概率统计答案(更新至第五章).pdf

Solutions for Homework Huaming King 2012 12 23 King If I mean such probability exists some of the HW appear in the fi nal test are you sure you are capable of solving them 1 2 3 If F x Rx Ce t dt x is a distribution function what should C be 1 Solution Since F x is a distribution function then Z Ce t dt 2C Z 0 Ce tdt 2C 1 which implying that C 1 2 4 Suppose that random variable X has the following distribution function F x 0 x 5 1 5 5 x 2 3 10 2 x 0 1 2 0 x 2 1x 2 Find the probability distribution of X Solution It is easy to see that F x is a purely jumping function Therefore X is a discrete type random variable Since P X x F x F x one follows that P x 5 1 5 0 1 5 P x 2 3 10 1 5 1 10 P x 0 1 2 3 10 2 10 P x 2 1 1 2 1 2 We come to conclusion that the probability density matrix of X is 5 202 2 10 1 10 2 10 5 10 5 4 4 3 Solution 8 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 n P n 3 P n 4 P n 5 P n 6 P n 7 1 P n 3 2 8 3 A3 8 n 3 1 2 3 4 3 A3 4 P n 3 A3 4 A3 8 4 3 3 8 7 6 1 14 2 P n 4 8 4 A4 8 n 4 C1 3 A4 8 A1 4A13A14A12 A1 4A13A14A12 P n 4 3A1 4A13A14A12 A4 8 3 4 3 4 2 8 7 6 5 6 35 3 P n 5 8 5 A5 8 n 5 C2 4 6 A5 8 A1 4A13A14A13A12 A1 4A13A14A13A12 3 P n 5 6A1 4A13A14A13A12 A5 8 6 4 3 4 3 2 8 7 6 5 4 9 35 4 P n 6 2 7 5 P n 7 3 14 34567 1 14 6 35 9 35 2 7 3 14 7 Suppose that random variable X has distribution P X k C 2 3 k k 0 1 2 3 Find 1 the value of C 2 P 1 X 2 3 P 0 X 2 5 Solution 1 One knows that 4 X k 0 C 2 3 k C 1 2 3 4 9 8 27 1 implying that C 27 65 So P X k 27 65 2 3 k 2 P 1 X 2 P X 1 P X 2 27 65 2 3 2 3 2 6 13 3 P 0 X 0 0y 0 Proof Let Fy x be the distribution function of Y Then Fy x P Y x P X x P X x2 Z x2 0 p t dt let t y2 Z x 0 2yp y2 dy 1 Then we conclude that Y is a continuous random variable with density function py y 2yp y2 y 0 22 Suppose X p and P X 1 2P X 2 Find EX DX and P X 3 Solution Since X is a Poisson random variable we suppose that P X k k k e Then one follows that P X 1 1 1 e 2 2 2 e 2P X 2 Solving this equation one has that 1 Therefore P X k 1 k e 1 Then EX DX 1 and P X 3 1 3 e 1 1 6e 27 Suppose that X N 5 4 Find 3 P 1 X 9 Solution One knows 5 and 2 Then it follows that X 5 2 N 0 1 7 Therefore we have P 1 X 9 P 1 5 2 X 5 2 9 5 2 P 2 X 