Calculating the expectancy












0














The distance between two numbers will be set as $ | i - j | $ . I pick two numbers without replacement from $ 0,1,2, dots ,n $ let $ X $ be the distance between the numbers. What is the expectancy of $ X $ ?



there are $ {n+1 choose 2} $ options. so $ {n+1 choose 2} = frac{n(n+1)}{2} $ and the probability will be $$ P{X = i} = frac{2(n+1-i)}{n(n+1)}$$



Then the expectancy $$ E[X] = sum_{i=1}^n x_i cdot p(x_i) = sum_{i=1}^n x_i frac{2(n+1-i)}{n(n+1)} $$



I don't really know how to proceed from here?






probability probability-theory







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  • 1




    I don't know why you are switching to $x_i cdot p(x_i)$. Your probability $P{X = i}$ is correct and the expectation is $E[X] = sum_{i = 1}^n i cdot P{X = i}$
    – Lee David Chung Lin
    Dec 2 at 12:12










  • I need to somehow get to this answer : $ frac{n+2}{3} $
    – bm1125
    Dec 2 at 13:07










  • You have to set $x_i=i$ and then apply the usual techniques to evaluate the sum.
    – Ingix
    Dec 2 at 13:53
















0














The distance between two numbers will be set as $ | i - j | $ . I pick two numbers without replacement from $ 0,1,2, dots ,n $ let $ X $ be the distance between the numbers. What is the expectancy of $ X $ ?



there are $ {n+1 choose 2} $ options. so $ {n+1 choose 2} = frac{n(n+1)}{2} $ and the probability will be $$ P{X = i} = frac{2(n+1-i)}{n(n+1)}$$



Then the expectancy $$ E[X] = sum_{i=1}^n x_i cdot p(x_i) = sum_{i=1}^n x_i frac{2(n+1-i)}{n(n+1)} $$



I don't really know how to proceed from here?






probability probability-theory







share|cite|improve this question




















  • 1




    I don't know why you are switching to $x_i cdot p(x_i)$. Your probability $P{X = i}$ is correct and the expectation is $E[X] = sum_{i = 1}^n i cdot P{X = i}$
    – Lee David Chung Lin
    Dec 2 at 12:12










  • I need to somehow get to this answer : $ frac{n+2}{3} $
    – bm1125
    Dec 2 at 13:07










  • You have to set $x_i=i$ and then apply the usual techniques to evaluate the sum.
    – Ingix
    Dec 2 at 13:53














0












0








0


1





The distance between two numbers will be set as $ | i - j | $ . I pick two numbers without replacement from $ 0,1,2, dots ,n $ let $ X $ be the distance between the numbers. What is the expectancy of $ X $ ?



there are $ {n+1 choose 2} $ options. so $ {n+1 choose 2} = frac{n(n+1)}{2} $ and the probability will be $$ P{X = i} = frac{2(n+1-i)}{n(n+1)}$$



Then the expectancy $$ E[X] = sum_{i=1}^n x_i cdot p(x_i) = sum_{i=1}^n x_i frac{2(n+1-i)}{n(n+1)} $$



I don't really know how to proceed from here?






probability probability-theory







share|cite|improve this question















The distance between two numbers will be set as $ | i - j | $ . I pick two numbers without replacement from $ 0,1,2, dots ,n $ let $ X $ be the distance between the numbers. What is the expectancy of $ X $ ?



there are $ {n+1 choose 2} $ options. so $ {n+1 choose 2} = frac{n(n+1)}{2} $ and the probability will be $$ P{X = i} = frac{2(n+1-i)}{n(n+1)}$$



Then the expectancy $$ E[X] = sum_{i=1}^n x_i cdot p(x_i) = sum_{i=1}^n x_i frac{2(n+1-i)}{n(n+1)} $$



I don't really know how to proceed from here?






probability probability-theory




probability probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 2 at 12:30









Bernard

118k639112




118k639112










asked Dec 2 at 11:17









bm1125

60516




60516








  • 1




    I don't know why you are switching to $x_i cdot p(x_i)$. Your probability $P{X = i}$ is correct and the expectation is $E[X] = sum_{i = 1}^n i cdot P{X = i}$
    – Lee David Chung Lin
    Dec 2 at 12:12










  • I need to somehow get to this answer : $ frac{n+2}{3} $
    – bm1125
    Dec 2 at 13:07










  • You have to set $x_i=i$ and then apply the usual techniques to evaluate the sum.
    – Ingix
    Dec 2 at 13:53














  • 1




    I don't know why you are switching to $x_i cdot p(x_i)$. Your probability $P{X = i}$ is correct and the expectation is $E[X] = sum_{i = 1}^n i cdot P{X = i}$
    – Lee David Chung Lin
    Dec 2 at 12:12










  • I need to somehow get to this answer : $ frac{n+2}{3} $
    – bm1125
    Dec 2 at 13:07










  • You have to set $x_i=i$ and then apply the usual techniques to evaluate the sum.
    – Ingix
    Dec 2 at 13:53








