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Can we say that

$E \left(\frac{n}{n-1}X \right) = E(X)$

because $n/(n-1)$ is basically equal to 1. Or can we not say this? I am just factoring out the n/(n-1).

$n$ is the sample size.

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    It is of course not equal not if you meant almost equal to E(X) the answer is yes for$n$large since$n/(n-1)$approaches$1$from above assuming E(X)>0.2012-05-21

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The proposed identity is not true, but $ \lim_{n\to \infty}E \left(\frac{n}{n-1}X \right) = E(X), $ and sometimes people might have a reason to care about that.