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Statistics 116 - Fall 2004 Theory of Probability Assignment # 6 Due Friday, November 05 Solution prepared by Xiaohu Zhang, [email protected] Nov 5, 2004 Q. 1) Q. 2) (Ross #5.19) Answer Let X be a normal random variable with mean 12 and variance 4. Find the value of c such that P (X > c) = 0:10. : Letting Z = (x 12)=2 then Z is a standard normal. Now, :10 = P fZ > (c 12)=2g. From table 5.1, we nd P fZ < 1:28g = :90 and so (c 12)=2 = 1:28 or c = 14:56 (Ross # 5.31) (a) A re station is to be located along a road of length A, A < 1. If res will occur at points uniformly chosen on (0; A), where should the station be located so as to minimize the expected distance from the re? That is, choose a so as to minimize E (jX aj): (b) Now suppose that the road is of innite length { stretching from point 0 outward to 1. If the distance of a re from point 0 is exponentially distributed with rate , where should the re now be located? That is, we want to minimize E (jX aj) where X is now exponential with rate : Answer (a): 1 Z A 1 aj dx A 0 Z a Z A 1 (a x) dx + (x = A 0 a 1 a2 A2 a2 = ( + aA ( A 2 2 2 a2 A = a+ A 2 E jX aj = jx 1 a) dx A a2 )) Let the derivative be zero, we get 2Aa 1 = 0, thus a = A2 . To check this is the minimizer, one can check the second derivative which is always positive, so it is indeed minimizer. : Answer (b) E jX Z A aj = jx aje x dx (a x)e x dx + 0 Z a = 0 = a Z a 0 e x dx Z a 1 a 1 e + ) 2 a 1 = a+ e 0 Z A a xe ( = a( (x a)e x dx x dx + a a e 1 Z a xe x dx a 1 Z a 1 a 1 a 1 e + 2 ) + ( e a + 2 e a ) 2 Let the derivative be zero, we get 1 2e a = 0, thus a = (log 2)=. To check this is the minimizer, one can check the second derivative which is always positive, so it is indeed minimizer. Q. 3) (Ross #5.35) The lung cancer hazard rate of a t-year-old male smoker, (t) is such that (t) = 0:027 + 0:00025 (t 40)2 ; t 40: Assuming that a 40-year-old male smoker survives all other hazards, what is the probability that he survives to (a) age 50 without contracting lung cancer; (b) age 60 without contracting lung cancer. Answer (a): This problem is very similar to Example 5f of the textbook (pp216-217). The general result is as follows. 2 e x dx 1 a( e a ) Suppose the hazard rate function is (t), then probability for an A-year person to survive until age B (B > A) is P fA-year-old person reaches ageB g = P fhis life time > B jhis life time > Ag = P fhis life time > B g=P fhis life time > Ag = (1 F (B ))=(1 F (A)) Z B = expf 0 Z B = expf A (t)dtg= expf Z A 0 (t)dtg (t)dtg Applying to this problem, the probability h R for him ito survive to age 50 50 :353 = 0:7026 without contracting lung cancer is exp 40 (t)dt = e Answer (b): Similarly, the probability h R for him ito survive to age 60 with60 out contracting lung cancer is exp Q. 4) (t)dt = e 1:207 = 0:2991 (Ross #5.5 { Theoretical) Use the fact that for any non-negative ran- dom variable Y , E (Y ) = to prove that E (X n ) = Answer: E [X n ] = = = Q. 5) 40 Z Z0 1 Z0 1 0 0 1 Z 1 1 Z 0 P (Y > t) dt ntn 1 P (X > t) dt: P fX n > sgds P fX n > tn gdtn by s = tn P fX > tgntn 1 dt (Ross # 5.28 Theoretical) Let X be a continuous random variable having cumulative distribution function F . Dene a new random variable Y by Y = F (X ). Show that Y is uniformly distributed over (0,1). : Let FY () be the cumulative distribution function of Y . It suces to show FY (y ) = y for y 2 (0; 1) since this is the cumulative distribution function for uniform random variable over (0,1). In the following we use FX () to denote the cumulative distribution function of X . Answer 3 P fY y g P fFX (X ) y g P fX FX 1 (y )g FX (FX 1 (y )) y FY (y ) = = = = = 4