Wikipedia:Reference desk/Archives/Mathematics/2017 June 13

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June 13

Bessel's correction for sampling a finite population

Consider the numbers $(X_{1}\cdots X_{N})$. The variance is

$S^{2}=(N^{-1}\sum X_{i}^{2})-(N^{-1}\sum X_{i})^{2}$

Pick a number $n$ such that $0<n<N$. Consider the variance of each of the ${\binom {N}{n}}$ n-samples. The mean value of these sample variances, ${\bar {s^{2}}}$, is a little less than the population variance $S^{2}$.

Conjecture:

${\bar {s^{2}}}=(1-(n^{-1}-N^{-1})(1-N^{-1})^{-1})S^{2}$.

I obtained this result by numerical experimentation. I need your help to prove (or disprove) it.

The limiting case $N^{-1}=0$ gives Bessel's correction

${\bar {s^{2}}}=(1-n^{-1})S^{2}$.

Bo Jacoby (talk) 11:42, 13 June 2017 (UTC).

First, simplify the factor to rewrite the formula which needs to be proved:

${\bar {s^{2}}}={\frac {N(n-1)}{n(N-1)}}S^{2}$.

Next, instead of using the definition, use the formula given at the bottom of the section "Discrete random variable" in the the article.

$S^{2}={\frac {1}{2N^{2}}}\sum _{i,j\in X}(x_{i}-x_{j})^{2}$

Where $X=\{1,2,\dots N\}$. The average of the variances for the smaller samples is then

${\frac {1}{\binom {N}{n}}}\sum _{|Y|=n}{\frac {1}{2n^{2}}}\sum _{i,j\in Y}(x_{i}-x_{j})^{2}$

where the first summation runs over all subsets Y of X having n elements. For a given i and j with i≠j (we don't care about i=j since those terms don't contribute to the sum), the number of sets Y containing i and j is ${\binom {N-2}{n-2}}$. So each of the terms $(x_{i}-x_{j})^{2}$ occurs that many times in the double sum. Therefore we can pull out that factor and make it a single sum, then simplify to get

${\frac {\binom {N-2}{n-2}}{2n^{2}{\binom {N}{n}}}}\sum _{i,j\in X}(x_{i}-x_{j})^{2}$

Plugging in the original sum, this becomes

${\frac {2N^{2}{\binom {N-2}{n-2}}}{2n^{2}{\binom {N}{n}}}}S^{2}={\frac {N(n-1)}{n(N-1)}}S^{2}$.

Nice problem. It kind of makes sense since you're applying the reverse correction for N and then applying a new correction for n. --RDBury (talk) 13:29, 13 June 2017 (UTC)

Yes, very interesting and a very elegant result. So the estimate using the unbiased infinite-population formula needs to be multiplied by the reciprocal of the above (by ${\tfrac {N-1}{N}}$) to account for a finite population. There's an interesting related result stated without proof in Standard error#Correction for finite population. The standard error is the standard deviation of the sample mean. In that context the infinite-population estimate needs to be multiplied by ${\sqrt {\frac {N-n}{N-1}}}$ for a finite population (the square root sign is just because we're dealing with a square root of a variance). Loraof (talk) 00:17, 14 June 2017 (UTC)

Thank you, both of you! The formula can be written symmetrically

$(1-n^{-1})^{-1}{\bar {s^{2}}}=(1-N^{-1})^{-1}{\bar {S^{2}}}$

showing that a Bessel correction must be applied to the population variance too. RDBury, which article are you referring to? Bo Jacoby (talk) 06:03, 14 June 2017 (UTC).

Sorry, I was referring to the article on variance you linked to in the original post. --RDBury (talk) 11:22, 14 June 2017 (UTC)

Thanks! Bo Jacoby (talk) 15:18, 14 June 2017 (UTC).

Square roots etc.

Are all square roots of integers either an integer or an irrational number?

What about cube roots? Are they too either integers or irrational numbers?

And what about fourth roots, fifth, sixth etc? — Preceding unsigned comment added by 83.226.140.123 (talk) 19:09, 13 June 2017 (UTC)

Yes in all cases. This is proven for square roots in Proof by infinite descent#Irrationality of √k if it is not an integer. The proof for cube roots and higher works the same way but is a bit more tedious. Loraof (talk) 20:03, 13 June 2017 (UTC)

The general case can also be proven by the rational roots theorem. See: [1] CodeTalker (talk) 20:10, 13 June 2017 (UTC)

(ec) And Irrational number#General roots gives a simple proof for the general case, relying on the fact the integers have a unique prime factorization: if, say, the 17th root of 25 was the fully reduced rational non-integer number ${\tfrac {a}{b}}$ (b>1) then $\left({\tfrac {a}{b}}\right)^{17}=25$ (still fully reduced), so $\left({\tfrac {a^{17}}{b^{17}}}\right)={\tfrac {25}{1}},$ but this impossible since the denominators don't match up ($b^{17}\neq 1$). Loraof (talk) 20:52, 13 June 2017 (UTC)

Note that if you're taking a square, fourth, sixth, etc. root of a negative integer, your result will be a complex number (a + bi). a and b will not be non-integer rationals. —Guanaco 20:22, 13 June 2017 (UTC)

Wouldn't the root just be an imaginary number, with no real component ? (I suppose you can consider that to be a complex number, where a=0, if you prefer, but then the roots of positive numbers could also be considered to be complex numbers, where b=0.) In the case of sqrt(-3), that gives us sqrt(3)i, where b=sqrt(3), which is irrational, right ? StuRat (talk) 15:27, 14 June 2017 (UTC)

Right for square roots, since i is one of the square roots of −1. In general the n-th root of K, where K < 0 and n is even, is the n-th root of |K| times any n-th root of −1. One of the latter is, for example when n = 4, ${\tfrac {\sqrt {2}}{2}}+{\tfrac {\sqrt {2}}{2}}i.$ Loraof (talk) 17:58, 14 June 2017 (UTC)

See our article roots of unity. CodeTalker (talk) 18:06, 14 June 2017 (UTC)