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# Chapter3_all - Continuous Random Variables Reading Chapter...

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1 Continuous Random Variables Reading: Chapter 3.1 – 3.8 Homework: 3.1.2, 3.2.1, 3.2.4, 3.3.2, 3.3.7, 3.4.4, 3.4.5, 3.4.9, 3.5.3., 3.5.6, 3.6.1, 3.6.6, 3.7.1, 3.7.3, 3.7.11, 3.8.1. G. Qu ENEE 324 Engineering Probability 2 Cumulative Distribution Function CDF of a random variable X is: F X (x) = P[X≤x] X is a continuous random variable if F X (x) is continuous. (the range of X contains a continuous interval) Example: X: a random integer between 0 and 4. S X ={0,1,2,3,4} P X (x) = 0.2 for x=0,1,2,3,4 and 0 otherwise F X (x) is an non-decreasing piece-wise constant function that is not continuous at x=0,1,2,3,4 Y: a random real number between 0 and 4. S Y =[0,4] is a continuous region, not a countable set. F Y [y] = P[Y≤y] = ? P Y [Y=y] = ? > < = 4 1 4 0 4 / 0 0 ) ( y y y y y F Y

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2 G. Qu ENEE 324 Engineering Probability 3 Properties of CDF Theorem 3.1: for any random variable X F X (-∞) = 0, F X (∞) = 1 F X (x) = 0 for x<x min , F X (x) = 1 for x ≥x max F X (x) ≥ F X (x’) if x≥x’ F X (x) – F X (x’) = P[x’< X ≤ x] Example: Quiz 3.1 P[Y≤-1] P[Y≤1] P[2<Y≤3] P[Y>1.5] > < = 4 1 4 0 4 / 0 0 ) ( y y y y y F Y G. Qu ENEE 324 Engineering Probability 4 Probability Density Function The PDF of a continuous random variable X is Δ - Δ + = = Δ ) ( ) ( lim ) ( ) ( 0 x F x F dx x dF x f X X X X What does PDF tell directly? large f X (x) implies that F X (x) increases fast at x. f X (x) is the slope of CDF at x, it measures how fast CDF F X (x) increases. large f X (x) implies that higher probability that X has value close to x. Similar to P X (x). ] [ ) ( ) ( ) ( Δ + = - Δ + = Δ x X x P x F x F x f X X X
3 G. Qu ENEE 324 Engineering Probability 5 Properties of PDF Theorem 3.2: for any random variable X f X (x) ≥ 0 Theorem 3.3: Example: Problem 3.2.5 Find the conditions for a and b to make f X (x) a valid PDF. - = x X X dt t f x F ) ( ) ( 1 ) ( ) ( = = - X X F dt t f ] [ ) ( 2 1 2 1 x X x P dt t f x x X = + = otherwise x bx ax x f X 0 1 0 ) ( 2 G. Qu ENEE 324 Engineering Probability 6 Expected Values For discrete r.v. X, E[X] = ∑ xP X (x) For continuous r.v X, Theorem 3.4: Theorem 3.5: E[X-μ X ] = 0 E[aX+b] = aE[X]+b Var[X] = E[X 2 ] – E[X] 2 Var[aX+b] = a 2 Var[X] - = dx x xf X E X ) ( ] [ - = dx x f x g X g E X ) ( ) ( )] ( [

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4 G. Qu ENEE 324 Engineering Probability 7 In-class Quiz Quiz 3.3: Given the PDF of r.v. Y below, find Expected value: E[Y]=? Second moment: E[Y 2 ]=? Variance: Var[Y]=? Standard deviation: σ Y =? - = otherwise x y y f Y 0 1 1 2 / 3 ) ( 2 G. Qu ENEE 324 Engineering Probability 8 Continuous E[X] = (a+b)/2 Var[X] = (b-a) 2 /12 Uniform Random Variable Discrete E[X] = (a+b)/2 Var[X] = (b-a)(b-a+2)/12 + = + - = otherwise b a a x a b x P X 0 ,..., 1 , 1 1 ) (   > + - + - < = b x b x a a b a x a x x F X 1 1 1 0 ) ( - = otherwise b x a a b x f X 0 1 ) ( >
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