Lec12 - IOE/S 265 IOE/Stat 265, Fall 2009 2009 Lecture...

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Unformatted text preview: IOE/S 265 IOE/Stat 265, Fall 2009 2009 Lecture Lecture #12: Joint, Marginal, and Conditional Probability Probability 5-1 5-2 5-3 5-4 5-5 Jointly Distributed Random Variables Expected Values, Covariance & Correlation Statistics and Their Distributions Th Di Distribution of the Sample Mean Distributions of a Linear Combination Distributions of Linear Combination 1 Jointly Distributed Variables Distributed Variables Most real problems involve more than single Most real problems involve more than a single random variable. It is necessary to analyze and make inference It is necessary to analyze and make inference about several random variables simultaneously. X 1 0 0 1 Z Y Y 0 1 1 0 Z 0 1 0 1 X X = height (1 = in spec, 0 = out of spec) Y = width (1 = in spec, 0 = out of spec) width (1 in spec out of spec) Z = depth (1= in spec, 0 = out of spec) 2 5.1 Definitions Definitions Joint Probability Mass Function – pg 185 Marginal Probability Mass Function – pg 186 Joint Probability Density Function – pg 186 Marginal Probability Density Function – pg 188 Probability Density Function pg 188 Independence – pg 190, 192 More than Two Random Variables – pg 191 Example 1: (Revisit) TV Repairs 1: (Revisit) TV Repairs TV Repair Example Repair Example (continued from Lectures (continued from Lectures 4 & 5) 5) Event A: sell a TV from one of 3 brands (A1, A2 or A3); Event B: repair TV Event B: repair TV Suppose Selling Mix: Likelihood to Repair Given Model A1 = 25% Likelihood to Repair Given Model A2 = 20% Likelihood to Repair Given Model A3 = 10% A1 = 50%, A2 = 30% and A3 = 20% Conditional Prob Density Function – pg 193 First Convert marginal to joint probabilities table. Example: f(x=A1, y=repair) = 0.5*0.25 = 0.125 3 4 Joint Probability Table Probability Table Wh What are some requirements of joint probability table? Sum of all cells = 1, all table values ≥ 0. Joint, Marginal, and Conditional probabilities each sum to 1. What is the marginal PMF of Y = repair? fY(x,repair) What is the probability of having a non-model A1? 0 (no repair) 0.375 0.24 0.18 1 (repair) 0.125 0.06 0.02 f(x=A2,y) or f(x=A3,y) = f(x=A2, y=0) + f(x=A2, y = 1) + f(x=A3, y=0) or f(x=A f(x=A y=0) f(x=A 1) f(x=A y=0) + f(x=A3, y = 1) Y X Spreadsheet: Lec12.xls available on CTools 5 A1 A2 A3 6 Example 1: TV Repair (continued) 1: TV Repair (continued) Are and independent? Are X and Y independent? Does f(X=A1,Y=1) = fX(X=A1) * fY(Y=1) ? fX(X=A1) = f(A1,0) + f(A1,1) = 0.375 + .125 = 0.5 fY(Y=1) = f(A1,1) + f(A2,1) + f(A3,1) = 0.205 f(A1,1) = 0.125 0.125 fx(A1)*fy(1) = 0.5*0.205 = 0.1025 -- So, Dependent (i.e. repair rates are not the same for all brands) Example 2: CD Case Dimensions 2: CD Case Dimensions Lengths (X) and Widths (Y) of rectangular CD covers Lengths (X) and Widths (Y) of rectangular CD covers in personal computers are distributed as follows: Y: 129 m m 130 m m 131 m m 15 m m 12.0% 42.0% 6.0% 16 m m 8.0% 28.0% 4.0% x A1 A2 A3 y 0 (no repair) 1 (repair) 0.375 0.125 0.24 0.06 0.18 0.02 7 X: 8 Example (Continued) Example 2 (Continued) Find P(X>129 and Y>15) = ? Example 2 (Continued) (Continued) Find marginal density functions for and Find marginal density functions for X and Y Are X and Y Independent ? Find the conditional pdf of X given Y = 15 9 14 Example 2: CD Case Dimensions 2: CD Case Dimensions Lengths (X) and Widths (Y) of rectangular CD covers Lengths (X) and Widths (Y) of rectangular CD covers in personal computers are distributed as follows: Example 2 (Continued) (Continued) Find the conditional expectation of given Find the conditional expectation of X given Y=15 Y: 129 m m 130 m m 131 m m 15 m m 12.0% 42.0% 6.0% 16 m m 8.0% 28.0% 4.0% Find the Find the value of E(X). Does it differ from of E(X) Does it diffe E(X | Y=15) ? X: 19 22 Example (Continued) Example 2 (Continued) Find the probability the length (X) is less than or equal to 130 mm given the width (Y) is 15 mm. Example 2: Does the Variance of depend on X? Does the Variance of Y depend on X? 26 28 Example 3: CD Case Dimensions (Modified) (Modified) Lengths (X) and Widths (Y) of rectangular CD covers Lengths (X) and Widths (Y) of rectangular CD covers in personal computers are distributed as follows: Example 3 (Continued) (Continued) Find P(X>129 and Y>15) = ? Y: 129 m m 130 m m 131 m m 15 m m 12.0% 40.0% 8.0% 16 m m 8.0% 30.0% 2.0% Are and Independent Are X and Y Independent ? X: Notice: Marginal probabilities are same as before! 33 35 Example (Continued) Example 3 (Continued) Find marginal density functions for and Find marginal density functions for X and Y Example 3 (Continued) (Continued) Find the conditional expectation of given Find the conditional expectation of X given Y=15 Find the conditional pdf of X given Y = 15 Find the Find the value of E(X). Does it differ from of E(X) Does it diffe E(X | Y=15) ? 38 43 Example 3 (Continued) (Continued) Find the probability the length (X) is less than Find the probability the length (X) is less than or equal to 130 mm given the width (Y) is 15 mm mm. Example 3: Does the Variance of depend on X? Does the Variance of Y depend on X? 48 51 Example 4: Washer Dimensional Quality Quality Hole diameter(X) and thickness(Y) varies from washer diameter(X) and thickness(Y) varies from washer to washer according to ⎧1 ⎪ (x + y) f(x, y) = ⎨ 6 ⎪ 0 ⎩ 1 ≤ x ≤ 2, 4 ≤ y ≤ 5 otherwise Find P(1≤X≤1.5 and 4.5≤Y≤5.0) 55 59 Example 4 (Continued) (Continued) Find marginal density functions for and Find marginal density functions for X and Y Example 4 (Continued) (Continued) Find the conditional expectation of given Find the conditional expectation of Y given X=1.2 Find the conditional pdf of Y given X=1.2 Find the value of E(Y). Does it differ from E(Y X=1 E(Y | X=1.2) ? 60 63 Example (Continued) Example 4 (Continued) Are and independent? Are X and Y independent? Example 4 (Continued) (Continued) Find the probability the hole thickness is less Find the probability the hole thickness is less than or equal to 4.8 mm given the diameter is is 1.2 mm. mm 67 69 Example 5: Mixture Experiments 5: Mixture Experiments Useful application related to Mixture Experiments Mi Let X - proportion of a mix that is component A Let Let Y - proportion of a mix that is not A: y = 1 – x of mix that is not A: The sum of X and Y represents 100% of the mixture f(x,y) = k(xy) where 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 subject to x+y = 1 What is k if f(x,y) is pdf?: 71 ...
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