Functions of Several Random Variables (Ch. 5.5)

Transcription

1 (Ch. 5.5) Iowa State University Mar 7, 2013 Iowa State University Mar 7, / 37

2 Outline Iowa State University Mar 7, / 37

3 several random variables We often consider functions of random variables of the form: U = g(x, Y,..., Z) where X, Y,..., Z are random variables. U is itself a random variable. Iowa State University Mar 7, / 37

4 Example: connecting steel parts Suppose that a steel plate with nominal thickness.15 in. is to rest in a groove of nominal width.155 in., machined on the surface of a steel block. X = plate thickness Y = slot width U = Y X, the wiggle room of the plate Iowa State University Mar 7, / 37

5 The distributions of X, Y, and U Determining the distribution of U is difficult in the continuous case. Iowa State University Mar 7, / 37

6 Outline Iowa State University Mar 7, / 37

7 Approximating E(U) and Var(U) when determining f U (u) is too hard If X, Y,..., Z are independent, g is well-behaved, and the variances Var(X ), Var(Y ),..., Var(Z) are small enough, then U = g(x, Y,... Z) has: E(U) g(e(x ), E(Y ),..., E(Z)) ( ) g 2 ( ) g 2 ( ) g 2 Var(U) Var(X ) + Var(Y ) + + Var(Z) x y z These formulas are often called the propagation of error formulas. Iowa State University Mar 7, / 37

8 Example: an electric circuit R is the total resistance of the circuit. R 1, R 2, and R 3 are the resistances of resistors 1, 2, and 3, respectively. E(R i ) = 100, Var(R i ) = 2, i = 1, 2, 3. R = g(r 1, R 2, R 3 ) = R 1 + R 2R 3 R 2 + R 3 Iowa State University Mar 7, / 37

9 Example: an electric circuit E(R) g(100, 100, 100) = (100)(100) = 150Ω g r 1 = 1 g = (r 2 + r 3 )r 3 r 2 r 3 r3 2 r 2 (r 2 + r 3 ) 2 = (r 2 + r 3 ) 2 g = (r 2 + r 3 )r 2 r 2 r 3 r2 2 r 3 (r 2 + r 3 ) 2 = (r 2 + r 3 ) 2 ( Var(R) (1) 2 (2) 2 (100) 2 ) 2 ( + ( ) 2 (2) 2 (100) 2 ) 2 + ( ) 2 (2) 2 = 4.5 SD(R) Ω Iowa State University Mar 7, / 37

10 Outline Iowa State University Mar 7, / 37

11 X 1, X 2,..., X n are independent random variables and Y = a 0 + a 1 X 1 + a 2 X a n X n then: E(Y ) = E(a 0 + a 1 X 1 + a 2 X a n X n ) = a 0 + a 1 E(X 1 ) + a 2 E(X 2 ) + + a n E(X n ) Var(Y ) = Var(a 0 + a 1 X 1 + a 2 X a n X n ) = a 2 1 Var(X 1 ) + a 2 2 Var(X 2 ) + + a 2 n Var(X n ) Iowa State University Mar 7, / 37

12 Your turn: linear Say we have two independent random variables X and Y with E(X ) = 3.3, Var(X ) = 1.91, E(Y ) = 25, and Var(Y ) = 65. Find: E(3 + 2X 3Y ) E( 4X + 3Y ) E( 4X 6Y ) Var(3 + 2X 3Y ) Var(2X 5Y ) Var( 4X 6Y ) Iowa State University Mar 7, / 37

13 Answers: linear E(3 + 2X 3Y ) = 3 + 2E(X ) 3E(Y ) = = 65.4 E( 4X + 3Y ) = 4E(X ) + 3E(Y ) = = 61.8 E( 4X 6Y ) = 4 E(X ) 6 E(Y ) = = Iowa State University Mar 7, / 37

14 Answers: linear Var(3 + 2X 3Y ) = 2 2 Var(X ) + ( 3) 2 Var(Y ) = = Var(2X 5Y ) = 2 2 Var(X ) + ( 5) 2 Var(Y ) = = Var( 4X 6Y ) = ( 4) 2 Var(X ) + ( 6) 2 Var(Y ) = = Iowa State University Mar 7, / 37

15 Your turn: more linear Say X Binomial(n = 10, p = 0.5) and Y Poisson(λ = 3). Calculate: E(5 + 2X 7Y ) Var(5 + 2X 7Y ) Iowa State University Mar 7, / 37

16 Answer: more linear First, note that: E(X ) = np = = 5 E(Y ) = λ = 3 Var(X ) = np(1 p) = 10(0.5)(1 0.5) = 2.5 Var(Y ) = λ = 3 Now, we can calculate: E(5 + 2X 7Y ) = 5 + 2E(X ) 7E(Y ) = = 6 Var(5 + 2X 7Y ) = 2 2 Var(X ) + ( 7) 2 Var(Y ) = = 157 Iowa State University Mar 7, / 37

17 iid random variables. Identically Distributed: Random variables X 1, X 2,..., X n are identically distributed if they have the same probability distribution. iid : Random variables X 1, X 2,..., X n are iid if they are Independent and Identically Distributed. Iowa State University Mar 7, / 37

18 Your turn: averages of iid random variables X 1, X 2,..., X n are iid with expectation µ and variance σ 2. Derive: where: E(X ) Var(X ) X = X 1 + X X n n the mean of the X i s. Iowa State University Mar 7, / 37

