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• 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 ?.

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

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

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

For the second part, you can proceed similarly: $X_i = sum_k=1^n Y_k^(i)$ and $X_j = sum_ell=1^n Y_ell^(j)$, so:
$$
X_i X_j = sum_k=1^n sum_ell=1^n Y_k^(i) Y_ell^(j) implies
mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n mathbbE(Y_k^(i) Y_ell^(j)).
$$
We will first treat the case where $i ne j$. Then, for each term in the sum such that $k = ell$, we must have $Y_k^(i) Y_ell^(j) = Y_k^(i) Y_k^(j) = 0$ since it impossible for ball $k$ to be placed both in box $i$ and in box $j$. On the other hand, if $k ne ell$, then the events corresponding to $Y_k^(i)$ and $Y_ell^(j)$ are independent since the placement of balls $k$ and $ell$ are independent, which implies that $Y_k^(i)$ and $Y_ell^(j)$ are independent random variables. Therefore, in this case,
$$mathbbE(Y_k^(i) Y_ell^(j)) = mathbbE(Y_k^(i)) mathbbE(Y_ell^(j)) = frac1r cdot frac1r.$$
In summary, if $i ne j$, then
$$mathbbE(X_i X_j) = sum_k=1^n sum_ell=1^n delta_k ne ell cdot frac1r^2 = fracn(n-1)r^2$$
where $delta_k ne ell$ represents the indicator value which is 1 when $k ne ell$ and 0 when $k = ell$.

For the case $i = j$, I will leave the similar computation of $mathbbE(X_i^2)$ to you, with just the hint that the difference is in the expected value of $mathbbE(Y_k^(i) Y_ell^(j))$ for the case $k = ell$.

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
$$E[X_i]=nfrac1r=fracnr$$