1




1




I don't know why you are switching to $x_i cdot p(x_i)$. Your probability $P{X = i}$ is correct and the expectation is $E[X] = sum_{i = 1}^n i cdot P{X = i}$
– Lee David Chung Lin
Dec 2 at 12:12




I don't know why you are switching to $x_i cdot p(x_i)$. Your probability $P{X = i}$ is correct and the expectation is $E[X] = sum_{i = 1}^n i cdot P{X = i}$
– Lee David Chung Lin
Dec 2 at 12:12












I need to somehow get to this answer : $ frac{n+2}{3} $
– bm1125
Dec 2 at 13:07




I need to somehow get to this answer : $ frac{n+2}{3} $
– bm1125
Dec 2 at 13:07












You have to set $x_i=i$ and then apply the usual techniques to evaluate the sum.
– Ingix
Dec 2 at 13:53




You have to set $x_i=i$ and then apply the usual techniques to evaluate the sum.
– Ingix
Dec 2 at 13:53










1 Answer
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You are almost there.



$x_i = i$ in your formula for the expectation.



$$E[X]=sum_1^nfrac{2i(n+1-i)}{n(n+1)}$$



Then, use the identities:



$$sum_1^n i =frac{n(n+1)}{2}$$



and



$$sum_1^n i^2 =frac{n(n+1)(2n+1)}{6}$$



We get



$$E[X]=frac{2}{n(n+1)}(n+1)sum i -frac{2}{n(n+1)}sum i^2$$



$$E[X]=frac{2}{n}frac{n(n+1)}{2} - frac{2}{n(n+1)}frac{n(n+1)(2n+1)}{6}$$



Therefore



$$E[X]=n+1-frac{2n+1}{3}=frac{n+2}{3}$$






share|cite|improve this answer





















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    You are almost there.



    $x_i = i$ in your formula for the expectation.



    $$E[X]=sum_1^nfrac{2i(n+1-i)}{n(n+1)}$$



    Then, use the identities:



    $$sum_1^n i =frac{n(n+1)}{2}$$



    and



    $$sum_1^n i^2 =frac{n(n+1)(2n+1)}{6}$$



    We get



    $$E[X]=frac{2}{n(n+1)}(n+1)sum i -frac{2}{n(n+1)}sum i^2$$



    $$E[X]=frac{2}{n}frac{n(n+1)}{2} - frac{2}{n(n+1)}frac{n(n+1)(2n+1)}{6}$$



    Therefore



    $$E[X]=n+1-frac{2n+1}{3}=frac{n+2}{3}$$






    share|cite|improve this answer


























      1














      You are almost there.



      $x_i = i$ in your formula for the expectation.



      $$E[X]=sum_1^nfrac{2i(n+1-i)}{n(n+1)}$$



      Then, use the identities:



      $$sum_1^n i =frac{n(n+1)}{2}$$



      and



      $$sum_1^n i^2 =frac{n(n+1)(2n+1)}{6}$$



      We get



      $$E[X]=frac{2}{n(n+1)}(n+1)sum i -frac{2}{n(n+1)}sum i^2$$



      $$E[X]=frac{2}{n}frac{n(n+1)}{2} - frac{2}{n(n+1)}frac{n(n+1)(2n+1)}{6}$$



      Therefore



      $$E[X]=n+1-frac{2n+1}{3}=frac{n+2}{3}$$






      share|cite|improve this answer
























        1












        1








        1






        You are almost there.



        $x_i = i$ in your formula for the expectation.



        $$E[X]=sum_1^nfrac{2i(n+1-i)}{n(n+1)}$$



        Then, use the identities:



        $$sum_1^n i =frac{n(n+1)}{2}$$



        and



        $$sum_1^n i^2 =frac{n(n+1)(2n+1)}{6}$$



        We get



        $$E[X]=frac{2}{n(n+1)}(n+1)sum i -frac{2}{n(n+1)}sum i^2$$



        $$E[X]=frac{2}{n}frac{n(n+1)}{2} - frac{2}{n(n+1)}frac{n(n+1)(2n+1)}{6}$$



        Therefore



        $$E[X]=n+1-frac{2n+1}{3}=frac{n+2}{3}$$






        share|cite|improve this answer












        You are almost there.



        $x_i = i$ in your formula for the expectation.



        $$E[X]=sum_1^nfrac{2i(n+1-i)}{n(n+1)}$$



        Then, use the identities:



        $$sum_1^n i =frac{n(n+1)}{2}$$



        and



        $$sum_1^n i^2 =frac{n(n+1)(2n+1)}{6}$$



        We get



        $$E[X]=frac{2}{n(n+1)}(n+1)sum i -frac{2}{n(n+1)}sum i^2$$



        $$E[X]=frac{2}{n}frac{n(n+1)}{2} - frac{2}{n(n+1)}frac{n(n+1)(2n+1)}{6}$$



        Therefore



        $$E[X]=n+1-frac{2n+1}{3}=frac{n+2}{3}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 at 13:58









        ip6

        54839




        54839






























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