I'm having a little trouble with the proof that the expected value of [tex] x_i [/tex] is [tex] \bar{X} [/tex](adsbygoogle = window.adsbygoogle || []).push({});

What I have is

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

Then

[tex] Pr(x_i=X_j) = 1/N [/tex]

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 ?

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

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