SOR201 Solutions to Examples 3

Transcription

1 page 0SOR0(00) SOR0 lutions to Examples (a) An outcome is an unordered sample {i,, i n }, a subset of {,, N}, where the i j s are all different The random variable X can take the values n, n +,, N If X x, the other numbers selected have values in the range,, (x ) If X x, (n ) numbers have been selected from the set {,, (x )} and 0 numbers from the set {(x + ),, N} This can be done in ( x n ) ways The total number of possible selections is ( N P(X x) ( x n ) / ( N, x n,, N (b) If the largest number in the sample is y, then all numbers in the sample are y, and vice versa; also y n and y need not be an integer Consider y to be one of the values n, n +,, N P(X y) P(all numbers in sample y) No of ways of selecting n numbers from {,, y}, without replacement No of ways of selecting n numbers from {,, N}, without replacement ( ( y / N, y n, n +,, N P(X x) P(X x) P(X x ) [( ) ( x n x )] ( n / N ( ( x n ) / N, x n,, N (using the identity ( ( y r ) + y ) ( r y+ ) r ) (i) (a) Y X or X ± Y P(Y y) P(X y) + P(X y) p( y) + p( y), y,, 9, and P(Y 0) p(0) { X, when X 0 (b) W X X, when X < 0 P(W w) p(w) + p( w), w,, and P(W 0) p(0) {, when X > 0 (c) Z sgn(x), when X < 0 0, when X 0 P(Z ) p(x), x>0 P(Z 0) p(0), P(Z ) x<0 p(x) /continued overleaf

2 page 0SOR0(00) (ii) Let g(a) E[(X a) ] E[X ax + a ] E(X ) ae(x) + a dg E(X) + a 0 when a E(X) da d g > 0 when a E(X) da E[(X a)] is minimized when a E(X) [Note: E( X b ) is minimized when b median of X] (i) (a) We have P(X > m) (b) No memory property: xm+ P(X x) pq m + pq m+ + pq m [ + q + q + ] pqm q qm since p + q P((X > m + (X > m)) P(X > m + n X > m) P(X > m) P(X > m + since (X > m + (X > m) P(X > m) qm+n q m qn P(X > [using result in part (a)] (ii) We can write out E(X) as follows: E(X) xp(x x) x0 xp(x x) x P(X ) + P(X ) + P(X ) + P(X ) + P(X ) + P(X ) Since the series is convergent and the terms are positive, we can re-arrange the order of the terms, summing vertically, we obtain E(X) P(X > x) x0 For the geometric distribution in part (i): P(X > 0) ; P(X > x) q x (x a positive integer) E(X) + q + q + q p

3 page 0SOR0(00) We are given that X and Y are independent random variables with (a) P(X Y ) (b) We have that P(X k) P(Y k) pq k, k 0,, ; p + q P(X k, Y k) k0 P(X k)p(y k) k0 ( ) pq k k0 p (q ) k P(X + Y k0 p [since X and Y are independent] [since 0 < q < ] ( q ) p [since p q] ( + q) P(X x, Y n x) x0 P(X x)p(y n x) x0 pq x pq n x x0 p q n (n + )p q n x0 [independence] P(X x X + Y P(X x and X + Y P(X + Y P(X x, Y n x) P(X + Y pqx pq n x [using independence and (*)] (n + )p q n n +, x 0,,, n the discrete uniform distribution on (0,,,, (c) U min(x, Y ) takes the values 0,,, The event (U u) can be decomposed into mutually exclusive events thus: (U u) (X u, Y u) (X u, Y u + ) (X u, Y u + ) (X u +, Y u) (X u +, Y u) ( ) /continued overleaf

