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Iam 530 Elements of Probability and Statistics

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Iam 530 Elements of Probability and Statistics IAM 530 ELEMENTS OF PROBABILITY AND STATISTICS LECTURE 3-RANDOM VARIABLES VARIABLE • Studying the behavior of random variables, and more importantly functions of random variables is essential for both the theory and practice of statistics. Variable: A characteristic of population or sample that is of interest for us. Random Variable: A function defined on the sample space S that associates a real number with each outcome in S. In other words, a numerical value to each outcome of a particular experiment. • For each element of an experiment’s sample space, the random variable can take on exactly one value TYPES OF RANDOM VARIABLES We will start with univariate random variables. • Discrete Random Variable: A random variable is called discrete if its range consists of a countable (possibly infinite) number of elements. • Continuous Random Variable: A random variable is called continuous if it can take on any value along a continuum (but may be reported “discretely”). In other words, its outcome can be any value in an interval of the real number line. Note: • Random Variables are denoted by upper case letters (X) • Individual outcomes for RV are denoted by lower case letters (x) DISCRETE RANDOM VARIABLES EXAMPLES • A random variable which takes on values in {0,1} is known as a Bernoulli random variable. • Discrete Uniform distribution: 1 P(X x) ; x 1,2,..., N; N 1,2,... N • Throw a fair die. P(X=1)=…=P(X=6)=1/6 DISCRETE RANDOM VARIABLES • Probability Distribution: Table, Graph, or Formula that describes values a random variable can take on, and its corresponding probability (discrete random variable) or density (continuous random variable). • Discrete Probability Distribution: Assigns probabilities (masses) to the individual outcomes. PROBABILITY MASS FUNCTION (PMF) • Probability Mass Function – 01 p i and p 1 i i – Probability : P() X xii p Example Consider tossing three fair coins. • Let X=number of heads observed. • S={TTT, TTH, THT, HTT, THH, HTH, HHT, HHH} • P(X=0)=P(X=3)=1/8; P(X=1)=P(X=2)=3/8 7 CUMULATIVE DISTRIBUTION FUNCTION (CDF) Cumulative Distribution Function (CDF): F()() y P Y y b F()()() b P Y b p y y FF( ) 0 ( ) 1 F( y ) is monotonically increasing in y Example X= Sum of the up faces of the two die. Table gives value of y for all elements in S 1st\2nd 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 PMF and CDF y p(y) F(y) 2 1/36 1/36 # of ways 2 die can sum to y py() 3 2/36 3/36 # of ways 2 die can result in 4 3/36 6/36 y 5 4/36 10/36 F()() y p t t 2 6 5/36 15/36 7 6/36 21/36 8 5/36 26/36 9 4/36 30/36 10 3/36 33/36 11 2/36 35/36 12 1/36 36/36 PMF-Graph Dice Rolling Probability Function 0,18 0,16 0,14 0,12 0,1 p(y) 0,08 0,06 0,04 0,02 0 2 3 4 5 6 7 8 9 10 11 12 y Example 2 • Machine Breakdowns – Sample space : S{,,} electrical mechanical misuse – Each of these failures may be associated with a repair cost – State space : {50,200,350} – Cost is a random variable : 50, 200, and 350 – P (cost=50)=0.3, P (cost=200)=0.2, P (cost=350)=0.5 – 0.3 + 0.2 + 0.5 =1 50 200 350 xi 0.3 0.2 0.5 pi fx() 0.5 0.3 0.2 50 200 350 Cost($) • Cumulative Distribution Function x50 F ( x ) P (cost x ) 0 50x 200 F ( x ) P (cost x ) 0.3 200x 350 F ( x ) P (cost x ) 0.3 0.2 0.5 350x F ( x ) P (cost x ) 0.3 0.2 0.5 1.0 Fx() 1.0 0.5 0.3 0 50 200 350 x($cost) CONTINUOUS RANDOM VARIABLES • When sample