5.Random Variables and Expectations

7/28/2019 5.Random Variables and Expectations

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RANDOM VARIABLES

AND EXPECTATION


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Random Variables

A random variable is a function or rule thatassigns a numerical value to each simpleevent in a sample space.

A random variable reflects the aspect of arandom experiment that is of interest for us.

There are two types of random variables:

Discrete random variable Continuous random variable.


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Random Variables

IfXis a function that assigns a realnumbered value to every possible event ina sample space of interest,Xis called a

random variable. It is denoted by capital letters such asX, Y

andZ.

The specified value of the random variableis unknown until the experimental outcomeis observed.


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EXAMPLES

The experiment of flipping a fair coin.Outcome of the flip is a random variable.

S={H,T}X(H)=1 andX(T)=0

Select a student at random from allregistered students at METU. We want toknow the weight of these students.

X= the weight of the selected student

S: {x: 45kg X 300kg}


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A random variable is discrete if it can assumea countable number of values.

A random variable is continuous if it can

assume an uncountable number of values.

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Continuous random variable

After the first value is defined

the second value, and any valuethereafter are known.

Therefore, the number of

values is countable

After the first value is defined,

any number can be the next one

Discrete random variable

Therefore, the number of

values is uncountable

0 1 2 3 ...

Discrete and Continuous RandomVariables


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A table, formula, or graph that lists all possiblevalues a discrete random variable can assume,together with associated probabilities, is calleda discrete probability distribution.

To calculate the probability that the randomvariable X assumes the valuex,P(X = x), add the probabilities of all the simple events for which

X is equal tox, or Use probability calculation tools (tree diagram),

Apply probability definitions

Discrete Probability Distribution


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If a random variable can assume valuesxi, then the following must be true:

1)x(p.2

xallfor1)p(x0.1

ixall

i

ii

Requirements for a DiscreteDistribution


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EXAMPLE

Consider an experiment in which a fair coin istossed 3 times.

X= The number of heads

Lets assign 1 for head and 0 for tail. The samplespace is

S: {TTT,TTH,THT,HTT,THH,HTH,HHT,HHH}

Possible values ofXis 0, 1, 2, 3. Then, the

probability distribution ofXis

x 0 1 2 3 Total

p(x) 1/8 3/8 3/8 1/8 1


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EXAMPLE

An automobile service facility specializing inengine tune-ups knows that 45% of all tune-ups are done on 4-cylinder automobiles,

40% of 6-cylinder automobiles and 15% on8-cylinder automobiles.

LetX= The number of cylinders on the next car

to be tuned.


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Example (Contd.)

a) What is the pmf ofX?

b) Draw the line graph for the pmf.

c) What is the probability that the next car tuned has at least6-cylinder? More than 6-cylinder?

d) Draw a probability histogram.


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Distribution and Relative Frequencies

In practice, often probability distributions areestimated from relative frequencies.

ExampleA survey reveals the following frequencies(1,000s) for the number of color TVs per household.

Number of TVs Number of Households x p(x)0 1,218 0 1218/Total = .0121 32,379 1 .3192 37,961 2 .374

3 19,387 3 .1914 7,714 4 .0765 2,842 5 .028Total 101,501 1.000


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Determining Probability of Events

The probability distribution can be used tocalculate the probability of different events

Example continued

Calculate the probability of the followingevents:

P(The number of color TVs is 3) = P(X=3)

=.191

P(The number of color TVs is two or more) =P(X2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)=

.374 +.191 +.076 +.028 = .669


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EXAMPLE

Find a constant c so thatp(x) satisfies theconditions of being a pmf ofX.

2, 1,2,3,....

( ) 30 ,elsewhere

x

c xp x

Letp(x)=x/15, x=1,2,3,4,5 be a pmf ofX.

FindPr(X=1 or X=2).

FindPr(1X3).


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The Cumulative DistributionFunction

IfXis a random variable, then thecumulative distribution function (cdf),denoted byF(x) is given by

for all real numbersx. It is a non-decreasingstep function ofx and it is right continuous.

:

( ) ( ) ( )

y y x

F x P X x p y


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The Cumulative DistributionFunction

For any two numbers a and b, a b

P (a Xb) = F (b)F (a-)

where a- represents the largest possibleXvalue which is less than a.

Ifa and b are integers,

P (a Xb) = F (b)F(a-1)

Taking a=b,P( X=a ) = F (a)-F (a-1)


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EXAMPLE

Let X is the number of days of sick leave takenby a randomly selected employee of a largecompany during a particular year. If the max.number of allowable sick days per year is 14,

possible values ofXare 0,1,2,,14. WithF(0)=0.58, F(1)=0.72,F(2)=0.76,F(3)=0.81,F(4)=0.88 and F(5)=0.94,find

P(2 X5) =

P(X = 3) =

P(X2) =


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Describing the Population/

Probability Distribution The probability distribution represents a

population

Were interested in describing the population

by computing various parameters.

Specifically, we calculate the population

mean and population variance.


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EXPECTED VALUES

LetXbe a rv with pdffX(x) andg(X) be afunction ofX. Then, the expected value (orthe mean or the mathematical expectation)

ofg(X)

X

x

X

g x f x , if X is discrete

E g X

g x f x dx, if X is continuous

providing the sum or the integral exists, i.e.,


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EXPECTED VALUES

E[g(X)] is finite ifE[| g(X) |] is finite.

