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Say we have to randomly pick k integral numbers out of n. The numbers are from the range < a; b >. What is the expected value of average absolute deviation from the mean for that random subset of k-numbers as the number of drawings approaches infinity?

Sorry if didn't make myself clear. Could you explain the answer so that it is understandable for a not so bright high school student?

EDIT: This is not homework :) Somobody asked me to program a vizualization of Lotto lottery results and I just got curious about the statistics of that.

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    Are the numbers picked with or without replacement? Is the 'mean' the observed mean or the 'true' mean?2011-09-30

2 Answers 2

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I do not have a complete answer; I am posting this just in case someone finds it interesting or useful. I assume sampling with replacement; the “without replacement” variation seems much harder.

We are interested in the quantity $ \mathbf E \left[\frac{1}{k} \sum_{i=1}^k \left| X_i - \frac{X_1 + X_2 + \cdots + X_k}{k} \right| \right] $ where $X_1, \ldots, X_k$ are iid and drawn from a distribution $\mathcal D$. By linearity of expectation and symmetry, this is equal to $ \begin{align*} \mathbf E \left[\left| X_k - \frac{X_1 + X_2 + \cdots + X_k}{k} \right| \right] &= \mathbf E \left[\left| \frac{(k-1)X_k - (X_1 + X_2 + \cdots + X_{k-1})}{k} \right| \right] \\ &= \frac{k-1}{k} \cdot \mathbf E \left[\left| X_k - \frac{X_1 + X_2 + \cdots + X_{k-1}}{k-1} \right| \right] \\ &= \frac{k-1}{k} \cdot \mathbf E \left[\left| X_k - Y \right| \right] \end{align*} $ where $Y = \frac{X_1 + \cdots + X_{k-1}}{k-1}$ is independent of $X_k$.

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    In the question, the distribution is given (a uniform in the integers, in the $[a,b]$ range). As $k$ tend to infinity, $Y$ tends to the mean, so the quantity of interest tends to the median absolute deviation, which is $n/4$ (assuming $N$ is even). Isn't this the complete answer?2011-12-30
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Since this may be homework, here are some hints:

  1. When the number of drawings approaches infinity, what will the mean be close to?
  2. Can you work out the deviation from the mean for each of the $n = b-a+1$ possible values?
  3. Can you take the absolute values of these deviations, and then take the average (i.e. add them up and divide by $n$)?

You may find yourself doing the calculations twice, once for $n$ odd and once for $n$ even.

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    If $k$ is not the number of drawings which approaches infinity then your original comment may be right. Incidentally, I think my answer to (3) is that these converge to $\frac{n}{4}$ for even $n$ and $\frac{n^2-1}{4n}$ for odd $n$ as the number of drawings approaches infinity.2011-08-31