Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-1 Lesson 5: Continuous Probability Distributions

Lesson5-1 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data

Lesson 5:

Continuous Probability Distributions


Outline

Continuous probability distributions

Features of univariate probability distribution

Features of bivariate probability distribution

Marginal density and Conditional density

Expectation, Variance, Covariance and Correlation Coefficient

Importance of normal distribution

The normal approximation to the binomial


Types of Probability Distributions

Number of random variables

Joint distribution

1 Univariate probability distribution

2 Bivariate probability distribution

3 Trivariate probability distribution

… …

n Multivariate probability distribution

Probability distribution may be classified according to the number of random variables it describes.


Continuous Probability Distributions

A continuous random variable is a variable that can assume any value in an interval thickness of an item time required to complete a task temperature of a solution height, in inches

These can potentially take on any value, depending only on the ability to measure accurately.


Cumulative Distribution Function

The cumulative distribution function, F(x), for a continuous random variable X expresses the probability that X does not exceed the value of x

Let a and b be two possible values of X, with a < b. The probability that X lies between a and b is

x)P(XF(x)

F(a)F(b)b)XP(a


Probability Density Function

Let X be a random variable that takes any real values in an interval between a and b. The number of possible outcomes are by definition infinite.

The main features of a probability density function f(x) are: P(X (-, +)) = P(X (a,b)) = 1. P(X = x) = 0. f(x) 0 for all x and may be larger than 1. The probability that X falls into an subinterval (c,d) is

and lies between 0 and 1.

d

c

dxxfdcXP )()),((


The probability concepts for the discrete case is not readily applicable….

What is the probability that a dart randomly thrown will end up exactly in segment A (which lies on a straight line)?Suppose the dart has equal chance to land on any point of the line.

A

1.

a b




A

0.5.

a b1/2




A

0.25.

a b1/21/4




A

0.125.

a b1/21/41/8



What is the probability that a dart randomly thrown will end up exactly at a point A (which lies on a straight line)?Suppose the dart has equal chance to land on any point of the line.

A

0!!

a b1/21/41/8



What is the probability that a dart randomly thrown will end up exactly at a point A or a point B (which lie on a straight line)?Suppose the dart has equal chance to land on any point of the line.

A

0!!Since A & B are mutually exclusive, P(A or B) = P(A) + P(B) =0.

a b1/21/41/8

B



What is the probability that a dart randomly thrown will end up exactly at one of the single point on the line?Suppose the dart has equal chance to land on any point of the line.

A

0!!Since for distinct points A & B are mutually exclusive, P(A or B) = P(A) + P(B) =0.P (one of the single point on the line) = 0 ?????

a b1/21/41/8

B



Would like to modify the concept of discrete probability to fit into the case of continuous random variable!

What do we know? P(X<k) should be proportional to (k-0)

Try P(X<k) = (k-a)*c where c is a constant. P(X<b) = (b-a)*c implies c= 1/(b-a)




A

0.5 = (b-a)/2 * c = 1/2.

a b1/2

C=1/(b-a)

c is called the probability density.

Probability is simply the area



In discrete case, E(X) = ∑X X P(X).In continuous case, P(X) =0 for any point of X. How do we compute E(X) then?

Split the domain into n equal parts, so that the width of these n interval is dx=(b-a)/n. In each of these n intervals, the probability is well define. Then, take the left boundary value of each of this n intervals, multiply by the probability for the interval. Then, we have a weighted average similar to the discrete case.E(X)= a*dx*c + (a+dx)*dx*c+ (a+2dx)*dx*c+ … +(a+(n-1)dx)*dx*c

a b

c

dx



E(X)= a*dx*c + (a+dx)*dx*c+ (a+2dx)*dx*c+ … +(a+(n-1)dx)*dx*c

a b

c

dx

However, it is an approximation because a is only an approximate of the points within the interval (a, a+dx). Approximation improves if dx is made smaller, or n larger. That is, when dx is very very very closed to zero (but still positive), E(X)= limn-> [a*dx*c + (a+dx)*dx*c +… +(a+(n-1)dx)*dx*c]= limn-> ∑i [xi*dx*c ]=limdx->0 ∑i [xi*dx*c ]

][b

a

cdxxE(X)

xi xi+dx



What if we want to allow it more likely for the dart to end up in some segment of the line (say, the middle segment)?

a b

c2

c1 c1

1/4 3/4

We can do it as long as we have the areas added up to 1: ¼*c1 + ½*c2 + ¼*c1 = 1.



