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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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Can you tell us where you came across this question? I would be surprised if it is homework, since you are in high school. – Srivatsan Aug 31 '11 at 11:52
Well, I'm not at high school anymore, my math skills are though :) – user467799 Aug 31 '11 at 13:33
Oops, sorry. I presumed this is not homework. But the current answer feels this could be homework, and hence it is given hint-style. Now that we know this isn't, perhaps a more descriptive answer is appropriate. (I, for one, think that the question is actually fairly non-trivial and interesting.) – Srivatsan Aug 31 '11 at 13:37
+1 on non-trivial. There's a reason why one usually uses variance and standard deviation instead of mean absolute deviation -- it's much easier to work with, despite looking more complicated at first glance. – Henning Makholm Aug 31 '11 at 14:53
Are the numbers picked with or without replacement? Is the 'mean' the observed mean or the 'true' mean? – leonbloy Sep 30 '11 at 13:47

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? – leonbloy Dec 30 '11 at 0:02

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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I'm not sure that this is valid. It seems that in step (2) you're envisioning using the expected mean instead of the observed mean, and it's not obvious that this will work, because the variation in the observed mean is not independent of the individual observations. (The eventual answer is likely right, though, but it will require more subtle reasoning than this sketch reveals to show it). – Henning Makholm Aug 31 '11 at 14:52
@Henning: I would expect the result to be similar to the $\frac{k-1}{k}$ bias effect in the sample/population variance, which tends to 1 as the number of drawings approaches infinity. – Henry Aug 31 '11 at 19:18
Yes, yes, I have convinced myself that it doesn't matter for the final result, but doing so was not trivial. (And I'm still not 100% sure it would be the same for a continuous distribution). – Henning Makholm Aug 31 '11 at 19:28
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. – Henry Aug 31 '11 at 20:45

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