Computing the expectation of the number of balls in a box The 2019 Stack Overflow Developer...

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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












4












$begingroup$



  • 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 $mathbb{E}left[X_{i}right], mathbb{E}left[X_{i}X_{j}right]$.


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











$endgroup$












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










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










  • $begingroup$
    Computationally, the answer to the second part appears to be $frac{n^2}{r^2}$
    $endgroup$
    – Sean Lee
    2 hours ago
















4












$begingroup$



  • 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 $mathbb{E}left[X_{i}right], mathbb{E}left[X_{i}X_{j}right]$.


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











$endgroup$












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










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










  • $begingroup$
    Computationally, the answer to the second part appears to be $frac{n^2}{r^2}$
    $endgroup$
    – Sean Lee
    2 hours ago














4












4








4





$begingroup$



  • 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 $mathbb{E}left[X_{i}right], mathbb{E}left[X_{i}X_{j}right]$.


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











$endgroup$





  • 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 $mathbb{E}left[X_{i}right], mathbb{E}left[X_{i}X_{j}right]$.


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




probability-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Felix Marin

68.9k7110147




68.9k7110147










asked 3 hours ago









631631

465




465












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










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










  • $begingroup$
    Computationally, the answer to the second part appears to be $frac{n^2}{r^2}$
    $endgroup$
    – Sean Lee
    2 hours ago


















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










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










  • $begingroup$
    Computationally, the answer to the second part appears to be $frac{n^2}{r^2}$
    $endgroup$
    – Sean Lee
    2 hours ago
















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




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












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




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












$begingroup$
Computationally, the answer to the second part appears to be $frac{n^2}{r^2}$
$endgroup$
– Sean Lee
2 hours ago




$begingroup$
Computationally, the answer to the second part appears to be $frac{n^2}{r^2}$
$endgroup$
– Sean Lee
2 hours ago










3 Answers
3






active

oldest

votes


















1












$begingroup$

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



$$ mathbb{E}[X_i] = frac{n}{r} $$



Now, we would like to know what is $mathbb{E}[X_i X_j] $.



We begin by making the following observation:



$$X_i = n - sum_{jneq i}X_j $$



Which gives us:



$$ X_isum_{jneq i}X_j = nX_i - X_i^2$$



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



begin{align}mathbb{E}[X_i X_j] &= frac{1}{r}Big(mathbb{E}[X_i sum_{jneq i} X_j] + mathbb{E}[X_i^2]Big) \
&= frac{1}{r} mathbb{E}[nX_i] \
&= frac{n^2}{r^2}
end{align}






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    For the first part, you can use linearity of expectation to compute $mathbb{E}[X_i]$.
    Specifically, you know that for a fixed box, the probability of putting a ball in it
    is $frac{1}{r}$. Let



    $$
    Y_k^{(i)} = begin{cases}
    1 &, text{ if ball $k$ was placed in box $i$} \
    0 &, text{ otherwise}
    end{cases},
    $$
    which satisfies $mathbb{E}[Y_k^{(i)}] = mathbb{P}(Y_k^{(i)} = 1) = frac{1}{r}.$
    Then you can write



    $$
    X_i = sum_{j=1}^n Y_j^{(i)} Rightarrow mathbb{E}X_i = sum_{j=1}^n frac{1}{r} = frac{n}{r}.
    $$






    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.
      $$
      P(X=k) = frac{{K choose k} {N- Kchoose n - k}}{{N choose n}}.
      $$



      $N$ 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 $nfrac{K}N$, hence the mean of your variable
      $$E[X_i]=nfrac{1}{r}=frac{n}{r}$$






      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









        1












        $begingroup$

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



        $$ mathbb{E}[X_i] = frac{n}{r} $$



        Now, we would like to know what is $mathbb{E}[X_i X_j] $.



        We begin by making the following observation:



        $$X_i = n - sum_{jneq i}X_j $$



        Which gives us:



        $$ X_isum_{jneq i}X_j = nX_i - X_i^2$$



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



        begin{align}mathbb{E}[X_i X_j] &= frac{1}{r}Big(mathbb{E}[X_i sum_{jneq i} X_j] + mathbb{E}[X_i^2]Big) \
        &= frac{1}{r} mathbb{E}[nX_i] \
        &= frac{n^2}{r^2}
        end{align}






        share|cite|improve this answer











        $endgroup$


















          1












          $begingroup$

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



          $$ mathbb{E}[X_i] = frac{n}{r} $$



          Now, we would like to know what is $mathbb{E}[X_i X_j] $.



