discrete random variables lecture
TRANSCRIPT
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Random Variables
A random variable (rv) associates anumber with each outcome in thesample space
We denote random variables with uppercase letter, XThe observed numerical value once theexperiment is run is denoted by thecorresponding lower case letter, x
In mathematical terms, a rv is a functionwhose domain is the sample space andthe range is the set or real numbers
2007 Thomson Brooks/Cole, a part of TheThomson Corporation.
Random Variables
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Random Variable Example
A tall antenna is built on a mountain topwhere an extreme wind event occurs.Either the antenna fails (F) or survives (S)
If the rv X is associated withthe outcomes,
1 indicates that the antennasurvived, 0 indicates that theantenna failed
S,Fs
0FX,1SX
Types of Random Variables
Discrete Random Variable:takes a finite number of values
e.g. the number of cars lined up at the MassPike toll plaza #8 (Palmer)
Continuous Random Variable:takes all values in an interval
e.g. the time each car mustwait at the toll plaza
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Probability Mass Function
ExampleSuppose you flip two coins, a rv, X, isthe number of heads in the experiment
What is the sample space?What is the probability of each outcome inthe sample space?
Probability Mass Function
A Probability Mass Function (pmf), alsocalled probability distribution , is afunction p(x) that assigns to eachpossible value x that the random variableX can take, its probability
p(x i) 0 for each possible value x i of X
xsX:SsallP
xXPxp
ixall
i 1)x(p
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pmf histogram
0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2
p ( x
)
pmf:P (X = 0) = 0.5P (X = 1) = 0.3P (X = 2) = 0.2
Cumulative DistributionFunction Example
From the previous example, let F(x) denotethe cumulative distribution function (cdf) ofthe rv X
X pmf, P(X=x) cdf, F(X=x)
0 0.5 0.5
1 0.3 0.5+0.3=0.8
2 0.2 0.5+0.3+0.2=1
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Cumulative Distribution
Function Any probability distribution must followthe axioms of probability
F(- )=0; F( ) = 1F(x) 0 and is weakly increasingIt is continuous in x
Any function that satisfies these axiomsis a cdf
Expected Value Example
Let X = number of working bulldozersafter 6 months. Assume the probabilitythat a bulldozer is working after 6months is 0.8, and there are 3 dozers.
Find the pmf Find the cdf What is the expected value of the number ofdozers working after 6 months?
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Expected Value
A random variable has probabilities passociated with outcomes x with set ofpossible values D and pmf p(x)
The expected value is the long runexpected mean, if you were to see Xover and over again
Dx
x xpxXE
Example:Expected ValuesIn previous example, at least 2 dozers areneeded to finish a $100K job. Every dozerthat was brought in after 6 months costs$10K. What is your expected profit if youstart with 3 dozers?
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Expected Value of a Function
Consider that there might be a functionalrelationship with X with a set of possiblevalues D and pmf p(x) such that we havea probability of h(X)
D
)x(p)x(h)X(hE
Properties of Expected Value
bXEabaXE
]X[EXEXXE 2121
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Example Solution
E[X2] = 0 .008 + 1 .096 + 4 .384 + 9 .512= 6.24
E[X] = = 2.4(E[X])2 = 2 = 2.4 2 = 5.76
2 = 6.24 5.76 = 0.48
P(x) .008 .096 .384 .512x 0 1 2 3x2 0 1 4 9
222 ]X[E
Variance
The variance is defined as
22
D
2
2
2
]X[E
)x(p)x(
])X[(E
)X(V
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Properties of Variance &
Standard Deviation
Xa)baX(V 22
XbaX |a|
Example:Variance Properties
Your profit is equal to $6000 plus $10 forevery dozer that is working at 6 months.
What is the variance of your profit?What is the STD of your profit?