# Parametric function or statistic?

1. Jan 17, 2007

### EvLer

Which one would you say this is :

E(|x1-x2|)
x1, x2, xn - a sample of n values on the underlying random variable...
I was thinking this is a statistic

2. Jan 17, 2007

### EnumaElish

"E" of something is a probability-weighted average outcome, so it involves the entire set of outcomes rather than just a sample. A statistic involves only a sample. The x's in E[f(X1,x2)] are not sample values, they are shorthands for two random variables or distributions (e.g., "Normal distribution 1" and "Normal distribution 2").

3. Jan 18, 2007

### EvLer

so... parametric function?

4. Jan 18, 2007

### EnumaElish

"Probabilistic representation of the set of outcomes." I guess para. func. is acceptable for cases where the prob. rep. reduces to a parametric function.

5. Jan 18, 2007

### EvLer

so, then what about |x1-x2|? would you say it is a statistic?

6. Jan 18, 2007

### HallsofIvy

Yes, that's a statistic. The initial confusing factor was your use of "E". Did you intend that to be the expectation? If so, E(u) is the expected value of a random variable or function of a random variable. It doesn't make sense to talk about the "expected" value of specific values from a sample.

7. Jan 18, 2007

### EvLer

thanks a lot for explaining.
so to make sure I understand, it would be wrong to "do" expected value on a sample? since E's domain is data from entire distribution?
edit: would you call E() a function? if not, what is it formally? sorry if I don't get this right away...
Thank you again.

8. Jan 18, 2007

### EnumaElish

Let c be a (deterministic) constant. Then the expected value of c is always c.

You can take the expectation of a sample, but the result will be the sample itself. That's because a sample {x1, ..., xn} (differently from random vars. {X1, ..., Xn}) is a set of values that have already been determined (by the very act of selecting them from a set of probable outcomes). There is nothing inherently probabilistic or random about a sample, once you have the sample.

And yes, in "standard" theory and applications, E[f(O)] (where "E" is a shorthand for "EP") is a function given by the integral of f(O)dP(O) where f is any (continuous, to be safe) function defined over the set of outcomes O, and P is the cumulative probability distribution defined over O. For discrete cases, it is the inner product f(O).p(O) where p is the density defined over O.

E[f(O)] is the P-measure of the set f(O), and hence the technically correct notation "EP."

Last edited: Jan 19, 2007
9. Jan 21, 2007

### EnumaElish

Examples:
Continuous P, specific parameters: Let U[0,1] be the uniform prob. dist. over the unit interval I=[0,1]: U[0,1](v) = v for v in I. Then E[U[0,1]] = EU[0,1][I] = $\int_0^1 v dU[0,1](v) = \int_0^1 v dv = \frac{v^2}{2}|_0^1 = 1/2$.
Continuous P, general parameters: Let U[a,b] be the uniform prob. dist. over any interval J = [a,b]: U[a,b](v) = (v-a)/(b-a) for v in J. Then
$$E[U[a,b]] = E_{U[a,b]}[J] = \int_a^b v dU[a,b](v) = \int_a^b \frac v{b-a} dv =\frac{v^2}{2(b-a)}|_a^b = \frac{b^2-a^2}{2(b-a)} = \frac{a+b}2$$

Discrete P, specific params.: Let p be the uniform density function p(v) = 1/2 for all v in O = {0,1}. Then EP[{0,1}] = (1/2,1/2}.{0,1} = (1/2) 1 + (1/2) 0 = 1/2.
Discrete P, gen. params.: Let p be the uniform density function p(v) = 1/n for all v in W = {1, ..., n}. Then EP[W] = (1/n, ..., 1/n}.{1, ..., n} = (1 + ... + n)/n = n(n+1)/(2n) = (n+1)/2.

Last edited: Jan 21, 2007