Computing the expectation of the number of balls in a box The 2019 Stack Overflow Developer Survey Results Are InThere is two boxes with one with 8 balls and one with 4 ballsdrawing balls from box without replacemntRandom distribution of colored balls into boxes.Optimal Number of White BallsCompute possible outcomes when get balls from a boxPoisson Approximation Problem involving putting balls into boxesCompute expected received balls from boxesput n balls into n boxesA question of probability regarding expectation and variance of a random variable.Distributing 5 distinct balls into 3 distinct boxes

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Computing the expectation of the number of balls in a box



The 2019 Stack Overflow Developer Survey Results Are InThere is two boxes with one with 8 balls and one with 4 ballsdrawing balls from box without replacemntRandom distribution of colored balls into boxes.Optimal Number of White BallsCompute possible outcomes when get balls from a boxPoisson Approximation Problem involving putting balls into boxesCompute expected received balls from boxesput n balls into n boxesA question of probability regarding expectation and variance of a random variable.Distributing 5 distinct balls into 3 distinct boxes










5












$begingroup$


  • There are $r$ boxes and $n$ balls.

  • Each ball is placed in a box with equal probability, independently of the other balls.

  • Let $X_i$ be the number of balls in box $i$,
    $1 leq i leq r$.

  • Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.

I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Are there any restrictions on $j$?
    $endgroup$
    – Sean Lee
    4 hours ago










  • $begingroup$
    @SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
    $endgroup$
    – 631
    4 hours ago










  • $begingroup$
    Computationally, the answer to the second part appears to be $fracn^2r^2$
    $endgroup$
    – Sean Lee
    3 hours ago















5












$begingroup$


  • There are $r$ boxes and $n$ balls.

  • Each ball is placed in a box with equal probability, independently of the other balls.

  • Let $X_i$ be the number of balls in box $i$,
    $1 leq i leq r$.

  • Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.

I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Are there any restrictions on $j$?
    $endgroup$
    – Sean Lee
    4 hours ago










  • $begingroup$
    @SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
    $endgroup$
    – 631
    4 hours ago










  • $begingroup$
    Computationally, the answer to the second part appears to be $fracn^2r^2$
    $endgroup$
    – Sean Lee
    3 hours ago













5












5








5





$begingroup$


  • There are $r$ boxes and $n$ balls.

  • Each ball is placed in a box with equal probability, independently of the other balls.

  • Let $X_i$ be the number of balls in box $i$,
    $1 leq i leq r$.

  • Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.

I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.










share|cite|improve this question











$endgroup$




  • There are $r$ boxes and $n$ balls.

  • Each ball is placed in a box with equal probability, independently of the other balls.

  • Let $X_i$ be the number of balls in box $i$,
    $1 leq i leq r$.

  • Compute $mathbbEleft[X_iright], mathbbEleft[X_iX_jright]$.

I am preparing for an exam, and I have no idea how to approach this problem. Can someone push me in the right direction ?.







probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Felix Marin

68.9k7110147




68.9k7110147










asked 4 hours ago









631631

565




565











  • $begingroup$
    Are there any restrictions on $j$?
    $endgroup$
    – Sean Lee
    4 hours ago










  • $begingroup$
    @SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
    $endgroup$
    – 631
    4 hours ago










  • $begingroup$
    Computationally, the answer to the second part appears to be $fracn^2r^2$
    $endgroup$
    – Sean Lee
    3 hours ago
















  • $begingroup$
    Are there any restrictions on $j$?
    $endgroup$
    – Sean Lee
    4 hours ago










  • $begingroup$
    @SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
    $endgroup$
    – 631
    4 hours ago










  • $begingroup$
    Computationally, the answer to the second part appears to be $fracn^2r^2$
    $endgroup$
    – Sean Lee
    3 hours ago















$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
4 hours ago




$begingroup$
Are there any restrictions on $j$?
$endgroup$
– Sean Lee
4 hours ago












$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
4 hours ago




$begingroup$
@SeanLee In the question, no. I'm guessing it would have the same restrictions as i.
$endgroup$
– 631
4 hours ago












$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
3 hours ago




$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
3 hours ago










3 Answers
3






active

oldest

votes


















2












$begingroup$

Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:



$$ mathbbE[X_i] = fracnr $$



Now, we would like to know what is $mathbbE[X_i X_j] $.