          We begin by making the following observation:



          $$X_i = n - sum_{jneq i}X_j $$



          Which gives us:



          $$ X_isum_{jneq i}X_j = nX_i - X_i^2$$



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



          begin{align}mathbb{E}[X_i X_j] &= frac{1}{r}Big(mathbb{E}[X_i sum_{jneq i} X_j] + mathbb{E}[X_i^2]Big) \
          &= frac{1}{r} mathbb{E}[nX_i] \
          &= frac{n^2}{r^2}
          end{align}






          share|cite|improve this answer











          $endgroup$
















            1












            1








            1





            $begingroup$

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



            $$ mathbb{E}[X_i] = frac{n}{r} $$



            Now, we would like to know what is $mathbb{E}[X_i X_j] $.



            We begin by making the following observation:



            $$X_i = n - sum_{jneq i}X_j $$



            Which gives us:



            $$ X_isum_{jneq i}X_j = nX_i - X_i^2$$



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



            begin{align}mathbb{E}[X_i X_j] &= frac{1}{r}Big(mathbb{E}[X_i sum_{jneq i} X_j] + mathbb{E}[X_i^2]Big) \
            &= frac{1}{r} mathbb{E}[nX_i] \
            &= frac{n^2}{r^2}
            end{align}






            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:



            $$ mathbb{E}[X_i] = frac{n}{r} $$



            Now, we would like to know what is $mathbb{E}[X_i X_j] $.



            We begin by making the following observation:



            $$X_i = n - sum_{jneq i}X_j $$



            Which gives us:



            $$ X_isum_{jneq i}X_j = nX_i - X_i^2$$



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



            begin{align}mathbb{E}[X_i X_j] &= frac{1}{r}Big(mathbb{E}[X_i sum_{jneq i} X_j] + mathbb{E}[X_i^2]Big) \
            &= frac{1}{r} mathbb{E}[nX_i] \
            &= frac{n^2}{r^2}
            end{align}







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours ago









            Sean LeeSean Lee

            791214




            791214























                3












                $begingroup$

                For the first part, you can use linearity of expectation to compute $mathbb{E}[X_i]$.
                Specifically, you know that for a fixed box, the probability of putting a ball in it
                is $frac{1}{r}$. Let



                $$
                Y_k^{(i)} = begin{cases}
                1 &, text{ if ball $k$ was placed in box $i$} \
                0 &, text{ otherwise}
                end{cases},
                $$
                which satisfies $mathbb{E}[Y_k^{(i)}] = mathbb{P}(Y_k^{(i)} = 1) = frac{1}{r}.$
                Then you can write



                $$
                X_i = sum_{j=1}^n Y_j^{(i)} Rightarrow mathbb{E}X_i = sum_{j=1}^n frac{1}{r} = frac{n}{r}.
                $$






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  For the first part, you can use linearity of expectation to compute $mathbb{E}[X_i]$.
                  Specifically, you know that for a fixed box, the probability of putting a ball in it
                  is $frac{1}{r}$. Let



                  $$
                  Y_k^{(i)} = begin{cases}
                  1 &, text{ if ball $k$ was placed in box $i$} \
                  0 &, text{ otherwise}
                  end{cases},
                  $$
                  which satisfies $mathbb{E}[Y_k^{(i)}] = mathbb{P}(Y_k^{(i)} = 1) = frac{1}{r}.$
                  Then you can write



                  $$
                  X_i = sum_{j=1}^n Y_j^{(i)} Rightarrow mathbb{E}X_i = sum_{j=1}^n frac{1}{r} = frac{n}{r}.
                  $$






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    For the first part, you can use linearity of expectation to compute $mathbb{E}[X_i]$.
                    Specifically, you know that for a fixed box, the probability of putting a ball in it
                    is $frac{1}{r}$. Let



                    $$
                    Y_k^{(i)} = begin{cases}
                    1 &, text{ if ball $k$ was placed in box $i$} \
                    0 &, text{ otherwise}
                    end{cases},
                    $$
                    which satisfies $mathbb{E}[Y_k^{(i)}] = mathbb{P}(Y_k^{(i)} = 1) = frac{1}{r}.$
                    Then you can write



                    $$
                    X_i = sum_{j=1}^n Y_j^{(i)} Rightarrow mathbb{E}X_i = sum_{j=1}^n frac{1}{r} = frac{n}{r}.
                    $$






                    share|cite|improve this answer









                    $endgroup$



                    For the first part, you can use linearity of expectation to compute $mathbb{E}[X_i]$.
                    Specifically, you know that for a fixed box, the probability of putting a ball in it
                    is $frac{1}{r}$. Let



