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### 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?

```

```
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/
```
Associated Topics:
College Statistics

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