19 Answers: averages of iid random variables ( ) X1 + X X n E(X ) = E n = 1 n E(X 1) + 1 n E(X 2) n E(Xn) = 1 n µ + 1 n µ n µ }{{} n times = n 1 n µ = µ Remember E(X ) = µ: it s an important result. Iowa State University Mar 7, / 37

20 Answers: averages of iid random variables ( ) X1 + X X n Var(X ) = Var n ( ) 1 2 = Var(X 1 ) + n = 1 n 2 σ2 + 1 n 2 σ n 2 σ2 }{{} n times 1 = n n 2 σ2 ( ) 1 2 Var(X 2 ) + + n ( ) 1 2 Var(X n) n = σ2 n Remember Var(X ) = σ2 : it s another important result. n Iowa State University Mar 7, / 37

21 Example: length of seeds A botanist has collected a sample of 10 seeds and measures the length of each. The seed lengths X 1, X 2,..., X 10 are supposed to be iid with mean µ = 5 mm and variance σ 2 = 2 mm 2. E(X ) = µ = 5 Var(X ) = σ 2 /n = 2/10 = 0.2 Iowa State University Mar 7, / 37

22 Outline Iowa State University Mar 7, / 37

23 If X 1, X 2,..., X n are any iid random variables with mean µ and variance σ 2 <, and if n 25, ) X Normal (µ, σ2 n (CLT) one of the most important and useful results in statistics. Iowa State University Mar 7, / 37

24 Example: tool serial numbers W 1 = last digit of the serial number observed next Monday at 9 AM W 2 = last digit of the serial number the Monday after at 9 AM W 1 and W 2 are independent with pmf: { 0.1 w = 0, 1, 2,..., 9 f (w) = 0 otherwise W = 1 2 (W 1 + W 2 ) has the pmf: Iowa State University Mar 7, / 37

25 Example: tool serial numbers What if W = 1 8 (W 1 + W W 8 ), the average of 8 days of initial serial numbers? Iowa State University Mar 7, / 37

26 Example: excess sale time S = sample mean excess sale time (over a 7.5 s threshold) for 100 stamp sales. Each individual excess sale time should have an Exp(α = 16.5 s) distribution. That means: E(S) = α = 16.5 s SD(S) = Var(S) = α = 1.65 s By the Central Limit, S N(16.5, ) We want to approximate P(S > 17). Iowa State University Mar 7, / 37

27 Example: excess sale time P(S > 17) = P( S > ) P(Z > 0.303) (Z N(0, 1)) = 1 P(Z 0.303) = 1 Φ(0.303) = from the standard normal table = 0.38 Iowa State University Mar 7, / 37

28 Example: net weight of baby food jars Individual jar weights are iid with unknown mean µ and standard deviation σ = 1.6 g ( ) V = sample mean weight of n jars N. µ, 1.62 n We want to find µ. One way to hone in on µ is to find n such that: P(µ 0.3 < V < µ + 0.3) = 0.8 That way, our measured value of V is likely to be close to µ. Iowa State University Mar 7, / 37

29 Example: net weight of baby food jars 0.8 = P(µ 0.3 < V < µ + 0.3) = P( / n < V µ 1.6/ n < / n ) P( 0.19 n < Z < 0.19 n) (by CLT) = 1 2Φ( 0.19 n) (look at the N(0,1) pdf) Φ 1 (0.1) = 0.19 n n = Φ 1 (0.1) 2 ( 0.19) 2 = ( 1.28)2 ( 0.19) 2 = (standard normal table) Hence, we ll need a sample size of n = 47. Iowa State University Mar 7, / 37

30 Example: cars Suppose a bunch of cars pass through certain stretch of road. Whenever a car comes, you look at your watch and record the time. Let X i be the time (in hours) between when the i th car comes and the (i + 1) th car comes, i = 1,..., 44. Suppose you know: X 1, X 2,..., X 44 iid f (x) = e x x 0 Find the probability that the average time gap between cars exceeds 1.05 hours. Iowa State University Mar 7, / 37

31 Example: cars µ = E(X 1 ) = = 0 xf (x)dx xe x dx = e x (x + 1) 0 integration by parts = 1 Iowa State University Mar 7, / 37

32 Example: cars E(X1 2 ) = = 0 x 2 f (x)dx x 2 e x dx = e x (x 2 + 2x + 2) 0 integration by parts = 2 σ 2 = Var(X 1 ) = E(X1 2 ) E 2 (X 1 ) = = 1 Iowa State University Mar 7, / 37

33 Example: cars X approx. N(µ, σ 2 /n) = N(1, 1/44) Thus: X 1 1/44 N(0, 1) Iowa State University Mar 7, / 37

34 Example: cars Now, we re ready to approximate: P(X > 1.05) = P( X 1 1/44 > /44 ) = P( X 1 1/44 > 0.332) P(Z > 0.332) = 1 P(Z 0.332) = 1 Φ(0.332) = = Iowa State University Mar 7, / 37

35 Example: cars Density of X_ Density of X_ x_1 Iowa State University Mar 7, / 37

36 Example: cars Density Density of Average(X) x Iowa State University Mar 7, / 37

37 Example: cars Density Densities of and Average(X) and N(1,1/44) average(x) N(1, 1/ x Iowa State University Mar 7, / 37