The expected value of the sum of two random variables E(X + Y) = E(X) + E(Y). Why is the average of the sum of two sets of data not equal to the sum of the averages. avg(X + Y) != avg(X) + avg(Y)

Answer 1

Proof:

Given /assuming there are the same number of stats in each set, suppose there are K statistics.

X = {x1,x2, ..., xk}

Y = {y1,y2,...,yk}

avg(X) = (x1+x2+...+xk)/k

avg(Y) = (y1+y2+...+yk)/k

avg(x+y) = (x1+x2+...+xk+y1+y2+...+yk)/(2k)

But

avg(X)+avg(Y) = (X1+x2+...+xk)/k + (y1+y2+...+yk)/k

= (x1+x2+...+xk+y1+y2+...+yk)/k

So no, they are not the same , because the averages are off by a factor of 2 as you can see

by the denominators highlighted in bold. This is assuming, as stated, there are the SAME number of stats in each set.

Otherwise you will have k statistics in the first set and j statistics in the second set, where k and j are different. Then the equations become:

Avg(x) = (x1+x2+...+xk)/k and Avg(y) = (y1+y2+...+yj)/j

Avg(x+y) = ( x1+x2+...+xk+y1+y2+...+yj)/(j+k) while

Avg(x) + avg(y) = (x1+x2+...+xk)/k + (y1+y2+...+yj)/j

= j(x1+x2+...+xk) + k(y1+y2+..yj)/ (jk)

Not even close!!!