Help with proof that E[x_i] is X bar

[sid9221]

I'm having a little trouble with the proof that the expected value of x_i is \bar{X}

What I have is

E[x_i]=\sum_{j=1}^{N}X_j Pr(x_i=X_j)

Then

Pr(x_i=X_j) = 1/N

This is the bit I can't understand, how does that probability evaluate to that value.

I know the denominator is how many ways you can choose n out N. I think that the numerator should be how many ways you can choose (n-1) out of (N-1). But I seem to have an extra n.

Any advice ?

 

[FactChecker]

I am used to \bar{X}being defined as E(X) or as a sample mean. That would make the identity either true by definition or false. Can you define that notation, please?

 

[sid9221]

I am used to \bar{X}being defined as E(X) or as a sample mean. That would make the identity either true by definition or false. Can you define that notation, please?

I'm trying to show that E[x_i]=\bar{X}
where \bar{x} is the sample mean and \bar{X} is the population mean

 

[mathman]

By definition E(x_i) is the population mean, assuming all x_i have the same mean.