Probability HW#5 - 國立臺灣大學

1 Probability HW#5 Due: May 20, (6 points)ComputeE[X] ifXhas a density function given by(a)f(x) ={14xe x/2x >00otherwise;(b)f(x) ={c(1 x2) 1< x <10otherwise;(c)f(x) ={5x2x >50x (a) Note that limx xne x/2= 0 for alln [X] = xf(x) dx= 014x2e x/2dx= 12x2e x/2 0 0 xe x/2dx= 12x2e x/2 0+ ( 2xe x/2) 0 0 2e x/2dx= 4e x/2 0= 4(b)E[X] = xf(x) dx= 1 1c(x x3) dx= 0 The last equality holds sincex x3is an odd function (c)E[X] = xf(x) dx= 55xdx= 5 lnx| 5= (4 points)The density function ofXis given byf(x) ={a+bx20 x 10otherwiseIfE[X] =35, ,f(x) is a Probability density function. f(x) dx= 10a+bx2dx=ax+b3x3 10=a+b3= 1 And we can computeE[X],E[X] = xf(x) dx= 10ax+bx3dx=a2x2+b4x4 10=a2+b4=35 Hence{3a+b= 310a+ 5b= 12We havea= 3/5 andb= 6 (4 points)The lifetime in hours of an electronic tube is a random variable having aprobability density function given byf(x) =xe x,x 0 Compute the expected lifetime of such a xf(x) dx= 0x2e xdx= x2e x 0 0 2xe xdx= x2e x 0+ ( 2xe x) 0 0 2e xdx= 2e x 0= 2 The expected lifetime of such a tube is (6 points)}}}}}

2 Suppose that the height, in inches, of a 25-year-old man is a normal randomvariable with parameters = 71 and 2= What percentage of 25-year-old men are over 6 feet, 2 inches tall? What percentage of men in the6-footer club are over 6 feet, 5 inches? the normal random variable with parameters = 71 and 2= , andZbe the standard normal random variable. Note that 1 inchequals to 12 feets, = = Then the first problem is to computeP(X >6 12 + 2).P(X >74) =P(X >74 )=P(Z > ) = 1 ( ) the second problem, we need to computeP(X >6 12 + 5|X >6 12) =P(X >77|X >72)Hence,P(X >77|X >72) =P(X >77)P(X >72)=P(Z > )P(Z > )=1 P(Z < )1 P(Z < )=1 ( )1 ( ) (4 points)The lifetimes of interactive computer chips produced by a certain semicon-ductor manufacturer are normally distributed with parameters = 106hours and = 3 105hours.

3 What is the approximate Probability thata batch of 100 chips will contain at least 20 whose lifetimes are less the normal random variable with parameters = 106hours and = 3 (X < 106) =P(X 1063 105< 106 1063 105)=P(Z <43)= (43) the number of chips that will have a lifetime less than a binomial random variable with parametersn= 100 andp= The normal approximation yields thatP(N 20) =P(N )=P(N (100)( ) 100( )( ) (100)( ) 100( )( ))=P(N ) 1 ( ) (6 points)Jones figures that the total number of thousands of miles thatan auto can bedriven before it would need to be junked is an exponential random variablewith parameter 1/20.

4 Smith has a used car that he claims has been drivenonly 10,000 miles. If Jones purchases the car, what is the Probability thatshe would get at least 20,000 additional miles out of it? Repeat under the as-sumption that the lifetime mileage of the car is not exponentially distributed,but rather is (in thousands of miles) uniformly distributedover (0,40). the exponetial random variable with parameter 1/20,Udenote the uniform random variable with parameters 0 and prob-ability function of an exponential random variable with parameter 1/20 isF(x) = 1 e x/20,x 04 The desired Probability isP(X >30|X >10). Since exponential randomvariables have memoryless property, we haveP(X >30|X >10) =P(X >20) = 1 P(X 20) =e 1 Probability function of an uniform random variable withparameter 0and 40 isF(x) =140 0x=x40,0 x 40 HenceP(U >30|U >10) =P(U >30)P(U >10)=1 P(U 30)1 P(U 10)=1 30/401 10/40=13is the desired (6 points)The lung cancer hazard rate (t) of at-year-old male smoker is such that (t) =.

5 027 +.00025(t 40)2t 40 Assuming that a 40-year-old male smoker survives all other hazards, whatis the Probability that he survives to (a) age 50 and (b) age 60withoutcontracting lung cancer? the lifetime distribution. We can compute the prob-ability function by1 F(t) = exp( t0 (u) du)={0t <40exp( t40 (u) du)t 40where (u) du= + (t 40)3 HenceP(X 40) = 1 F(40) = exp( 0) = 1(a) The Probability that he survives to age 50 isP(X 50) = 1 F(50)= exp( (t 40)3 5040)= exp( ) (b) The Probability that he survives to age 60 isP(X 60) = 1 F(60)= exp( (t 40)3 6040)= exp( ) (6 points)Suppose that the life distribution of an item has the hazard rate function (t) =t3,t >0.}

6 What is the Probability that(a) the item survives to age 2?(b) the item s lifetime is between .4 and (c) a 1-year-old item will survive to age 2? the life distribution. First we compute the probabil-ity function of the life (t) = 1 exp( t0 (u) du)= 1 exp( 14t4)(a) It is to computeP(X 2)P(X 2) = 1 P(X <2) = 1 F(2) =e 4 (b) The Probability is given byP( X ) =P(X ) P(X )=F( ) F( )= (1 e ) (1 e )=e e (c) The Probability that a 1-year-old item will survive to age 2 isP(X 2|X 1) =P(X 2)P(X 1)=1 F(2)1 F(1)=e 4e 1/4=e 15/4 (9 points)LetZbe a standard normal random variableZ, and letgbe a differnetiablefunction with derivativeg.

