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I could use more comments in the binomial example.

For example, how does this follow? "the asymptotic variance of ${\displaystyle {\displaystyle \log \left({\frac {X_{n}}{n}}\right)}\log \left({\frac {X_{n}}{n}}\right)}$ does exist and is equal to .."

For example, how does this follow? "since p>0, ${\displaystyle {\displaystyle \Pr \left({\frac {X_{n}}{n}}>0\right)\rightarrow 1}}$ ${\displaystyle {\displaystyle \Pr \left({\frac {X_{n}}{n}}>0\right)\rightarrow 1}}$ as ${\displaystyle {\displaystyle n\rightarrow \infty }n\rightarrow \infty ..}$"

For example, how does this follow? "the logarithm of the estimated relative risk ${\displaystyle {\displaystyle {\frac {\hat {p}}{\hat {q}}}}{\frac {\hat {p}}{\hat {q}}}}$ has asymptotic variance equal to .."

I'm sure these are trivial to someone who can already see it, but I don't.

Also, more text about the practical implications of this example would be good. The relative risk thing at the end is a start, but what about the proportion itself? What is a concrete, real-world example?

dfrankow (talk) 16:48, 21 December 2018 (UTC)Reply

You do not need normality to use the delta method —Preceding unsigned comment added by 75.73.53.186 (talk) 01:28, 2 March 2011 (UTC)Reply

At the beginning of the proof, it says "..since ${\displaystyle X_{n}\to \theta }$..". Where does this come from? — Preceding unsigned comment added by 128.40.213.241 (talk) 14:29, 5 November 2015‎ ${\displaystyle \forall \epsilon >0,P(|X_{n}-\theta |<\epsilon )=P({\sqrt {n}}|X_{n}-\theta |<{\sqrt {n}}\epsilon )\to 1}$ as ${\displaystyle n\to \infty }$ since ${\displaystyle {\sqrt {n}}(X_{n}-\theta )\to N(0,\sigma ^{2})}$ and ${\displaystyle {\sqrt {n}}\epsilon \to \infty }$ Vinzklorthos (talk) 20:08, 5 August 2020 (UTC)Reply

It does exist provided you are dealing with positive support ${\displaystyle X\in (0,\infty )}$. Now the average of ${\displaystyle X_{n}}$ will need n to be large enough for the density at 0 to become vanishingly small. Limit-theorem (talk) 20:57, 8 July 2015 (UTC)Reply

Perhaps to avoid the subtleties of almost sure convergene, it is best to insert another example with C-infinity transformation on the real line. I can do so later. Limit-theorem (talk) 13:57, 15 July 2015 (UTC)Reply

I was the one who added that comment. The support is not positive since a binomial random variable can be zero. Btyner (talk) 17:49, 23 August 2015 (UTC)Reply

The example doesn't state that it is an approximation of the variance of a binomial, rather that it is an approximation of the variance of the limiting normal distribution whose variance does exist. I'd recommend removing the disclaimer. 2602:306:3844:5400:90C5:53FB:DA9D:CDDA (talk) 21:41, 20 April 2016 (UTC)Reply

I went ahead and reworded it in this edit, and removed the disclaimer. Btyner (talk) 00:49, 28 April 2016 (UTC)Reply