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# Solucionario Walpole 8 ED

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One of the basic assumptions for the exponential distribution centers around the \u201clack- of-memory\u201d property for the associated Poisson distribution. Thus the drill bit of problem 6.80 is assumed to have no punishment through wear if the exponential dis- tribution applies. A drill bit is a mechanical part that certainly will have significant wear over time. Hence the exponential distribution would not apply. 6.83 The chi-squared distribution is a special case of the gamma distribution when \u3b1 = v/2 and \u3b2 = 2, where v is the degrees of the freedom of the chi-squared distribution. So, the mean of the chi-squared distribution, using the property from the gamma distribution, is µ = \u3b1\u3b2 = (v/2)(2) = v, and the variance of the chi-squared distribution is \u3c32 = \u3b1\u3b22 = (v/2)(2)2 = 2v. 6.84 Let X be the length of time in seconds. Then Y = ln(X) follows a normal distribution with µ = 1.8 and \u3c3 = 2. (a) P (X > 20) = P (Y > ln 20) = P (Z > (ln 20\u2212 1.8)/2) = P (Z > 0.60) = 0.2743. P (X > 60) = P (Y > ln 60) = P (Z > (ln 60\u2212 1.8)/2) = P (Z > 1.15) = 0.1251. (b) The mean of the underlying normal distribution is e1.8+4/2 = 44.70 seconds. So, P (X < 44.70) = P (Z < (ln 44.70\u2212 1.8)/2) = P (Z < 1) = 0.8413. Chapter 7 Functions of Random Variables 7.1 From y = 2x\u2212 1 we obtain x = (y + 1)/2, and given x = 1, 2, and 3, then g(y) = f [(y + 1)/2] = 1/3, for y = 1, 3, 5. 7.2 From y = x2, x = 0, 1, 2, 3, we obtain x = \u221a y, g(y) = f( \u221a y) = ( 3\u221a y )( 2 5 )\u221ay ( 3 5 )3\u2212\u221ay , fory = 0, 1, 4, 9. 7.3 The inverse functions of y1 = x1 + x2 and y2 = x1 \u2212 x2 are x1 = (y1 + y2)/2 and x2 = (y1 \u2212 y2)/2. Therefore, g(y1, y2) = ( 2 y1+y2 2 , y1\u2212y2 2 , 2\u2212 y1 )( 1 4 )(y1+y2)/2(1 3 )(y1\u2212y2)/2( 5 12 )2\u2212y1 , where y1 = 0, 1, 2, y2 = \u22122,\u22121, 0, 1, 2, y2 \u2264 y1 and y1 + y2 = 0, 2, 4. 7.4 Let W = X2. The inverse functions of y = x1x2 and w = x2 are x1 = y/w and x2 = w, where y/w = 1, 2. Then g(y, w) = (y/w)(w/18) = y/18, y = 1, 2, 3, 4, 6; w = 1, 2, 3, y/w = 1, 2. In tabular form the joint distribution g(y, w) and marginal h(y) are given by y g(y, w) 1 2 3 4 6 1 1/18 2/18 w 2 2/18 4/18 3 3/18 6/18 h(y) 1/18 2/9 1/6 2/9 1/3 85 86 Chapter 7 Functions of Random Variables The alternate solutions are: P (Y = 1) = f(1, 1) = 1/18, P (Y = 2) = f(1, 2) + f(2, 1) = 2/18 + 2/18 = 2/9, P (Y = 3) = f(1, 3) = 3/18 = 1/6, P (Y = 4) = f(2, 2) = 4/18 = 2/9, P (Y = 6) = f(2, 3) = 6/18 = 1/3. 7.5 The inverse function of y = \u22122 ln x is given by x = e\u2212y/2 from which we obtain |J | = | \u2212 e\u2212y/2/2| = e\u2212y/2/2. Now, g(y) = f(ey/2)|J | = e\u2212y/2/2, y > 0, which is a chi-squared distribution with 2 degrees of freedom. 