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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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What is $n$? ${}$ – user21436 Apr 24 '12 at 4:47
For $X$ that does not depend on $n$, and the constant $n$, $$E \left(\frac{n}{n-1}X \right) = \frac{n}{n-1}E(X)$$ The rest depends on what is $n$. – penartur Apr 24 '12 at 4:51
Basically equal to 1? Would you willingly pay 25 percent more tax than you owe because, after all, 5/4 is basically equal to 1? – Gerry Myerson Apr 24 '12 at 5:42
Do you mean the expectation value as written, or its limit for $n\to\infty$? In that limit, there's a precise sense in which "$n/(n-1)$ is basically equal to $1$", but if that's not what you mean, I laud your sense of commitment to the community and hope your extra taxes will be spent wisely. – joriki Apr 24 '12 at 7:17
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. – Michael Chernick May 21 '12 at 17:54

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.

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