimum is also a stopping time,
τ = min(τ 1, τ 2) = τ 1 ∧
τ 2,
is a stopping time. We use this result mainly when one of the
stopping times is a constant, τ = N .
Clearly, any constant N is a stopping time. Then
τ ∧ N is a stopping time, which is
bounded by N . For example, if τ
is the first time one wins $5 in a game of coin tossing then
τ ∧ 10 is the time of winning $5 if it happens before 10
tosses, or time 10 if $5 were not won by toss 10.
Note that also max(τ 1, τ 2) = τ 1 ∨ τ 2
and τ 1 + τ 2 are also
stopping times. But we don’t use these properties.
6.2 Optional Stopping Theorem
Prove that a martingale has a constant mean for any deterministic
time. For example, if (M n) is a martingale, then prove
that
E (M 5) = E (M 4) = E (M 3)
= E (M 2) = E (M 1)
= M 0.
There is nothing special about time 5. We can prove the same for
all fixed times. What if we substite a fixed deterministic time
with a random one?
It turns out that the mean of the stopped martingale does not
change also for some random times, such as a bounded stopping time.
What we observed above might not be true! This is why we need the
following theorem.
Theorem 17 (Optional Stopping Theorem) Let M n
be a martingale.EMtau=M0
1. If τ ≤ K < ∞ is a bounded
stopping time then
E (M τ ) = E (M 0).
2. If M n are uniformly bounded, |M n| ≤
C for any n, then for any stopping time
τ (even non finite),
E (M τ ) = E (M 0).
The proof of this theorem is outside this course.
27
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
32/89
6.3 Hitting probabilities in a simple Random Walk Ruin Prob.
Unbiased RW
Suppose that you are playing a game of chance by betting on the
outcomes of
tosses of a fair coin ( p = 0.5). You win $1 if heads
come up and lose $1 if tails come up. You start with $20. Find the
probability of winning $10 before losing all your initial capital
of $20.
Solution. Denote the required probability by a. X 0,
X 1, . . . , X n, . . . denote the capital at times
0, 1, . . . , n , . . .. X 0 = 20, for any n,
X n+1 = X n + Y n+1,
where Y n+1 is the outcome of the n +
1
toss, Y n+1 = ±1 with probabilities 0.5.
X n is an unbiased a random walk. Denote by
τ the time when you either win 10 or lose 20. In
terms of the process
X n τ = min{n : X n =
30 or 0}.
Denote by a the probability that you win 10 before losing
20, ie X τ = 30, then 1 − a is the
probability X τ = 0 if you lose 20 before
winning 10.
Thus the distribution of X τ is:
X τ = 30 and X τ = 0 with
probability 1 − a. We have seen that the process X n
is a martingale. Applying the optional
stopping theorem (without proving that we can)
E (X τ ) = E (X 0)
= X 0 = 20.
On the other hand, calculating the expectation
directly E (X τ ) = 30×a+0×(1−a). Thus we have
from these equations 30a = 20, and a = 2/3.
Thus the probability of winning $10 before losing the initial
capital of $20 is 2/3.
The same calculation gives that the probability of the process
X n hitting level b before it hits level
c, having started at x, b < x < c is
given byunbiased RW
a = c − x
Biased RWuin biased RW
Let a simple random move to the right with probability
p and to the left with probability q = 1
− p. We want to find the probability that it hits level
b before it hits level c, when started at x,
b < x < c. Let τ be the stopping time
of random walk hitting b or c.
28
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
33/89
Stopping the exponential martingale M n =
(q/p)X n we have
E (q/p)X τ = (q/p)x.
But X τ = b with probability
a and X τ = c with
probability 1 − a. Hence
a(q/p)b + (1 − a)(q/p)c = (q/p)x.
a = (q/p)x − (q/p)c
6.4 Expected duration of a game Unbiased RW
We use the martingale M n = X 2n − n
and stop it at τ . Assuming it is allowed
EX 2τ − Eτ = x2. Eτ
= ab2 + (1 − a)c2 − x2, where a is the hitting
probability
in unbiased RW. Biased RW
We use the martingale M n = X n − nµ.
Here µ = p − q = 2 p − 1. Stopping
it at τ gives
EX τ − µEτ = x. Eτ =
(ab + (1
− a)c
biased RW.
Exercise: Give a proof that Optional stopping applies in the
martingale above.
6.5 Discrete time Risk Model
Time is discrete n = 0, 1, 2, . . . (years). The
insurer charges a premium of ck > 0 in
the kth year. Let X k denote the
aggregate claim amount (sum total of all claims) in the kth
year. The insurer has funds x at the start of year
1.
