Finding Covariance and Expectation

Date: 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/