What is we want to allow it is more likely for the dart to end up in some segment of the line (say, the middle segment)?

a b

c2

c1 c1

1/4 3/4It is better to define the density “function” to allow the density to vary with x in a general way. f(x) = c1 if x in (a, a+1/4) = c2 if x in (a+1/4, a+3/4) = c3 if x in (a+3/4, b)

])([b

a

dxxfxE(X)


c d x

f(x)

Shaded area under the curve is the probability that X is between c and d

Probability as an area

d

c

f(x)dxdxcP )(

xf(x)dxXE )(


Probability Density Function

The cumulative density function F(x0) is the area under the probability density function f(x) from the minimum x value (a) up to x0

00

0

xx

a

f(x)dxf(x)dx)F(x


Expectations for Continuous Random Variables

The mean of X, denoted μX , is defined as the expected value of X

The variance of X, denoted σX2 , is defined as the

expectation of the squared deviation, (X - μX)2, of a random variable from its mean

-

X xf(x)dx E(X) μ

f(x)dx)μ(x])μE[(Xσ XXX222


Linear Functions of Variables Let W = a + bX , where X has mean μX and variance

σX2 , and a and b are constants

Then the mean of W isE(W) = E(a+bX) = a + bE(X) = a + b μX

the variance isVar(W) = Var(a+bX) = b2Var(X) = b2σX

2

the standard deviation of W is|b|σX


An important special case of the previous results is the standardized random variable

Z =( X- μX ) /σX

which has a mean 0 and variance 1

Linear Functions of Variables


The Univariate Uniform Distribution

If c and d are numbers on the real line, the random variable X ~ U(c,d), i.e., has a univariate uniform distribution if

otherwise 0

dxcfor c-d

1=f(x)

The mean and standard deviation of a uniform random variable x are

122

cdand

dcXX


The Uniform Density


8 10 12 14 16 18 20 22

Learning exercise 4: Part-time Work on Campus

A student has been offered part-time work in a laboratory. The professor says that the work will vary from week to week. The number of hours will be between 10 and 20 with a uniform probability density function, represented as follows:

How tall is the rectangle? What is the probability of

getting less than 15 hours in a week?

Given that the student gets at least 15 hours in a week, what is the probability that more than 17.5 hours will be available?


8 10 12 14 16 18 20 22

Learning exercise 4: Part-time Work on Campus

How tall is the rectangle? (20-10)*h = 1 h=0.1

What is the probability of getting less than 15 hours in a week? 0.1*(15-10) = 0.5

Given that the student gets at least 15 hours in a week, what is the probability that more than 17.5 hours will be available? 0.1*(20-17.5) = 0.25 0.25/0.5 = 0.5P(hour>17.5)/P(hour>15)


The Normal Distribution

Continuous Probability

Distributions

Probability Distributions

Uniform

Normal

Exponential


‘ Bell Shaped’ Symmetrical Mean, Median and Mode

are Equal

Location is determined by the mean, μ

Spread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + to

Mean = Median = Mode

x

f(x)

μ

σ

Normal Distribution N(,2)

The normal probability distribution is asymptotic. That is the curve gets closer and closer to the X-axis but never actually touches it.


The Normal Distribution N(,2)

The normal distribution closely approximates the probability distributions of a wide range of random variables

Distributions of sample means approach a normal distribution given a “large” sample size

Computations of probabilities are direct and elegant

The normal probability distribution has led to good business decisions for a number of applications

Sum of normal random variables remain normal.

