Fin500J_topic10_Probability_2010_0000000

1
Fin500J Topic 10 Fall 2010 Olin Business School
Fin500J: Mathematical Foundations in Finance
Topic 10: Probability and Statistics
Philip H. Dybvig
Reference: Probability and Statistics, DeGroot and Schervish, Chapter 3, 4, 5
Slides designed by Yajun Wang

Outline
 Definition of a Random Variable
 Discrete Random Variables
 Continuous Random Variables
 Expectations, Variances
 Exponential Distributions
 Joint Probability Distributions
 Marginal Probability Distributions
 Covariance
 Bivariate Normal Distributions
Fall 2010 Olin Business School 2
Fin500J Topic 10

Definition of a Random Variable
 A random variable is a real valued function defined on a sample space
S. In a particular experiment, a random variable X would be some
function that assigns a real number X(s) for each possible outcome
 A discrete random variable can take a countable number of values.
 Number of steps to the top of the Eiffel Tower*
 A continuous random variable can take any value along a given
interval of a number line.
 The time a tourist stays at the top
once s/he gets there
3
Fall 2010 Olin Business School
S
s
Fin500J Topic 10
* The answer ranges from 1,652 to 1,789. See Great Buildings

Probability Distributions, Mean and Variance for Discrete
Random Variables
 The probability distribution of a discrete random variable is
defined as a function that specifies the probability associated
with each possible outcome the random variable can assume.
 p(x) ≥ 0 for all values of x
 p(x) = 1
4
Fin500J Topic 10
 The mean, or expected value, of a discrete random variable is
( ) ( ).
E x xp x
   
 The variance of a discrete random variable x is
2 2 2
[( ) ] ( ) ( ).
E x x p x
  
   


The Binomial Distribution
 A Binomial Random
Variable
 n identical trials
 Two outcomes: Success or
Failure
 P(S) = p; P(F) = q = 1 – p
 Trials are independent
 x is the number of S’s in n
trials
Flip a coin 3 times
Outcomes are Heads or Tails
P(H) = .5; P(F) = 1-.5 = .5
A head on flip i doesn’t change
P(H) of flip i + 1
5
Fin500J Topic 10

The Binomial Distribution (Example 1)
Results of 3 flips Probability Combined Summary
HHH (p)(p)(p) p3 (1)p3q0
HHT (p)(p)(q) p2q
HTH (p)(q)(p) p2q (3)p2q1
THH (q)(p)(p) p2q
HTT (p)(q)(q) pq2
THT (q)(p)(q) pq2 (3)p1q2
TTH (q)(q)(p) pq2
TTT (q)(q)(q) q3 (1)p0q3
6
Fin500J Topic 10

The Binomial Distribution Probability Distribution
x
n
x
q
p
x
n
x
P 






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
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7
Fin500J Topic 10
 Example: Binomial tree model in option pricing.

Mean and Variance of Binomial Distribution
8
Fin500J Topic 10
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variance
,
mean 


The Binomial Distribution Probability Distribution
9
Fin500J Topic 10
 Example 2: Say 40% of the class is female.
What is the probability that 6 of the first 10 students
walking in will be female?
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The Poisson Distribution
 Evaluates the probability of a (usually small) number of occurrences
out of many opportunities in a …
 period of time, area, volume, weight, distance and other units of
measurement
10
Fin500J Topic 10
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x 
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
  = mean number of occurrences in the given unit of
time, area, volume, etc.
 Mean µ = , variance: 2 = 
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The Poisson Distribution (Example 3)
11
Fin500J Topic 10
 Example 3: Say in a given stream there are an average of 3 striped
trout per 100 yards. What is the probability of seeing 5 striped
trout in the next 100 yards, assuming a Poisson distribution?
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Continuous Probability Distributions
 A continuous random variable can take any numerical value within
some interval.
 A continuous distribution can be characterized by its probability
density function.
For example: for an interval (a, b],
Fin500J Topic 10
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a
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• The function f (x) is called the probability density function of X. Every
p.d.f. f (x) must satisfy
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all
for
x
f

