CEE11_s11_ch3

CEE11_s11_ch3 - Discrete random variables CEE 11 ch 3...

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Discrete random variables CEE 11 – ch 3 Copyright Dr. Brigitte Baldi 2011 © Random variables A random variable X is a function that associates a unique numerical value with every outcome of a random experiment. ± A discrete random variable can take only a finite (or a countably infinite, i.e. ordered list) number of values. ² 0 for head, 1 for tail, when flipping a coin ² Number of items inspected before a defective item is found ² Number of cracks in a portion of a highway during the winter months ± A continuous random variable can take any real value within a finite or infinite interval. ² Life time, pH, concentration of pollutants, . ..
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Probability mass function (PMF) The probability mass function, p(x) = P(X = x) ,of±a ± discrete random variable X gives the probability that the random variable will take on each of its possible values. p(x) = 1/16 x = 0 4/16 x = 1 6/16 x = 2 4/16 x = 3 1/16 x = 4 0 otherwise X is the number of heads in 4 successive coin flips
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Cumulative distribution function If p(x) = P(X = x) is the probability mass function (pmf) of a discrete random variable, then its cumulative distribution function (cdf) is F(x) = P(X x) = Σ p(y) for y x If a and b are any two numbers such that a b then F(a X b) = F(b) – F(a ־ ) where F(a ־ ) evaluates F at the point immediately to the left of a. p(x) = 1/16 x = 0 4/16 x = 1 6/16 x = 2 4/16 x = 3 1/16 x = 4 0 otherwise X is the number of heads in 4 successive coin flips F(x) = 0x < 0 1/16 x = 0 5/16 x = 1 11/16 x = 2 15/16 x = 3 16/16 x = 4 1x > 4 0 0.2 0.4 0.6 0.8 1 01234 X F(x)
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Expected value (mean) The expected value E(X) or mean μ of a random variable X represents its central value. It is the weighted sum of the possible values of X in the domain D using the probabilities as weights. p(x) = 1/16 x = 0 4/16 x = 1 6/16 x = 2 4/16 x = 3 1/16 x = 4 0 otherwise X is the number of heads in 4 successive coin flips E(X) = (0)(1/16) + (1)(4/16) + (2)(6/16) + (3)(4/16) + (4)(1/16) = 32/16 = 2 Variance The variance V(X) or σ 2 of a random variable X is a measure of spread about its expected value μ . The standard deviation σ is the square root of σ 2 p(x) = 1/16 x = 0 4/16 x = 1 6/16 x = 2 4/16 x = 3 1/16 x = 4 0 otherwise X is the number of heads in 4 successive coin flips E(X) = 2 V(X) = (0-2) 2 (1/16) + (1-2) 2 (4/16) + (2-2) 2 (6/16) + (3-2) 2 (4/16) + (4-2) 2 (1/16) = (4+4+0+4+4)/16 = 1
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Calculations rules for the mean Let X be a random variable taking values over D and with pmf p(x). If Y is a function of X the such that Y = h(X) , then If a and b are two constants, then E(aX) = aE(X) E(X + b) = E(X) + b E(aX + b) = aE(X) + b Also note that, if X and Y are two random variables, then E(X + Y) = E(X) + E(Y) E(X – Y) = E(X) – E(Y) Calculations rules for the variance Let X be a random variable taking values over D and with pmf p(x). If Y is a function of X the such that Y = h(X) , then If a and b are two constants, then V(aX) = a 2 V(X) V(X + b) = V(X) V(aX + b) = a 2 V(X) Also note that, if X and Y are two independent random variables, then V(X + Y) = V(X) + V(Y) V(X – Y) = V(X) + V(Y) {} 2 [( ) ] () [( ) ] () xD VhX hx EhX px =−
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Let X be a discrete random varible with the following pmf and cdf: 0.10 x = 0 0.25 x = 1 0.35 x = 2 ()= 0.20 x = 3 0.10 x = 4 0 otherwise px Let h(x) = x 2 +2 () x p 0(0.1)+1(0.25)+2(0.35)+3(0.20)+4(0.10)=1.95
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This note was uploaded on 07/21/2011 for the course CEE 11 taught by Professor Bhatti during the Spring '11 term at UC Irvine.

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CEE11_s11_ch3 - Discrete random variables CEE 11 ch 3...

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