Assign5.dvi - Statistics 265 Assignment 5 Solutions 5.2 Three balanced coins are tossed independently One of the variables of interest is Y1 the

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Unformatted text preview: Statistics 265 Assignment 5 Solutions 5.2 Three balanced coins are tossed independently. One of the variables of interest is Y1; the number of heads. Let Y2 denote the amount of money won on a side bet in the following manner. If the rst head occurs on the rst toss, you win $1: If the rst head occurs on toss 2 or on toss 3 you win $2 or $3; respectively. If no heads appear you lose $1 (that is, win $1). a. Find the joint probability function for Y1 and Y2 : b. What is the probability that fewer than three heads will occur and you will win $1 or less? That is, nd F(2,1)]. Solution: a. The sample points and the values of the random variable Y1 and Y2 are given by s HHH HHT HTH HTT THH THT TTH TTT The joint probability function of Y1 and Y2 is p(0; (y1 ; y2 ) (3; 1) (2; 1) (2; 1) (1; 1) (2; 2) (1; 2) (1; 3) (0; 1) p( s 1 8 1 8 1 8 1 8 1 8 1 8 1 8 1 8 f g) p(0; 1) = 1) = 1 8 0 0 0 p(1; p(0; 2) = p(0; 3) = 1) = 0 1 p(1; 1) = 8 1 p(1; 2) = 8 1 p(1; 3) = 8 p(2; 1) = 0 1 p(2; 1) = 4 1 p(2; 2) = 8 p(2; 3) = p(3; 1) = 0 1 p(3; 1) = 8 p(3; 2) = 0 p(3; 3) = 0 0 b. The probability that fewer than three heads will occur and you will win $1 or less is given by P (Y1 2; Y2 1) = F (2; 1) = p(0; 1) + p(1; 1) + p(2; 1) =1+1+1 884 = 1: 2 5.4 Let Y1 and Y2 have the joint probability density function given by f (y1 ; y2 ) = ky1 y2 ; 0; 0 y1 1; 0 elsewhere. y2 1 a. Find the value of k that makes this a probability density function. b. Find the joint distribution function for Y1 and Y2 : 1 c. Find P (Y1 2 ; Y2 3 ): 4 Solution: a. Since the total probability must be 1; we need Z 1 1 Z 1 1 f (y1 ; y2 ) dy1dy2 = 1; and since the density fuction is only nonzero on the square 0; 1] 0; 1]; then Z1 0 Z1 0 ky1y2 dy1 Z1 21 y dy2 = k 1 y2 dy2 = Z1 k 0 0 2 0 = k y22 2 =4 and we need k = 4: b. The joint distribution function of Y1 and Y2 is F (y1; y2 ) = Z y2 Z y1 k 21 0 2 y2 dy2 1 1 f (t1 ; t2) dt1dt2: case 1: If 0 y1 1; 0 y2 1; then = Z y2 Z y1 0 0 2 4t1 t2 dt1 dt2 = 2y1 F (y1; y2 ) Z y2 0 t2 dt2 22 = y1 y2 case 2: If 0 y1 1; y2 > 1; then F (y1; y2 ) = Z 1 Z y1 0 0 4t1t2 dt1dt2 = 2 Z y1 0 t1 dt1 2 = y1 case 3: If y1 > 1; 0 y2 1; then F (y1; y2 ) = Z y2 Z 1 1 0 4t1 t2 dt1 dt2 = 2 Z y2 0 t2 dt2 2 = y2 2 case 4: If y1 > 1; y2 > 1; then F (y1; y2 ) = Z 1Z 1 0 0 4t1 t2 dt1 dt2 = 1 =0 case 5: If y1 0 or y2 0; then F (y1; y2 ) = Z y2 Z y1 0 0 f (t1 ; t2) dt1dt2 1; 0 y2 1 1; y2 > 1 2 F (y1 ; y2) = y y1 > 1; 0 y2 1 >2 > >1 y1 > 1; y2 > 1 > > : 0 otherwise. 