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1. ## Geometric distribution question

The question is:

A fair die is rolled repeatedly. Let X be the number of rolls needed to obtain a 5 and Y the number of rolls needed to obtain a 6. Calculate E(X given Y=2).

So I thought since the number of rolls needed to obtain a 5 is independent from the number of rolls needed to obtain a 6. And since p=1/6 I thought maybe the answer is just 1/(1/6) which the expected value of a geometric distribution. But that's not correct.The actual answer is 6.6. Can someone please tell me what I am doing wrong?

3. Originally Posted by desi2000
The question is:

A fair die is rolled repeatedly. Let X be the number of rolls needed to obtain a 5 and Y the number of rolls needed to obtain a 6. Calculate E(X given Y=2).

So I thought since the number of rolls needed to obtain a 5 is independent from the number of rolls needed to obtain a 6. And since p=1/6 I thought maybe the answer is just 1/(1/6) which the expected value of a geometric distribution. But that's not correct.The actual answer is 6.6. Can someone please tell me what I am doing wrong?
Yeah I think this was a confusion from other posters before. You could probably search here or in AO.

4. ## Alternate Solution

Originally Posted by desi2000
The question is:

A fair die is rolled repeatedly. Let X be the number of rolls needed to obtain a 5 and Y the number of rolls needed to obtain a 6. Calculate E(X given Y=2).

So I thought since the number of rolls needed to obtain a 5 is independent from the number of rolls needed to obtain a 6. And since p=1/6 I thought maybe the answer is just 1/(1/6) which the expected value of a geometric distribution. But that's not correct.The actual answer is 6.6. Can someone please tell me what I am doing wrong?
You are missing the fact that once you are told that the second toss resulted
in a 6, the expected number of tosses required to obtain your first 5 changes.

What makes you think X and Y are independent? If they were,
Pr[X=2 and Y=2] would = P[X=2]P[Y =2].
The former probability is zero; the latter is ((5/36)^2);

What follows is a solution derived from first principles; an alternate solution appears below this one.

Let Px(X) be the conditional distribution of X given Y=2
Px(1) = 5 on first roll; 6 on second =((1/6)^2)
Px(2) = 5 on second roll(impossible since Y=2) =0
Px(3) = no 5 or 6 on the first roll; 6 on the second roll; 5 on third roll
=(2/3)((1/6)^2)
Px(4) = no 5 or 6 on first roll; 6 on second roll; no 5 on third roll; 5 on the fourth roll
=(2/3)((1/6)^2)(5/6)

E[X given Y=2]
=1((1/6)^2) + 2(0) +3[(2/3)(1/6)^2] + 4[(2/3)((1/6)^2)(5/6)] +5[(2/3)((1/6)^2)((5/6)^2)] + ........../P(Y=2)

=((1/6)^2)[1 + 3(2/3) +4(2/3)(5/6) + 5(2/3)((5/6)^2) + ........../P(Y=2)
=((1/6)^2)[1+ (2/3)[3+4(5/6) +5((5/6)^2) +............../P(Y=2)
=((1/6)^2[1+ (2/3)[48]] / (5/36)
=(33/36)/(5/36) = 33/5 = 6.6

For the alternate solution, we partition the conditional expectations into two mutualy exclusive parts. The first one imposes the restriction that X<2 and Y
=2. Of course,the only way this can happen is if X=1, in which case, our conditional expected value is one, since X represents the number of rolls needed fo the first 5: P(X=1 and Y=2)/P(Y=2) = 1/5 = .2. Now the second partition considers the case where E[(X) given X > 2and Y =2)]. Because of the memoryless property of the geometric distribution, E[X] given X > 2 has the same ditribution as E[X + 2]=
E[X] +2 = 6 + 2 = 8. So this is our conditional Expectatin of X given X>2 and Y=1. The probobility for this conditional expectation is just the complement of the first probability we calculated . Now remember that the first conditional expectation is 1 and we multiply that by the associated probability. We multiply 1(.2)
to obtain our second part, we multiply 8(.8) and then add the two products: 1(.2) + 8(.8) = 6.6.

Good Luck To You

5. Thank you so much R_T!!! your solution makes perfect sense!!! you are wonderful!!! :laugh:

6. Originally Posted by desi2000
Thank you so much R_T!!! your solution makes perfect sense!!! you are wonderful!!! :laugh:
You're very welcome. Good Luck with The Exam.

7. Why does weighting the conditional expectations on the conditional probabilities work ?

I.E.
Why is it that
E(X|Y = 2) = E[X|X=1]*P(X=1|Y=2) + E[X|X>=3]*P(X>=3|Y=2)
?

How is this justified mathematically ( or otherwise ) ? From what Theorem or formula can this be derived ? For me this is a big hole in the solution to this one .

I have searched the forums and nobody has addressed this issue . Is it because the answer is obvious ? I don't think so !

Why does weighting the conditional expectations on the conditional probabilities work ?

I.E.
Why is it that
E(X|Y = 2) = E[X|X=1]*P(X=1|Y=2) + E[X|X>=3]*P(X>=3|Y=2)
?

How is this justified mathematically ( or otherwise ) ? From what Theorem or formula can this be derived ? For me this is a big hole in the solution to this one .

I have searched the forums and nobody has addressed this issue . Is it because the answer is obvious ? I don't think so !
The Law of Total Probability.

9. In what way ? I don't see the connection .

In what way ? I don't see the connection .
Do you know the Law of Total Probability? This is something you are expected to know for the test, and it is at the very beginning of the material on the test. Here is the Wikipedia reference:
http://en.wikipedia.org/wiki/Law_of_total_probability