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Let $\lbrace x_i,y_i \rbrace_{i=1}^{n}$ be a random sample. Let $\bar{X}$ and $\bar{Y}$ be the sample means.

I want to rewrite the statement

$$\sum_{i=1}^n x_i y_i - \bar{X}\bar{Y}$$ in terms of the standard error $S_{xy}$

My friend claims I can simply say

$$\frac{\sum_{i=1}^n x_i y_i}{n} = \Bbb E[XY]$$ and that

$$\Bbb E [XY] - \bar{X}\bar{Y} = S_{xy}$$

I don't follow why this should be true in general. But I can't explain why.

My Question

Does the sample mean equal the population mean in general and if so why? How does this relate to the variance?

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  • $\begingroup$ If the random variables have a continuous distribution, then the answer is never. With something like the binomial distribution, there's a small chance for equality only if $pn$ is an integer. $\endgroup$
    – Alex R.
    Commented Apr 18, 2016 at 18:56
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    $\begingroup$ Your friends argument is correct if you interpret the expectation $\mathbb E$ as being relative to the empirical distribution of the data. In that case, it is trivially true that $\mathbb E X = \bar X$ for example. $\endgroup$
    – guy
    Commented Apr 17, 2019 at 15:01
  • $\begingroup$ @AlexR. Excepting, perhaps, a degenerate case where all values are identical? (I.e. variance is 0) $\endgroup$
    – Alexis
    Commented Sep 8, 2021 at 17:29

1 Answer 1

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As your sample size gets closer to the actual population size, you expect to get a "true" mean- that is there is always some errors associated with the statistics ( derived from the sample) when compared to the actual parameters (derived from the population)- that said,there is a trade-off between sample size and the error you might get when comparing the sample mean with the population mean.This error is known as the "margin of error" when you comprise the population size by selecting a sample of the population. The more error you can tolerate, the smaller sample you need. The less error you can tolerate, the bigger sample (closer to the actual population) you will need.

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