Let X be the number of 1's and Y the number of 2's that occur in n rolls of a fair die. Compute Cov(X,Y).

Do I use ?

A fair die is rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find E[X|Y=5].

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- April 21st 2009, 11:08 PMnoob mathematician[SOLVED] 2 questions
Let X be the number of 1's and Y the number of 2's that occur in n rolls of a fair die. Compute Cov(X,Y).

Do I use ?

A fair die is rolled. Let X and Y denote, respectively, the number of rolls necessary to obtain a 6 and a 5. Find E[X|Y=5]. - April 22nd 2009, 12:10 AMMoo
Hello,

Why not ?

You can compute E[X] and E[Y] I guess ^^ (same binomial distributions)

For the distribution , note that it's equivalent to saying "what is the probability to get l 2's, while there already had k 1's ?

You can also see that l can get values only between 0 and n-k.

And that it follows a binomial distribution (n-k,1/5) ~ Z

So we actually have :

But looks like a lot , doesn't it ? :D

Can you try to do this ? - April 22nd 2009, 08:12 AMnoob mathematician
- April 22nd 2009, 10:48 AMMoo
Yes, but the sum isn't exactly this !

There is the sum , which depends on k, so you can't just calculate the two sums separately.

I'm sorry I should've added brackets :

Quote:

However I'm stuck with , especially the (n-k) part

Do it with words : P(Y=l|X=k) is the probability that there are l 2's, while there are k 1's, among n numbers. So it means that there**already**have k 1's, leaving us with n-k 'undefined' numbers.

But, there can't be any 1's left in these n-k, since they're included in the k numbers (X=k is the total of 1's among the n numbers). So there are 5 remaining possibilities : {2,3,4,5,6}

Thus, P(Y=l|X=k) is equal to P(Z=l), where Z follows a binomial distribution (n-k,1/5)

Does it look better this way ? (Worried)

Now the sum :

Then you're left with :

But

And

Finally,

And you get the desired result ;)