Chapter4_427

Chapter4_427 - Chap 4: Continuous RV Continuous RV can take...

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1 Chap 4: Continuous RV Continuous RV can take on any number an interval. (Not isolated values like discrete case) X= height to nearest inch is discrete; Y = true height is continuous. U = time till failure of a product is continuous.
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2 Sect. 4.1: pdf ’s Probability density function (pdf) f(x) of a continuous RV X has two key properties: 1. f(x) > 0 for all x. 2. . “Prob. X is in A = integral of pdf over A” i.e., area under the curve A dx x f A X P ) ( ) (
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3 Key points: Prob. X is equal to a specific value is zero (area is zero). Hence, equalities don’t matter for cont. RV. ex) P(X < c) = P(X < c) pdf is NOT a probability. It can even be greater than 1 for some values. We must have 1 ) ( dx x f
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One View: Cont. RV as Approx. to Disc. RV Idea: from prob. histograms for disc. RV to pdf Compare: integral = limit of sum of areas of rectangles
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Ex) X has pdf f(x) = c x 3 if 0 x 1 = 0 elsewhere. 1. Find c. 2. Find P(0.7 < X < 1.5) . f(x) x
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Find c: reduces to = (c/4) x 4 hence c=4. . ( ) 1 f x dx  0 1 1 1 0 3 dx cx
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P(0.7 < X < 1.5)= = (4/4) x 4 = .76 . 1 .7 dx x x x f 1 7 . 0 3 5 . 1 1 1 7 . 0 3 5 . 1 7 . 0 4 0 4 ) (
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8 Last Time: Sect. 4.1: Continuous RV The pdf f(x) > 0 of a continuous RV X: “Prob. X is in A = integral of pdf over A” i.e., area under the curve prob. X is equal to a specific value is 0 pdf is not a probability and A dx x f A X P ) ( ) ( 1 ) ( x f
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Cumul. Dist. Functions (cdf): F(x) = P(X < x) 0 < F(x) < 1, for all x As As F(x) is non-decreasing. Regions where F is flat are regions of prob. = 0 For a continuous RV, F(x) is a continuous function of x. Wherever the derivative exists, f(x) = dF(x) / dx Same as discrete RV ,  x 0 ) ( x F 1 ) ( x F , x
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Ex) X has with pdf f(x) = .25 x 3 if 0 x 2 = 0 elsewhere. 1. P(0.5 < X < 1) 2. Dist. func. of X. Note: pdf is positive only for 0 < x < 2. Hence, we know F(x) = 0 for all x < 0 and F(x) = 1 for all x > 2. . x 0 059 . | 16 / 25 . 1 5 . 4 1 5 . 3 x dx x
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( f(x) = .25 x 3 if 0 x 2 ) For 0 < x < 2, Final answer: F(x) = 0, x < 0 = x 4 / 16, 0 < x < 2 = 1, x > 2 “endpoints” match-up (F is continuous) Check dF(x)/ dx . x 0 16 / 25 . ) ( 4 0 3 x dy y x F x
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Ex) X is continuous and has dist. func. (cdf) F(x) = 1 – 9 / x 2 if x > 3, and = 0 for x < 3. 1. P(X < 5 ) = F(5) = 1 – 9 / 25 = 0.64 2. P(4 < X < 5 ) = F(5) – F(4) = .64 – ( 1 - 9 / 16) = 0.2025. 3. pdf: For x < 3: dF(x) / dx = 0 For x > 3 : dF(x) / dx = (-9) (-2) / x 3 = 18 / x 3 (no derivative at x = 3, but no problem) So f(x) = 18 / x 3 if x > 3, and = 0 for x < 3 or f(x) = 18 / x 3 if x > 3, and = 0 for x < 3. They give the same results; i.e., we get the same prob’s. .
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Remark: RV of “mixed type” RV that has both discrete and continuous parts. Ex) R is measured rainfall on a summer day, so R > 0. R = 0 with positive prob., say P(R = 0) = p and, when R > 0, it is continuous. I will not ask you about these F(r) 1 p 0 r
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14 Expected Values Similar idea as discrete case: Summaries of prob. dist.
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This note was uploaded on 03/22/2011 for the course STAT 427 taught by Professor Staff during the Spring '08 term at Ohio State.

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Chapter4_427 - Chap 4: Continuous RV Continuous RV can take...

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