space is uncountable (continuous) • For a continuous random variable P(X = x) = 0. Examples: • Continuous Uniform(a,b) 1 f (X) a x b. b a • Suppose that the random variable X is the diameter of a randomly chosen cylinder manufactured by the company. PROBABILITY DENSITY FUNCTION (PDF) • Probability Density Function – Probabilistic properties of a continuous random variable fx( ) 0 f( x ) dx 1 statespace Example – Suppose that the diameter of a metal cylinder has a p.d.f 2 f( x ) 1.5 6( x 50.2) for 49.5 x 50.5 fx( ) 0, elsewhere fx() 49.5 50.5 x • This is a valid p.d.f. 50.5 2 3 50.5 (1.5 6(x 50.0) ) dx [1.5 x 2( x 50.0) ]49.5 49.5 [1.5 50.5 2(50.5 50.0)3 ] [1.5 49.5 2(49.5 50.0)3 ] 75.5 74.5 1.0 • The probability that a metal cylinder has a diameter between 49.8 and 50.1 mm can be calculated to be 50.1 2 3 50.1 (1.5 6(x 50.0) ) dx [1.5 x 2( x 50.0) ]49.8 49.8 [1.5 50.1 2(50.1 50.0)3 ] [1.5 49.8 2(49.8 50.0)3 ] fx() 75.148 74.716 0.432 49.5 49.8 50.1 50.5 x CUMULATIVE DISTRIBUTION FUNCTION (CDF) x F()()() x P X x f y dy dF() x fx() dx P()()() a X b P X b P X a F()() b F a P()() a X b P a X b x F( x ) P ( X x ) (1.5 6( y 50.0)2 ) dy 49.5 3 x [1.5yy 2( 50.0) ]49.5 [1.5xx 2( 50.0)33 ] [1.5 49.5 2(49.5 50.0) ] 1.5xx 2( 50.0)3 74.5 PXFF(49.7 50.0) (50.0) (49.7) (1.5 50.0 2(50.0 50.0)3 74.5) (1.5 49.7 2(49.7 50.0)3 74.5) 0.5 0.104 0.396 PX(49.7 50.0) 0.396 1 PX(Fx() 50.0) 0.5 PX( 49.7) 0.104 49.5 49.7 50.0 50.5 x Example • Suppose cdf of the random variable X is given as: F(x) 4 x32 6 x 3 x Find the pdf for X. dF(x) 1 1 2 12x22 12 x 3 12 x x 12 x dx 42 THE EXPECTED VALUE Let X be a rv with pdf fX(x) and g(X) be a function of X. Then, the expected value (or the mean or the mathematical expectation) of g(X) g x fX x , if X is discrete x E g X g x fX x dx, if X is continuous providing the sum or the integral exists, i.e., <E[g(X)]< . EXPECTED VALUE (MEAN) AND VARIANCE OF A DISCERETE RANDOM VARIABLE • Given a discrete random variable X with values xi, that occur with probabilities p(xi), the population mean of X is E()() X xii p x all x i • Let X be a discrete random variable with possible values xi that occur with probabilities p(xi), and let E(xi) = The variance of X is defined by 2 2 2 V()()()() X E X xii p x all xi The stan dard deviation is 2 Mean: EX( ) VXEXEXEX( )2 ( ( )) 2 ( ) 2 (x )2 p ( x ) x 2 2 x 2 p ( x ) all x all x x22 p( x ) 2 xp ( x ) p ( x ) all x all x all x EXEX22 ( ) 2 (1) 2 2 Example – Rolling 2 Dice y p(y) yp(y) y2p(y) 2 1/36 2/36 4/36 3 2/36 6/36 18/36 12 4 3/36 12/36 48/36 E( Y ) yp ( y ) 7.0 5 4/36 20/36 100/36 y 2 6 5/36 30/36 180/36 12 2E Y 2 2 y 2 p() y 2 7 6/36 42/36 294/36 y 2 8 5/36 40/36 320/36 54.8333 (7.0)2 5.8333 9 4/36 36/36 324/36 5.8333 2.4152 10 3/36 30/36 300/36 11 2/36 22/36 242/36 12 1/36 12/36 144/36 Sum 36/36= 252/36 1974/36=5 1.00 =7.00 4.833 Example 2 • The pmf for the number of defective items in a lot is as follows 0.35,x 0 0.39,x 1 p( x ) 0.19, x 2 0.06,x 3 0.01,x 4 Find the expected number and the variance of defective items. Results: E(X)=0.99, Var(X)=0.8699 28 EXPECTED VALUE (MEAN) AND VARIANCE OF A CONTINUOUS RANDOM VARIABLE • The expected value or mean value of a continuous random variable X with pdf f(x) is E()() X xf x dx all x • The variance of a continuous random variable X with pdf f(x) is 2Var()()()() X E X 2 x 2 f x dx all x E()()() X2 2 x 2 f x dx 2 all x Example • In the flight time example, suppose the probability density function for X is 42 f( x ) , 0 x 0.5; f( x ) , 0.5 x 1.
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