Hints and Answers to Selected Exercises in Fundamentals of Probability: A First Course, Anirban DasGupta, Springer, 2010

Transcription

1 Hints and Answers to Selected Exercises in Fundamentals of Probability: A First Course, Anirban DasGupta, Springer, 00

2 Chapter The number of sample points are: a) 3760; b) 560; c) 6840; d) 9; e) / 5 Yes, because the total possible number of initials is less than 0,000 6 a) P (AB)+P(BC)+P(AC) 3P (ABC) 7 a) EF c G c ;b)egf c ;d)e F G 9 a) Mutually exclusive; c) Not mutually exclusive 0 a) Head on the first toss; c) Same outcome on the first two tosses 3/4 3 / 4 5/6 5 (3!) 4 /9! 6 a) 0 0 ; b) 0!/ /6 /3 Use inclusion-exclusion formula with A i as the event that the handyman gets day i of the week off, i =,,, 7 3 Just count the sample points 5 Use inclusion-exclusion formula with A i as the event that the ith child gets chosen everyday of the week For example, P (A )=(/3) 7 6 / (8!) /8; /; 9/3; 43/64 33 (a) is more likely 35 (7 6) ( 0 6!!(3 6 ) 8! 6 ) Think of the complement of the event, which is that the m shoes are of m different colors, and from each of these m colors, you would choose either the left or the right shoe 40 4 (n r+)r!(n r)! n! (n k+) (n+)(n+)

3 3) 3) 43 a) (6 ( 5 3) ; b) ( 6 ( 5 3)( 39 3) 44 Use inclusion-exclusion formula with A i as the event that the hand of the ith player is void in the particular suit For example, P (A )= (39 47 /4 Chapter 3) ( 5 3) =05 withm = (365) 3 Use Stir- n 5 Solve the equation m(m ) (m n+) m n ling s approximation to find the root Chapter 3 3 5/6 3 5/6 33 /3 34 P (B A) 8/9 36 5% of days 38 0/9 39 Consider the events A i = North has i aces and South has 4 i, and look at P (A 4 i= A i )andp(a 3 4 i= A i ) ( 3 ) ( 59 ) 3 The winning probability of Sam is j=0 ( ) j Find one of the other two probabilities, and then the third one by subtraction 35 Consider the events A i = The last face to show up is face i, and observe that P ( 6 i=a i )=6P(A ) 36 First solve the equation p 4 +( p) 4 = / weeks 30 Both strategies are equally good 3 /8; use Bayes theorem 3 x/(x + y) 34 p/(6 5p) 36 a) 000; b)

4 Consider the probability that the first cell remains empty and the other two do not, and prove that this probability is 3y( x)( x y) Then find the required conditional probability 39 List the sample points and find P (b <ac), where b is the off-diagonal element, by direct counting The probability of this event is about 008, which is small 334 Chapter 4 (N r)(n r ) (N )(N ) 4 p(0) = p(4) = /6; p() = p(3) = /4; p() = 3/8 F (x) =0(x<0); = /6(0 x<); = 5/6( x<); = /6( x< 3); = 5/6(3 x<4); = (x 4) 4 p(0) = p(3) = /6; p() = 5/8; p() = /9; p(4) = /9; p(5) = /8 43 p(0) = /; p() = /4; p() = /8; p(3) = p(4) = /6 44 Use the total probability formula; eg, P (X =)= 6 E(X) =7/ (a) P (h(x) =±) = 4/3; P (h(x) =0)=5/3; E(h(X)) = 0 48 First prove that the variance of X must be zero 6n= n n,etc; 40 The expected value is 3n(n )5n 6 n 4 (a) /5; (b) 4/ n Consider indicator random variables I i of the events that the pair of cards marked as i remain in the jar Show that E(I i )= (N m ) ( N m ) 40 Show that P (X x) = 0 for x < 5 and is equal to (x 5) ( 0 5 ) for x =5, 6,, 0 From here, find the pmf of X 4 Show that P (X >n)= (366 n) for n =, 3,, 365, and (365) n then apply the tailsum formula log 4

