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Practice problems 1. What is exponential, γ, Binimial, Bernoulli, negative binomial distribution? 2. Explain the memoryless property of the exponential distribution. P rob(X > b|X > a) = P rob(X > (b − a)) for b > a. This means just because you managed to be 70, that is no evidence that your exceptionally healthy in this model (but it also means it is a fresh start!). 3. What do you think if you here central limit theorem? Approximation of Binomial coefficients, universal property of independent events, moment generating function. It means for example that lim Ee √t n Pn j=1 ( Xj −EXj σj n ) 2 /2 = et for independent random variables with uniformly bounded moments. 4. Estimate the probability of winning 10 times out of 20, 100 times out of 200, where you flip a coin with P rob(up) = p? Does a tabular of the Errorfunction for N (0, 1) help? We get P10 20 = p (1 − p) ≈ 0.176 10 10 10 and 200 p100 = p (1 − p) ≈ 0.05 . 100 200 I think 20 a calculator can still manage, is probably okay. The approxi10 100 100 100 mated values are calculate using p = 1/2. Now the central limit theorem, tells us that P rob(| n X j=1 Xj − E j | > √ a2 e− 2 nσa) ≺ P rob(|g| > a) ≤ √ . 2πa √ nσa = 1 in order to calculate 1 − P10 and 1 − P100 . p √ For 10 trials and σ = 1/4 = 1/2 we get a = 2/ 20 in the first case and √ √ a = 2/ 200 = 2/10 ≈ 0.15 for n = 200. In our case we need For small numbers a the estimate P rob(|g| > a) ≤ 1 2 −a /2 e√ 2πa is not helpful. Thus instead, we look at P rob(| n X Xj − Ej | ≤ j=1 √ 2a 4 nσa) ≈ P rob(|g| ≤ a) ≈ √ = √ . 2π 2πn √ Thus we expect the probability to go like 1/ n for n large. Actually the value for 1000 is 0.01783 . (Maybe I forgot 2π). 5. Find the distribution of Z given that moment generating function is i) 1/3 + 2/3et . 2 /3+et2 −1+et −1 ii) et/3+t iii) e √ 1 −1 1−t . . ad i) P rob(Z = 0) = 1/3, P rob(Z = 1) = 2/3 ad ii) Z = Z1 + Z2 + Z3 + Z4 all independent. We need MZ1 (t) = et/3 2 /3 MZ2 (t) = et . This Z1 = 1/3. If Z2 is N (0, σ 2 ) distributed then Z2 = √ σY , Y N (0, 1) and hence MZ2 (t) = EetσY = e σ 2 t2 2 . Thus σ 2 = 2/3 does the job. For Z3 we choose Z3 = PN j Vj where N is 1-Poisson and Vj is N (0, 2) (to kill the 2). Finally Z4 is 1-Poisson. √ P as iii) For ξ 2 we have 1/ 1 − 2t thus Z = N j=1 Yj , N is 1-Poisson and Yj are iid 1/2g 2 with g N (0, 1). 6. Find the moment generating function for Z = PN j=1 Xj where X is exponential with EX = 3. For exponential we have for µ = EX that Z ∞ dx 1 1 1 tX Ee = etx e−x/µ = = , µ µ 1/µ − t µ−t 0 provided that t < µ. Thus 1 MZ (t) = e µ−t −1 . 2 7. Flip a coin with P rob(up) = p. When you see ‘up’ you another flip a coin with P rob(up) = 1/2. When you see ‘down’ you another flip a coin with P rob(up) = 1/4. What is the expected time to see the second coin up? We have to condition 1 P rob(X2 = up) = pP rob(X2 = up|X1 = up) + P rob(X2 = up|X1 = down) 2 1 1 1 p = p + (1 − p) = + . 2 4 4 4 8. Calculate E(X|Y ) and E(Y |X) in the following examples: a) fX,Y (x, y) = c(x + 1 + y)2 , 0 ≤ x, y ≤ 1. b) fX,Y (x, y) = c((x + 1)2 + y 2 ), 0 ≤ x, y ≤ 1. Why don’t we have to calculate c? Recall that E(X|Y ) = g(Y ) where Z g(y) = E(Xy ) = xfX|y (x)dx . In order to find that we need fX|y (x) = fXY (x, y) . fY (y) This means we have to calculate f˜Y (y) = Z 1 f˜X,Y (x, y)dx . 