5 2 2 2 2 2 2 1 28 Suppose that X U 0 5 Find the probability that the equation 4x2 4Xx X 2 0 has real solution Solution The equation 4x2 4Xx X 2 0 has real solutions if and only if 4X 2 4 4 X 2 16X2 16X 32 0 Equivalently one has X 2 X 1 0 which implies that X 1 X 2 Since X U 0 5 P X 1 0 Then one conclude that P equation has real solution P X 2 5 2 5 0 3 5 30 Suppose that random variable X has density function p x 1 1 x2 x 0 y 0 1 k 2 F x y 3 P 0 X 1 0 Y 0 y 0 2 F x y Z x 0 Z y 0 p s t dsdt Z x 0 Z y 0 12e 3s 4tdsdt Z x 0 3e 3sds Z y 0 3e 3tdt 1 e 3x 1 e 3y 3 P 0 X 1 0 Y 2 F 1 2 F 0 2 F 1 0 F 0 0 11 1 e 3 1 e 8 0 1 e 8 0 1 e 3 0 F 1 2 1 e 3 1 e 8 4 X Y p x y 1 2 0 x 1 0 y 2 X Y 1 2 Solution X Y 1 2 min X Y 1 2 P X Y 1 2 P min X Y 1 P Y X 4 P Y 1 2 X 1 2 Solution 1 X p1 x Z 2 0 p x y dy Z 2 0 x2 1 3xydy 2x2 2 3x 0 x 1 Y p2 y Z 1 0 p x y dx Z 1 0 x2 1 3xydx 1 3 1 6y 0 y 2 2 X Y y pX Y x y p x y p2 y x2 1 3xy 1 3 1 6y 0 x 1 0 y 2 14 Y X x pY X y x p x y p1 x x2 1 3xy 2x2 2 3x 0 x 1 0 1 ZZ x y x y 1 p x y dxdy Z 1 0 Z 2 1 x x2 1 3xydy dx Z 1 0 x2 1 x 1 3x 2 1 x 2 2 dx 65 72 P Y X ZZ x y y x p x y dxdy Z 1 0 Z 2 x x2 1 3xydy dx Z 1 0 x2 2 x 1 3x 2 x2 2 dx 17 24 4 P Y 1 2 X 1 2 P Y 1 2 X 1 2 P X 1 2 R1 2 0 R1 2 0 p x y dxdy R1 2 0 p1 x dx R1 2 0 R1 2 0 x2 1 3xydxdy R1 2 0 2x2 2 3xdx R1 2 0 1 2x 2 1 24xdx R1 2 0 2x2 2 3xdx 5 32 13 X Y p x y e y 0 x 0 p2 y Z y 0 e ydx ye y y 0 15 p x y e y6 ye x y p1 x p2 y X Y 2 X Y y pX Y x y p x y p2 y e y ye y 0 x y Y X x pY X y x p x y p1 x e y e x 0 x y 14 X 13 0 30 7 Y 24 0 60 4 X Y X Y Solution X Y P X Y 3 P X 1 Y 2 P X 1 P Y 2 0 3 0 6 0 18 P X Y 5 P X 1 Y 4 P X 3 Y 2 P X 1 P Y 4 P X 3 P Y 2 0 3 0 4 0 7 0 6 0 54 P X Y 7 P X 3 Y 4 P X 3 P Y 4 0 7 0 4 0 28 X Y X Y 357 0 180 540 28 15 X U 0 1 Y e 1 X Y Z X Y Solution p1 x 1 0 x 1 p2 y e y y 0 z 1 pX Y z Z 1 0 p1 x p2 z x dx Z 1 0 p1 x p2 z x dx Z 1 0 e z x dx 16 e z Z 1 0 ex dx e z e 1 pX Y z 0z 1 z 0 1 x y z 0 x 1 x 0 1 x z 16 X Y p x y e x y x 0 y 0 1 2 X Y Solution 1 2 X Y FZ z FZ z P 1 2 X Y z P X Y 2z ZZ x y x y 2z p x y dxdy Z 2z 0 Z 2z x 0 e x y dydx Z 2z 0 e x e 2zdx 1 e 2z 2ze 2z 1 2 X Y pZ z F Z z 2e 2z 2e 2z 4ze 2z 4ze 2z z 0 17 X Y G p Z max X Y solution P X n P Y n qn 1p P Z n P max X Y n P X n