5 Differentiate E(X a) with respect to a 48 Write n= np (X >n)= n= n( j=n+ p(j)) as j= p(j)( j n= n), and simplify 49 Mean = (n +)/; variance = (n )/ Find medians separately for odd and even n 43 (a) At most, people; (b) at most 55,555 people 433 First show that the maximum possible value of E(X )ife(x) =µ is µm Then apply the Paley-Zygmund inequality 434 Look at random variables whose pmf go to zero at the rate x p+ 437 Show that {f(x ) c} and {g(x ) d} are independent events for any c, d 438 A sufficient condition is that E(X )=E(X )=0 Chapter 5 5 pgf is s/( s), s < ; the mgf is e t /( e t ),t<log 5 For example, G(s) = s, s 0 54 e bt ψ(at) 55 Use the fact that x x for all x 57 Use the fact that E[X(X ) (X k +)]=G (k) () For example, σ = G () () + G () () [G () ()] 59 Use the fact that e tx is a convex function of x for any t 5 κ = p; κ = p( p); κ 3 = p 3p +p 3 ; κ 4 = p 7p +p 3 6p 4 Chapter (n = 0); 0445(n = 30); 03(n = 50) The limit can be proved to be zero 6 p(0) = p(3) = 0; p() = 038 The distribution is not Binomial ; 075 The most likely value is zero (n =);0687(n =);05973(n =3);5845(n =4);05757(n = 5) 68 For p /

6 6 (k/n) n ((k )/N ) n,k =,,,N for with replacement sampling; ( n) ( k k n ),k = n,,n for without replacement sampling ( N n) 6 Compute a Poisson approximation with n = 00 and p = (00 00) = The final answer is The probability of getting an A is (0) 5 ; the probability of getting a B is ( ) 5 4 (0) 5 (08) + ( ) 6 4 (0) 5 (08),etc 66 Binomial(00, 75) 69 Find the sum ( k=0 P (X) = Y = k) andsimplifyittoe λ ( p) n n n! k=0( λp p )k /[(k!) (n k)!] 60 Solve two appropriate equations to first show that p = 0988,n = 86, and so the variance is np( p) = Use the skewness and kurtosis formulas for a general Poisson distribution; the answers are 0447; No You have to first prove that if three such values exist, they must be consecutive integers 630 Compute a Poisson approximation to P (X ) The answer is No +e 634 λ 637 (a) 0839; (b) 00803; (c) Expand (q+p) n and (q p) n by using the Binomial expansion theorem and add, with q = p Chapter 7 7 It is a density for k = 7 c = 73 (a) c =;(b)forx>0,f(x) =/+x x 4 /; (c) 085; 085;

7 74 F (x) =0(x<0), = p( cos(x))+( p)sinx(0 x π), =(x> π) Median solves p( cos(x)) + ( p)sinx =05 If p =05, the median is π 4 76 (a) c = 4/6875; (b) 4074%; %; (c) 9 minutes 77 F (x) =0(x<0), = x/5(0 x 5), =(x>5); expected payout is $87500 This distribution is a mixed distribution 79 Density function of the ratio of the two lengths is, 0 <v<; +v expected value is log = 0693 You should find the density by first finding the CDF 70 One example is f(x) =, 0 <x< x 7 One example is f(x) =,x x 73 Quantile function is p 76 (a) Mean = 0; variance = / 77 c =/5; mean is 3π 5π and the variance is 69π 00 ] Quantile function is tan [(p )π 75th percentile equals 70 Use the trigonometric identity arctan x + arctan y = π + arctan x+y xy for all x, y > 0,xy >, and plug in suitable special values of x, y 7 The density function of Y = is e y y 3/ X π,y >0 73 First prove that E[f(X)] must belong to the range space of f and then use the intermediate value theorem In the standard normal case, x 0 = log Use the formula for the area of a triangle in pp 439 of the text 77 α/ Γ( α+ ) π 730 µ = x 0 [e h(t)dt 0 ]dx n 73 n+ 733 n 735 Use Lyapounov s inequality suitably 736 Use Jensen s inequality suitably 737 I(x) =0(x ), = x log x (x >) Chapter 8 8 a =/,b=/4 8 a =,b=5 7

