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The basket contains N numbered balls $1; 2; ...; N$. From the basket without returns $n<=N$ balls are retrieved. Let S denote the sum of numbers a moat of taken out balls. Find the mathematical expectation E (S) and the variance this Var (S) of this sum.

Solution:

I wanted to clarify, to solve this problem, you need to write a distribution and then start counting and if you need a distribution, then which one?

Ben
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    The expectation is easy. The variance is harder, but will probably be a polynomial function of $N$ and $n$. Personally I would experiment to find that polynomial – Henry Nov 26 '21 at 10:01
  • @Henry Can you tell me how to find this polynomial, where and what to read, since unfortunately I cannot imagine it – Ben Nov 26 '21 at 10:03
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    Try small N and n first. – Ivan Neretin Nov 26 '21 at 10:10
  • @Ivan Neretin Can you explain in a little more detail what you mean? – Ben Nov 26 '21 at 10:12
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    For example if $n=1$ the variance is $\frac1{12}(N^2-1)$ – Henry Nov 26 '21 at 10:12
  • Where did $1/12$ come from? – Ben Nov 26 '21 at 10:14
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    When n=1, the distribution of S is rather obvious and well known. – Ivan Neretin Nov 26 '21 at 10:17
  • Do I need to set the distribution correctly now, so that I can then calculate everything that is needed? – Ben Nov 26 '21 at 10:25
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    Let $X_i$ be the number on the $i$-th ball picked, for $i=1,2,\ldots, n$. Note that by symmetry, $X_{i}\sim Uniform({1,2,\ldots,N})$ for all $i=1,2,\ldots, n$ and for any $k\in {1,2,\ldots,N}$, we have that conditional on $X_{i}=k$, $X_{j}$ is uniformly distributed on ${1,2,\ldots,N} \backslash {k}$ (for $j\ne i$). You can use these facts (and the law of total expectation where necessary) to work out $\Bbb{E}[X_{i}], \mathrm{Var}(X_{i}), \Bbb{E}[X_{i}X_{j}], \mathrm{Cov}(X_{i}, X_{j})$, and thus ultimately the mean and variance of the sum $S_{n} := X_{1} + \cdots + X_{n}$. – Minus One-Twelfth Nov 26 '21 at 10:25
  • Thanks for the answer, but I don't understand at all what this is about. I understood the train of your thoughts, but I don’t understand how to count. – Ben Nov 26 '21 at 10:32
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    I think that if $n=2$ the variance might be $\frac16(N^2-N-2)$. So my heroic guess before I look any further is that the general expression may be something like $\frac1{12}n(N+1)(N-n)$. The $\frac1{12}$ looks familiar as the variance of a continuous uniform distribution on $[0,1]$. The $(N-n)$ term looks familiar from the finite population correction for the standard error. – Henry Nov 26 '21 at 10:41
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    @Henry Also, your expression is symmetric with respect to $n\leftrightarrow N-n$, which is a good sign. – Ivan Neretin Nov 26 '21 at 10:43
  • Can you please show how you calculated the mathematical expectation, perhaps then it will become clearer – Ben Nov 26 '21 at 11:00
  • This is answered at https://math.stackexchange.com/q/2813390/321264 and its linked posts. – StubbornAtom Nov 26 '21 at 11:34

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