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Statistics 116 - Fall 2004 Theory of Probability Assignment # 9 Solutions Q. 1) (Ross # 7.51) The joint density of X and Y is given by ( f(x; y) = e y y 0 < x < y; 0 < y < 1 otherwise. 0 Compute E[X 3 jY = y]. Solution: Since fX jY (xjy) = Ry 31 3 0 x y dx = y =4. R ye 0 e y =y y =ydx = y1 for 0 < x < y, then E[X 3 jY = y] = Q. 2) (a) If X Poisson(), compute MX (t). (b) Show that if Xn Binomial(n; =n) then MXn (t) converges to MX (t) for every t, where X Poisson(). Solution: (a) MX (t) = E[etX ] = (b) MXn (t) = E[etXn ] = 1 X k=0 etk 1 k etk e = exp( + et ) k! k=0 X n k lambda t p (1 p)n k = (pet +1 p)n = (1+ (e 1))n k n , which in the limit, when n goes to 1, gives us exp((et Q. 3) (Ross # 8.9) 1)). If X is a Gamma random variable with parameters (n; 1). Approximately how large need n be so that X P 1 > 0:01 < 0:01? n Answer: 1 (a) using Chebyshev's inequality; (b) the Central Limit Theorem. Solution: (a) Since E[X=n] = 1, by Chebyshev inequality we have : P fj X n 1j > 0:01g < Var(X=n) = 104 =n (0:01)2 . Then 104 =n < 0:01 if and only if n > 106 . P (b) Let Xi Gamma(1; 1) for i = 1; :::n. Then X = ni=1 Xi Gamma(n; 1) P fj X X + X2 + ::: + Xn pn 1j > 0:01g = P fj 1 n n j > 0:01png = P fjzj > 0:01png . By central limit theorem z has standard normal distribution. So, p look at the table to nd that 0:01 n > 2:58, therefore n > (258)2 ' 66564. Q. 4) (Ross # 8.11) Many people believe that the daily change of price of a company's stock on the stock market is a random variable with mean 0 and variance 2 . That is, if Yn represents the price of the stock on the n-th day, then Yn = Yn 1 + Xn ; n1 where X1 ; X2 ; : : : ; are independent and identically distributed random variables with mean 0 and variance 2 . Suppose that the stock's price today is 100. If 2 = 1, what can you say about the probability that the stock's price will exceed 105 after 10 days, without using the Central Limit Theorem? Solution: Note, that Y10 Y0 = X1 + X2 + ::: + X10 , and E[Y10 Y0 ] = 0 as well as var(Y10 Y0 ) = 10. Moreover, Y10 Y0 has normal distribution with the mean and variance above. You can solve this problem by three dierent ways now. 1. Using the fact of standard normal distribution : P fY10 Y0 > 5g = P Y10 Y0 5 >p 10 10 p 2. Using ONE sided Chebyshev inequality : P fY10 Y0 > 5g 2 2 2 = ; 2 2 +a 7 ' 0:057: where 2 is the variance of Y10 sided Chebyshev inequality : P fY10 Y0 and equals to 10. 3. Using TWO Y0 > 5g P fjY10 = 3 Var 10 : 25 Y0 j > 5g P10 i=1 Xi 25