8 83 /3 85 (a) x 3 3x is a strictly monotone function of x on (0, ), taking values in (, 0) The equation x 3 3x = y has a unique real root, say x = g(y) The density function of Y = X 3 3X is g (y), <y<0 (b) f Y (y) = y, 0 <y</4; (c) f Y (y) = πy 3/4, 0 <y< y 86 (a) X can be simulated as X = U /5 ; ( ) ; (b) X can be simulated as X = sin πu ( ) (c) X can be simulated as X = log ( U) if U > and X = ( ) +log U if U 87 Substitute β =6 α into the equation F (0) = 0, and solve for α numerically Then find β =6 α n j=0 ( 89 The nth moment equals n j) (j+)(n j+) n 8 Write the mean absolute deviation as E( X µ ) =c m m+n 0 ( m m+n x)x m ( x) n dx + c m (x m m+n m+n )xm ( x) n dx, and then integrate term by term by expanding ( x) n Here c is the normalizing constant of the Beta density 8 α = 0985,β = 08548,P(X >) = 0086 This problem requires numerical work 84 e / The density is e (y )/,y 87 Use the fact that n i= X i G(n, ) and then follow the same steps as in e 8 The mean is n i= m i and the second moment is n i= m i (m i +) 83 The number of centuries is log( 5/N ) log,wheren =0 5 Thisworks out to approximately 836 centuries 84 P (X >λ) =e, which does not depend on λ 85 (a) The expected profit is E[(c c )min(x, t)] c 3 t; (b) The optimum value of t satisfies F (t) = c 3 c c,provided0< c 3 c c < 88 The density function of Y = e X is a standard exponential density 8

9 89 The density function of Y =loglog is X e ey e y, <y< 830 The density function of Y = θ is X αyα, 0 <y 83 (a) 00465; (b) 00009; (c) this is the probability that X>Y where X, Y are independent Poisson variables with means 4/3 and ; see Theorem 63 in text for more details on how to compute it 83 s t 833 (a) The answer is yes; (b) the answer is no 834 The distribution is Bin(n, u) t Chapter ; ; 0357; 05 9 ; 5 93, The density of Y = is Z y π e y, <y< It is uniformly bounded 94 Z +:all;z 3 : all -3; Z 3 : all zero 95 The density of Y = φ(z) is, 0 <y log(πy ) π Φ(x) = +erf(x/ ) 90 A :668%; B :47%; C :383%; D :857%; F :8% 9 If X denotes the diameter of a ball bearing, then we want E(X 48 X 5) and Var(X 48 X 5) These are the mean and the variance of a truncated distribution (section 4) For example, E(X X 5) equals xf(x)dx This is equal to 5 (why?) The variance P (48 X 5) calculation should be done by first calculating E(X 48 X 5) 93 Y = g(z) has a mixed distribution Its CDF is F Y (y) =0(y< a), = Φ( a)(y = a), =Φ(y)( a <y<a), =(y a) oz 95 First prove that E[Φ(X)] = P (Z < X), where Z is an independent standard normal variable Since Z X N( µ, +σ ), P (Z < X)= µ P (Z X<0) = Φ( +σ ) 98 Median = e µ ;mode=e µ σ 90 0, 0, 0,

10 9 (a) 00037; (b) About at x = 65 Chapter 0 0 Exact = 03770; normal approximation without continuity correction = 0643; normal approximation with continuity correction = Exact = 088; normal approximation = The first person has to make at least 44 correct predictions and the second person has to make at least 434 correct predictions Find the probability of the intersection of these two events 06 P (Z )= This equals P ( n i= (X i Xi ) > 0) Use the central limit theorem on these new variables X i Xi The answers are: n =0:00339; n =0: 00049; n =30: reservations can be allowed rolls should suffice 05 First show that the mean and the variance of log X are - and The final answers are 038 for (a) and for (b) Formula for 99% confidence interval is (X +335)± X Formula for 95% confidence interval is (X +9) ± X For X =5, 95% confidence interval is 69 ± 478, and 99% confidence interval is 835 ± (a) n; (b)99n; (c) Use the CLT for each of the totals scored by Tom and Sara and then find a normal approximation for the difference of these two totals The final answer is (a) P (Z > 4) 0; (b) use normal approximation for Binomial with n =5,p= 054 6x( x)dx; the final answer is To find the third moment of S n,expand(x + X + + X n ) 3 and 0

