# Help with proof that E[x_i] is X bar

1. Apr 5, 2014

### 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.

2. Apr 5, 2014

### 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?

3. Apr 5, 2014

### sid9221

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

4. Apr 5, 2014

### mathman

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