19 9 c. The probability we want is P Y1 1 ; Y2 3 = F 1 ; 3 = 4 16 = 64 : 2 4 24 5.11 Let (Y1 ; Y2) denote the coordinates of a point chosen at random inside a unit circle whose center is at the origin. That is, Y1 and Y2 have a joint density function given by (1 2 2 ; y1 + y2 1 f (y1 ; y2 ) = 0 elsewhere. Find the probability P (Y1 Y2 ): Solution: Since the density function is only nonzero on the circular region shown 22 y1 y2 2 y1 Therefore, we have 8 > > > > > < 0 0 y1 y1 y2 y2 + y2 =1 1 2 0 y 1 and the joint probability density is given by ( f (y1 ; y2 ) = 1 0 2 + y2 1 otherwise, 2 y1 then P (Y1 Y2 ) = = ZZ f (y1 ; y2) dy1 dy2 4 y1 y2 Z5 4 1 r drd 4 r2 1 0 = 1 54 = 1: 2 3 2 5.14 Refer to Exercise 5.2. a. Derive the marginal probability distribution for your winnings on the side bet. b. What is the probability that you obtained three heads, given that you won $1 on the side bet? Solution: a. The marginal probability function for Y2 is p2 ( 1) = P (Y2 = 1) = 1 8 1+1+1=1 p2(1) = P (Y2 = 1) = 8482 111 p2(2) = P (Y2 = 2) = + = 884 1 p2(3) = P (Y2 = 3) = 8 1 Y= 1= = 1) = P (YP (Y 3;= 21) 1) = 8 = 2 = 1 : 1 84 2 2 b. The probability that you obtained three heads, given that you won $1 on the side bet is P (Y1 =3 Y2 5.16 In Exercise 5.4, we derived the fact that 4y1 y2 ; 0 y1 1; 0 y2 1 0 elsewhere is a valid joint probability density function. a. Find the marginal density functions for Y1 and Y2: b. Find P (Y1 1 Y2 3 ): 2 4 c. Find the conditional density function of Y1 given Y2 = y2 : d. Find the conditional density function of Y2 given Y1 = y1 : 3 e. Find P (Y1 4 Y2 = 1 ): 2 Solution: The joint density function for Y1 and Y2 is given by 4y1 y2 0 y1 1; 0 y2 1 f (y1 ; y2 ) = 0 otherwise. f (y1 ; y2 ) = a. The marginal density function for Y1 is given by f1 (y1 ) = Z 1 1 f (y1 ; y2) dy2 = Z1 0 4y1 y2 dy2 = 2y1 ; for 0 1; and f (y1 ; y2 ) = 0 otherwise, that is, 2y1 0 y1 1 f1 (y1 ) = 0 otherwise. Similarly, the marginal density function for Y2 is given by 2y2 0 y2 1 f2 (y2 ) = 0 otherwise. y1 4 b. By de nition of the conditional probability, we have P Y1 1 2 Y2 3 4 = 1; Y P Y1 22 3 P Y2 4 3 4 Now, the numerator can be found from P Y1 1; Y 22 3 = Z 2 Z 1 4y y dy dy = Z 2 2y y2 1 dy = Z 2 2y 1 9 12 21 1 12 1 3 3 4 16 0 0 0 4 4 7Z2 7 22 7 7 = 16 2y1 dy1 = 16 y1 = 16 1 = 64 ; 4 0 0 1 1 1 1 1 dy1 while the denominator can be found from P Y2 3 = Z 1 f (y ) dy = Z 1 2y dy = y2 1 = 1 9 = 7 2 2 2 23 322 3 4 16 16 4 4 4 3 4 and therefore, P Y1 1Y 2 2 = c. The conditional density of Y1 given Y2 = y2 is by de nition 8 < f (y1 ; y2) f2 (y2 ) > 0 f2 (y2 ) f (y1 j y2 ) = : 0 otherwise. and since f (y1 ; y2) = 4y1 y2 for 0 0 < y2 1; we have d. Similarly, if 0 < y1 1; we have f (y2 y1) y1 7 64 7 16 = 16 = 1 : 64 4 1; 0 = 2y1 0 2y2 0 y2 1; and f2 (y2 ) = 2y2 for 0 0 y1 1 otherwise. 