Normal distribution is completely characterized by two parameters, mean and variance.


N(,2)

x

x

x

(a)

(b)

(c)

Changing shifts the location of the distribution.Changing 2 changes the dispersion.


The Normal Probability Distribution

The random variable X ~ N(,2), i.e., has a univariate normal distribution if for all x on the real line (-,+ )

e2

1=f(x)

2-x

21

-

and are the mean and standard deviation, = 3.14159 … and e = 2.71828 is the base of natural or Naperian logarithms.


Normal Distribution Probability

Probability is the area under the curve!

c dX

f(X) A table may be constructed to help us find the probability


Moments of Standard Normal Random Variables N(0, )

Mean=0 Variance =1 Skewness = 0 Kurtosis = 3 Excess kurtosis =0


Infinite Number of Normal Distribution Tables

Normal distributions differ by mean & standard deviation.

Each distribution would require its own table.

X

f(X)

Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data

The Standard Normal Probability Distribution -- N(0,1)

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

It is also called the z distribution.

A z-value is the distance between a selected value, designated X, and the population mean , divided by the population standard deviation, . The formula is:

X

Z

0])([1

)(1

)(

XEXE

XEZE

1)(1

)(1

)(22

XVarXVarX

VarZVar

Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data

The Standard Normal Probability Distribution

Any normal random variable can be transformed to a standard normal random variable

Suppose X ~ N(µ, 2) Z=(X-µ)/ ~ N(0,1)

P(X<k) = P [(X-µ)/ < (k-µ)/ ]


Standardize the Normal Distribution

0

= 1

Z

Because we can transform any normal random variable into standard normal random variable, we need only one table!

Normal Distribution

Standardized Normal Distribution

X

XZ


Standardizing Example

Z 0 .12

Normal distribution N(5,100)= 5, = 10

Standardized Normal Distribution N(0,1) = 0, = 1

X5 6.2

12.010

52.6

XZ

010

55

XZ


Obtaining the Probability

Z0

= 1

0.12

Z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

0.0478

.02

0.1 .0478

Standardized Normal Probability Table (Portion)

ProbabilitiesShaded Area Exaggerated


Example P(3.8 X 5)

Z0-0.12

Normal Distribution

0.0478


Shaded Area Exaggerated

X = 5

= 10

3.8

12.010

58.3

XZ


Example (2.9 X 7.1)

0-.21 Z.21

Normal Distribution

.1664

.0832.0832


5

= 10

2.9 7.1 X


21.010

59.2

XZ

21.010

51.7

XZ


Example P(X 8)

Z0 .30

Normal Distribution


.1179

.5000 .3821

X = 5

= 10

8


30.010

58

XZ


Example P(7.1 X 8)

0 .30 Z.21

Normal Distribution

.0832

.1179 .0347


= 5

= 10

87.1 XShaded Area Exaggerated

21.010

51.7

XZ

3.010

58

XZ


Normal Distribution Thinking Challenge

You work in Quality Control for GE. Light bulb life has a normal distribution with µ= 2000 hours & = 200 hours. What’s the probability that a bulb will last between 2000 & 2400 hours? less than 1470 hours?


Solution P(2000 X 2400)

Z0 2.0

Normal Distribution

.4772


X = 2000

= 200

2400

P(2000<X<2400) = P [(2000-µ)/ <(X-µ)/ < (2400-µ)/ ]= P[(X-µ)/ < (2400-µ)/ ] – P [(X-µ)/ < (2000-µ)/ ]= P[(X-µ)/ < (2400-µ)/ ] – 0.5


2.0200

20002400

σμXZ


Solution P(X 1470)

Z 0-2.65

Normal Distribution

.4960 .0040

.5000


X = 2000

= 200

1470

P(X<1470) = P [(X-µ)/ < (1470-µ)/ ]


2.65200

20004701

σμXZ


Finding Z Values for Known Probabilities

Z .00 .02

0.0 .0000 .0040 .0080

0.1 .0398 .0438 .0478

0.2 .0793 .0832 .0871

.1179 .1255

Z Z = 0

Z = 1

.31

.1217.01

0.3 .1217


What Is Z Given P(Z) = 0.1217?