11 take expectations term by term Then simplify 07 The roundoff error on one transaction is uniformly distributed on { 50, 49, 48,, 0,, 49} Thisgivesµ = 05 Now find σ,andthen use a normal approximation to P ( 5 S n 5), with n = It is about 45 times more likely that you will get exactly 50 heads than that you will get more than sixty heads 03 It is just a bit more likely that you will get exactly 0 sixes Chapter The joint pmf of (X, Y )isp(, 0) = /7,p(3, 0) = p(3, ) = p(4, ) = p(6, ) = p(6, ) = /7, and p(x, y) = 0 otherwise E(X) =6/7; Var(X) = /49; E(Y )=;Var(Y )=6/7; ρ X,Y 059 The joint pmf of (X, Y )isp(0, 4) = p(3, ) = p(4, 4) = /6,p(, ) = /4,p(, 0) = 3/8,p(3, ) = 3/6, and p(x, y) = 0 otherwise E(Y )=3/6 3 (a) c = /36; (b) They are independent; (c) E(X) = E(Y ) =,E(XY )=4 4 (a) c = /5; (b) They are not independent; (c) E(X) = 53/5; E(Y ) = 67/5; E(XY ) = 47/5 6 The joint pmf of (X, Y )isp(, 3) = p(, 4) = p(, 5) = p(, 3) = p(, 5) = p(, 6) = p(3, 4) = p(3, 5) = p(3, 7) = p(4, 5) = p(4, 6) = p(4, 7) = /, and p(x, y) = 0 otherwise X, Y are not independent 7 The direct method of calculating E(X Y = y) is to use the formula x xp(x,y) )( 39 )( 3 x 3 x y )( 6+x 3 y E(X Y = y) = In this problem, p(x, y) x p(x,y) =( 3 x 0,x + y 3 However, you can avoid the calculation by logically arguing how many clubs South (or any other player) should get, if y clubs have already been picked up by North You can also logically see that E(X Y =3) and E(Y X = 3) must be the same 9 E(Y )=0 0 Var(X) = λ + λ Use the fact that P (X >Y)andP (Y > X) are equal in this problem, and the fact that P (X >Y)+P (Y > X)+P (X = Y )= Nowfind P (X = Y ) and then conclude that P (X Y )= p ; P (X >Y)= p p ),x,y,

12 3 pmf of X + Y is P (X + Y =)=P (X + Y =8)=004; P (X + Y = 3) = P (X+Y =7)=0; P (X+Y =4)=P (X+Y =6)=0; P (X+Y = 5) = 06 4 (a) They are not independent; (b) 0; (c) 0 7 X Poi(λ + η); Y Poi(µ + η) X and Y are not independent The joint pmf is found by simplifying the sum P (X = x, Y = y) = min(x,y) w=0 e λ λ x w (x w)! e µ µ y w (y w)! e η η w w! 8 (a) 35+y; (b)35y; (c) y 0 The joint pmf of (X, Y )isp(, 0) = /; p(3, 0) = p(3, ) = p(4, 0) = p(4, ) = /8 E(X Y =0)=35 You can argue logically what E(X Y =) should be You can prove this by using the result in problem 6 Show, by using problem 6, that E(XY )=E Y [E(XY Y = y)] E(Y )E(X), and so Cov(X, Y ) 0 3 Use the two iterated expectation formulas E(X) =E Y [E(X Y = y)] and E(XY )=E Y [E(XY Y = y)] 4 Without loss of generality, you may assume that each of X, Y takes thevalues0, Now try to represent the covariance between X and Y in terms of P (X =0,Y =0) P (X =0)P (Y = 0) Finally show that if P (X =0,Y =0)=P (X =0)P (Y = 0), then that would force X, Y to be independent 6 First show that for any two random variables U, V with finite variance, Cov(U, V )= [P (U u, V v) P (U u)p (V v)]dudv Apply this to U = g(x),v = h(x) 7 The joint mgf is ( 6 et + 6 et + 3 )4 The value of the covariance is 9 8 The joint mgf is ψ(t,t )=e λ µ η+λet +µe t +ηe t +t Find E(XY ) by evaluating t t ψ at (t,t )=(0, 0) Chapter (a) c = 4; (b) They are independent; (c) f X (x) =x, 0 x ; f Y (y) =y, 0 y,e(x) =E(Y )=/3; E(XY )=4/9 (a) c = 4; (b) They are not independent; (c) f X (x) =x( x), 0 x ; f Y (y) =y( y), 0 y ; E(X) =E(Y )=/5; E(XY )=/5

