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Making notation consistent[edit]

I'm going to make some changes late in the article to establish consistent notation. Right now in the section Population variance and sample variance we have in the subsection Sample Variance:

Taking directly the variance of the sample gives:
\sigma_y^2 = \frac 1n \sum_{i=1}^n \left(y_i - \overline{y} \right)^2

and then

Correcting for this bias yields the unbiased sample variance:
\sigma_S^2 = \frac{1}{n-1} \sum_{i=1}^n \left(y_i - \overline{y} \right)^2

with no definition given for S, though \sigma_S^2 is usually (and subsequently here) notated as s2. Then in Distribution of the sample variance we have an analysis of s2 with no definition having been given for it. Then in Samuelson's inequality we have

Values must lie within the limits m ± s (n − 1)1/2 .

with m and s undefined (and s meaning something different from previously). Finally, in Relations with the harmonic and arithmetic means we have

... where m is the minimum of the sample

in which m means something different from in the previous subsection.

So I'm about to rationalize the notation -- please give me a chance to finish before revising! Duoduoduo (talk) 16:34, 16 May 2013 (UTC)

Done! Duoduoduo (talk) 17:34, 16 May 2013 (UTC)

Poisson variance derivation[edit]

Right now the Poisson section says:

The variance is equal to:
 \operatorname{Var}(X) = \sum_{k=1}^{n} \frac{\lambda^k}{k!} e^{-\lambda} (k-\lambda)^2 = \lambda,

It seems to me that this can't be right, since the value of the undefined n would affect the sum. Is n supposed to be infinity? Duoduoduo (talk) 14:03, 10 July 2013 (UTC)

True, and easy to check that the formula given above is false. Also the zero term is missing. Since all terms in the sum are non-negative, and the zero term is positive,
 \sum_{k=1}^n \frac{\lambda^k}{k!} e^{-\lambda}(k-\lambda)^2 <
\sum_{k=0}^\infty \frac{\lambda^k}{k!} e^{-\lambda} (k-\lambda)^2 = E(X-E[X])^2 = Var(X) = \lambda.

Mathstat (talk) 20:14, 10 July 2013 (UTC)

Formula for binomial variance in same section is also wrong. Mathstat (talk) 20:19, 10 July 2013 (UTC)

Meaning (interpretation) of Variance[edit]

While reading the article on variance, I found that a paragraph on how variance should be interpreted (in an intuitive way) was somehow missing. I don't feel confident to write that bit myself, but if someone could add this part it would I believe make the article more interesting and complete. — Preceding unsigned comment added by Marc saint ourens (talkcontribs) 18:29, 1 September 2013 (UTC)

Done. Thanks very much for the suggestion! Duoduoduo (talk) 16:56, 2 September 2013 (UTC)

Variance for six-sided die[edit]

Isn't the variance given in the article only correct for an infinite number of rolls of a die? For one roll, the variance is zero. For two rolls, a quick calculation suggests the variance is 5/8. Grover cleveland (talk) 16:27, 4 June 2014 (UTC)

Variance is not dependent on an expeiment, but only on properties of the die. You are confused with sample variance. Nijdam (talk) 20:06, 4 June 2014 (UTC)