Finding X Values for Known Probabilities

Z Z = 0

Z = 1

.31X = 5

= 10

?

Normal Distribution Standardized Normal Distribution

.1217 .1217

1.810)31.0(5 ZXShaded Area Exaggerated


EXAMPLE 1

The bi-monthly starting salaries of recent MBA graduates follows the normal distribution with a mean of $20,000 and a standard deviation of $2,000. What is the z-value for a salary of $24,000?

00.22000$

000,20$000,24$

X

z


EXAMPLE 1 continued

A z-value of 2 indicates that the value of $24,000 is 2 standard deviation above the mean of $20,000.

A z-value of –1.50 indicates that $17,000 is 1.5 standard deviation below the mean of $20,000.

50.12000$

000,20$000,17$

X

z

What is the z-value of $17,000 ?


Areas Under the Normal Curve

About 68 percent of the area under the normal curve is within one standard deviation of the mean.

± P( - < X < + ) = 0.6826

About 95 percent is within two standard deviations of the mean. ± 2 P( - 2 < X < + 2 ) = 0.9544

Practically all is within three standard deviations of the mean. ± 3 P( - 3 < X < + 3 ) = 0.9974


EXAMPLE 2

The daily water usage per person in New Providence, New Jersey is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons.

About 68 percent of those living in New Providence will use how many gallons of water?

About 68% of the daily water usage will lie between 15 and 25 gallons.


EXAMPLE 2 continued

What is the probability that a person from New Providence selected at random will use between 20 and 24 gallons per day?

00.05

2020

X

z

80.05

2024

X

z

P(20<X<24)=P[(20-20)/5 < (X-20)/5 < (24-20)/5 ] =P[ 0<Z<0.8 ]

The area under a normal curve between a z-value of 0 and a z-value of 0.80 is 0.2881. We conclude that 28.81 percent of the residents use between 20 and 24 gallons of water per day.


How do we find P(0<z<0.8)

P(0<z<0.8) = P(z<0.8) – P(z<0)=0.7881 – 0.5=0.2881

P(z<c)

c

P(0<z<c)

c0

P(0<z<0.8) = 0.2881


EXAMPLE 2 continued

What percent of the population use between 18 and 26 gallons of water per day?

40.05

2018

X

z

20.15

2026

X

z

Suppose X ~ N(µ, 2) Z=(X-µ)/ ~ N(0,1)

P(X<k) = P [(X-µ)/ < (k-µ)/ ]


How do we find P(-0.4<z<1.2)

P(z<c)

c

P(0<z<c)

c0

P(-0.4<z<1.2) = P(-0.4<z<0) + P(0<z<1.2)=P(0<z<0.4) + P(0<z<1.2)=0.1554+0.3849=0.5403

P(-0.4<z<1.2) = P(z<1.2) - P(z<-0.4)= P(z<1.2) - P(z>0.4) = P(z<1.2) – [1- P(z<0.4)]=0.8849 – [1- 0.6554]=0.5403

P(-0.4<z<0) =P(0<z<0.4) because of symmetry of the z distribution.


Steps to find the X value for a known probability:1. Find the Z value for the known probability2. Convert to X units using the formula:

Finding the X value for a Known Probability

ZσμX


Finding the X value for a Known Probability

Example: Suppose X is normal with mean 8.0 and standard

deviation 5.0. Now find the X value so that only 20% of all values are

below this X

X? 8.0

.20

Z? 0


Find the Z value for 20% in the Lower Tail

20% area in the lower tail is consistent with a Z value of -0.84


X? 8.0

.20

Z-0.84 0

1. Find the Z value for the known probability

z F(z)

.82 .7939

.83 .7967

.84 .7995

.85 .8023

.80


2. Convert to X units using the formula:

Finding the X value

803

0584008

.