13 3 (a) c = ; (b) They are not independent; (c) f X (x) =e x,x > 0; f Y (y) =ye y,y > 0; E(X) =;E(Y )=;(d)e(x Y = y) =y/; (e) E(Y X = x) =x + 5 (a) c = 3 ;(b)no;(c)f 4π X(x) = 3( 4 x ), x ; you can now find f Y (y),f Z (z) by using a suitable symmetry in the formula for the joint density; (d) E(X Y = y) = 0 for any y; (e) Similarly, E(Y X = x) = 0 for any x; (f)0 6 y y + 8 (you can get the answer without doing any hard calculations) 9 E(X X + Y )= 5 8 π; use the fact that X and X + Y are independent X+Y 0 3 8/55 The joint distribution of (X, Y ) does not have a density The joint distribution is a distribution on the line segment y =x, 0 x You can write the joint CDF of (X, Y )asp(x x, Y y) =P (X x, X y) = P (X x, X y ) and simplify this last line 4 E(X n )= 0 n a, b satisfy a(σ + ρσ σ )+b(σ + ρσ σ )=0 8 /4 0 /4 Note the similarity of this problem to Example 7 in the text 048 If we denote X +Y = U, X Y = V,Y = W,then(U, V, W )doesnot have a joint density The joint distribution of (U, V, W ) is a distribution on the plane w = u v You can characterize the joint distribution of (U, V, W ) by writing the joint distribution of (U, V ) which is a bivariate normal, and the conditional distribution of W given U = u, V = v is a one point distribution at w = u v 3 The correlation coefficient between X and X is µ µ +σ 5 (a) Y N(, ); (b) ;(c)x Y = y N( y, ) 6 The mean is 3/4, and the variance is 3/80 7 (a) 3 π x y,x + y ; (b) 3 4 ( x ), x 3

14 σφ( x µ σ ) Φ( x µ σ ) 9 The mean residual life is E(X X >x)=µ + 30 You have to find E(Y X >40) This is different from E(Y X = 40 40) E(Y X > 40) equals yp(x,y)dxdy p(x, y) is a bivariate normal P (X>40) density, and you will be able to simplify the integral in the numerator The denominator is easy because X is marginally normal 3 (a) The densities of U, V, W are 3( u) (0 <u<), 6v( v)(0 < v < ), 3w (0 < w < ) (b) U U[0, ] and T = V has the density V W U t(0 <t<) and V are independent (c) E( U )=/; E( V )=/3 V W V W 33 U has an Exponential density with mean The densities of V,W 3 are 6e v ( e v )(v >0) and 3e w ( e w ) (w>0) T = W U has the density e t ( e t )(t >0) 34 (a) The epoch T of the last departure has the density 4 λ (e t/λ e t/λ t λ e t/λ ); (b) /; (c) The total time T spentbymaryatthepost office has the density λ e t/λ ( e t/λ ) 36 (a) (09) n ( + n); (b) Find the smallest n such that 9 (099)n (098) n 0 This gives n = 57 Chapter 3 3 U = XY has the density f U (u) =( u), 0 u ; V = X has the Y density f V (v) = (0 v ), = (v>) 3 3v 3 3 Z = X + Y has the density f Z (z) =z (0 z ), =z z ( <z ) V = X Y has the density f V (v) = v ( v 0), =( v) (0 < v ) W = X Y has the density f W (w) =( w), 0 w 33 (a) No; (b) c =/; (c) Z = X + Y has the density f Z (z) = z e z,z >0, which is a Gamma density u log u 34 (a) No; (b) c =8;(c)U = XY has the density f U (u) =, 0 < u< 35 T = NXY has a mixed distribution T can be equal to zero with a positive probability; P (T =0)= Conditional on T>0,T has a density 4 You have to find this by conditioning separately on N =andn =,and take the mixture of the two densities that you will get for (XY )=XY and (XY )=XY To complete this problem, first show that V = XY has the density f V (v) = log v, 0 <v<, and then proceed as outlined 4