                    $$
                    Y_k^{(i)} = begin{cases}
                    1 &, text{ if ball $k$ was placed in box $i$} \
                    0 &, text{ otherwise}
                    end{cases},
                    $$
                    which satisfies $mathbb{E}[Y_k^{(i)}] = mathbb{P}(Y_k^{(i)} = 1) = frac{1}{r}.$
                    Then you can write



                    $$
                    X_i = sum_{j=1}^n Y_j^{(i)} Rightarrow mathbb{E}X_i = sum_{j=1}^n frac{1}{r} = frac{n}{r}.
                    $$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 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.
                        $$
                        P(X=k) = frac{{K choose k} {N- Kchoose n - k}}{{N choose n}}.
                        $$



                        $N$ 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 $nfrac{K}N$, hence the mean of your variable
                        $$E[X_i]=nfrac{1}{r}=frac{n}{r}$$






                        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.
                          $$
                          P(X=k) = frac{{K choose k} {N- Kchoose n - k}}{{N choose n}}.
                          $$



                          $N$ 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 $nfrac{K}N$, hence the mean of your variable
                          $$E[X_i]=nfrac{1}{r}=frac{n}{r}$$






                          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.
                            $$
                            P(X=k) = frac{{K choose k} {N- Kchoose n - k}}{{N choose n}}.
                            $$



                            $N$ 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 $nfrac{K}N$, hence the mean of your variable
                            $$E[X_i]=nfrac{1}{r}=frac{n}{r}$$






                            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.
                            $$
                            P(X=k) = frac{{K choose k} {N- Kchoose n - k}}{{N choose n}}.
                            $$



                            $N$ 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 $nfrac{K}N$, hence the mean of your variable
                            $$E[X_i]=nfrac{1}{r}=frac{n}{r}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            RScrlliRScrlli

                            751114




                            751114






























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                                Nässjö kommun Tettstader | Kjelder | NavigasjonsmenyeVIAFISNIGeoNamesMusicBrainz (area)

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                                Nässjö kommunKommunar i Jönköpings län svenskeJönköpings länAdministrasjonssenterNässjöAneby kommunEksjö kommunVetlanda kommunSävsjö kommuneVaggeryd kommuneJönköping kommunejernbaneknutepunktBackyard BabiesEidsbergtettstader31. desember2005 Nässjö kommun kommune Kommunevåpen Land   Sverige Len Jönköpings län Landskap Småland Areal 988,49 km²  - land 930,29 km²  - vatn 58,2 km² Folketal 29 375 Wikimedia Commons: Nässjö Municipality Nässjö kommune ligg i det svenske länet Jönköpings län. Administrasjonssenter i kommunen er Nässjö. Kommunen grensar til Aneby kommun i nord, Eksjö kommun og Vetlanda kommun i aust, Sävsjö kommune og Vaggeryd kommune i sør og Jönköping kommune i vest. Kommunen er kjend som jernbaneknutepunkt, og som heimstaden til rockebandet Backyard Babies. Nässjö er venskapskommune med Eidsberg. Nässjö sentrum sett frå vasstårnet. Tettstader | Nässjö kommune har tol...
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                                Kvitkval Innhaldsliste Taksonomi og utvikling | Utsjånad og levevis | Utbreiing | Åtferd |...

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                                Nær truga artarPattedyr i NoregNarkvalfamilien tannkvalarktiskenarkvalenmiocenkvalendelfinanetannkvalarLatinbiosonarenekkolokaliseringisbjørnenspekkhoggarinuittarArktisNew YorkNord-AmerikaEuropaJapan Kvitkval Status i verda: Nær truga Status i Noreg: Datamangel Kvitkval Systematikk Domene: Eukaryota Rike: Animalia Rekkje: Chordata Underrekkje: Vertebrata Klasse: Mammalia Orden: Cetaceae Familie: Monodontidae Slekt: Delphinapterus Art: D. leucas Vitskapleg namn Delphinapterus leucas Pallas, 1776 Kvitkval er ein tannkval som lever i arktiske og subarktiske område. Innhaldsliste 1 Taksonomi og utvikling 2 Utsjånad og levevis 3 Utbreiing 4 Åtferd 5 Bestandsstorleik og menneskeleg påverknad 6 Kjelder Taksonomi og utvikling | Kvitkvalen vart første gongen skildra av Peter Simon Pallas i 1776. Han er medlem av Monodontidae -familien saman med narkvalen. D...
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