Continuous Probability Distributions
 There are an infinite
number of possible
outcomes
 P(x) = 0
 Instead, find P(a<x≤b)
Table
Software
Integral calculus
Fin500J Topic 10
 If a random variable X has a continuous distribution for which the
p.d.f. is f(x), then the expectation E(X) and variance Var(X) are
defined as follows:
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The Uniform Distribution on an Interval
 For two values a and b
 Mean and Variance
d
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Fin500J Topic 10
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The Normal Distribution
 The probability density function f(x):
µ = the mean of x,  = the standard deviation of x
2
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Fin500J Topic 10
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The Normal Distribution (Cont.)
Fin500J Topic 10




x
z
 Example 4: Say a toy car goes an average of 3,000 yards between
recharges, with a standard deviation of 50 yards (i.e., µ = 3,000 and  =
50) .
What is the probability that the car will go more than 3,100 yards
without recharging?
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 A popular model for the change in the price of a stock over a period of
time of length u is:
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on with
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The Exponential Distribution
 Probability Distribution for an Exponential Random Variable x
 Probability Density Function
 Mean: Variance:
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Fin500J Topic 10
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The Exponential Distribution (Example 5)
Fin500J Topic 10
• Example 5: Suppose the waiting time to see the nurse at the student
health center is distributed exponentially with a mean of 45 minutes.
What is the probability that a student will wait more than an hour to get
his or her generic pill?
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e
x
P
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a
x
P
a


Normal, Exponential Distribution (Matlab)
 >p = normcdf([-1 1],0,1);
>P(2)-p(1)
P = normcdf(X,mu,sigma) computes the
normal cdf at each of the values in X using
the corresponding parameters in mu and
sigma. X, mu, and sigma can be vectors,
matrices, or multidimensional arrays that
all have the same size.
Example 4:
>p=1-normcdf(3100,3000,50)
>p =
0.0228
 P = expcdf(X,mu)
P = expcdf(X,mu) computes the exponential
cdf at each of the values in X using the
corresponding parameters in mu. The
parameters in mu must be positive.
Example 5:
>mu=45;
>> p=1-expcdf(60,45)
p =
0.2636
Fin500J Topic 10 Fall 2010 Olin Business School 19

Joint Probability Distributions
 In general, if X and Y are two random variables, the probability
distribution that defines their simultaneous behavior is called a joint
probability distribution.
 For example: X : the length of one dimension of an injection-molded
part, and Y : the length of another dimension. We might be interested
in
 P(2.95  X  3.05 and 7.60  Y  7.80).
Fin500J Topic 10 20

Discrete Joint Probability Distributions
 The joint probability distribution of two discrete random variables X,Y
is usually written as fXY(x,y)= Pr(X=x, Y=y). The joint probability
function satisfies
 Example 6: X can take only 1 and 3; Y can take only 1,2 and 3 ; and the
joint probability function of X and Y is:
 

x y
XY
XY y
x
f
and
y
x
f .
1
)
,
(
0
)
,
(
Joint distribution of X and Y
(1) Compute P(X≥2, Y≥2)
P(X≥2, Y≥2)=P(X=3,Y=2)+P(X=3,Y=3)=0.2+0.3=0.5
(2) Compute Pr(X=3)
P(X=3)=P(X=3,Y=1)+P(X=3,Y=2)+P(X=3,Y=3)=0.2+0.2+0.3=0.7
Fin500J Topic 10 21

Continuous Joint Distributions
Fin500J Topic 10 22
 A joint probability density function for the continuous
random variables X and Y, denotes as fXY(x,y), satisfies the
following properties:

Continuous Joint Distributions (Example 7)
?
)
Pr(
)
2
(
?
)
1
(
.
0
,
1
)
,
(
2
2