U n denotes the insurance company surplus at time
n. U 0 = x is the initial fund.
The premium in year n is c. The payout at time
n is X n (X n is the
aggregate
claim, ). Then the equation for surplus at the end of year n
is
U n = U 0 + cn − n
k=1
X k
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
34/89
Assumptions.
c − E (X n) > 0. The premiums are greater
than the expected payout. X 1, X 2, . . . are
identically distributed and independent.
Exercise 1. Find the expected surplus and the sd of the
surplus U n.
2. Use the Law of Large Numbers to give an approximate value for
U n.
3. Use the Central Limit Theorem to give the approximate
distribution of U n
6.6 Ruin Probability
The probability of ruin is the probability that surplus becomes
negative. More precisely, the time of ruin T ,
T = min{n : U n < 0},
where T = ∞ if
U n ≥ 0 for all
n = 1, 2, · · · . The probability that ruin has occurred by
time n is
P (T
P (T < ∞).
This probability is the central question of study in Actuarial
mathematics/ Insurance.
7 Applications in Insurance Insurance is an agreement where, for an
upfront payment (called premium) the company agrees to pay the
policyholder a certain amount if a specific loss occurs. The
individual transfers this risk to an insurance company in exchange
for a fixed premium.
30
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
35/89
Theorem 18 Assume that {c−X k, k = 1, 2, · ·
· } are i.i.d. random variables, and there exists a
constant R > 0 such that uin exp
bound
Ee−R(c−X 1) = 1.
Then for all n P x(T ≤ n|U 0
= x) ≤ e−Rx
Proof. Step 1. Show that M n = e−RU n is
a martingale. Step 2. Use the Martingale stopping Theorem with the
stopping time min(T, n) = T
∧ n
Step 3. Extract information from the resulting equation. Step 1.
Finite expectation. E |e−RU n|
= E (e−RU n) = e−Rx
∏n k=1 E (e−R(c−X k)) = e−Rx < ∞.
Proof of the martingale property. Since
U n+1 = U n + c − X n+1,
we have E (M n+1|U 1, · · · , U n)
= = E (e−RU n+1|U 1, · · · , U n)
by definition of M n+1
= E (e−RU n−R(c−X n+1)|U 1, · · ·
, U n) by definition of U n+1
= e−RU nE (e−R(c−X n+1)|U 1, · · ·
, U n) since U n is known
= e−RU nE (e−R(c−X n+1))
= e−RU nE (e−R(c−X n+1)) by independence
= e−RU nby definition of R.
This together with finite expectation implies that M n
= e−RU n, n = 0, 1, . . . is a
martingale.
We have seen that T is a stopping time.
T ∧ n is a stopping time bounded by n,
min(T, n) ≤ n.
We can apply the Martingale Stopping
E (M T ∧n) = E (M 0)
E (e−RU T ∧n) = E (e−RU 0)
= e−Rx.
We now “open” the T ∧n by using the indicators,
T ∧n = T I (T ≤ n) +
nI (T > n). Thus
e−RU T ∧n = e−RU T I (T
≤
n) + e−RU nI (T > n).
31
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
36/89
+ E
≥ E (e−RU T I (T
≤ n) since E (e−RU nI (T >
n)) > 0
≥ E (I (T ≤ n))
since U T < 0 and
e−RU T > 1
= P (T ≤ n) as required .
7.1 The bound for the ruin probability. Constant R.
Bound on the ruin probability is e−Rx.
We now turn to finding constant R. The constant R
is found from the equation
E (
( eRX
= 1.
Recall that the second term is the moment generating function
of X
E
( eRX
7.2 R in the Normal model
Example. Suppose that the aggregate claims have N (µ,
σ2) distribution. We give a bound on the ruin probability.
The mgf of N (µ, σ 2
) is given by
mX (R) = E (eRX )
= E (eN (Rµ,R2σ2)) = eRµ+ 1 2 R2σ2
(or use the formula for Normal moment generating function). Thus
the equation for R becomes
mX (R) = eRc
eRµ+ 1 2R
2
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
37/89
σ2
Remark The aggregate claims in consecutive years
X 1, X 2, . . . , X n, . . . are as-
sumed to have the same distribution, say as X 1.
Suppose that there are n insured individuals. Then
each has individual claim distribution Y . So that in
one year the aggregate claim is
X 1 = n
i=1
Y i,
where Y i is the claim of person i. If the
individual claim has mean µY and sd σ2
Y , then the
nσY ≈ N (0, 1).
