# Properties of an Average

1. Apr 25, 2004

### 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.

2. Apr 26, 2004

### 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)

3. Apr 26, 2004

### 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
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?

4. Apr 26, 2004

### 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

-Ray.

Last edited: Apr 27, 2004
5. Apr 26, 2004

### Ebolamonk3y

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

6. Apr 27, 2004

### 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.