0 Here f˜ means everything without the annoying c. Thus we get Z 1 Z 1 1 2 ˜ x2 +1+y 2 +2x+2y+2xydx = +y 2 +1+2y+y . fY (y) = (x+1+y) dx = 3 0 0 Thus (x + 1 + y)2 . fX|y (x) = 4/3 + 3y + y 2 Similarly Z Z 2 x(x+1+y) dx = 1 x3 +x+xy 2 +2x2 +2xy+2x2 ydx = 0 3 1 1 y2 2 2 17 + + + +y+ y = +4/3y 4 2 2 3 3 12 Hence g(y) = 2 17 +4/3y+ y2 12 2 4/3+3y+y . For the other example, we get f˜Y (y) = Z (x + 1)2 + ydx = 1/3(8 − 1) + y and Z xf˜X,Y (x, y)dx = Z Z = x(x + 1)2 + ydx x3 + 2x2 + xdx + Thus g(y) = y y = 1/4 + 2/3 + 1/2 + . 2 2 17 + y2 12 7 +y 8 . 9. When Alice sends a signal, then Bob receives N (x/2−1, 1) distributed random variable. Now Alice sends a N (0, 1) signal. Calculate E(X|Y ) and E(Y |X) as far as you can. First things first E(Y |X) = g(X), X is N (0, 1) and g(x) = EYx = x/2 − 1 because Yx is N (x/2 − 1, 1) distributed. For the other one we complete the square fX|y (x) = fY |x (y)fX (x) fX,Y (x, y) = . fY (y) fY (y) Thus we have to consider (y − (x/2 − 1))2 + x2 = x2 /4 − x + 1 + y 2 − 2(x/2 − 1)y + x2 = 5/4x2 − x − xy + 1 + y 2 + 2y 2 − 2y 2 − 2y 2 2 − 2y 2 = 5/4(x2 − 2x( )+( ) −( ) + 1 + y 2 + 2y 5 5 5 5 2 − 2y 2 4 = (x − ) − (1 − 2y + y 2 ) + 1 + y 2 + 2y 4 5 25 5 2 − 2y 2 21 2 58 21 = (x − ) + y + y+ . 4 5 25 25 25 Thus Xy is normal N ( 2−2y , 25 ) distributed and EXy = 6 4 2−2y 6 = g(y). 10. Let (X, Y ) be a joint gaussian random variable with EX = EY = 0 and ! 1 1 cov = 0 1 4 Find the joint distribution function. Solution: Let X0 , Y0 be independent N (0, 1). Then X = X0 + Y0 and Y = Y0 has exactly the same covariance matrix, and we determined the distribution. As for the joint probability density we recall that for cjk = E(Yj − EYj )(Yk − EYk ) and EYj = 0 = EYk we have pY1 ,...,Yk = e−(y,Ay)/2 1 (2π)n/2 | det A|1/2 . In two dimensions matrices are easy to invert. The inverse of ! c11 c12 C = c21 c22 is given by C −1 1 = det(C) c22 −c12 −c21 c11 In our case C −1 1 −1 = 0 1 ! . ! . Moreover, det(C) = 1. 11. Out of four balls which are drawn with replacement, a coupon collecting problem requires to get three out of four. The probabilities 1/3 for type 1 and type 2, and 1/6 for type 3 and type 4. What is the expected waiting time. Give two solutions. Solutions: Let us consider the possible orders of getting three different coupons (1, 2, 3), (1, 2, 4), (1, 3, 2), (1, 3, 4), (1, 4, 2), (1, 4, 3) 5 are possible solutions starting with 1. There are 18 such solutions. Assuming the the first possibility we get EX1A(1,2,3) = = X k1 ,k2 (k1 + k2 )P (1, ...., 1, 2, ...., 2, 3) | {z } | {z } X k1 k1 x k2 k2 (k1 + k2 )(1/3) (1/3) (1/6) k1 ,k2 1X k(1/3)k 6 k=2 1 X k(1/3)k−1 . = 18 k=2 = We know that P∞ k=1 kq k−1 = X 1 . (1−q)2 Thus k(1/3)k−1 = k=2 5 1 −1 = . 2 (2/3) 4 This means EX1A(1,2,4) = EX1A(1,2,3) = 5 . 72 We also have EX1A(1,3,2) = EX1A(1,4,2) , and EX1A(1,4,3) = EX1A(1,3,4) . Let us calculate the latter. EX1A(1,4,3) = = X k1 ,k2 (k1 + k2 )P (1, ...., 1, 4, ...., 4, 3) | {z } | {z } X k1 k1 x k2 k2 (k1 + k2 )(1/3) (1/6) (1/6) k1 ,k2 = X 1X k1 (1/3)k1 ( (1/6)k2 ) 6 k =1 k =1 1 + ∞ X 1X k1 (1/6)k2 ( 6 k =1 k 2 2 (1/3)k1 ) . 1 =1 Then we note that ∞ X k=1 k q = ∞ X k=0 qk − 1 = 1 1 − (1 − q) q −1 = = . 1−q 1−q 1−q 6 Hence 1 36 1/3 1 9 1/6 + 6 4 1 − 1/6 6 25 2/3 1 9 1 1 36 1 = + 6 4 5 6 25 2 1 1 + 24 = . = 8 × 25 8 EX1A(1,4,3) = All the other cases are similar. 7