Y ).
Example Consider a car owner who has an 80% chance of no
accidents in a year,
a 20% chance of being in a single accident in a year, and no chance
of being in morethan one accident in a year. For simplicity, assume
that there is a 50% probability that after the accident the car
will need repairs costing 500, a 40% probability that the repairs
will cost 5000, and a 10% probability that the car will need to be
replaced, which will cost 15,000. Hence the distribution of the
random variable Y, loss due to accident:
f (x) =
0.80 if x = 0
0.10 if x = 500
0.08 if x = 5000
0.02 if x = 15000
The car owners expected loss is the mean of this distribution,
E (Y ) = 750. The standard deviation of the loss
σY = 2442.
An insurance company that will reimburse repair costs resulting
from accidents for 100 such car owners
For the company the loss in one year is sum of losses for each car.
If the loss to car i is Y i, then
X 1 = 100 i=1
Y i,
33
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
38/89
Note that most of Y is are zero. This fact is
taken into account in the loss (claim) distribution.
For the company, the expected loss in one year is sum of expected
losses
µ = µX = E
100 i=1
100
car = 596, 336, 400.
So that the aggregate loss in one year X has
approximately Normal distribution with this parameters.
Suppose the premium is set to be 30% higher than the expected
claim, c = 1.3µ Then
R = 2(c − µ)
596, 336, 400 = 7.55 × 10−5
So, if the company has initial fund of x = 100, 000
= 105, then the ruin proba- bility is less than e−7.55 =
0.0005.
Note that initial fund of only x = 10, 000 = 104 is not
enough, the ruin proba- bility is less than e−0.755 =
0.47.
7.3 Simulations
Suppose you want to simulate from a strictly increasing c.d.f.
F . Let U be a
uniform random variable. We have that
Y = F −1(U ),
has c.d.f. F . In fact
P (Y ≤ x) = P (F −1(U ) ≤ x)
= P (U ≤ F (x)) = F (x).
Example: simulation of an exponential.
34
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
39/89
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
40/89
8 Brownian Motion
Botanist R. Brown described the motion of a pollen particle
suspended in fluid in
1828. It was observed that a particle moved in an irregular, random
fashion. In 1900 L. Bachelier used the Brownian motion as a model
for movement of
stock prices in his mathematical theory of speculation. A. Einstein
in 1905 explained Brownian motion as a result of molecular
bom-
bardment by the molecules of the fluid. Mathematical foundation for
Brownian motion as a stochastic process was done
by N. Wiener in 1931, hence it is also called the Wiener
process.
8.1 Definition of Brownian Motion
Defining Properties of Brownian Motion {Bt} .
Time t, 0 ≤ t ≤ T .Properties BM
1. (Normal or Gaussian increments) For all s < t,
Bt − Bs has N (0, t − s)
distribution, Normal distribution with mean 0 and variance t
− s.
2. (Independent increments) Bt − Bs is independent of
the past, that is, of Bu, 0 ≤ u ≤ s.
3. (Continuity of paths) Bt, t
≥ 0 are continuous functions of t.
The initial point B0 is a constant, often 0.
If B0 = x then Bt is BM
started at x. We explain these properties below.
Defining Property 1 of Brownian Motion
Bt − Bs is N (0, t − s) for s < t. By
Theorem 2 with σ =
√ t − s, the distribution of Bt − Bs is the
same as the
distribution of √
E (Bt − Bs) = 0.
E (Bt − Bs) = EBt − EBs = 0.
Thus for all s and t EBt = E Bs.
In particular EBt = EB0 = B0.
36
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
41/89
The last equality, because expectation of a constant is that
constant. Next for a random variable X with zero
mean, EX = 0, we have
V ar(X ) = E (X −
EX )2 = E (X 2).
Since (Bt − Bs) has zero mean, and by a property
of N (0, σ2) distribution
E (Bt − Bs)2 = V ar(Bt − Bs) = t − s,
SD(Bt − Bs) = √
t − s.
If we take s = 0 then we obtain E (Bt − B0) =
0 and E (Bt − B0)2 = t.
8.2 Independence of Increments For any times s and
t, s < t, the random variable Bt − Bs is
independent of all the variables Bs and Bu,
u < s.
BMGauss Theorem 19 Brownian motion has covariance
function min(t, s).
Proof: Take t > s. Then Bt can be
written as a sum of Bs and increment (Bt −
Bs),
Bt = Bs + (Bt − Bs).
Hence E (BsBt) = EB2
.