Xx

X

g x f x < , if X is discrete

E g X

g x f x dx< , if X is continuous


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Population Mean (Expected Value)

Given a discrete random variable X withvaluesxi, that occur with probabilitiesp(xi),the population mean ofX is.

ixall

ii)x(px)X(E


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Let X be a discrete random variable withpossible valuesxi that occur withprobabilitiesp(x

i

), and letE(xi

) = . Thevariance ofX is defined by

ixall

i

2

i

22 )x(p)x()X(E)X(V

Population Variance

2

isdev iationdardtansThe


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EXAMPLE

The pmf for the number of defective itemsin a lot is as follows

0.35, 0

0.39, 1

( ) 0.19, 2

0.06, 3

0.01, 4

x

x

p x x

x

x

Find the expected number and the variance of

defective items.


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EXAMPLE

What is the mathematical expectation if wewin $10 when a die comes up 1 or 6 andlose $5 when it comes up 2, 3, 4 and 5?

X= amount of profit


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EXAMPLE

A grab-bay contains 6 packages worth $2each, 11 packages worth $3, and 8packages worth $4 each. Is it reasonable

to pay $3.5 for the option of selecting oneof these packages at random?

X= worth of packages


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Laws of Expected Value

LetXbe a rv and a, b, and c be constants.Then, for any two functionsg1(x) andg2(x)whose expectations exist,

1 2 1 2)a E ag X bg X c aE g X bE g X c

1 10 , 0.b) If g x for all x then E g X

1 2 1 2) .c If g x g x for all x, then E g x E g x

1 1)d If a g x b for all x, then a E g X b


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Laws of Expected

Value

E(c) = c

E(X + c) = E(X) + c

E(cX) = cE(X)

Laws of

Variance

V(c) = 0

V(X + c) = V(X)

V(cX) = c2V(X)

Laws of Expected Value and Variance


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SOME MATHEMATICAL

EXPECTATIONS

Population Mean: = E(X)

Population Variance:

2 22 2

0Var X E X E X (measure of the deviation from the population mean)

Population Standard Deviation: 2 0

Moments:* k

k E X the k-th moment

k

k E X the k-th central moment


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SKEWNESS

Measure of lack of symmetry in the pdf.

3

3

3 3/2

2

E XSkewness

If the distribution ofXis symmetric around itsmean ,

3=0 Skewness=0


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KURTOSIS

Measure of the peakedness of the pdf. Describesthe shape of the r,v.

4

4

4 2

2

E X

Kurtosis

Kurtosis=3 Normal

Kurtosis >3 Leptokurtic

(peaked and fat tails)

Kurtosis


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EXAMPLE

An appliance dealer sells 3 different modelsof upright freezers having 13.5, 15.9 and 19.1cubic feet of storage space, respectively.

X= the amount of storage space purchased bythe next customer.

The pmf ofX0.2, 13.5

( ) 0.5, 15.9

0.3, 19.1

x

p x x

x


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EXAMPLE (contd.)

a) E(X)=?

b) If the price of a freezer having capacityXcubic feet is25X-8.5, what is the expected price paid by the next

customer to buy a freezer?

c) Var (X)=?

d) Suppose that although the rated capacity of a freezer isX,the actual capacity is h (X)=X-0.01X2. What is the expectedactual capacity of the freezer purchased by the nextcustomer?


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A continuous random variable has anuncountably infinite number of values inthe interval (a,b).

Continuous Probability Distributions

The probability that a continuous variableX will assume any particular value is zero.Why?

0 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 1

1/4 + 1/4 + 1/4 + 1/4 = 1The probability of each value


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0 11/21/3 2/3

1/2 + 1/2 = 11/3 + 1/3 + 1/3 = 1

1/4 + 1/4 + 1/4 + 1/4 = 1

As the number of values increases the probability of each

value decreases. This is so because the sum of all the

probabilities remains 1.

When the number of values approaches infinity (becauseX

is continuous) the probability of each value approaches 0.

The probability of each value

Continuous Probability

Distributions


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To calculate probabilities we define aprobability density functionf(x).

The density function satisfies the following

conditionsf(x) is non-negative,

The total area under the curve representing f(x)equals 1.

x1 x2

Area = 1P(x1


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CUMULATIVE DISTRIBUTIONFUNCTION

min

( ) ( ) ( )x

X

F x P X x f x dx

Example: LetXis the amount of time for which abook on two-hour reserve at a college library is

checked out by a randomly selected student.Suppose thatXhas density function

0.5 ,0 2( )

0 ,otherwise

x xf x

P(X1)=?

P(0.5 X1.5)=?P(X


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EXPECTED VALUE

The expected value or mean value of acontinuous random variableXwith pdff(x) is

( ) ( )

all x

E X xf x dx

The variance of a continuous randomvariableXwith pdff(x) is

2 2 2

all x

2 2 2 2

all x

( ) ( ) ( ) ( )

( ) ( ) ( )

Var X E X x f x dx

E X x f x dx


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EXAMPLE

LetXbe a random variable and it is a lifelength of light bulb. Its pdf is

f(x)=2(1-x), 0< x < 1

FindE(X) and Var(X).


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EXPECTED VALUE OF SUMS OF

RANDOM VARIABLES

IfXand Yare random variables, then

E(X+Y) = E(X) + E(Y)


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Example At computer store, the annual demand for a particular software

package is a discrete random variable X. The store owner ordersfour copies of the package at 10$ per copy and charges customers$35 per copy. At the end of the year, the package is obsolete andthe owner loses the investment on unsold copies. The pdf ofXisgiven by the following table:

FindE(X) and Var(X).

Express the owners net profit Yas a linear function ofX, and findE(Y) and Var(Y).

39

x 0 1 2 3 4

P(x) .1 .3 .3 .2 .1

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5.Random Variables and Expectations

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