Re: The mean of the sum of m-subset

On Aug 12, 5:18 pm, "Stephen J. Herschkorn" <sjhersc...@xxxxxxxxxxxx>
wrote:
Stephen J. Herschkorn wrote:
Stephen J. Herschkorn wrote:

Mohammad Nabil wrote:

On Aug 11, 8:46 am, "Stephen J. Herschkorn" <sjhersc...@xxxxxxxxxxxx>
wrote:

Chip Eastham wrote:

On Aug 10, 8:13 pm, Mohammad Nabil <mohammad.nabi...@xxxxxxxxx>
wrote:

If we have a set of n real numbers, and we pick x from it
according to the uniform distribution ( a = 0, b = n-1 ). If we
picked m numbers, m less than or equal n, then what is the mean
of their sum ? And how would we approach calculating the variance ?
[...]

Are you picking m numbers (from the given set of n real numbers)
with replacement or without replacement? The latter prevents a
number from being picked more than once in the m summands.

The method of selection affects the variance, but not the mean.

I calculated it and the mean is equal, but may I ask what's the
basis behind that?

Let X_i be the ith number picked, Y = sum(i=1..m, X_i). EY =
sum(i=1..m, EX_i), and EX_i (= EX1) is the same regardless of
whether we select with or without replacement.

To determine the variance,

Var(Y) = sum(i=1..m, Var(X_i)) + sum((i,j): i!=j, Cov(X_i, X_j)
= m Var(X1) + m (m-1) Cov(X1, X2).

Alternatively, for j = 1, 2, 3,..., n, let I_j = 1 if the jth
number, whose value is x_j, is picked; I_j = 0 otherwise. Then
Y = sum(j=1..n, x_j I_j).

Using this last formulation, I get

Var(Y) = [m (n-m) / n^2] {sum(j=1..n, (x_j)^2) - [2 / (n-1)]
sum(j=1..n-1, k= j+1..n; x_j x_k)},

= m (n-m) {n sum(j=1..n, (x_j)^2) - [sum(j=1..n, x_j)]^2} / [n^2 (n-1)]



but I strongly urge the interested reader to verify my algebra.

--
Stephen J. Herschkorn sjhersc...@xxxxxxxxxxxx
Math Tutor on the Internet and in Central New Jersey and Manhattan

Thanks very much :) Your help is really appreciated :)

.

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