Now Brownian motion has independent increments: (Bt −Bs)
and Bs are indepen- dent, therefore expectation of their
product is the product of their expectations (Theorem 8) so
that
E
= E BsE (Bt − Bs).
Brownian motion has Normal increments: (Bt − Bs) is
N (0, t − s). Therefore its mean is zero,
E (Bt
− Bs) = 0. So that
E (BsBt) = E (B2 s).
Next, writing Bs = B0 + (Bs − B0) and using
independence of terms, we have
E (B2 s ) = E (B2
0 + (Bs − B0)2 + 2B0(Bs − B0)) = E (B2 0)
+ s = B2
0 + s.
Here we used that E (Bs − B0)2 = s as the
variance of N (0, s) distribution, and that
B0 is non-random, E (B2
0) = B2 0 .
− B0) = E (B0) = B0.
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
42/89
Cov(Bs, Bt = E (BtBs− EBtEBs = B 2 0 +
s − B
2 0 = s.
If t < s, then similarly (or exchanging roles
of s and t) E (BtBs)
= t. Therefore
Cov (
9 Brownian Motion is a Gaussian Process
The distributions of B (t) for any time t
are called marginal distributions of Brow- nian motion.
The joint distributions of the vector (B(t1), B(t2)) of Brownian
motion sampled at two arbitrary times t1 < t2
are called bivariate distributions.
Similarly for any n the joint distributions of the
vector (B(t1), B(t2), . . . , B(tn)) of Brownian motion sampled at
n arbitrary times t1 < t2 < . .
. < tn are called n- dimensional
distributions.
Finite dimensional distributions are the joint distributions
when n = 1, 2, 3, . . .. To describe a random process it
is not enough to know the distributions of its
values at any time t, but also joint distributions.A
stochastic (random) process is called Gaussian if all its finite
dimensional distributions are multivariate Normal.
In this lecture we prove that Brownian motion is a Gaussian
process.
BMGaussPr Theorem 20 Brownian Motion is a Gaussian
process.
9.1 Proof of Gaussian property of Brownian Motion Proof: of
Theorem 20.
We need to show that all joint distributions of BM at time points
t1, t2, . . . , tn for all n = 1, 2, . .
. are Multivariate Normal. Take BM started at 0, B0
= 0.
Start with n = 1. By the property of increments of BM,
with s = 0 we have Bt − B0 has N (0, t)
distribution. Hence Bt has N (0, t)
distribution.
Now take n = 2. Write
(B(t1), B(t2)) = (B(t1), B(t1) + (B(t2) − B(t1)).
38
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
43/89
Denote X = B(t1), and Y =
B(t2) − B(t1). By the property of independence of
increments of BM X and Y are
independent, X ∼ N (0, t1),
Y ∼ N (0, t2 − t1).
Then (B(t1), B(t2)) = (X, X + Y ).
Write X = √
t1Z 1, Y = √
t2 − t1Z 2, where Z 1, Z 2 are
independent standard Nor- mal. Denote σ1 =
√ t1, σ2 =
(X, X + Y ) = (σ1Z 1,
σ1Z 1 + σ2Z 2) = AZ,
where matrix A =
σ2
.
Similarly, for n = 3, the joint distribution of the
vector (B(t1), B(t2), B(t3)) is
trivariate normal with mean (0, 0, 0), and covariance matrix t1
t1 t1
t1 t2 t2 t1 t2 t3
For a general n can complete the proof by induction.
Alternatively, write directly
(B(t1), B(t2), . . . , B(tn))
= (B(t1) Y 1
, B(t1) Y 1
+ (B(tn)) − B(tn−1) Y n
),
Denote Y 1 = B (t1), and for k
> 1 Y k = B (tk) − B(tk−1). Then by
the property of independence of increments of Brownian
motion, Y k’s are independent. They also have normal
distribution, Y 1 ∼ N (0, t1),
and Y k ∼ N (0, tk − tk−1). B(t2)
= Y 1 + Y 2, etc, B(tk)
= Y 1+Y 2+. . . Y k.
Z 1 = Y 1/
√ t1, and Z k = Y k/
√ tk − tk−1 are independent
standard normal. Thus
with
A =
, σ1 = √ t1, σk = √ tk −
tk−1.
(B(t1), B(t2), . . . , B(tn))
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
44/89
BMGauss Corollary Brownian motion is a Gaussian
process with constant mean function, and covariance
function min(t, s).
Example Find the distribution of B(1) + B(2)
+ B(3) + B(4). Consider X = (B(1), B(2),
B(3), B(4)). Since Brownian motion is a Gaussian process, all its
finite dimensional distributions are Normal, in particular X
has a multivariate Normal distribution with mean vector zero
and covariance matrix given by σij = C
ov(X i, X j). For example, Cov(X 1,
X 3) = C ov((B(1), B(3)) = 1.
