Help with proof that the expected value of $x_i$ is $\bar{X}$ 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 ?
 A: Every $X_j$ is equally likely to be selected (this isn't true for every population, but it is a common situation [for example, a die, where 1, 2, 3, 4, 5, 6 are all equally likely to be thrown]). So since there are $N$ possible values, the probability of each is ${1 \over N}.$ 
So you have \begin{align}\mathrm E[x_i]&=\sum_{j=1}^N X_j\,\mathrm P[x_i=X_j]\\ &={1 \over N} \sum_{j=1}^N X_j\\ &={1 \over N} (N \bar{X})\\&= \bar{X}\;.\end{align}
A: If, as some of the comments have suggested, that $x_i$ is a single (discrete) random variable, and the $X_j$ are all the potential values that $x_i$ can take, then basically the definition of expectation of $x_i$ is $\mathbb{E}(x_i) := \sum_{j=1}^n X_j \mathbb{P}(x_i = X_j)$. However, in no way does this imply anything about the values that $\mathbb{P}(x_i = X_j)$ can take (other than that they are non-negative and must sum to unity over $j$)!! The probabilities $\mathbb{P}(x_i = X_j)$ define the distribution of the discrete random variable $x_i$, and since the expectation of $x_i$ is just a quantity associated with its distribution, then $\mathbb{E}(x_i)$ changes as the probabilities $\mathbb{P}(x_i = X_j)$ change.
Perhaps you're missing something from your question - maybe you've been given a question where you know that $\mathbb{E}(x_i)$ is fixed to be equal to a given value, and you're also given the values $X_j$, then it may be that you're asked to solve for all possible values of $\mathbb{P}(x_i = X_j)$?
Several comments:
1) Your use of capitals to represents values and lower-case to represent random variables is very confusing - mathematicians use the opposite convention.
2) Your use of the index $i$ is very confusing - it is not used anywhere (usefully) in your question. Come to think of it, your question mentions some combinatorial statement - perhaps you have written your question badly, in which case I urge you to rephrase your question. If this question was prompted by some other question that you're trying to solve, I would ask that original question.
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EDIT in response to comment:
Ah, if you're trying to prove that the sample mean is unbiased estimator of the population mean, then let's start with only two random variables $x_1$ and $x_2$ (using your notation). We need to prove that the expectation operator is linear i.e. $\mathbb{E}(x_1 + x_2) = \mathbb{E}(x_1) + \mathbb{E}(x_2)$ (note that if this is the case, then it's trivial to extend this to the case of there being $n$ random variables - use induction). So to prove linearity (and letting the $X_1$ and $X_2$ are the possibles values that $x_1$ and $x_2$ can take, respectively):
$\begin{align}
\mathbb{E}(x_1 + x_2) &= \sum_{X_1, X_2} (X_1+X_2) \mathbb{P}(x_1 = X_1, x_2 = X_2) \\
                      &= \sum_{X_1} \sum_{X_2} \left[X_1 \mathbb{P}(x_1 = X_1, x_2 = X_2) + X_2 \mathbb{P}(x_1 = X_1, x_2 = X_2)\right] \\
&= \sum_{X_1} \sum_{X_2} X_1 \mathbb{P}(x_1 = X_1, x_2 = X_2) + \sum_{X_2} \sum_{X_1} X_2 \mathbb{P}(x_1 = X_1, x_2 = X_2) \\
&= \sum_{X_1} \left[ X_1 \sum_{X_2}  \mathbb{P}(x_1 = X_1, x_2 = X_2)\right] + \sum_{X_2} \left[ X_2\sum_{X_1}  \mathbb{P}(x_1 = X_1, x_2 = X_2)\right] \\
&= \sum_{X_1}X_1\mathbb{P}(x_1 = X_1) + \sum_{X_2}X_2\mathbb{P}(x_2 = X_2)\\
&= \mathbb{E}(x_1) + \mathbb{E}(x_2) \,\,\,\,\,\text{QED}
\end{align}$
From this, it should be easy to see that $\mathbb{E}(\bar{x})$ is equal to the population mean.