Y n P X n P Y n P X n 2 n X k 1 qn 1p 2 p 1 q qn 1 2 p2 1 qn 2 1 q 2 1 qn 2 P Z n P Z n P Z n 1 1 qn 2 1 qn 1 2 I feel very tired and have to stop here You must do all the homework left by yourself 17 4 1 Xn p x 1 x 3 x 1 0 X 1 Xn Z x p x dx Z x 1 x 1 x 3 dx Z x 1 1 x 2 dx 2 Z 1 1 x2 dx 0 lim n P Pn i 1Xi n 0 2 200 0 6 1kw 0 999 Solution i 1 2 200 Xi 1 i 0 i Xi P Xi 1 0 6 P X 0 0 4 E Xi 0 6 D Xi 0 6 0 4 0 24 200 P200 i 1Xi 1kw 200 x P 200 x 0 999 P 200 x P 200 200 0 6 200 0 6 0 4 x 200 0 6 200 0 6 0 4 x 200 0 6 200 0 6 0 4 0 999 18 x 200 0 6 200 0 6 0 4 3 1 x 141 5 141 5 0 999 3 U 0 5 0 5 1 300 15 2 n 10 0 9 n Solution i Xi Xi U 0 5 0 5 E Xi 0 D Xi 0 5 0 5 2 12 1 12 P Pn i 1Xi n 0 pn 12 x x 1 P 300 X i 1 Xi 15 P 15 300 X i 1 Xi 15 P 15 p300 12 P300 i 1Xi p300 12 15 p300 12 15 p300 12 15 p300 12 2 3 1 2 0 99865 1 0 9973 2 P P300 i 1Xi 10 0 9 P n X i 1 Xi 10 P 10 n X i 1 Xi 10 P 10 pn 12 Pn i 1Xi pn 12 10 pn 12 10 pn 12 10 pn 12 2 10 pn 12 1 0 9 19 10 pn 12 0 95 10 pn 12 1 64 n 440 4 0 01 0 95 100 Solution n Xi 1 i 0 i Xi P Xi 1 0 99 P Xi 0 0 01 E Xi 0 99 DXi 0 99 0 01 n Pn i 1Xi n P n 100 0 95 P n 100 1 P n 100 1 P n n 0 99 n 0 99 0 01 0 8 P n 0 8n 1 P n 0 8n 1 P n n 0 9 n 0 09 0 8n 0 9n 0 09n 1 0 8n 0 9n 0 09n 0 9n 0 8n 0 09n n 3 0 95 n 3 1 65 n 25 6 X P 100 P 80 X 100 Solution Poisson i i 1 2 100 P 1 100 X i 1 i P 100 E i 1 D i P P100 i 1 i 100 100 x x P 80 X 100 P 80 100 X i 1 i 100 P 80 100 100 P100 i 1 i 100 100 100 100 100 100 100 100 80 100 100 0 2 0 5 1 2 0 5 0 9773 0 4773 7 lim n n X k 0 nk k e n 1 2 Proof i i 1 2 P 1 n X k 1 k P n E n X k 1 k n D n X k 1 k 21 lim n P Pn k 1 k n n x x x 0 lim n P Pn k 1 k n n 0 0 1 2 lim n P n X k 1 k n 1 2 lim n n X k 0 nk k e n 1 2 5 2 2 4 X 100 Chebyshev X 90 Solution D X D X n 2 100 4 100 Chebyshev P X DX 2 4 100 2 P X 1 4 100 2 0 4 P X 0 4 1 4 100 0 42 0 9 P 0 4 X 0 4 0 9 4 m n F 1 F1 m n 1 F n m 22 Solution m n X Y X 2 m Y 2 n X m Y n F m n F m n X m Y n F m n 1 Y n X m F n m 1 n m F F m n P F1 m n 1 P 1 1 F1 m n 1 2 1 F n m P 1 F n m P 1 F n m 1 3 2 3 P 1 1 F1 m n P 1 F n m 1 F n m 1 F1 m n 5 N 40 52 1 36 X 38 43 2 64 X 40 1 3 n P X 40 1 0 95 Solution E X 40 D X 2 n 52 n 1 P 38 X 43 P 38 40 q 25 36 X 40 q 25 36 43 40 q 25 36 23 P 2 4 X 40