4 page 0SOR0(00), invoking the most general ( countably additive ) form of Axiom, we have: P(U u) P(U u, Y u)+ P(X u, Y y)+ P(Y u, X x) yu+ (pq u ) + pq u pq y + yu+ p q u + p q u+y yu+ p q u + p q u+ i0 pq u pq x xu+ p q u + p q u+ /( q) p q u + pq u+ pq u (p + q) pq u (q + ), u 0,, q i xu+ (a) Sample Points X Y Z H H H H H H H H H T H H H T H H H T H H H T H H H T H H H H H H H T T H H T H T H T H H T T H H H T H H T T H H T H T H T H H T H H T T H H T H T H H T T H H H H H T T T H T H T T T H H T T H T T H T T H T H T T T H H T H T T T H T H T T H T T H T H T T T H H H T T T T T H T T T T T H T T T T T H T T T T T H T T T T T /continued overleaf

5 page 0SOR0(00) (b) Let p(x, y, z) P(X x, Y y, Z z) From the listing of the sample space in (a), we deduce that All other probabilities of the form p(0, 0, 0) p(,, ) p(,, ) p(,, ) 6 p(,, ) p(,, ) 6 p(,, ) p(,, ) p(,, ) p(,, ) p(,, ) p(x, y, z), 0 x, 0 y, 0 z are zero [Check: p(x, y, z) ] x,y,z The joint probability function of (X, Y ) is given by P(X x, Y y) P(X x, Y y, Z z) z0 p(x, y, z) for 0 x, 0 y z0 If this function is tabulated in a two-way table, the row and column totals give the probability function values for the random variables Y and X respectively, since P(X x) and P(Y y) P(X x, Y y) p(x, y, z), 0 x y0 P(X x, Y y) y0 z0 x0 x0 z0 p(x, y, z), 0 y Thus: 0 X Y Similarly, the joint probability function of (Y, Z) and the probability functions of Y and Z are Z Y /continued overleaf

6 page 6 0SOR0(00) (c) E(Y ) (0 E(Y ) (0 ) + ( ) + ( ) + ( ) 8 ) + ( Var(Y ) E(Y ) {E(Y )} E(Z) (0 E(Z ) (0 ) + ( )+( ) + ( ) + (9 ) 8 8 Var(Z) E(Z ) {E(Z)} 79 ( 6 6 ) + ( ) + ( ) + ( ) + ( ) 6 6 )+( )+(9 )+(6 )+( ) ) 0 6 E(Y Z) ( ) + ( ) + ( ) + ( ) + ( +( 6 ) + ( 7 ) + ( ) + ( ) 90 Hence Cov(Y, Z) 90 6 and ρ(y, Z) (d) P(X x Y ) P(X x, Y )/P(Y ), 0 x P(Y y, Z z X ) Thus x 0 P(X x Y ) 0 P(X, Y y, Z z), 0 y, 0 z P(X ) P(Y, Z X ) 0 P(Y, Z X ) 6 0 P(Y, Z X ) all other probabilities are zero (e) The random variable E(Y X) takes the values E(Y X x), x 0,,, where Thus: E(Y X x) 0 ; yp(y y X x) y0 P(X x, Y y) y P(X x) y0 E(Y X 0) 0 / / E(Y X ) / / E(Y X ) / 0/ 0/ 0 E(Y X ) / 0/ + 6/ 0/ + / 0/ 8 0 E(Y X ) / + / 6 / / 0 E(Y X ) / / The probabilities are derived from {P(X x)} Thus: 6 E(Y X) 0 6 Prob and E[E(Y X)] (0 ) + ( 6 ) + ( 6 0 ) + ( ) 8 E(Y ) )