0 y2 1 otherwise. Z3 0 3 4 0 y2 1; then if f (y1 y2) j j = e. The probability we want is P Y1 3 4 Y2 = 1 2 = Z3 4 0 f y1 y2 j= 1 dy 1 2 = 4 2y1 dy1 = 2 y1 9 = 16 : 5 5.20 In Exercise 5.8, we were given the following joint probability density function for the random variables Y1 and Y2; which were the proportions of two components in a sample from a mixture of insecticide: 2; 0 y1 1; 0 y2 1; 0 y1 + y2 1 f (y1 ; y2) = 0 elsewhere, a. Find P (Y1 b. Find P (Y1 Solution: 1 2 Y2 1Y 2 2 ) =) 1 4: 1: 4 a. Now, P (Y1 1 2 ; Y2 1 4 )= Z 1 Z 1 y2 4 0 1 4 0 1 2 2 y2 2 dy1dy2 = Z1 4 0 21 y2 1 dy 2 2 =2 Z1 4 0 1 2 y2 dy2 = y2 1 4 0 13 = 1 16 = 16 : 4 = Z 1 y2 0 The marginal density function for Y2 is given by f2 (y2 ) = Z 1 1 f (y1 ; y2 ) dy1 2 dy1 = 2(1 1 4 0 1 4 y2 ) for 0 y2 1; so that P Y2 1 4 = Z1 0 4 2(1 1 4 y2 )dy2 = 2y2 1 4 0 2 y2 1 =2 = 3 16 7 16 1 = 7: 16 16 = 3: 7 Therefore, P Y1 1Y 2 2 = P (Y1 1; Y 22 1 P Y2 4 ) b. If 0 y2 < 1; then the conditional density function of Y1 given Y2 = y2 is f (y1 ; y2) f y1 y2 = = 2(1 2 y ) ; f2 (y2 ) 2 f y1 y2 that is, for 0 y1 1 1 y2 : Therefore, if Y2 = 4 ; then f y1 y2 = 1 1y 2 =1 =1=4 4 3 3 4 for 0 y1 3 ; and we have 4 P Y1 1 2 Y2 =1 = 4 6 Z3 4 1 2 4 dy = 4 1 = 1 : 3 1 34 3 5.30 Refer to Exercise 5.2. The number of heads in three coin tosses is binomially distributed with n = 3; p = 1 : 2 Are the total number of heads and your winnings on the side bet independent? Examine your answer to Exercise 5.14(b).] Solution: We have the following: P (Y1 1 = 3; Y2 = 1) = 8 1 P (Y1 = 3) = 8 1 P (Y2 = 1) = 2 so that 1 = 3; Y2 = 1) = 8 6= 1 1 = P (Y1 = 3) P (Y2 = 1); 82 and the random variables Y1 and Y2 are not stochastically independent. P (Y1 f (y1 ; y2 ) = 5.32 In Exercise 5.4, we derived the fact that 4y1 y2 ; 0; 0 y1 1; 0 elsewhere y2 1 is a valid joint probability density function. Are Y1 and Y2 independent? Solution: The joint density function of Y1 and Y2 is f (y1 ; y2) = 2y1 2y2 ; 0; = = 2y1 0 0 y1 1; 0 elsewhere 0 y1 1 otherwise y2 1 and the the marginal density functions were found in problem 5.16 to be f1 (y1 ) and 2y2 0 y2 1 0 otherwise therefore, f (y1 ; y2) = f1 (y1 ) f2 (y2 ) for all (y1 ; y2); and the random variables Y1 and Y2 are stochastically independent. f2 (y2 ) 7 5.46 In Exercise 5.4, we derived the fact that f (y1 ; y2 ) = 4y1 y2 ; 0; 0 y1 1; 0 elsewhere y2 1 is a valid joint probability density function. a. Find E (Y1 ): b. Find V (Y1 ): c. Find E (Y1 Y2 ): Solution: a. The expected value of Y1 is E (Y1 ) = Z 1 1 Z 1 1 y1 f (y1 ; y2 ) dy1dy2 Z1 0 = Z 1Z 1 0 0 2 4y1 y2 dy2dy1 =4 b. The variance of Y1 is V (Y1 ) = Z Z1 0 2 y1 dy1 y2 dy2 4 = 3 1 = 2: 23 1 3 2y1 dy1 4 = 2 4 = 