Properties of an Average

[Ebolamonk3y]

What are the properties of an average (mean) of something? Like... is it communative, associative with other averages? I duffed up this one on a test. :frown:

[matt grime]

Things like, for rvs, E(aX+bY)=aE(X)+bE(Y) for a and b constants, and if they are independent E(XY)=E(X)E(Y)

[Ebolamonk3y]

Here is the original question...


Of the following five statements, I to V, about the binary operation of averaging (arithmetic mean), those which are always true are...

I. Averaging is associative
II. Averaging is commutative
III. Averaging distributes over addition
IV. Addition distributes over Averaging
V. Averaging has an identity


multiple choice...

A) II Only B) I + II only C) II and III Only D) II and IV only E) II and V only


which one grime?

[rgoudie]

I'll jump in here with my first posting.

I The arithmetic mean is not associative:
Let m be the function that yields the arithmetic mean of its two parameters.
m(m(a, b), c) = m((a+b)/2, c) = (a+b)/4 + c/2
m(a, m(b, c)) = m(a, (b+c)/2) = a/2 + (b+c)/4.

II The arithmetic mean is commutative since addition is commutative:
m(a, b) = (a+b)/2 = (b+a)/2 = m(b, a).

III The arithmetic mean does not distribute over addition:
m(a, b+c) = (a+b+c)/2.
m(a, b) + m(a, c) = (a+b)/2 + (a+c)/2 = (2a+b+c)/2.

IV Addition does not distribute over averaging:
a + m(b, c) = a + (b+c)/2.
m(a+b, a+c) = (a+b+a+c)/2 = (2a+b+c)/2.

V The arithmetic mean does not have an identity:
m(a, i) = a
(a+i)/2 = a
a+i = 2a
i = a

Your only choice for a correct answer is A.

-Ray.

[Ebolamonk3y]

Woah... neato! Stuff I have no clue about...

[matt grime]

"binary operation of averaging (arithmetic mean),"

who the hell wrote that? the arithmetic mean is not a binary operation. They could at least have included the words "of two numbers" explicitly so it made sense.