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

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

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}

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

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

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:
$$
X_i X_j = sum_{k=1}^n sum_{ell=1}^n Y_k^{(i)} Y_{ell}^{(j)} implies
mathbb{E}(X_i X_j) = sum_{k=1}^n sum_{ell=1}^n mathbb{E}(Y_k^{(i)} Y_{ell}^{(j)}).
$$
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,
$$mathbb{E}(Y_k^{(i)} Y_{ell}^{(j)}) = mathbb{E}(Y_k^{(i)}) mathbb{E}(Y_{ell}^{(j)}) = frac{1}{r} cdot frac{1}{r}.$$
In summary, if $i ne j$, then
$$mathbb{E}(X_i X_j) = sum_{k=1}^n sum_{ell=1}^n delta_{k ne ell} cdot frac{1}{r^2} = frac{n(n-1)}{r^2}$$
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 $mathbb{E}(X_i^2)$ to you, with just the hint that the difference is in the expected value of $mathbb{E}(Y_k^{(i)} Y_{ell}^{(j)})$ 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.
$$
P(X=k) = frac{{K choose k} {N- Kchoose n - k}}{{N choose n}}.
$$

$N$ 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
$$E[X_i]=nfrac{1}{r}=frac{n}{r}$$