1 15.math-review friday 8/18/00. 15.math-review2 yevent: xin this setting we are talking about some...
Post on 22-Dec-2015
218 views
TRANSCRIPT
1
15.Math-Review
Friday 8/18/00Friday 8/18/00
15.Math-Review 2
Event: In this setting we are talking about some uncertain event. The
outcome of which is uncertain
Outcome: The result of an observation of the event once the uncertainty has
been resolved.
Probability: The likelihood that certain outcome is realized for the event.
Example:To roll a balanced 6-sided die is an event. The number that appears
on top of the die once its rolled it’s the outcome. And any outcome (any of the 6 sides) has a probability of 1/6.
Random Variables
15.Math-Review 3
We have an uncertain event. If the outcome of the uncertainty is a number then it is
called a random variable:
Random Variables
Example:The result of the flip of a coin is uncertain event. We can obtain
heads or tails. This is not a random variable.If we associate the variable X a value equal to 1 if the coin flip is
head and 0 if the coin flip is tails, then X is a random variable.
15.Math-Review 4
A random process (or event) is one whose outcome cannot be specified in advance (except in probabilistic terms).
A random variable is a number that reflects the outcome of a random process.
Random Variables
15.Math-Review 5
A random variable can be discrete or continuous. Discrete: The values the random variable can take are fixed
discrete amounts.
Example: The number on top after the roll of a die can be 1, 2, 3, 4, 5 or 6.
Continuous: The random variable can take any value in some interval.
Example: If we draw a student at random in the class and record their height in principle we can obtain any number between .5 meters and 2.5 meters. (Very unlikely in the extremes)
Random Variables
15.Math-Review 6
The behavior of a discrete random variable can be described with a probability distribution of the form:
Probability Distributions
x P(x=ai)
a1 p1
a2 p2
: :: :
an pn
The die example: x P(x=ai)
1 1/62 1/63 1/64 1/65 1/66 1/6
15.Math-Review 7
The probability that a continuous random variable takes exactly one value =0. There are infinite possible values the variable can take.
For a cont. r.v. we ask what is the probability that the r.v. falls within a certain interval.
If X is a cont. r.v. its cumulative distribution function F(x) is defined by:
F(x) = P(Xx)
Probability Distributions
15.Math-Review 8
What is P(a Xb) for a cont. r.v.? Since the events {X<a} and {aXb} are disjoint
(mutually exclusive) we have:
P(X b) = P({X<a}{aXb})=P(X<a)+P(aXb)
= P(X a)+P(aXb) This means that
P(aXb)= P(X b) - P(X a)=F(b)-F(a)
Probability Distributions
15.Math-Review 9
We define f (x) the probability density function of the cont. r.v. X as:
Probability Distributions
)(
)()(
xd
xdFxf
By construction we have that:
b
a
x
dttfaFbFbXaP
xXPdttfxF
)()()()(
)()()(
But what is this density function?
15.Math-Review 10
The density function is the function f( ) such that the probability over a tiny interval of length y around point x is yf(x).
Probability Distributions
x
f(x)
1
x
F(x)
Interval of length yP(Xshaded area) yf(x)
15.Math-Review 11
Probability Distributions
Cumulative Chart
.000
.250
.500
.750
1.000
0
10000
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 6
Frequency Chart
.000
.043
.086
.129
.172
0
428.7
857.5
1715
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 6
Density function of a continuous r.v. These graphs show the
empirical frequency and cumulative frequency of the ‘die’ r.v.
What happens if we allow the r.v. to take more values between 1 and 6?
