Measurements - How can it be that precise?

[Omega0]

Guys, could you give me an online source where the derivation of the factor $$1/\sqrt{N}$$ for the standard derivation for one data sample is derived correctly?
Thanks.

 

[Dale]

http://scipp.ucsc.edu/~haber/ph116C/iid.pdf

 

[FactChecker]

http://scipp.ucsc.edu/~haber/ph116C/iid.pdf

I'm afraid that that reference refers to other charts that prove the essential facts. I could not find those charts. I think this is more self-contained and complete: http://www.milefoot.com/math/stat/rv-sums.htm
(For continuous random variables, the summations signs are replaced by integrals.)

 

[Omega0]

Dale, that's it. Now I got it. Thanks to you and the other guys, Stephen, FastChecker etc.

To justify my stupid fault a bit: In the (very good) book it says, see the derivation above, my translation: "Here the $s^2\left(y_i\right)$ are all the same because they are built each from the same input values. "
What I absolutely didn't understand is that, naturally, they don't need to be the same input values but the same $s^2$.
This was my big failure.
Having said this, I would have found it fair if the professors would have remarked this. Seems they expected a knowledge in statistics above my level.

Thanks and sorry for my ignorance.

 

[Stephen Tashi]

Which formulae for population parameters also work for sample statistics that estimate them?

If we have a random samples $S_x$, $S_y$ of N things taken from each of two random variables $X$ and $Y$, we can imagine the values in $S_x$ and $S_y$ to define an empirical joint probability distributionj. So the sample mean of the pairwise sums of values in in $S_x$ and $S_y$ should be the sum of their sample means - analagous (but not identical) to the fact that $E(X+Y) = E(X) + E(Y)$

However if $X$ and $Y$ are independent random variables, there is no guarantee that the empirical distribution of N pairs of numbers of the form $(x,y), x \in S_x, y\in S_y$ will factor as an distribution of x-value times an (independent) distribution of y-values. So we can't conclude the sample variance of $x+y$ is the sample variance of the x-values plus the sample variance of the y-values.

If, instead of N pairs of values, we looked at all the possible $N^2$ values $(x,y)$ we would get an empirical distribution where the x and y values are independent. Then something analgous to $Var(X+Y) = Var(X) + Var(Y)$ should work out. We would have to be specific about what definition of "sample variance" is used. For example, can we used the unbiased estimators of the population variances?

 

[FactChecker]

I prefer to think that (using variables with zero mean for simple equations):
$$\sigma^2_{X+Y} = E( (X+Y)^2 ) = E( X^2 + 2XY + Y^2) = E(X^2) + 2E(XY) + E(Y^2)$$ $$ = \sigma^2_X + 2*cov(XY) + \sigma^2_Y$$
So the independence of $X$ and $Y$ implies the desired result.

 

[Stephen Tashi]

So the independence of $X$ and $Y$ implies the desired result.

$E(X^2) = \sigma_X^2$ in the case of a random variable with zero mean. For a set of data $x_i$ realized from such a random variable, we don't necessarily have a zero sample mean.

 

[FactChecker]

$E(X^2) = \sigma_X^2$ in the case of a random variable with zero mean. For a set of data $x_i$ realized from such a random variable, we don't necessarily have a zero sample mean.

Good point. But in terms of expected values, it is correct.