15 above 36 P (m = X) =+ a 3a(b+c) ; P (m = Z) = a(3b a) ; P (m = Y )canbe 6bc 6bc found by subtraction 38 E( XY XY )=E( X +Y X )=0 +Y 39 It is better to first find the CDF of X + Y This is easy for t, but must be done carefully for t> The CDF of X + Y is P (X + Y t) = πt(0 <t ), and P 4 (X + Y t) =t( π arctan t ) + t for 4 <t Of course, for t,p(x + Y t) = To evaluate E( X + Y ), apply its definition, and find the answer by numerical integration 3 The point P Q has the planar coordinates (U, V )=(X Y,Z W ) First show that the joint density of (U, V ) in polar coordinates is (arccos( r ) r r ), 0 r, π <θ<π This will lead to the π 4 density of r as 4 r(arccos( r ) r r ), 0 r We need E(r), which is π 4 4 π 0 r (arccos( r ) r r 8 )dr Thisintegralequals =090545, which 4 45π is the needed average distance 3 /6 33 P ( X < ) = /4; P (X <Y)=/ They would have been the Y same if Y was a nonnegative random variable 34 The answer is arctan( σ) π τ 36 By using the general formula for the density of a product in the text, first show that the density of U = XY is of the form c c u γ u xα δ γ ( x) β (x u) δ dx, wherec,c are the normalizing constants of the Beta densities for X, Y Now simplify this expression in terms of the F Hypergeometric function 38 The density of U = XY is f U (u) = log u, <u< π u 30 This can be proved by simply using the Jacobian formula You have to be a bit careful and take the cases Y and Y> separately, and then put the two cases together 3 The density of W = X + Y + Z is w g(w),w >0 3w 3 The required density is,w >0 (+w) 4 34 The density of Z = X + Y is f Z (z) = e z (0 z ), = (e )e z,z > 35 The density of Z = X + Y is f Z (z) =Φ( z µ z µ ) Φ( ) σ σ 5

16 37 Given z 0, let n be such that n z<n+ Then the density of Z = X + Y at z is f Z (z) = e λ λ n n! 38 (a) c = r ;(b)no;(c)rhas the density,r >0, θ U[ π, π], π (+r ) 3/ and r, θ are independently distributed; (d) 330 The correlation is The correlation is 0 33 The density of Z = X + Y is f Z (z) = e z/µ e z/λ,z >0; the density µ λ of W = X Y is f W (w) = λ+µ ew/µ (w<0), and = λ+µ e w/λ (w>0) 335 If X U[0, ], then n,n, are each uniform on {0,,,, 9}, and they are independent 336 The integer part of X and the fractional part of X are independent in this case The integer part has the pmf P ( X = n) =e n ( e ),n = 0,,,, and the fractional part has the CDF P ({X} y) = ey, 0 e y (e ) y Note the interesting fact that the integer part therefore has a Geometric distribution 338 The density of R = X + X + + Xn n+3 4Γ( is f R (r) = ) πγ( n ) r n (+r ) (n+3)/,r > 0 You can find the density of X + X + + X n from this by a simple transformation 6

17 Hints and Answers to Supplementary Problems Appendix I: Word Problems I 7576; 85 I 00; 07; 07 I3 /6 I4 /3 I5 /6; /3; / I I8 The winning probabilities of A, B, C are 36/9, 30/9, 5/9 respectively I I 9/9 I I6 For with replacement sampling, the probability is /9; for without replacement sampling, the probability is 3/55 ( I7 4 ( 39 3) ( 5 3) ( 4 ( 6 3) ( 5 3) + ( 4 ( 3) 3 ( 5 3) I9 0 x )( 3 ) I0 ( 39 x=5 ( 5 x ) ( 6 4) 6 x=3 ( 6 x) I I3 4! 4 4 [ 0 ( 5 x )( 0 x) 5 x=0 ( 50 0) I6 ] I8 (06) 5 I9 /5 I30 Denoting the selection of a green ball by G and that of a white ball by W, the favorable sample points are GGGGG, GGGW G, GW GGG, GW GW G Compute the probability of each of these sample points and add I3 /3 I ; 058; 0559 I35 (a) ; (b) all numbers in [, ] are medians of the distribution; (c) First prove that n(n ) n= =4 n I36 Var( X ) =E(X ) [E( X )] E(X ) E(X) (since E( X ) 7

