Given a set of N objects and associate a non-negative integer value k_i to each object in the set. Consider the following computation problem: let k'_i = \sum_i {k'_i (1/N)} = E';
Can we conclude that:
V = \sum_i {k_i^2 (1/N)} - (\sum_i {k_i (1/N)})^2 >= \sum_i {k'_i^2 (1/N)} - E'^2 =V' ?
If not, what is the relation between V and V'? I am wondering whether their relation is independent of the exact difference between k_i and k'_i, thanks!
If I understand well, you assume a set of sample objects O = {o_1, ... o_N}, and a uniform probability distribution p : O -> [0,1]; that is, forall (i=1,... N). p(o_i) = 1/N.
Next, you consider two random variables X and Y, defined over
your space of sample objects O. For any object o_i \in O,
where i=1,...,n you let X(o_i) = k_i, and Y(o_i) = k'_i.
Finally, you assume that Y
To see that the answer is "no", even when the numbers are constrained to be non-negative, just take (10, 10), with mean 10 and variance 0, and (0, 10), with mean 5 and variance 25.
To Andrea, thanks for the answer. Your understanding is correct except for that I assume k_i and k'_i non-negative. My original thought was, since the difference k_i^2 - k'_^2 seems grow faster than k_i-k'_i, possiblly V always larger than V'.
To Adam, good example, thanks. But it is strange why it is not as my original thought, never mind...
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