Σ =
1 2 3 31 2 3 4
Now, let a = (1, 1, 1, 1). Then
aX =
X 1 + X 2 + X 3 + X 4
= B(1) + B(2) + B(3) + B(4).
aX has a Normal distribution with mean zero and variance
aΣaT , the sum of the elements of the covariance matrix
in this case.
Thus B(1) + B(2) + B(3) + B(4) has a Normal
distribution with mean zero and variance 30. Alternatively, we can
calculate the variance of the sum by the covariance formula
V
ar(X 1 + X 2 + X 3 + X 4)
=
Cov(X 1 + X 2 + X 3 + X 4,
X 1 + X 2 + X 3 + X 4)
=
i,j
40
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
45/89
9.2 Processes obtained from Brownian motion Two process used in
applications are the Arithmetic and Geometric Brownian
motion.
Arithmetic Brownian motion X t = µt + σBt,
where µ and σ are constants. This is also
known as Brownian motion with drift.
Theorem 21 If X t is Brownian motion
with drift above then X t−µt σ
is a standard Brownian motion.
It is easy to show that X t is a Gaussian
process. Calculation of its mean and covariance functions is left
as exercise.
Geometric Brownian motion
S t = S 0eµt+σBt .
What is the distribution of S t? Compute its mean
and variance.
In particular, we have a
Corollary X = E (X |Y )
and X − X = X −
E (X |Y ) are uncorrelated.
Proof: of the Theorem that best predictor is
X = E (X |Y ). Take
any Z , which is a function of Y . We
need to show that
E (X − X )2 ≤ E (X −
Z )2.
E (X − Z )
(X − X )( X − Z )
= E (X − X )2
+ E ( X − Z )2, by the previous
result
≥ E (X − X )2.
Thus X = E (X |Y ) is the
optimal, best predictor/estimator.
41
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
46/89
9.3 Conditional expectation with many predictors
Let X, Y 1, Y 2, . . . , Y n be random
variables. By definition, the optimal predictor
X minimizes the mean square error, i.e.
for any Z function of Y ’s
E (X − X )2 ≤ E (X −
Z )2.
Theorem 22 The best predictor X based
on Y 1, Y 2, . . . , Y n is given
by
X = E (X |Y 1, Y 2, . . . Y
n).
Conditional expectation given many random variables is defined
similarly as the mean of the conditional distribution. It is
denoted by
E (X |Y 1, Y 2, . . . Y n) Notation: If
we denote the information generated by Y 1, Y 2, .
. . , Y n by F n then
E (X |Y 1, Y 2, . . . Y n)
= E (X |F n).
Note that often it is hard to find a formula for the conditional
expectation. But
in the multivariate Normal case it is known and is established by
direct calculations.
Cond.Exp.MvN Theorem 23 (Normal Correlation)
Suppose X and Y jointly form
a multi- variate normal distribution. Then the vector of
conditional expectations is given by the
following
E (X|Y) = E (X) + Cov(X, Y)Cov−1(Y, Y)(Y −
E (Y)).
Cov(X, Y) denotes the matrix with elements
Cov(X i, Y j), Cov−1(Y, Y)
denotes the inverse of the covariance matrix
of Y .
Example Best predictor of X based
on Y in Bivariate Normal best Pred
Direct application of the formula
E (X |Y ) = E X + Cov(X,
Y )
V ar(Y ) (Y − EY ).
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
47/89
Example Best predictor of future value of Brownian motion
based on the present value.best PredBM
Consider best predictor of Brownian motion Bt+s at
the future time t + s if we know the
present value Bt.
Since (Bt, Bt+s) is a Bivariate Normal
with V ar(Bt) = t and Cov(Bt,
Bt+s) = min(t, t + s) = t, EBt = 0
we obtain
E (Bt+s|Bt) = Bt.
Further one can check that even if we know many past values of
Brownian motion at times t1 < t2
< . . . < tn = t
E (Bt+s|Bt1 , Bt2, . . . , Bt) = Bt.
This is known as the martingale property of Brownian motion.
43
8/19/2019 Financial Economics pt 2 Modelling in Insurance and
Finance
http://slidepdf.com/reader/full/financial-economics-pt-2-modelling-in-insurance-and-finance
48/89
A process M t, t ≥ 0, is a martingale if
• for all t, E |M t| < ∞ •
for all t and s > 0,
E (M