18 E(X) ) =Var(X) I38 This is the negative hypergeometric distribution (see Exercise 65) (a) To find the mass function, show that P (X >x)= (39 x ),x =,,, 39, ( 5 x ) and then find the mass function by subtraction; (b) By the tailsum formula, E(X) =+ 39 ( 39 x ) x= =53/4 = 3786 ( 5 x ) I40 Consider a two valued random variable X with the mass function 0005 P (X = µ 0005) = δ, P (X = N) =δ If δ =,thenx has N+0005 µ mean µ NowtakeN to be sufficiently large I4 Since E(X) = µ = E(X ), and E(X ) (E(X)) for any random variable X, it follows that µ µ, ie, 0 µ Therefore, Var(X) =E(X ) (E(X)) = µ µ = µ( µ) 4 I44 (a) E(X) = E(Y ) = 4/3; (b) Var(X) = Var(Y ) = 8/9; (c) P (Z =0)=4/9; P (Z =)=4/7; P (Z =)=/7; (d) E(Z) =6/7 I45 The variance is 67 I47 Mean = 4405; variance = 363 I49 For example, take X such that P (X =0)=,P(X = 000) = ) ( 5 3) I50 4 ( I5 (a) 0; (b) Show that P (N >n)= 0 9 ( n),n =,,, 0 0 n Then, by the tailsum formula, E(N) =+ 0 n= P (N >n) I5 9/6 I54 log I55 The mgf equals ψ(t) = +et +e t +e 3t +e 4t +e 5t The mean is zero 5e t I56 E(Y )= (n )+4( n 4)+ = n;themgfofy equals ψ(t) = + +(n )e t +( n 4)e 4t + n 4 n I57 The mgf of XY is p + pe λ(et ) I59 The mgf is n i= ( p i + p i e t ); the variance is n i= p i ( p i ) I60 X takes the values ± with probability each, I6 (a) The mgf is e λ n (λe t ) x x=0 + P (X >n); (b) The limit equals the x! mgf of X itself, ie, e λ(et ) I64 The nth factorial moment is λ n I65 (a) c = ;(b)p(0) = /8; p() = /; p() = /4; p(3) = /8; (c) 8 /8 8

19 I66 e (s )λ I68 Yes, it is possible Take a random variable X with the stated property and take Y = X I70 66 I7 For sending one parcel, the expected cost is $950 For sending two parcels, the expected cost is $50 It is better to send one parcel I73 The distribution of X is a binomial with parameters 3 and Therefore, the mean is and the variance is I75 e 5/6 p I77 θ I79 (a) 00835; (b) 0007; (c) k = 5 in each case I80 (a) (39 3) ; (b) 03664; (c) ( 5 3) I8 P (max(x, Y ) n) = 4 n 5 n + n,n 0 By using the tailsum formula now, E(max(X, Y )) = n=0 (4 n + 5 n n )= 9 =4 I83 The mgf of X Y is e λ µ+λet +µe t I85 The first four moments are, 6,, 94 I86 The first four moments are 5, 75, 65, 075 I87 0 I88 Use the tailsum method to find the mean I90 Suppose the couple has N children First prove that P (N > n)= ( r n x=n r+ x) p x ( p) n x,r n r Then find the pmf of N by subtraction 587 I9 =4 630 I9 No; because by problem I93, Var(XY ) > Var(X)Var(Y ) = E(X)E(Y ) = E(XY ) I94 X must be ± with probability each I96 06 I97 (a) The density is f(x) =0(x<), =(x )/3( x ), = /3( x 4), =(5 x)/3(4 x 5), =0(x>5); (b) The CDF is F (x) =0(x ), =(x ) /6( <x ), = x/3 /( <x 4), = 5x/3 (x + 9)/6(4 <x<5), =(x 5); (c) 3; (d) 3 3 I98 (a) c = 4; (b) 0; 075; 075; P (C A B) = 0 and is undefined 0 I00 F (x) =0(x<0), =09(x =0), = 0 e x (x >0) I0 (a) 0; (b) ; (c) prove that x 3 x x > 0ifx> So the 9

20 required probability is P (X >); (d) ( + e 3 ); (e) 0 I03 The density of Y = e X is f(y) = y ( y<), and 0 log y e otherwise I05 The density of Y = g(x) isf(y) =e y + e y, 0 <y<and0 y otherwise I07 Y has the CDF F (y) = +y, ( y ), =0(y< ), =(y>) In other words, Y has the same distribution as Z I09 (a) σ equals 53; (b) 696; (c) I0 The shortest interval with probability 5 under each of these three distributions is [, ] I 6 I4 Mean equals log,medianequals, and variance equals n n n I5 3 I6 4 I7 log( arctan X) is one such transformation π I8 (a) c = ;(b) π 4 π π I0 square inches I π log I3 n 770 log(φ(4)) I4 a< I5 The function has a minima of zero and a maxima of I7 Hazard rates with a bathtub shape can be generated by using densities of the form cx d e xp,x > 0,p,d > 0 Of course, this is not the only possible way to generate bathtub hazard rates I9 (09) 0 I30 (09544) 0 I3 The deciles are e 8,e 084,e 055,e 053,,e 053,e 055,e 084,e 8 I3 The name is lognormal I33 No, because the mean µ would be zero, and then the three percentile values become inconsistent I35 For a 90% confidence interval, n approximately 7σ ; for a 95% confidence interval, n approximately 384σ ; for a 99% confidence interval, n approximately 663σ I