.).(.

ZσμX

So 20% of the values from a distribution with mean 8.0 and standard deviation 5.0 are less than 3.80


EXAMPLE 3

Professor Mann has determined that the scores in his statistics course are approximately normally distributed with a mean of 72 and a standard deviation of 5. He announces to the class that the top 15 percent of the scores will earn an A.

What is the lowest score a student can earn and still receive an A?


Example 3 continued

To begin let k be the score that separates an A from a B. 15 percent of the students score more than k, then 35 percent

must score between the mean of 72 and k. Write down the relation between k and the probability:

P(X>k) = 0.15 and P(X<k) =1-P(X>k) = 0.85 Transform X into z:

P[(X-72)/5) < (k-72)/5 ] = P[z < (k-72)/5] P[0<z < s] =0.85 -0.5 = 0.35

X72 k

Z0 ?

0.150.35


Example 3 continued

Find s from table: P[0<z<1.04]=0.35

Compute k: (k-72)/5=1.04 implies K=77.2

Those with a score of 77.2 or more earn an A.

X72 77.2

Z0 1.04

0.150.35


How do we know that the data are likely drawn from normal?

1. Check the moments Skewness =0 Excess Kurtosis = 0

(For instance, refer to section 16.7 “Tests for Skewness and Excess Kurtosis”, p.567 of Estimation and Inference in Econometrics by Davidson and MacKinnon)

2. Normal probability plot1. Suppose we have n observations in the sample.

Sort them in ascending order. Compute the empirical z value (i.e., (x-mx)/sx)

2. Generate a column 0.5, 1.5, …..,[ 0.5+(n-1)]. Call this column U.3. Generate another column p(z) = U/n. 4. Generate another column theoretical z = NORMSINV(p(z))5. Plot empirical z against the theoretical z.6. If the data has normal distribution, the plot should be a straight

line.


Normal Probability Plot (The data are generated from a normal distribution.)

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5

Theoretical z value

z valu

e from

data


Normal Probability Plot (The data are generated from a uniform distribution.)

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5

Theoretical z value

z valu

e from

data


The Normal Approximation to the Binomial

The normal distribution (a continuous distribution) yields a good approximation of the binomial distribution (a discrete distribution) for large values of n (number of trials).

The normal probability distribution is generally a good approximation to the binomial probability distribution when n and n(1- ) are both greater than 5.

Why can we approximate binomial by normal?Because of the Central Limit Theorem.


Normal Distribution Approximation for Binomial Distribution

Recall the binomial distribution: n independent trials probability of success on any given trial = P

Random variable X: Xi =1 if the ith trial is “success”

Xi =0 if the ith trial is “failure”

nPμE(X)

P)nP(1-σVar(X) 2



The shape of the binomial distribution is approximately normal if n is large

The normal is a good approximation to the binomial when nP(1 – P) > 9

Standardize to Z from a binomial distribution:

P)nP(1

npX

Var(X)

E(X)XZ



Let X be the number of successes from n independent trials, each with probability of success P.

If nP(1 - P) > 9,

P)nP(1

nPbZ

P)nP(1

nPaPb)XP(a


Binomial Approximation Example

40% of all voters support ballot proposition A. What is the probability that between 76 and 80 voters indicate support in a sample of n = 200 ?

E(X) = µ = nP = 200(0.40) = 80 Var(X) = σ2 = nP(1 – P) = 200(0.40)(1 – 0.40) = 48

( note: nP(1 – P) = 48 > 9 )

0.21900.28100.5000

0.58)F(F(0)

0)Z0.58P(

0.4)200(0.4)(1

8080Z

0.4)200(0.4)(1

8076P80)XP(76


The Exponential Distribution

Used to model the length of time between two occurrences of an event (the time between arrivals)

Examples: Time between trucks arriving at an unloading dockTime between transactions at an ATM MachineTime between phone calls to the main operator