15.Math-Review 12
Probability Distributions
Frequency Chart
.000
.024
.048
.072
.096
0
238.7
477.5
716.2
955
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 11
Cumulative Chart
.000
.250
.500
.750
1.000
0
10000
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 11
Here the r.v can take 11 equally likely and equally spaced values between 1 and 6. Note that the frequency
over an interval of length 1 is 2 of these bars: 2(.088)=0.1761/6
15.Math-Review 13
Probability Distributions
Frequency Chart
.000
.013
.025
.038
.050
0
125.7
251.5
377.2
503
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 21
Cumulative Chart
.000
.250
.500
.750
1.000
0
10000
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 21
Here the r.v can take 21 equally likely and equally spaced values between 1 and 6. Note that the frequency
over an interval of length 1 is 4 of these bars: 4(.042)=0.1681/6
15.Math-Review 14
Probability Distributions
Cumulative Chart
.000
.250
.500
.750
1.000
0
10000
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 41
Frequency Chart
.000
.007
.015
.022
.030
0
74.25
148.5
222.7
297
0.00 1.75 3.50 5.25 7.00
10,000 Trials 0 Outliers
Forecast: 41 Here the r.v can take 41 equally likely and equally spaced values between 1 and 6. Note that the frequency
over an interval of length 1 is 8 of these bars: 8(.022)=0.1761/6
15.Math-Review 15
The probability of any one value decreases to 0.The cumulative frequency function is smoother. In the limit, when we pass to a continuous
distribution: the cumulative frequency function will become the
cumulative distribution function. The frequency function will go to the function g(x)=0. The frequency in an interval of length y will be y/6.
Probability Distributions
15.Math-Review 16
Mean, Variance, Covariance
The behavior of r.v. is expressed by their distribution.The mean and variance give some summary
information of what the distribution looks like.The covariance describes how two r.v. relate to each
other.
x
f(x)
.000
.043
.086
.129
.172
1 3 42 5 6
15.Math-Review 17
Mean, Variance, Covariance
The mean of r.v. X is an average of the possible values of X weighted by the probability. In the discrete case:
If X can take the values x1, x2,…, xn with probabilities p1, p2,…, pn respectively.
n
iiiX xpXE
1
)(
In the continuous case:Where f(x) is the probability density function:
dtttfXEX )()(
15.Math-Review 18
Mean, Variance, Covariance
The variance of r.v. X is the weighted square distance from the mean. In the discrete case:
n
iii
n
iii
XExpXDEVSTD
XExpXVAR
1
2
1
22
)().(.
)()(
In the continuous case:
dttfXEtXDEVSTD
dttfXEtXVAR
)()().(.
)()()(
2
22
15.Math-Review 19
Lets consider the random variables X, Y:
Mean, Variance, Covariance
xi pi yi pi
1 0.5 3 0.4
1.5 0.3 5 0.6
3 0.2
Compute the mean, variance and standard deviation.
X= 1.55, X2 =0.5725, X= 0.756637
Y= 4.2, Y2 =0.96, Y= 0.979796
15.Math-Review 20
Mean, Variance, Covariance
Graphically:
X
f(x)
These are continuous r.v. distributions with the same variance and with different means. The mean is the average value the r.v. takes.
X X X
Distributions that have the same mean but different variances. The ‘fatter’ distributions have greater variance.
15.Math-Review 21
Mean, Variance, Covariance
A city newsstand has been keeping records for the past year of the number of copies of the WSJ sold daily. Records were kept for 200 days.
Number of copies Frequency
0 24
1 52
2 38
3 16
4 37
5 18
6 13
7 2
15.Math-Review 22
Mean, Variance, Covariance
What is the distribution of the number of copies of the WSJ sold in one day?
What is the average number of WSJ sold in one day? What is the standard deviation of the number of WSJ
sold each day?
15.Math-Review 23
Mean, Variance, Covariance
X= 2.53, X2 =3.3791, X= 1.838233
# Copies Frequency Prob. xi-E(x) (xi-E(x))^20 24 0.12 -2.53 6.40091 52 0.26 -1.53 2.34092 38 0.19 -0.53 0.28093 16 0.08 0.47 0.22094 37 0.185 1.47 2.16095 18 0.09 2.47 6.10096 13 0.065 3.47 12.04097 2 0.01 4.47 19.9809
200 1
2.53 3.3791 = sumproducts(col, prob)1.83823285
15.Math-Review 24
Mean, Variance, Covariance
If we are given two r.v. X and Y the covariance and correlation of X and Y are defined by: Discrete case:
YX
YiXi
n
ii
YiXi
n
ii
yxpYXCORR
yxpYXCOV
))((),(
))((),(
1
1
Where X and X are the mean and standard deviation of r.v. X respectively.
And pi is the probability of the joint distribution. In other words: pi = P(X=xi,Y=yi)
The continuous case is analogous
15.Math-Review 25
Mean, Variance, Covariance
Interpretation: If COV(X,Y)>0, then if X is greater than its mean, Y is greater
than its mean.