We begin by making the following observation:



$$X_i = n - sum_jneq iX_j $$



Which gives us:



$$ X_isum_jneq iX_j = nX_i - X_i^2$$



Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:



beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
&= frac1r mathbbE[nX_i] \
&= fracn^2r^2
endalign






share|cite|improve this answer











$endgroup$




















    3












    $begingroup$

    For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
    Specifically, you know that for a fixed box, the probability of putting a ball in it
    is $frac1r$. Let



    $$
    Y_k^(i) = begincases
    1 &, text if ball $k$ was placed in box $i$ \
    0 &, text otherwise
    endcases,
    $$

    which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
    Then you can write



    $$
    X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
    $$






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      Think of placing the ball in box "$i$" as success and not placing it as a failure.



      This situation can be represented using the Hypergeometric Distribution.
      $$
      P(X=k) = fracK choose k N- Kchoose n - kN choose n.
      $$



      $N$ is the population size (number of boxes $r$)



      $K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)



      $n$ is the number of draws (the number of balls $n$).



      $k$ is the number of observed successes (the number of balls in box "$i$").



      The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
      $$E[X_i]=nfrac1r=fracnr$$






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:



        $$ mathbbE[X_i] = fracnr $$



        Now, we would like to know what is $mathbbE[X_i X_j] $.



        We begin by making the following observation:



        $$X_i = n - sum_jneq iX_j $$



        Which gives us:



        $$ X_isum_jneq iX_j = nX_i - X_i^2$$



        Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:



        beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
        &= frac1r mathbbE[nX_i] \
        &= fracn^2r^2
        endalign






        share|cite|improve this answer











        $endgroup$

















          2












          $begingroup$

          Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:



          $$ mathbbE[X_i] = fracnr $$



          Now, we would like to know what is $mathbbE[X_i X_j] $.



          We begin by making the following observation:



          $$X_i = n - sum_jneq iX_j $$



          Which gives us:



          $$ X_isum_jneq iX_j = nX_i - X_i^2$$



          Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:



          beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
          &= frac1r mathbbE[nX_i] \
          &= fracn^2r^2
          endalign






          share|cite|improve this answer











          $endgroup$















            2












            2








            2





            $begingroup$

            Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:



            $$ mathbbE[X_i] = fracnr $$



            Now, we would like to know what is $mathbbE[X_i X_j] $.



            We begin by making the following observation:



            $$X_i = n - sum_jneq iX_j $$



            Which gives us:



            $$ X_isum_jneq iX_j = nX_i - X_i^2$$



            Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:



            beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
            &= frac1r mathbbE[nX_i] \
            &= fracn^2r^2
            endalign






            share|cite|improve this answer











            $endgroup$



            Since there are $r$ boxes and $n$ balls, and each ball is placed in a box with equal probability, we have:



            $$ mathbbE[X_i] = fracnr $$



            Now, we would like to know what is $mathbbE[X_i X_j] $.



            We begin by making the following observation:



            $$X_i = n - sum_jneq iX_j $$



            Which gives us:



            $$ X_isum_jneq iX_j = nX_i - X_i^2$$



            Now, fix $i$ (we can do this because of the symmetry in the question), and thus we have:



            beginalignmathbbE[X_i X_j] &= frac1rBig(mathbbE[X_i sum_jneq i X_j] + mathbbE[X_i^2]Big) \
            &= frac1r mathbbE[nX_i] \
            &= fracn^2r^2
            endalign







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 3 hours ago

























            answered 3 hours ago









            Sean LeeSean Lee

            801214




            801214





















                3












                $begingroup$

                For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
                Specifically, you know that for a fixed box, the probability of putting a ball in it
                is $frac1r$. Let



                $$
                Y_k^(i) = begincases
                1 &, text if ball $k$ was placed in box $i$ \
                0 &, text otherwise
                endcases,
                $$

                which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
                Then you can write



                $$
                X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
                $$






                share|cite|improve this answer









                $endgroup$

















                  3












                  $begingroup$

                  For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
                  Specifically, you know that for a fixed box, the probability of putting a ball in it
                  is $frac1r$. Let



                  $$
                  Y_k^(i) = begincases
                  1 &, text if ball $k$ was placed in box $i$ \
                  0 &, text otherwise
                  endcases,
                  $$

                  which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
                  Then you can write



                  $$
                  X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
                  $$






                  share|cite|improve this answer









                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
                    Specifically, you know that for a fixed box, the probability of putting a ball in it
                    is $frac1r$. Let



                    $$
                    Y_k^(i) = begincases
                    1 &, text if ball $k$ was placed in box $i$ \
                    0 &, text otherwise
                    endcases,
                    $$

                    which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
                    Then you can write



                    $$
                    X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    For the first part, you can use linearity of expectation to compute $mathbbE[X_i]$.
                    Specifically, you know that for a fixed box, the probability of putting a ball in it
                    is $frac1r$. Let



                    $$
                    Y_k^(i) = begincases
                    1 &, text if ball $k$ was placed in box $i$ \
                    0 &, text otherwise
                    endcases,
                    $$

                    which satisfies $mathbbE[Y_k^(i)] = mathbbP(Y_k^(i) = 1) = frac1r.$
                    Then you can write



                    $$
                    X_i = sum_j=1^n Y_j^(i) Rightarrow mathbbEX_i = sum_j=1^n frac1r = fracnr.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    VHarisopVHarisop