The exponential random variable T (t>0) has a probability density function

Where is the mean number of occurrences per unit time t is the length of time until the next occurrence e = 2.71828

T is said to follow an exponential probability distribution

0 for t λef(t) λt



λteF(t) 1

Defined by a single parameter, its mean (lambda)

The cumulative distribution function (the probability that an arrival time is less than some specified time t) is

where e = mathematical constant approximated by 2.71828

= the population mean number of arrivals per unit

t = any value of the continuous variable where t > 0


Exponential Distribution Example

Example: Customers arrive at the service counter at the rate of 15 per hour. What is the probability that the arrival time between consecutive customers is less than three minutes?

The mean number of arrivals per hour is 15, so = 15

Three minutes is .05 hours

P(arrival time < .05) = 1 – e- X = 1 – e-(15)(.05) = 0.5276

So there is a 52.76% probability that the arrival time between successive customers is less than three minutes


Joint Cumulative Distribution Functions

Let X1, X2, . . .Xk be continuous random variables

Their joint cumulative distribution function, F(x1, x2, . . .xk)

defines the probability that simultaneously X1 is less than x1, X2 is less than x2, and so on; that is

)xX,...,xX,xP(X)x,...,x,F(x kk2211k21


Joint Cumulative Distribution Functions

The cumulative distribution functions F(x1), F(x2), . . .,F(xk)

of the individual random variables are called their marginal distribution functions

The random variables are independent if and only if

)F(x...))F(xF(x)x,...,x,F(x k21k21


Features of a Bivariate Continuous Distribution

Let X1 and X2 be a random variables that takes any real values in a region (rectangle) of (a,b,c,d). The number of possible outcomes are by definition infinite.

The main features of a probability density function f(x1,x2) are:

f(x1,x2) 0 for all (x1,x2) and may be larger than 1.

The probability that (X1,X2) falls into a region (rectangle) or (p,q,r,s) is

and lies between 0 and 1. P((X1,X2) (a,b,c,d)) = 1.

P((X1,X2) = (x1,x2) ) = 0.

q

p

s

r

dxdxxxfsrqpXXP 212121 ),()),,,(),((


The Bivariate Uniform Distribution

If a, b, c and d are numbers on the real line, , the random variable (X1,X2) ~ U(a,b,c,d), i.e., has a bivariate uniform distribution if

otherwise 0

dxc and bxa for c)-a)(d-(b

1

=)x,f(x 2121


The Marginal Density

The marginal density functions are:

f(x,y)dx f(y)

f(x,y)dy f(x)


The Conditional Density

The conditional density functions are:

y) f(x,y)/f(f(x|y)

x) f(x,y)/f(f(y|x)


The Expectation (Mean) of Continuous Probability Distribution

For univariate probability distribution, the expectation or mean E(X) is computed by the formula:

For bivariate probability distribution, the the expectation or mean E(X) is computed by the formula:

xf(x)dxE(X)

dyxf(x,y)dx E(X)


Conditional Mean of Bivariate Discrete Probability Distribution

For bivariate probability distribution, the conditional expectation or conditional mean E(X|Y) is computed by the formula:

Unconditional expectation or mean of X, E(X)

dxxf(x|y)y)E(X|Y

]E[μ

E[E(X|Y)]

dx f(y)dyxf(x|y) E(X)

X


Expectation of a linear transformed random variable

If a and b are constants and X is a random variable, then E(a) = aE(bX) = bE(X)E(a+bX) = a+bE(X)

bE(x)a

x f(x) dxbf(x) dxa

bx f(x) dxa f(x) dx

xbx) f(x) d(a

bx) dxbx) f(a(abx)E(a


The Variance of a Continuous Probability Distribution

For univariate continuous probability distribution

-

f(x)dxμ)(X ] μ)E[(XVar(X) 22

If a and b are constants and X is a random variable, then Var(a) = 0Var(bX) = b2Var(X)Var(a+bX) = b2Var(X)


The Covariance of a Bivariate Discrete Probability Distribution

dxdyyx)fμ)(Yμ(X)]μ)(YμE[(XCov(X,Y) YXYX ),(

Covariance measures how two random variables co-vary.

