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# f00soln6 - Statistics 265 Elements of Probability Theory...

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Unformatted text preview: Statistics 265 Elements of Probability Theory Fall Term 2000 Assignment 6 Due: Tuesday December 5, 2000 Solutions 6.2 Let Y be a random variable with a density function given by (3 2 ;1 y 1 f (y) = 2 y 0 otherwise. a. Find the density function of U1 = 3Y: b. Find the density function of U2 = 3 ; Y: c. Find the density function of U3 = Y 2 : Solution: The distribution function of Y is F (y) = P (Y y) = Zy ;1 f (t) dt = Zy3 ; t2 dt = 1 1 + y3 2 2 ;1 for ;1 y 1: a. If U1 = 3Y then FU1 (u) = P (3Y u) = P Y u = F u = 1 1 + u3 3 3 2 27 for ;3 u 3: Di erentiating, we have 82 <u fU1 (u) = : 18 0 ;3 u 3 otherwise. b. If U2 = 3 ; Y then FU2 (u) = P (3 ; Y u) = P (Y 1 3 ; u) = 1 ; P (Y < 3 ; u) = 1 ; F (3 ; u) = 2 1 ; (3 ; u)3 for ;1 3 ; u 1 that is, 2 u 4: Di erentiating, we have (3 2 2u4 fU2 (u) = 2 (3 ; u) 0 otherwise. c. If U3 = Y 2 then p p FU3 (u) = P (Y 2 u) = P (; u Y u) = P (Y p p 1u3 + 1u2 = u3 2 = F ( u) ; F (; u) = 2 2 2 3 for 0 u 1: Di erentiating, we have fU3 (u) = u 0 31 2 2 0u1 otherwise. pu) ; P (Y < ;pu) 6.8 The total time from arrival to completion of service at a fast{food outlet, Y1 and the time spent waiting in line before arriving at the service window, Y2 were given in Exercise 5.9 with joint density function e;y1 0 f (y1 y2) = 0 y2 y1 < 1 otherwise. Another random variable of interest is U = Y1 ; Y2 the time spent at the service window. a. Find the probability density function for U: b. Find E (U ) and V (U ): Compare your answers with the results of Exercise 5.68. Solution: a. The region for which y1 ; y2 u is shown in the gure, over the region for which y1 and y2 are positive and y2 y1 : y =y y 1 2 2 y =u+y 1 u 2 y 1 a. The distribution function of U = Y1 ; Y2 is given by FU (u) = P (Y1 ; Y2 u) = Z 1 Z y2 +u 0 y2 e;y1 dy1dy2 = Z 1; 0 e;y2 ; e;(y2 +u) dy2 = 1 ; e;u for 0 u < 1: Di erentiating, we have fU (u) = e;u 0 0 u<1 otherwise: b. The random variable U = Y1 ; Y2 has a gamma distribution with parameters Therefore, E (U ) = = 1 and V (U ) = 2 = 1: 6.12 In Exercise 4.7, we determined that 8b < f (y) = : y2 0 = 1 and = 1: yb otherwise is a bona de probability density function for a random variable, Y: Assuming b is a known constant and U has a uniform distribution on the interval 0 1] transform U to obtain a random variable with the same distribution as Y: Solution: The distribution function of Y is given by FY (y) = Zy ;1 f (t) dt = for y b: 2 Zyb b 2 dt = 1 ; y t b Since U has a uniform distribution on 0 1] then P (U u) = u b for 0 u 1 and we want to nd a function G such that Y = G(U ) with FY (y) = 1 ; y for y b: Now, FY (y) = P (Y y) = P (G(U ) y) = P (U G;1(y)) = P (U u) = u for 0 u 1 since U has a uniform distribution on 0 1]: Therefore, we want b u= 1; y or b y = 1;u b that is, Y = G(U ) = 1 ; U : 6.16 Let the random variable Y possess a uniform distribution on the interval 0 1]: a. Derive the distribution of the random variable W = Y 2 : p b. Derive the distribution of the random variable W = Y : Solution: a. If W = Y 2 then FW (w) = P (W w) = P (Y pw ) = Z ww p w ) = P (; 2 pw 0 for 0 w 1: Di erentiating, the density function for W = Y 2 is 1 w; 2 0 1 2 fW (w) = 0<w<1 otherwise. p b. If W = Y then FW (w) = P (W p w) = P ( Y w ) = P (Y w2 ) = p for 0 w 1: Di erentiating, the density function for W = Y is fW (w) = 2w 0 3 0<w<1 otherwise. Z w2 0 p 1 dy = w 1 dy = w2 6.22 The Weibull density function is given by ( 1 m;1 ; ym my e f (y ) = 0 0<y<1 otherwise where and m are positive constants. This density function is often used as a model for the lengths of life of physical systems. Suppose Y has the Weibull distribution just given. a. Find the density function of U = Y m : b. Find E (yk ) for any positive integer k: Solution: 1 a. Let U = Y m then using the transformation approach, we have Y = U m and ; dy = 1 u (1;m) = 1 u; (mm 1) m du m m so that ; 1 m;1 ; u 1 u; (mm 1) = 1 e; u gU (u) = 1 m u m e m for u > 0: b. If Y has the Weibull distribution given above, and k is a positive integer, then Z1 k u k k E (Y k ) = E U m = 1 u m e; du = 1 ; m + 1 0 k m +1 k =; m +1 k m: Note that the integrand is the density (except for constants) of a gamma variable with parameters k m + 1 and so that the integration can be done by choosing the necessary constants. 6.24 Let Y have a uniform 0 1] distribution. Show that U = ;2 ln Y has an exponential distribution with mean 2: Solution: The density function for Y is given by f (y ) = 1 for 0 y 1 and since u = ;2 ln y then 2 y = e; u : Now dy 1 ;u 2 du = ; 2 e so that dy 1 2 2 fU (u) = f (y) du = 1 ; 2 e; u = 1 e; u 2 for u > 0 which is the density function of an exponential distribution with = 2: 4 6.25 The speed of a molecule in a gas at equilibrium is a random variable V whose density function is given by f (v) = av2 e;bv 2 v>0 where b = 2m where k T and m denote Boltzmann's constant, the absolute temperature, and the mass kT of the molecule, respectively. a. Derive the distribution of W = 1 mV 2 the kinetic energy of the molecule. 2 b. Find E (W ): Solution: r 2 W a. If W = mV then V = 2m and 2 r dv = 1 2 w; 1 = p 1 2 dw 2 m 2mw Therefore, ; p p a 2w w w bw 2 fW (w) = p m e; 2m = a 2w e; kT = a 32 w 1 e; kT 3 2mw m2 m2 and since the above density must integrate to 1 and since the variable part of the density is that of a gamma variable with = 3 and = kT then a must be chosen so that 2 p a 2= ; 1 3 2 m 3 ; 3 (kT ) 2 2 and the density is 1 w fW (w) = ; 3 1 3 w 2 e; kT ; 2 (kT ) 2 for w 0: b. For a gamma{type random variable, we have E (W ) = 5 = 3 kT: 2 6.32 Suppose that Y1 and Y2 are independent, standard normal random variables. Find the density function of U = Y12 + Y22 : Solution: Since Y1 and Y2 are independent standard normal random variables, the moment{generating functions for Y12 and Y22 can be written as mY12 (t) = and therefore, 1 1 (1 ; 2t) 2 and mY22 (t) = 1 2 (1 ; 2t) 1 1 mU (t) = mY12 (t) mY22 (t) = 1 ; 2t which is the moment generating function for a gamma random variable with = 1 and = 2: By the uniqueness theorem, U has a gamma distribution with = 1 and = 2: Equivalently, U has a 2 distribution with 2 degrees of freedom. 6.45 Show that, if Y1 has a 2 distribution with 1 degrees of freedom and Y2 has a 2 distribution with 2 degrees of freedom, then U1 + U2 has a 2 distribution with 1 + 2 degrees of freedom, provided that Y1 and Y2 are independent. Solution: From Example 4.13 with = 2i and = 2 the generating functions are mYi (t) = (1 ; 2t) 2i for i = 1 2: Since Y1 and Y2 are independent, then mU (t) = mY1 (t) mY2 (t) = (1 ; 2t);( 1 + 2 ) 2 which is the moment generating function for a 2 random variable with 1 + 2 degrees of freedom. The result now follows by the uniqueness theorem for moment{generating functions. 6.46 Let Y1 and Y2 be independent normal random variables, each with mean 0 and variance 2 : De ne U1 = Y1 + Y2 and U2 = Y1 ; Y2 : Show that U1 and U2 are independent normal random variables, each with mean 0 and variance 2 2: Hint: If (U1 U2) has a joint moment{generating function m(t1 t2) then U1 and U2 are independent if and only if m(t1 t2) = mU1 (t) mU2 (t): Solution: We have i h ih E et1 (Y1+Y2 )+t2 (Y1;Y2 ) = E eY1 (t1+t2 )+Y2 (t1;t2 ) h i = E e(t1 +t2 )Y1 e(t1 ;t2)Y2 = mY1 (t1 + t2) mY2 (t1 ; t2 ) 2 2 2 2 = e 2 (t1 +t2 ) e 2 (t1 ;t2 ) 1 2 = e 2 t2 e 2 t2 = mU1 (t1 ) mU2 (t2) and since the joint moment{generating function is the product of the marginal moment{generating functions, then the random variables U1 and U2 are independent. 6 ...
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