PSTAT 120B Probability and Statistics
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1 - Week University of California, Santa Barbara April 10, 013
2 Discussion section for 10B Information about TA: Fang-I CHU Office: South Hall 5431 T Office hour: TBA [email protected] Slides will be available at each section Feel free to interrupt me if anything during section puzzles you. Office hour will be determined after I have more information on the regular due date for your weekly assignment.
3 Topic for review Transformation Using the cumulative distribution function approach Using the formula on probability density function Exercise 6.3 (page 307). Exercise 6.5 (page 308). Exercise 6.11 (page 308).
4 The method of transformation Available methods to do transformation: 1. Cumulative distribution function approach. Formula on probability density function (will be introduced next time) Note: the obtained results by using above two methods are identical. In some problem, using one approach is easier to compute than the other.
5 Transformation- cumulative distribution function approach (Review from lecture) Let U be a function of the random variables Y 1,..., Y n. 1. Find the region U u in the (Y 1,..., Y n ) space.. Find the cdf U, F U (u) = P(U u) by integrating the joint pdf of {Y 1,..., Y n } over the region U u. 3. Find the pdf of U as f U (u) = df U(u) du. Note: The cumulative distribution function approach can only be applied on continuous cases.
6 Exercise A supplier of kerosene has a weekly demand Y possessing a probability density function given by y 0 y 1 f (y) = 1 1 < y elsewhere with measurements in hundreds of gallons. (This problem was introduced in Exercise 4.13.) The supplier s profit is given by U = 10Y 4.
7 Exercise 6.3(a) 6.3(a) Find the probability density function for U. Solution: (a)the cumulative distribution function for Y should be written as y 0 y 1 F Y (y) = y 1 1 < y elsewhere (why?) 1. F U (u) = P(U u) = P(10Y 4 u) = p(y u+4 10 ). compute the range of u using given formula U = 10Y 4
8 Exercise 6.3(a) 6.3(a) Find the probability density function for U. Solution: (a)the cumulative distribution function for U can be obtained as ( u+4 10 ) = (u+4) 00 4 u 6 F U (u) = u = u < u 11 1 elsewhere The probability density function for U is followed as u+4 f U (u) = F U (u) = u < u 11 0 elsewhere
9 Exercise 6.3(b),(c) 6.3 (b)use the answer to part (a) to find E(U). (c)find E(U) by the methods of Chapter 4. Solution (b) (c) E(U) = 6 4 (Find E(Y ) on your own!) u u du u 1 10 du = 5.583(check this on your own!) E(U) = E(10Y 4) = 10E(Y ) 4 = 10( 3 4 ) 4 = 5.583
10 Exercise The waiting time Y until delivery of a component for an industrial operation is uniformly distributed over the interval from 1 to 5 days. The cost of this delay is given by U = Y + 3. Find the probability density function for U. Solution: Y Uniform(1, 5) i.e. f Y (y) = 1 4 for 1 < y < 5 cumulative distribution function for Y : F Y (y) = y 4 for 1 < y < 5
11 Exercise 6.5 Solution: F U (u) = P(U u) = P(Y + 3 u) u 3 = P(Y ) u 3 = F Y ( ) u 3 = 1 4 Probability density distribution for U: f U (u) = F U (u) = 1 16 ( u 3 ) 1, for 5 u 53(why?).
12 Exercise Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the (a)probability density function for the average length of life of the two components. (b)mean and variance of this average, using the answer in part(a). Check your answer by computing the mean and variance, using Theorem 5.1.
13 Exercise 6.11(a) 6.11(a) (a)probability density function for the average length of life of the two components. Solution: Known: 1. length Y i exp(1) for i = 1,. i.e. f Yi (y i ) = e y i. denote mean as U = Y1+Y Way to approach: 1. F U (u) = P(U u) = P( Y1+Y u 0 u) = P(Y 1 u Y ) = u y e y1 y dy 0 1 dy = 1 e u ue u (check this on your own!). f U (u) = F U (u) = 4ue u, u 0 3. recognize f U (u) is in form of gamma density with parameter (, 1 )
14 Exercise (b) (b)mean and variance of this average, using the answer in part(a). Check your answer by computing the mean and variance, using Theorem 5.1. Solution: Known: U Γ(, 1 ) Way to approach: using formula E(U) = αβ and Var(U) = αβ Obtained answer: E(U) = 1 1 and Var(U) =
15 Exercise (b) (b)mean and variance of this average, using the answer in part(a). Check your answer by computing the mean and variance, using Theorem 5.1. Solution: If you couldn t recognize U has a gamma distribution, use the definition to find mean and variance for U. i.e. E(U) = uf U (u)du, E(U ) = u f U (u)du and Var(U) = E(U ) E(U) Use Theorem 5.1 to check: E(U) = E( Y 1+Y ) = 1+1 and Var(U) = Var( Y 1+Y ) = 1 4 (1 + 1) = 1 (why?) Note Y 1 and Y are independent, this implies Cov(Y 1, Y ) = 0
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