18 : 949 1 1 Y2 2 y1 f1 (y1 ) dy1 2 3 2 = Z1 0 c. The expected value of Y1 since E (Y2 ) = Z is E (Y1 Y2 ) = E (Y1 ) E (Y2 ) = 0; 1 1 y2 f2 (y2 ) dy2 = Z1 0 2 2y2 dy2 = 2 = E (Y1 ): 3 8 5.73 A population of N alligators is to be sampled in order to obtain an approximate measure of the di erence between the proportions of sexually mature males and sexually mature females. Obviously this parameter has important implications for the future of the population. Assume that n animals are to be sampled without replacement. Let Y1 denote the number of mature females and Y2 the number of mature males in the sample. If the population contains proportions p1 and p2 of mature females and males, respectively (with p1 + p2 < 1), nd expressions for E Y1 n Y2 n and V Y1 n Y2 n : The situation here is analogous to drawing n balls from an urn containing N balls, r1 of which are red, r2 of which are black, and N r1 r2 of which are of another color. Using the argument given in class (see Example 5.29 in the text), we can show that Solution: E (Y1 ) = np1 V (Y1 ) = np1(1 p1) N N n 1 where p1 = r1 ; and N E (Y2 ) = np2 V (Y2 ) = np2(1 p2) N N n 1 where p2 = r2 : N De ne two new random variables for i = 1; 2; : : : ; n : Ui n P = 1 0 if alligator 1 is male if not n P Vi = 1 0 if alligator 1 is female if not The Y1 = Ui and Y2 = Vi : In order to nd the variance of a linear combination of Y1 and Y2 ; we i=1 i=1 will have to nd the covariance Cov(Y1 ; Y2); and this involves nding E (Y1 Y2 ) : " E (Y1 Y2 ) = E n X i=1 ! Ui n X i=1 !# Vi 2 = E4 E (Ui Vj ): n X i=1 Ui Vi + XX i6=j 3 Ui Vj 5 = n X i=1 E (Ui Vi ) + XX i6=j Now, the only situation in which Ui Vi 6= 0 is when U1 = 1 and Vi = 1; that is the ith alligator is both male and female. Since this event is impossible we have E (Ui Vi ) = 0 for all i: Also, Ui Vi is nonzero if and only if Ui = 1 and Vj = 1; which happens with probability P (Ui = 1; Vj = 1) = P (Vj = 1 Ui = 1) i6=j P (Ui = 1) = N r2 1 then N N r1 N = p1 N p2 N = NN 1 1 p1 p2; and since there are n(n 1) terms in PP E (Ui Vj ); E (Y1 Y2 ) = 0 + n(n 1) 9 1 p1p2 and Cov(Y1 ; Y2) = n(n 1) 2 = n p1 p1 N n = n(N N ) 1 2 p1p2 N1 nN p1 p2 N1 p1 p2 : N n2 p1 p2 Therefore, E Y1 n Y2 n 1 = n (np1 np2) = p1 p2; and V Y1 n Y2 n 1 = n2 V (Y1 Y2) 1 = n2 (V (Y1 ) + V (Y2 ) 2Cov(Y1 ; Y2)) 1 N N = n2 n(N 1n) p1(1 p1) + n(N 1n) p2 (1 N = n(N 1n) p1 (1 p1 ) + p2(1 p2) + 2p1 p2] N = n(N 1n) p1 + p2 (p1 p2)2 p2 ) 2n(n N 1 N) p1 p2 10 ...
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This note was uploaded on 12/06/2010 for the course MATH 3310 taught by Professor Frohmader during the Spring '10 term at Cornell University (Engineering School).

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