7 page 7 0SOR0(00) 6 Let X time to freedom (hours) Y number of door originally chosen (, or ) E(X) E(X Y )P(Y ) + E(X Y )P(Y ) + E(X Y )P(Y ) Now E(X) 0 days 7 (a) P(X x) E(X Y ) + E(X) E(Y Y ) + E(X) E(X Y ) E(X) 0[ + E(X)] + 0[ + E(X)] + 0[] + 08E(X) 0 N- N N+ N+ N P(X x) P(X N x + ), E(X N ) N x (x N )P(X x) Now st term ( N)P(X ) nd term ( N)P(X ) etc N x,, N N x nd last term (N )P(X N ) last term (N )P(X N) (N )P(X ) )P(X ) Summing st and last terms gives 0; nd and nd last terms gives 0; etc E(X N ) 0 E(X) (N + ), ie E(X) N + Since P(X N + ) P(X N + ), the median of X is N + (actually any value between N and N + can be considered the media /continued overleaf

8 page 8 0SOR0(00) (b) P(X x) 0 M- M M+ M M M x P(Y y) P(Y M y), y 0,, M By a similar procedure to that in part (a), we can show that E(Y M) 0 and hence E(Y ) M Since P(Y < M) P (Y > M), the median of Y is M 8 We have: ( n ) P(X X ) P [X k, X k] k0 P(X k, X k) k0 P(X k)p(x k) k0 [me events] [independence] But P(X k) P(X n k) since P(H) P(T ) P(X X ) P(X k)p(x n k) k0 P(X k, X n k) k0 ( n ) P [X k, X n k] k0 P(X + X [independence] [me events] Because the two people toss independently, the experiment can be regarded as a single Bernoulli process with n trials and p, X X + X being the total number of heads obtained the required probability is P(X ( ) ( n n ( n n n ) ) [binomial distribution] ( ) ( n n n )

9 page 9 0SOR0(00) 9 (a) (b) E(X) (0 ) + ( E(X ) (0 )+( X Z ) + ( 0 )+( Var(X) E(X ) {E(X)} ( ) 0 ) + ( ) + ( ) + ( ) 80 0 )+(9 )+(6 )+( ) 0 E(XY ) ( ) + ( ) + ( 6 ) + ( ) + ( 6 +( ) + ( ) + ( ) + ( ) 8 E(XZ) ( ) + ( 6 ) + ( ) + ( ) + ( +( ) + ( ) + ( ) + ( ) Cov(X, Y ), ρ(x, Y ) 0 8 and Cov(X, Z) 88, 6 ρ(x, Z) (c) P(X x Y, Z ) (d) Thus: P(X x, Y, Z ), 0 x P(Y, Z ) P(X Y, Z ) 6 7 P(X Y, Z ) 7 ; all other probabilities are zero E(Z X x) P(X x, Z z) z P(X x) z0 E(Z X 0) 0 / / E(Z X ) / / E(Z X ) 6/ 0/ 0/ 0 E(Z X ) / + 6/ + / 0/ 0/ 0/ 0 E(Z X ) / / + / / + / / 0 E(Z X ) / / /continued overleaf ) )

10 page 0 0SOR0(00) From {P(X x)} we deduce E(Z X) 0 Prob Let X k S E[E(Z X)] (0 ) + ( ) + ( 6 E(Z) return when critical value k is used value on first roll The probability distribution of S is: 0 0 ) + ( 0 0 ) + ( 0 ) + ( ) Now j P(S j) E(X k ) j 6 E(X k S j)p(s j) { 0, if j 7 E(X k S j) E(X k ), if j < k and j 7 (because we start agai j, if j k and j 7 E(X 6 ) [ ]E(X 6) + [(6 ) + (8 ) + (9 ) +(0 ) + ( ) + ( )] 0E(X 6) + 70 ie 6 E(X 6) 70 E(X 6 ) [E(X 7 ) ]E(X 8 ) [ ]E(X 8) + [(8 ) + (9 ) +(0 ) + ( ) + ( )] E(X 8) + 0 ie E(X 8) 0 E(X 8 ) E(X 9 ) [ ]E(X 9) + [(9 ) + (0 ) + ( ) + ( )] 0 E(X 9) ie E(X 9) 00 E(X 9 ) E(X k ) is maximised at k 8