If a and b are constants and X is a random variable, then Cov(a,b) = 0Cov(a,bX) = 0Cov(a+bX,Y) = bCov(X,Y)


Correlation coefficient

The strength of the dependence between X and Y is measured by the correlation coefficient:

Y)Var(X)Var(

Cov(X,Y)Corr(X,Y)


Variance of a sum of random variables

If a and b are constants and X and Y are random variables, then

Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)

Cov(X,Y)Var(Y)Var(X)

)]μ)(YμE[(X)μE[(Y] )μ E[ (X

)]μ)(Yμ(X)μ(Y )μE[ (X

)]μ(Y)μE[ (X

) ]μ (μYE[ XY)Var(X

YXYX

YXYX

YX

YX

2

2

22

2

2

2

2

2


Variance of a sum of random variables

If a and b are constants and X and Y are random variables, then

Var(aX+bY) =a2Var(X) + b2Var(Y) + 2abCov(X,Y)

abCov(X,Y)Var(Y)bVar(X)a

)]μ)(YμabE[(X)μE[(Yb] )μE[ (X a

)]bμ)(bYaμ(aX)μ(Yb )μ(XE[ a

)]bμ(bY)aμE[ (aX

) ]bμ (aμbYE[ aXbY)Var(aX

YXYX

YXYX

YX

YX

2

2

2

22

222

222

2

2

2

2


Sums of Random Variables

Let X1, X2, . . .Xk be k random variables with

means μ1, μ2,. . . μk and

variances σ12, σ2

2,. . ., σk2.

Then, the mean of their sum is the sum of their means

kk μμμ)XXE(X ...... 2121


Sums of Random Variables Let X1, X2, . . .Xk be k random variables with

means μ1, μ2,. . . μk and variances σ1

2, σ22,. . ., σk

2.

Then:

If the covariance between every pair of these random variables is 0, then the variance of their sum is the sum of their variances

However, if the covariances between pairs of random variables are not 0, the variance of their sum is

2k

22

21k21 σ...σσ)X...XVar(X

)X,Cov(X2σ...σσ)X...XVar(X j

1K

1i

K

1iji

2k

22

21k21


Differences Between Two Random Variables

For two random variables, X and Y

The mean of their difference is the difference of their means; that is

If the covariance between X and Y is 0, then the variance of their difference is

If the covariance between X and Y is not 0, then the variance of their difference is

YX μμY)E(X

2Y

2X σσY)Var(X

Y)2Cov(X,σσY)Var(X 2Y

2X


Linear Combinations of Random Variables

A linear combination of two random variables, X and Y, (where a and b are constants) is

The mean of W is

bYaXW

YXW bμaμbY]E[aXE[W]μ


Linear Combinations of Random Variables

The variance of W is

Or using the correlation,

If both X and Y are joint normally distributed random variables then the linear combination, W, is also normally distributed

Y)2abCov(X,σbσaσ 2Y

22X

22W

YX2Y

22X

22W σY)σ2abCorr(X,σbσaσ


Example

Two tasks must be performed by the same worker.

X = minutes to complete task 1; μx = 20, σx = 5

Y = minutes to complete task 2; μy = 20, σy = 5

X and Y are normally distributed and independent

What is the mean and standard deviation of the time to complete both tasks?


Example

X = minutes to complete task 1; μx = 20, σx = 5

Y = minutes to complete task 2; μy = 30, σy = 8

What are the mean and standard deviation for the time to complete both tasks?

Since X and Y are independent, Cov(X,Y) = 0, so

The standard deviation is

YXW

503020μμμ YXW

89(8)(5) Y)2Cov(X,σσσ 222Y

2X

2W

9.43489σW


- END -

Lesson 5: Lesson 5: Continuous Probability Distributions

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Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson5-1 Lesson 5: Continuous Probability Distributions