21 I37 For Z 5, the mean, median, and mode are all zero; for Z, themean is, the median is 0675, and the mode is zero; for Z, themeanis πe π +Φ(), the median is 05, and the mode is zero I39 The distribution is approximately a normal distribution with mean and variance 6 36 I4 n approximately 3393 I I44 A normal approximation cannot be used here, because this density does not have a finite mean I I49 The conditional expectation of the number of heads in the last 5 tosses given that the number of heads in the first 5 tosses is x is given by the formula 5 + x 3 I50 By using the iterated variance formula, you can easily prove that Var(Y ) Var(X) I5 (a) Each of X, Y, Z takes the values ± with probability each; (b) each of these joint pmfs assigns probability to the four points (±, ±); (c) 4 each correlation is zero; (d) I5 (a) The variance is 3n ; (b) the limit is zero by Chebyshev s inequality 4 I 53 The percentages of A, B, C, D,andF would be 8%, 359%, 686%, 359% and 8% So the required probability is 40! (008) 8 (0359) 8 (0686) 8 (0359) 8 (008) 8 (8!) 5 I55 0 I56 (a) ( k ); (b) min(x, k); (c) The covariance between X and min(x, k) is (k +) k The variance of min(x, k) is + k 4 k 4k k Using these, the correlation between X and min(x, k) is This converges to when k k k 4 k (k ) k ; (c) Show that the co- The correlation is I57 (a) The expected values are N+ variance between the two numbers drawn is N+ (f) 0 N(N+ n ;(b) N N ; I59 (a) ;(b) ;(c) 6 6 I60 E(X X X 3 )=n(n )(n )p p p 3 I6 For a general k, the conditional expectation is 4 k 3 I64 (a) E(X Y = y) = 3( y); (b) E(Y X = x) = 3 ( x); (c) 4 4

22 E(XY )= 9 56 ;(d)e(x Y )= 35 I65 (a) ;(b)0;(c) φ(c)φ(c); (d) arcsin( 3 4 π 0 ) I66 The density of Z = X Y is f Z (z) = z, z I67 A necessary and sufficient condition that you can make a triangle with three segments of lengths a, b, c is that the sum of any two of a, b, c is at least as large as the third one Using this, you can show that the required probability in this problem is log I68 I69 The joint density of U = X, V = XY,W = XY Z is f U,V,W (u, v, w) = e u u v w v,u,v,w >0 uv I7 (a) Show that Cov(X + X,X X ) = 0, and that it follows from this that X + X,X X are independent; (b) since X + X,X X are independent, any function of X + X and any function of X X are independent; (c) write X + X and X X in terms of the polar coordinates, from which it will follow that they are independent I73 (a) E(XY )=, Var(XY )= 5 4 ;(b)e(x Y )=0;(c)0;(d)c = I I76 I77 3 I78 Yes, the expectation is finite and equals I80 (a) A bound for the density of Z = X + Y is f Z (z) π for all z; (b) no, the density of XY may not be bounded Think of a random variable Y with a density converging to very sharply at y =0 4 I8 9 θ( p) I83 p+θ( p) I84 The mean equals 0( e 0 ) by applying the iterated expectation formula; the variance has to be found by applying the iterated variance formula, and the terms in the formula have to be found numerically The variance is approximately 90 I85

23 Appendix I: True-False Problems T 4 T 6 F 8 T 9 F F 3 F 4 T 5 F 6 T 8 F 9 F T 3 T 5 T 7 T 8 F 30 T 3 F 33 F 35 T 36 T 39 T 40 T 4 T 45 T 46 T 47 T 49 T 50 F 5 T 5 F 53 T 3

24 55 F 57 T 59 F 6 T 63 T 64 T 65 T 66 T 68 T 70 T 7 F 7 T 73 T 76 T 77 T 78 T 79 F 8 T 8 T 84 F 86 F 87 T 88 F 90 T 9 F 94 F 95 T 97 T 99 F 00 T 0 T (the expected value is 49) 03 T 04 T 06 F 4

25 07 T 09 T 0 F T 3 T 4 F (ρ can also be 08) 6 T 8 T 9 F T 5