                    1,218421




                    1,218421





















                        0












                        $begingroup$

                        Think of placing the ball in box "$i$" as success and not placing it as a failure.



                        This situation can be represented using the Hypergeometric Distribution.
                        $$
                        P(X=k) = fracK choose k N- Kchoose n - kN choose n.
                        $$



                        $N$ is the population size (number of boxes $r$)



                        $K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)



                        $n$ is the number of draws (the number of balls $n$).



                        $k$ is the number of observed successes (the number of balls in box "$i$").



                        The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
                        $$E[X_i]=nfrac1r=fracnr$$






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Think of placing the ball in box "$i$" as success and not placing it as a failure.



                          This situation can be represented using the Hypergeometric Distribution.
                          $$
                          P(X=k) = fracK choose k N- Kchoose n - kN choose n.
                          $$



                          $N$ is the population size (number of boxes $r$)



                          $K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)



                          $n$ is the number of draws (the number of balls $n$).



                          $k$ is the number of observed successes (the number of balls in box "$i$").



                          The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
                          $$E[X_i]=nfrac1r=fracnr$$






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            Think of placing the ball in box "$i$" as success and not placing it as a failure.



                            This situation can be represented using the Hypergeometric Distribution.
                            $$
                            P(X=k) = fracK choose k N- Kchoose n - kN choose n.
                            $$



                            $N$ is the population size (number of boxes $r$)



                            $K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)



                            $n$ is the number of draws (the number of balls $n$).



                            $k$ is the number of observed successes (the number of balls in box "$i$").



                            The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
                            $$E[X_i]=nfrac1r=fracnr$$






                            share|cite|improve this answer









                            $endgroup$



                            Think of placing the ball in box "$i$" as success and not placing it as a failure.



                            This situation can be represented using the Hypergeometric Distribution.
                            $$
                            P(X=k) = fracK choose k N- Kchoose n - kN choose n.
                            $$



                            $N$ is the population size (number of boxes $r$)



                            $K$ is the number of success states in the population (just $1$ because the success is defined as placing the ball in box "$i$".)



                            $n$ is the number of draws (the number of balls $n$).



                            $k$ is the number of observed successes (the number of balls in box "$i$").



                            The expectation of the Hypergeometric Distribution is $nfracKN$, hence the mean of your variable
                            $$E[X_i]=nfrac1r=fracnr$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 3 hours ago









                            RScrlliRScrlli

                            761114




                            761114



























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                                Тонконіг бульбистий Зміст Опис | Поширення | Екологія | Господарське значення | Примітки | Див. також | Література | Джерела | Посилання | Навігаційне меню1114601320038-241116202404kew-435458Poa bulbosaЭлектронный каталог сосудистых растений Азиатской России [Електронний каталог судинних рослин Азіатської Росії]Малышев Л. Л. Дикие родичи культурных растений. Poa bulbosa L. - Мятлик луковичный. [Малишев Л. Л. Дикі родичи культурних рослин. Poa bulbosa L. - Тонконіг бульбистий.]Мятлик (POA) Сем. Злаки (Мятликовые) [Тонконіг (POA) Род. Злаки (Тонконогові)]Poa bulbosa Linnaeus, Sp. Pl. 1: 70. 1753. 鳞茎早熟禾 lin jing zao shu he (Description from Flora of China) [Poa bulbosa Linnaeus, Sp. Pl. 1: 70. 1753. 鳞茎早熟禾 lin jing zao shu he (Опис від Флора Китаю)]Poa bulbosa L. – lipnice cibulkatá / lipnica cibulkatáPoa bulbosa в базі даних Poa bulbosa на сайті Poa bulbosa в базі даних «Global Biodiversity Information Facility» (GBIF)Poa bulbosa в базі даних «Euro + Med PlantBase» — інформаційному ресурсі для Євро-середземноморського розмаїття рослинPoa bulbosa L. на сайті «Плантариум»