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
    5 hours ago










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










  • $begingroup$
    Computationally, the answer to the second part appears to be $fracn^2r^2$
    $endgroup$
    – Sean Lee
    4 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
    5 hours ago










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










  • $begingroup$
    Computationally, the answer to the second part appears to be $fracn^2r^2$
    $endgroup$
    – Sean Lee
    4 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 5 hours ago









Felix Marin

68.9k7110147




68.9k7110147










asked 5 hours ago









631631

585




585











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










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










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
















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










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










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















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




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












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




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












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




$begingroup$
Computationally, the answer to the second part appears to be $fracn^2r^2$
$endgroup$
– Sean Lee
4 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











$endgroup$








  • 1




    $begingroup$
    If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
    $endgroup$
    – Daniel Schepler
    25 mins ago











  • $begingroup$
    Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
    $endgroup$
    – Sean Lee
    11 mins ago



















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











      $endgroup$








      • 1




        $begingroup$
        If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
        $endgroup$
        – Daniel Schepler
        25 mins ago











      • $begingroup$
        Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
        $endgroup$
        – Sean Lee
        11 mins ago
















      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











      $endgroup$








      • 1




        $begingroup$
        If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
        $endgroup$
        – Daniel Schepler
        25 mins ago











      • $begingroup$
        Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
        $endgroup$
        – Sean Lee
        11 mins ago














      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 4 hours ago

























      answered 5 hours ago









      Sean LeeSean Lee

      801214




      801214







      • 1




        $begingroup$
        If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
        $endgroup$
        – Daniel Schepler
        25 mins ago











      • $begingroup$
        Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
        $endgroup$
        – Sean Lee
        11 mins ago













      • 1




        $begingroup$
        If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
        $endgroup$
        – Daniel Schepler
        25 mins ago











      • $begingroup$
        Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
        $endgroup$
        – Sean Lee
        11 mins ago








      1




      1




      $begingroup$
      If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
      $endgroup$
      – Daniel Schepler
      25 mins ago





      $begingroup$
      If indeed $E(X_i X_j) = E(X_i) E(X_j)$ for $i in j$ then that implies zero correlation. I would expect a bit of negative correlation. (And indeed, my preliminary calculation based on the decomposition from VHarisop's answer seems to result in $E(X_i X_j) = fracn(n-1)r^2$ for $i ne j$ and $E(X_i^2) = fracnr + fracn(n-1)r^2$.)
      $endgroup$
      – Daniel Schepler
      25 mins ago













      $begingroup$
      Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
      $endgroup$
      – Sean Lee
      11 mins ago





      $begingroup$
      Yeah, it seemed a little strange to me initially, but its consistent with your results btw: $frac1r[(r-1)E(X_iX_j) + E(X_i^2)] = fracn^2r^2$
      $endgroup$
      – Sean Lee
      11 mins ago












      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 5 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 5 hours ago









                  RScrlliRScrlli

                  761114




                  761114



























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