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

Y
X
c
otherwise
y
x
for
y
cx
y
x
fXY
Calculating probabilities from a joint p.d.f.
.
20
3
4
21
)
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.
4
21
,
21
4
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,
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1
0
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1
1
1
2
2
2
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 
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




dydx
y
x
Y
X
c
c
dxdy
y
cx
dxdy
y
x
f
x
x
x
XY

Marginal Probability Distributions (Discrete)
Marginal Probability Distribution: the individual
probability distribution of a random variable computed
from a joint distribution.
Fin500J Topic 10 24

Compute fX(1), fX(3), fY(1), fY(2) and fY(3) in Example 6 .
fX(1)=P(X=1,Y=1)+P(X=1,Y=2)=0.1+0.2=0.3
fX(3)= P(X=3,Y=1)+P(X=3,Y=2)+ P(X=3,Y=3)=0.2+0.2+0.3=0.7
fY(1)= P(X=1,Y=1)+P(X=3,Y=1)=0.1+0.2=0.3
fY(2)=P(X=1,Y=2)+P(X=3,Y=2)=0.2+0.2=0.4
fY(3)= P(X=3,Y=3)=0.3
Marginal Probability Distributions (Discrete, Example)

Marginal Probability Distributions(Continuous)
 Similar to joint discrete random variables, we can find the
marginal probability distributions of X and Y from the
joint probability distribution.
Fin500J Topic 10 26

Compute fX (x) and fY(y) in Example 7
Marginal Probability Distributions(Continuous, Example)
.
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dy
y
x
f
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f
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y
XY
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x
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X
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
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




Independence
• In some random experiments, knowledge of the values of
X does not change any of the probabilities associated with
the values for Y.
• If two random variables, X and Y are independent, then
.
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B
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and
A
X
Y
X
XY 
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



Fin500J Topic 10 28

Independence (Example 8)
 Let the random variables X and Y denote the lengths of two dimensions of a
machined part, respectively.
 Assume that X and Y are independent random variables, and the distribution of
X is normal with mean 10.5 mm and variance 0.0025 (mm)2 and that the
distribution of Y is normal with mean 3.2 mm and variance 0.0036 (mm)2.
 Determine the probability that 10.4 < X < 10.6 and 3.15 < Y < 3.25.
 Because X,Y are independent
Fin500J Topic 10 29

Covariance and Correlation Coefficient
The covariance between two RV’s X and Y is
Properties:
The correlation Coefficient of X and Y is
Fin500J Topic 10 30
 
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Cov ,
X Y
X Y
X Y
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b
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X
Var
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a
X
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
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

Covariance and Correlation (Example 6 (Cont.))
Fin500J Topic 10 31

Covariance and Correlation
Example 9
Fin500J Topic 10 32

Example 9 (Cont.)
Fin500J Topic 10 33

Example 9 (Cont.)
Fin500J Topic 10 34

Zero Covariance and Independence
• However, in general, if Cov(X,Y)=0, X and Y may not be independent.
Example 10: X is uniformly distributed on [-1,1], Y=X2 . Then,
Cov(X,Y)= 0, but X determines Y, i.e., X and Y are not independent.
• If X and Y are independent, then Cov(X,Y)=0.
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Bivariate Normal Distribution
Fin500J Topic 10 36

Bivariate Normal Distribution
Example 11
Fin500J Topic 10 37

Bivariate Normal Distribution (Matlab)
 y = mvncdf(xl,xu,mu,SIGMA) returns the multivariate normal cumulative probability
with mean mu and covariance SIGMA evaluated over the rectangle with lower and upper
limits defined by xl and xu, respectively. mu is a 1-by-d vector, and SIGMA is a d-by-d
symmetric, positive definite matrix.
 Examples 11 (Cont.)
mu=[3.00 7.70]; SIGMA=[0.0016 0.00256; 0.00256 0.0064];
XL=[2.95 7.60];
XU=[3.05 7.80];
>> p=mvncdf(XL,XU, mu,SIGMA)
p =
0.6975

Fin500J_topic10_Probability_2010_0000000

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