Finding Covariance and ExpectationDate: 12/02/1999 at 15:54:11 From: Dan Kepner Subject: Probability, Covariance If x is the number of 1's and y is the number of 2's that occur in n rolls of a fair die, what is the Cov(x,y)? How many times would you expect to roll a fair die before all six faces appear at least once? Thank you for your help. Date: 12/03/1999 at 07:36:03 From: Doctor Anthony Subject: Re: Probability, Covariance >If X is the number of 1's and Y is the number of 2's that occur in n >rolls of a fair die, what is the Cov(X,Y)? Cov(x,y) = E(xy) - E(x).E(y) In any one trial, if x = 1, then y = 0 and xy = 0. Similarly, if y = 1, then x = 0 and again xy = 0. In any one trial we can also have x = 0, y = 0, so again xy = 0. This continues to apply for n trials, and so Cov(x,y) = -E(x).E(y) = -(n/6)(n/6) = - n^2/36 >How many times would you expect to roll a fair die before all six >faces appear at least once? The first throw will certainly produce a new number. We must now find the expected number of throws to the next new number. This has the probability 5/6 of being new, so we set up a difference equation as follows. Let a = the expected number of trials to the second new number. We must make one trial at least and we have a probability 1/6 of returning to 'a'. So a = 1 + 1/6 a 5/6 a = 1 a = 6/5 For the third new number, let b = expected number of further trials. This time there is a probability of 2/6 = 1/3 of returning to b. b = 1 + 1/3 b 2/3 b = 1 b = 3/2 = 6/4 and to the fourth new number we have c = 1 + 1/2 c 1/2 c = 1 c = 2 = 6/3 The pattern is now clear. The expected number of trials to 6 new numbers is: E(no. of trials) = 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 6[1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1/1] = 6 x 49/20 = 14.7 So on average you should get all six numbers by the 15th throw. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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