7.6 The inverse function of y = 8x3 is x = y1/3/2, for 0 < y < 8 from which we obtain |J | = y\u22122/3/6. Therefore, g(y) = f(y1/3/2)|J | = 2(y1/3/2)(y\u22122/3/6) = 1 6 y\u22121/3, 0 < y < 8. 7.7 To find k we solve the equation k \u222b\u221e 0 v2e\u2212bv 2 dv = 1. Let x = bv2, then dx = 2bv dv and dv = x \u22121/2 2 \u221a b dx. Then the equation becomes k 2b3/2 \u222b \u221e 0 x3/2\u22121e\u2212x dx = 1, or k\u393(3/2) 2b3/2 = 1. Hence k = 4b 3/2 \u393(1/2) . Now the inverse function of w = mv2/2 is v = \u221a 2w/m, for w > 0, from which we obtain |J | = 1/\u221a2mw. It follows that g(w) = f( \u221a 2w/m)|J | = 4b 3/2 \u393(1/2) (2w/m)e\u22122bw/m = 1 (m/2b)3/2\u393(3/2) w3/2\u22121e\u2212(2b/m)w , for w > 0, which is a gamma distribution with \u3b1 = 3/2 and \u3b2 = m/2b. 7.8 (a) The inverse of y = x2 is x = \u221a y, for 0 < y < 1, from which we obtain |J | = 1/2\u221ay. Therefore, g(y) = f( \u221a y)|J | = 2(1\u2212\u221ay)/2\u221ay = y\u22121/2 \u2212 1, 0 < y < 1. (b) P (Y < 1) = \u222b 1 0 (y\u22121/2 \u2212 1) dy = (2y1/2 \u2212 y)\u2223\u22231 0 = 0.5324. 7.9 (a) The inverse of y = x + 4 is x = y \u2212 4, for y > 4, from which we obtain |J | = 1. Therefore, g(y) = f(y \u2212 4)|J | = 32/y3, y > 4. Solutions for Exercises in Chapter 7 87 (b) P (Y > 8) = 32 \u222b\u221e 8 y\u22123 dy = \u2212 16y\u22122|\u221e8 = 14 . 7.10 (a) Let W = X. The inverse functions of z = x + y and w = x are x = w and y = z \u2212 w, 0 < w < z, 0 < z < 1, from which we obtain J = \u2223\u2223\u2223\u2223 \u2202x\u2202w \u2202x\u2202z\u2202y \u2202w \u2202y \u2202z \u2223\u2223\u2223\u2223 = \u2223\u2223\u2223\u2223 1 0\u22121 1 \u2223\u2223\u2223\u2223 = 1. Then g(w, z) = f(w, z \u2212 w)|J | = 24w(z \u2212 w), for 0 < w < z and 0 < z < 1. The marginal distribution of Z is f1(z) = \u222b 1 0 24(z \u2212 w)w dw = 4z3, 0 < z < 1. (b) P (1/2 < Z < 3/4) = 4 \u222b 3/4 1/2 z3 dz = 65/256. 7.11 The amount of kerosene left at the end of the day is Z = Y \u2212 X. Let W = Y . The inverse functions of z = y \u2212 x and w = y are x = w \u2212 z and y = w, for 0 < z < w and 0 < w < 1, from which we obtain J = \u2223\u2223\u2223\u2223 \u2202x\u2202w \u2202x\u2202z\u2202y \u2202w \u2202y \u2202z \u2223\u2223\u2223\u2223 = \u2223\u2223\u2223\u22231 \u221211 0 \u2223\u2223\u2223\u2223 = 1. Now, g(w, z) = g(w \u2212 z, w) = 2, 0 < z < w, 0 < w < 1, and the marginal distribution of Z is h(z) = 2 \u222b 1 z dw = 2(1\u2212 z), 0 < z < 1. 7.12 Since X1 and X2 are independent, the joint probability distribution is f(x1, x2) = f(x1)f(x2) = e \u2212(x1+x2), x1 > 0, x2 > 0. The inverse functions of y1 = x1 + x2 and y2 = x1/(x1 + x2) are x1 = y1y2 and x2 = y1(1\u2212 y2), for y1 > 0 and 0 < y2 < 1, so that J = \u2223\u2223\u2223\u2223\u2202x1/\u2202y1 \u2202x1/\u2202y2\u2202x2/\u2202y1 \u2202x2/\u2202y2 \u2223\u2223\u2223\u2223 = \u2223\u2223\u2223\u2223 y2 y11\u2212 y2 \u2212y1 \u2223\u2223\u2223\u2223 = \u2212y1. Then, g(y1, y2) = f(y1y2, y1(1\u2212y2))|J | = y1e\u2212y1 , for y1 > 0 and 0 < y2 < 1. Therefore, g(y1) = \u222b 1 0 y1e \u2212y1 dy2 = y1e\u2212y1 , y1 > 0, and g(y2) = \u222b \u221e 0 y1e \u2212y1 dy1 = \u393(2) = 1, 0 < y2 < 1. Since g(y1, y2) = g(y1)g(y2), the random variables Y1 and Y2 are independent. 