X
Y P(X=xi,,Y=yi )= pi
X
Y
COV(X,Y)>0 COV(X,Y)<0
Y P(X=xi,,Y=yi )= pi
X
Y
X
If COV(X,Y)<0, then if X is greater than its mean, Y is smaller than its mean.
15.Math-Review 26
Mean, Variance, Covariance
Example: In our old example: xi pi yi pi
1 0.5 3 0.4
1.5 0.3 5 0.6
3 0.2
The joint distribution is:1 1.5 3
3 0.2 0.12 0.08 0.4
5 0.3 0.18 0.12 0.6
0.5 0.3 0.2 1.0
y
x
We already know that: X= 1.55, X
2 =0.5725, X= 0.756637
Y= 4.2, Y2 =0.96, Y= 0.979796
15.Math-Review 27
Mean, Variance, Covariance
So...
The covariance is 0 because these r.v. are independent. What if we change the problem a little:
1 1.5 3
3 0.1 0.22 0.08 0.4
5 0.4 0.08 0.12 0.6
0.5 0.3 0.2 1.0
y
x
0
)2.45)(55.13(12.0)2.45)(55.15.1(18.0)2.45)(55.11(3.0
)2.43)(55.13(08.0)2.43)(55.15.1(12.0)2.43)(55.11(2.0
))((),(1
YiXi
n
ii yxpYXCOV
15.Math-Review 28
Mean, Variance, Covariance
Now:
Now X and Y are no longer independent, in fact they have a negative covariance.
Correlation is CORR(X,Y) = -0.1/(0.756637*0.979796) = -0.13489
1.0
)2.45)(55.13(12.0)2.45)(55.15.1(08.0)2.45)(55.11(4.0
)2.43)(55.13(08.0)2.43)(55.15.1(22.0)2.43)(55.11(1.0
))((),(1
YiXi
n
ii yxpYXCOV
15.Math-Review 29
Mean, Variance, Covariance
Consider r.v. X,Y of means X, Y and standard deviations X,
y respectively.
Let a,b be some fixed numbers. Define Z=aX+bY, then:
),( 2)(
)(22222 YXCOVabbaZVAR
baZE
YXZ
YXZ
15.Math-Review 30
Some Derivations
First we show that E(aX+bY)=aE(X)+bE(Y), we will use this formula in the remaining derivations.
)()()()(
),(),(
),()()()(
11
11
11
YbEXaEyYPybxXPxa
yYxXPybyYxXPxa
yYxXPbyaxpbyaxbYaXE
i
l
iii
k
ii
ii
n
iiii
n
ii
ii
n
iii
n
iiii
In the fourth equality it helps to think in terms of a table to link the joint distribution with the ‘marginal’ distributions.
15.Math-Review 31
Some Derivations
VAR(X) = E((X-E(X))2) = E(X2 -2E(X)X+ E(X)2) = = E(X2) -2E(X)E(X) + E(X)2 = = E(X2) -E(X)2
COV(X,Y) = E( (X-E(X))(Y-E(Y)) ) = E(XY- E(X)Y- E(Y)X+ E(X)E(Y))= = E(XY)- E(X)E(Y)- E(Y)E(X) + E(Y)E(X) = = E(XY)- E(X)E(Y)
VAR(aX+bY)= E((aX +bY)2) - (E(aX+bY))2 = = E(a2X2+2abXY+b2Y2)-(aE(X)+bE(Y))2 = = a2E(X2)+2abE(XY)+b2E(Y2)-a2E(X)2-2abE(X)E(Y)-b2E(Y)2 = = a2(E(X2)-E(X)2)+ b2(E(Y2)- E(Y)2)+2ab(E(XY)-E(X)E(Y)) = = a2VAR(X)+b2VAR(Y)+2abCOV(X,Y)
15.Math-Review 32
Mean, Variance, Covariance
We know that an investment in Snowboard Inc. has a return that is a random variable with mean .9 and standard deviation 0.075. Also an investment in Skiboots Inc. has a return that is a random variable that we know has mean .9 and standard deviation 0.27. Also the return for these stocks has a correlation of -0.75.
If you decide to invest 30% of your capital in Snowboard Inc, and 70% in Skiboots Inc. What is the mean and variance of the return of the resulting portfolio?
What if you invest 50% on each?