View Full Version : Geometric distribution question

desi2000

December 25th 2009, 05:18 PM

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?

NoMoreExams

December 25th 2009, 06:48 PM

Look at the link in the link http://www.actuarialoutpost.com/actuarial_discussion_forum/showthread.php?t=181444 then ask if you are still confused

mathrix

December 25th 2009, 07:44 PM

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. :)

R_T

December 26th 2009, 12:16 AM

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

desi2000

December 27th 2009, 04:26 PM

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

R_T

December 27th 2009, 07:23 PM

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

You're very welcome. Good Luck with The Exam.

reader

May 23rd 2010, 12:06 PM

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 !

krzysio

May 23rd 2010, 08:12 PM

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.

reader

May 23rd 2010, 08:35 PM

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

krzysio

May 23rd 2010, 11:12 PM

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

sohpmalvin

May 24th 2010, 01:16 PM

Law of total probability: P(A) = P(AB) + P(AC) + P(AD) + ...

A very critical law to know, for exam P/1.

reader

May 25th 2010, 04:07 PM

You guys are funny ! Of course I know that law . But it doesn't explain much for this problem because we have E[X|Y=2] and E[X|X=1] and E[X|X>=1] instead of A .

This problem is a lot more involved than that . It seems to me you guys haven't real thought about why this problem works out the way it does and how the double expectation formula is involved .

Forget about it anyway . I asked an expert about this and he explained it to me in detail .

somiwomi

June 1st 2010, 11:33 PM

You guys are funny ! Of course I know that law . But it doesn't explain much for this problem because we have E[X|Y=2] and E[X|X=1] and E[X|X>=1] instead of A .

This problem is a lot more involved than that . It seems to me you guys haven't real thought about why this problem works out the way it does and how the double expectation formula is involved .

Forget about it anyway . I asked an expert about this and he explained it to me in detail .

can u please share the details? while i understand the question a little more after reading the above responses, i'd like to understand it better.

somiwomi

June 2nd 2010, 01:02 AM

how do get 48 from 3+4(5/6) +5((5/6)^2) +.............? what is the formula to get the summation?

NoMoreExams

June 2nd 2010, 01:42 AM

how do get 48 from 3+4(5/6) +5((5/6)^2) +.............? what is the formula to get the summation?

Probably be more useful to start your own thread since I don't see how this relates to what everyone is asking...

\sum_{n=0}^{\infty} (3+n)*(5/6)^n = \sum_{n=0}^{\infty} 3*(5/6)^n + n*(5/6)^n

\sum_{n=0}^{\infty} 3*(5/6)^n = 3/(1-5/6) = 18

\sum_{n=0}^{\infty} n*(5/6)^n = \sum_{n=1}^{\infty} n*(5/6)^n = 1/(1-5/6)^2 * (5/6) = 36 * 5/6 = 30

30 + 18 = 48

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