88 Chapter 7 Functions of Random Variables 7.13 Since I and R are independent, the joint probability distribution is f(i, r) = 12ri(1\u2212 i), 0 < i < 1, 0 < r < 1. Let V = R. The inverse functions of w = i2r and v = r are i = \u221a w/v and r = v, for w < v < 1 and 0 < w < 1, from which we obtain J = \u2223\u2223\u2223\u2223\u2202i/\u2202w \u2202i/\u2202v\u2202r/\u2202w \u2202r/\u2202v \u2223\u2223\u2223\u2223 = 12\u221avw. Then, g(w, v) = f( \u221a w/v, v)|J | = 12v \u221a w/v(1\u2212 \u221a w/v) 1 2 \u221a vw = 6(1\u2212 \u221a w/v), for w < v < 1 and 0 < w < 1, and the marginal distribution of W is h(w) = 6 \u222b 1 w (1\u2212 \u221a w/v) dv = 6 (v \u2212 2\u221awv)\u2223\u2223v=1 v=w = 6 + 6w \u2212 12\u221aw, 0 < w < 1. 7.14 The inverse functions of y = x2 are given by x1 = \u221a y and x2 = \u2212\u221ay from which we obtain J1 = 1/2 \u221a y and J2 = 1/2 \u221a y. Therefore, g(y) = f( \u221a y)|J1|+ f(\u2212\u221ay)|J2| = 1 + \u221a y 2 · 1 2 \u221a y + 1\u2212\u221ay 2 · 1 2 \u221a y = 1/2 \u221a y, for 0 < y < 1. 7.15 The inverse functions of y = x2 are x1 = \u221a y, x2 = \u2212\u221ay for 0 < y < 1 and x1 = \u221ay for 0 < y < 4. Now |J1| = |J2| = |J3| = 1/2\u221ay, from which we get g(y) = f( \u221a y)|J1|+ f(\u2212\u221ay)|J2| = 2( \u221a y + 1) 9 · 1 2 \u221a y + 2(\u2212\u221ay + 1) 9 · 1 2 \u221a y = 2 9 \u221a y , for 0 < y < 1 and g(y) = f( \u221a y)|J3| = 2( \u221a y + 1) 9 · 1 2 \u221a y = \u221a y + 1 9 \u221a y , for 1 < y < 4. 7.16 Using the formula we obtain µ \u2032 r = E(X r) = \u222b \u221e 0 xr · x \u3b1\u22121e\u2212x/\u3b2 \u3b2\u3b1\u393(\u3b1) dx = \u3b2\u3b1+r\u393(\u3b1 + r) \u3b2\u3b1\u393(\u3b1) \u222b \u221e 0 x\u3b1+r\u22121e\u2212x/\u3b2 \u3b2\u3b1+r\u393(\u3b1+ r) dx = \u3b2r\u393(\u3b1+ r) \u393(\u3b1) , since the second integrand is a gamma density with parameters\u3b1 + r and \u3b2. Solutions for Exercises in Chapter 7 89 7.17 The moment-generating function of X is MX(t) = E(e tX) = 1 k k\u2211 x=1 etx = et(1\u2212 ekt) k(1\u2212 et) , by summing the geometric series of k terms. 7.18 The moment-generating function of X is MX(t) = E(e tX) = p \u221e\u2211 x=1 etxqx\u22121 = p q \u221e\u2211 x=1 (etq)x = pet 1\u2212 qet , by summing an infinite geometric series. To find out the moments, we use µ =M \u2032 X(0) = (1\u2212 qet)pet + pqe2t (1\u2212 qet)2 \u2223\u2223\u2223\u2223 t=0 = (1\u2212 q)p+ pq (1\u2212 q)2 = 1 p , and µ \u2032 2 =M \u2032\u2032 X(0) = (1\u2212 qet)2pet + 2pqe2t(1\u2212 qet) (1\u2212 qet)4 \u2223\u2223\u2223\u2223 t=0 = 2\u2212 p p2 . So, \u3c32 = µ \u2032 2 \u2212 µ2 = qp2 . 7.19 The moment-generating function of a Poisson random variable is MX(t) = E(e tX) = \u221e\u2211 x=0 etxe\u2212µµx x! = e\u2212µ \u221e\u2211 x=0 (µet)x x! = e\u2212µeµe t = eµ(e t\u22121). So, µ = M \u2032