Large Numbers and

Let x be as "Stochastic" or "Random" variable. This means that x occurs in "random sequences" x₁, x₂,..., x_i,... of numbers with values belonging to a definite range (EG., +1 for heads, -1 for tails in a coin toss) each value appearing with a specific frequency (E.G., half heads, half tails.)

Partition a sequence of x into groups of N terms. Let the sum of the terms in the be y_l . The object is to show that for long sequences, the averages (<>) of the squared deviations (d) of x and y are related by

<d²(y)>  =  N <d²(x)>  .

The mean of y is given by . In a sequence of x, refer to the i^th element (i = 1,2,...,N) of the l ^th group as x _{l , i .} Then the corresponding element of the y sequence is   . Notice that the mean of y is related to that of x by the expression

because Nm is the total number of x values.

The deviation of the terms in the y-sequence from their mean can be expressed in terms of the corresponding deviation for x as follows:

d_l (y) º  y_l - <y>

so that

Then the average of the squared deviation of y satisfies the equation

so that

or

.

Using the definition of independence sequences of random variables, easily justified using plausible arguments^*, the value of the second bracketed term above is seen to be zero.

The bracketed expression in the surviving (first) term on the right hand side of EQ. is just <d²(x)> , so that the equation yields the predicted relationship,

<d²(y)>  =  N <d²(x)> .

* In the sum

sort the terms so that they are listed in order of increasing value for . Since there are a large number of terms, many will be found for which is the same, e.g. . We collect those terms together. The constant value (e.g. ) multiplies the sum of many terms over .

But the sum of many 's is zero. Repeating this, we see that the entire bracketed term is zero.

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