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Name: ID: Homework for 1/8 1. [§4-6] Let X be a continuous random variable with probability density function f (x) = 2x, 0 ≤ x ≤ 1. (a) Find E[X]. (b) Find E[X 2 ]. (c) Find Var[X]. (a) We have ∞ Z E[X] = Z 1 xf (x) dx = −∞ x · 2x dx = 2x3 3 1 0 1 = 2. 3 0 (b) We have 2 Z ∞ E[X ] = Z 2 2 x · 2x dx = x f (x) dx = −∞ 0 x4 2 1 = 1. 2 0 (c) We have Var[X] = E[X 2 ] − E[X]2 = 1 − 2 2 2 1 . = 3 18 2. [§4-31] Let X be uniformly distributed on the interval [1, 2]. Find E[1/X]. Is E[1/X] = 1/E[X]? Since X is uniformly distributed on [1, 2], the pdf of X is ( 1, 1 ≤ x ≤ 2 f (x) = . 0, otherwise On one hand, we have Z ∞ Z E[X] = xf (x) dx = −∞ 2 x · 1 dx = 1 x2 2 2 = 3. 2 1 On the other hand, we have 2 Z ∞ Z 2 1 1 1 E = f (x) dx = · 1 dx = (ln x) = ln 2 − ln 1. X x x −∞ 1 1 It follows immediately that E[1/X] 6= 1/E[X]. 3. [§4-42] Let X be an exponential random variable with standard deviation σ. Find P(|X − E[X]| > kσ) for k = 2, 3, 4, and compare the results to the bounds from Chebyshev’s inequality. Since X is exponentially distributed and has variance σ 2 , the pdf of X is 1 e−x/σ , x > 0 f (x) = σ . 0, otherwise It follows immediately that E[X] = σ. Thus we have P(|X − E[X]| > kσ) = P(X > (k + 1)σ or X < (1 − k)σ) Z ∞ 1 −x/σ e dx we only consider k = 2, 3, 4 = σ (k+1)σ Z ∞ = e−y dy (change of variable y = x/σ) k+1 = e−(k+1) . Therefore, we have P(|X − E[X]| > 2σ) = e−3 = .0498, P(|X − E[X]| > 3σ) = e−4 = .0183, P(|X − E[X]| > 4σ) = e−5 = .0067. On the other hand, the Chebyshev’s inequality gives us P(|X − E[X]| > kσ) ≤ Var[X] 1 = 2. k2 σ2 k In particular, we have 1 1 = 0.25, P(|X − E[X]| > 3σ) ≤ 2 = 0.1111, 2 2 3 1 P(|X − E[X]| > 4σ) ≤ 2 = 0.0625. 4 P(|X − E[X]| > 2σ) ≤ 4. [§4-80] Let X be a continuous random variable with density function f (x) = 2x, 0 ≤ x ≤ 1. Find the moment-generating function of X, M (t), and verify that E[X] = M 0 (0) and that E[X 2 ] = M 00 (0). 2 The mgf of X is Z ∞ Z 1 M (t) = E[etX ] = etx f (x) dx = etx · 2x dx −∞ 0 " # 1 Z 1 tx 1 1 tx = 2 e x − e dx t 0 t 0 t e 1 t =2 − 2 e −1 (integration by parts) t t t 1 e et =2 − 2+ 2 . t t t Note that by L’Hospital rule, we have tet − et + 1 tet + et − et = 2 lim = lim et = 1. t→0 t→0 t→0 t2 2t lim M (t) = 2 lim t→0 Furthermore, we have 2et 2et 2 et − 2 + − 3 , and M (t) = 2 t t t3 t 2 t t t e − 2te + 2et − 2 lim M 0 (t) = 2 lim t→0 t→0 t3 2 t t e + 2tet − 2tet − 2et + 2et = 2 lim t→0 3t2 2 2 = lim et = , 3 t→0 3 0 and et 3et 6et 6et 6 − 2 + − 4 + 4 , and t t t3 t t 3 t 2 t t t e − 3t e + 6te − 6et + 6 lim M 00 (t) = 2 lim 4 t→0 t→0 t t3 et + 3t2 et − 3t2 et − 6tet + 6tet + 6et − 6et = 2 lim t→0 4t3 1 1 = lim et = . 2 t→0 2 M 00 (t) = 2 Combing with the first problem (§4-6), we see that E[X] = M 0 (0) and E[X 2 ] = M 00 (0). 3 5. [§4-91] Use the mgf to show that if X follows an exponential distribution, cX(c > 0) does also. Let the pdf of X be ( f (X) = λe−λx , x > 0 . 0, otherwise Then the mgf of X is Z ∞ etx · λe−λx dx = MX (t) = 0 λ λ−t Z ∞ (λ − t)e−(λ−t) dx = 0 λ , λ−t for all t < λ. (The trick here is to identify a pdf for Exp(λ − t)). Let Y = cX. Then the mgf of Y is MY (t) = E[etY ] = E[et(cX) ] = E[e(ct)X ] = MX (ct) = λ/c λ = , λ − ct λ/c − t for all t with ct < λ, that is, t < λ/c. Compare MY (t) to MX (t), we see that Y is also exponentially distributed and the parameter is λ/c. Note that the result McX (t) = MX (ct) is generally true as long as the mgf are defined (for example, we need to make sure ct < λ). More generally, we have MaX+b (t) = ebt MX (t). 4 Homework for 1/10 1. [§5-16] Suppose that X1 , . . . , X20 are independent random variables with density functions f (x) = 2x, 0 ≤ x ≤ 1 Let S = X1 + · · · + X20 . Use the central limit theorem to approximate P(S ≤ 10). From Problem 1 of Homeowrk 1/8, we have µ = E[X] = 2 , 3 and σ 2 = Var[X] = 1 . 18 Therefore, according the central limit theorem, we have 10 − 20 · 23 S − 20 · 23 10 − nµ S − nµ √ √ ≤ = P q P(S ≤ 10) = P √ ≤q1 √ σ n σ n 1 · 20 · 20 18 18 ≈ P(Z ≤ −3.16) = 0.0008, where Z ∼ N (0, 1). 2. [§5-17] Suppose that a measurement has mean µ and variance σ 2 = 25. Let X be the average of n such independent measurements. How large should n be so that P(|X − µ| < 1) = .95? By applying central limit theorem, we have |X − µ| 1 √ < √ 0.95 = P(|X − µ| < 1) = P σ/ n σ/ n 1 1 1 ≈ P |Z| < √ =P − √ <Z< √ , 5/ n 5/ n 5/ n where Z ∼ N (0, 1). Since 0.95 = P(−1.96 < Z < 1.96), we have 1 √ = 1.96, 5/ n which implies that n = (5 · 1.96)2 = 96.04 ≈ 96. 5 3. [§5-18] Suppose that a company ships packages that are variable in weight, with an average weight of 15 lb and a standard deviation of 10. Assuming that the packages come from a large number of different customers so that it is reasonable to model their weights as independent random variables, find the probability that 100 packages will have a total weight exceeding 1700 lb. Let X1 , X2 , . . . , X100 denote the weight of these 100 packages, and S100 = X1 +· · ·+X100 the total weight. Since µ = E[Xi ] = 15 and σ 2 = Var[Xi ] = 102 for 1 ≤ i ≤ 100, the central limit theorem gives 1700 − 100µ S100 − 100µ √ √ > P(S100 > 1700) = P σ 100 σ 100 S100 − 100 · 15 1700 − 100 · 15 √ √ =P > 10 · 100 10 · 100 ≈ P(Z > 2) = 0.0228. 6

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