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  • Conditional expectation question

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

    Thanks in advance guys!

  • #2
    If you're going to ask on here, wait a bit before asking on AO (and vice-a-versa). This way you wouldn't use up twice as much effort to answer the same question

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    • #3
      First of all, you were successful in proving that you know the two sites very well. Second, it was effortless asking the same question on both sites. : )
      I love this site because I am familiar with the users here but the other site seems to have more members so i get answers to my questions faster.

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      • #4
        Originally posted by ronaldy27 View Post
        First of all, you were successful in proving that you know the two sites very well. Second, it was effortless asking the same question on both sites. : )
        I love this site because I am familiar with the users here but the other site seems to have more members so i get answers to my questions faster.
        The effort I was referring to wasn't on your part, it's on the part of people who "waste" time answering the same question. This is simple posting etiquette.

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        • #5
          I think that because X and Y are independent geometric (or negative binomial with r=1, whichever you prefer) distributions, that:

          P(X|Y=2) = P(X,Y=2) / P(Y=2) = P(X) P(Y=2) / P(Y=2) = P(X).

          Then, E(X|Y=2) = Sum(X*P(X|Y=2)) = Sum(X*P(X)) = E(X) = q / p = [(5/6) / (1/6)] = 5.

          I don't know if this is correct but that's what I got. I might be misunderstanding something. Just let me know if it's right or not, please.

          Thank you.

          **********(Nevermind the above post I just realized the above was one die being rolled and these are NOT independent distributions)***********
          Last edited by bsd058; January 19 2012, 12:19 PM.

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          • #6
            I read the solution on the other forum. Seems to make sense.

            Essentially, because Y=2, we know that the first roll is not a 6, because it takes at least 2 rolls to roll a 6 according to the question.

            So, that means that since the first roll is not a 6, the first roll can only be a 1, 2, 3, 4, or 5. Then P(X=1|Y=2) = number of ways to roll a 5/total possibilities = 1/5 = .2.

            Since the P(X=1|Y=2) = .2, and we know that the second roll wasn't a 5 (because it was a 6, P(X>2|Y=2) = 1 - P(X=1|Y=2) = 1 - .2 = .8. This .8 is because there are only two possibilities for X. That is when X = 1 or X > 2.

            A part I don't understand is that the other site states that E(X) = 1/p, when, according to Probability for Risk Management, since X is geometric, E(X) = q/p.

            So I got, E(X|X>2) = (5/6) / (1/6) + 2 (the 2 is added because the experiment starts at X=3, and so there were 2 rolls before X=3) = 5 + 2 = 7. Then E(X|Y=2) = (.2)(1) + (.8)(7) = 5.8.

            The other site got (which I'm sure is correct because I know I'm missing something) E(X|X>2) = (1 / (1/6)) + 2 = 8, and so got E(X|Y>2) = (.2)(1) + (.8)(8) = 6.6.

            I'm waiting on a macro in excel to replicate E(X) empirically to see if and where there is a convergence.

            The SOA solutions sheet also shows 6.6. Can anyone explain if 1/p is the appropriate formula to use for E(X) for a geometric distribution or if q/p is? Also, please explain why 1/p is appropriate.

            Once the macro is done, I'll post what it concluded.

            Thank you.

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            • #7
              Read the question carefully... What is the expected number of rolls to obtain a 5? What is the expected number of rolls to obtain any number on a fair die? You may be thinking of the expected number of rolls BEFORE obtaining a 5.

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              • #8
                Excel concluded there was a convergence at 6, not 5. So, E(X), such that X is number of rolls it takes to roll a 5, does equal 6 which fits with 1/p, not q/p, according to the experiment.

                Thank you, Ross. I didn't catch that detail. That makes sense to me now. Also explains why the spreadsheet converges at 6.
                Last edited by bsd058; January 19 2012, 07:16 PM.

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                • #9
                  Originally posted by NoMoreExams View Post
                  The effort I was referring to wasn't on your part, it's on the part of people who "waste" time answering the same question. This is simple posting etiquette.
                  Hey buddy, not many spend their time looking through questions on both sites as much as you and I do, so you don't have to worry about them.

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                  • #10
                    Originally posted by bsd058 View Post
                    Excel concluded there was a convergence at 6, not 5. So, E(X), such that X is number of rolls it takes to roll a 5, does equal 6 which fits with 1/p, not q/p, according to the experiment.

                    Thank you, Ross. I didn't catch that detail. That makes sense to me now. Also explains why the spreadsheet converges at 6.
                    Yeah I got 6 but I definitely didn't clearly understand the question.

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                    • #11
                      Originally posted by ronaldy27 View Post
                      Hey buddy, not many spend their time looking through questions on both sites as much as you and I do, so you don't have to worry about them.
                      LOL I think you're still missing the point. If many were spending time looking at both websites, they wouldn't waste your time pointing out that a search feature exists or that for a geometric distribution the moments differ depending on how you define your random variable (hint: http://en.wikipedia.org/wiki/Geometric_distribution).

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                      • #12
                        I love how you mocked one person for using "u" instead of "you" and you used "LOL." I'm smelling no good hypocrite right now.
                        Last edited by Irish Blues; January 20 2012, 07:44 PM.

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                        • #13
                          Originally posted by ronaldy27 View Post
                          I love how you mocked one person for using "u" instead of "you" and you used "LOL." I'm smelling no good hypocrite right now.
                          Ronaldy27, don't worry about NoMoreExams. It's really not worth calling people names. Even if they are what you say they are. Best way to get him back is to ignore him. Let's just move on to the next thing. It's just a forum, afterall. Not real life. It's supposed to be fun.

                          Also, I don't think NoMoreExams meant any disrespect to you.

                          By the way, do you understand the solution from the other site, now?
                          Last edited by Irish Blues; January 20 2012, 07:44 PM.

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                          • #14
                            bsd058 yeah you're right I agree with everything you said. Yeah I understand the solution, but I'm still unsure. I'm going to just accept it for now and when I go back and review everything, I'm definitely going to revisit all my questions and try to understand them better.

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                            • #15
                              Hey, ronaldy27. On a side note, do you know how long it normally takes Actuarial Outpost to activate an account?

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