7 (a) Show thatE[g (Z)] =E[Zg(Z)](b) Show thatE[Zn+1] =nE[Zn 1](c) FindE[Z4]. (z) denote the Probability density function ofZ. Hencef(z) =1 2 e z2/2(a)E[g (Z)] = g (z)f(z) dz= 1 2 e z2/2g (z) dz=1 2 e z2/2g (z) dz=1 2 (e z2/2g(z) ze z2/2g(z) dz)Ifg(x) e x2/2, the last equation becomes1 2 ze z2/2g(z) dz=E[Zg(Z)]ThenE[g (Z)] =E[Zg(Z)] wheng(x) e x2/2.(b) Letg(x) =xn e x2/2, theng (x) =nxn 1. By (a), we haveE[Zg(Z)] =E[Zn+1] =E[g (Z)] =E[nZn 1] =nE[Zn 1](c) By (b), we haveE[Z4] =E[Z3+1] = 3E[Z3 1] = 3E[Z1+1] = 3E[Z1 1] = 3E[1] = 3 HenceE[Z4] = (4 points)IfXhas hazard rate function X(t), compute the hazard rate function ofaXwhereais a positive (t) =P(aX t) =P(X t/a) =FX(t/a)We havefaX(t) =ddtFaX(t) =ddtFX(t/a) =fX(t/a) 1aThen the hazard rate function ofaXis aX(t) =faX(t)1 FaX(t)=fX(t/a) 1a1 FX(t/a)=1a X(t/a) (4 points)Compute the hazard rate function of a gamma random variable with pa-rameters ( , )

8 And show it is increasing when 1 and decreasing when a gamma random variable with parameters ( , ). Thenf(x) ={ e x( x) 1 ( )x 00x <0 Hence the hazard function is X(t) =f(t)1 F(t)= e t( t) 1 ( ) t e t( t) 1 ( )dx=e t( t) 1 te x( x) 1dxWe can simplify it, then we have X(t) =1 te (x t)(xt) 1dxLety=x t, then we have X(t) =1 0e y(1 +yt) 1dy8If 1 0, fort1 t2,y >0, we have(1 +yt1) 1 (1 +yt2) 1 Hence X(t1) =1 0e y(1 +yt1) 1dy 1 0e y(1 +yt2) 1dy= X(t2)That is,t1 t2implies X(t1) X(t2) when 1 0. X(t) is increasingwhen 1. Similarly, X(t) is decreasing when (9 points)Consider the beta distribution with parameters (a, b).}

9 Show that(a) whena >1 andb >1, the density is unimodal (that is, it has a uniquemode) with mode equal to (a 1)/(a+b 2).(b) whena 1,b 1, anda+b <2, the density is either unimodal withmode at 0 or 1 or U-shaped with modes at both 0 and 1;(c) whena= 1 =b, all points in [0,1] are Probability density function of a gamma distribution with pa-rameters (a, b) isf(x) =1B(a, b)xa 1(1 x)b 10 x 1 Note that a mode of a Probability density function occurs at local maximum.(a) Whena >1 andb >1f (x) =1B(a, b)((a 1)xa 2(1 x)b 1 (b 1)xa 1(1 x)b 2)=1B(a, b)xa 2(1 x)b 2((a 1) (a+b 2)x) = 09We can attain thatf (x) = 0 whenx= (a 1)/(a+b 2).

10 Sincea >1andb >1, we note that0<a 1a+b 2=a 1(a 1) + (b 1)<a 1a 1= 1 When 0< x <(a 1)/(a+b 2), 1 x >0 and (a 1) (a+b 2)x >0,hencef (x)>0,0< x <a 1a+b 2 Similarly,f (x)<0 when (a 1)/(a+b 2)< x < 1a+b 21f (x)+0 f (x) 0 Hence,f(x) has a maximum atx= (a 1)/(a+b 2). Therefore,f(x)is unimodal with mode equal to (a 1)/(a+b 2).(b) Ifa= 1, then we havef(x) =1B(1, b)(1 x)b 10 x 1 The derivative off(x) isf (x) =1B(1, b)(1 b)(1 x)b 2 Thenf (x)>0 for 0< x <1. That is,f(x) is a strictly increasingfunction on (0,1). Therefore,f(x) has maximum atx= 1. Now ifb= 1,then the density function becomesf(x) =1B(a,1)xa 10 x 1 The dreivative off(x) isf (x) =1B(a,1)(a 1)xa 2 Thenf (x)<0 for 0< x <1.