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December 17[edit]

Confusion about the Central Limit Theorem[edit]

If a bunch of random variables $X_{i}$ are independently and identically distributed with an exponential distribution, their sum apparently follows a Gamma distribution.

But according to the central limit theorem, for $X_{i}$ with mean zero and variance $\sigma ^{2}$ , ${\sqrt {n}}\left({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\right)$ becomes normally distributed $N(0,\sigma ^{2})$ as n gets sufficiently large, regardless of the distribution of $X_{i}$ .

Doesn't this imply that the sum $\sum _{i=1}^{n}X_{i}$ becomes approximately normally distributed, $N(0,n\sigma ^{2})$ ?

So, what's the deal with the distribution of a sum of i.i.d. exponentials? Does the sum of i.i.d. exponential random variables follow a Gamma distribution, but the sum of samples drawn randomly from an exponential distribution converge to a normal distribution? I don't understand the distinction. What's going on here? Thorstein90 (talk) 00:09, 17 December 2012 (UTC)[reply]

Well, exponentially distributed random variables don't have zero mean, so you wouldn't expect the sum of exponentially distributed rvs to be approximately

N(0,{\text{anything}})

. The sum of n iid exponentially distributed random variables with parameter λ=1 is

S_{n}\sim \Gamma (n,1)

, whose mean and variance are both n. This escapes to infinity, so it obviously can't become normal. You have to subtract to remove the mean, and divide to normalize the variance. Indeed, you can check directly from the pdf of the gamma distribution that

\lim _{n\to \infty }{\frac {S_{n}-n}{\sqrt {n}}}\sim N(0,1)

. There is no need for the central limit theorem to show this. I just noticed, the graph at the top of Gamma distribution shows that something like this is true. As the parameter k of the Gamma distribution increases, the shape of the graph of the pdf looks more and more like a normal distribution—if you ignore the scale on the x-axis. That's intuitively what the central limit theorem is telling you. Sławomir Biały (talk) 13:24, 17 December 2012 (UTC)[reply]

The sum of n iid variables with finite mean and variance

\mu

and

\sigma ^{2}

is distributed approximately normally for large n, with mean

n\mu

and variance

n\sigma ^{2}

.

The sum of n iid exponential variables with mean

\mu

(and variance

\mu ^{2}

) follows the Gamma distribution with

k=n

and

\theta =\mu

. Which means that for large k, the Gamma distribution is roughly normal, with mean

k\theta

and variance

k\theta ^{2}

. -- Meni Rosenfeld (talk) 14:24, 17 December 2012 (UTC)[reply]

Rotundas[edit]

Look at the Rotunda (geometry) article. Because the pentagonal rotunda is the only one made entirely out of regular polygons, this question might be tough. But I want to know if anyone can add an image of a hexagonal rotunda (for which having non-regular polygons as lateral faces is a must) to the article?? Georgia guy (talk) 01:40, 17 December 2012 (UTC)[reply]

Check out User talk:Tomruen#Rotunda problem for info on what a hexagonal rotunda would require. Anyone know what it actually looks like (having non-regular polygons is indeed a must.) Georgia guy (talk) 02:23, 17 December 2012 (UTC)[reply]

Could you just cut a soccer ball in half ? That has hexagons and pentagons. StuRat (talk) 03:22, 17 December 2012 (UTC)[reply]

But is that the kind of figure we're talking about?? Go to rotunda (geometry) and, using the rule in the article, guess what faces a hexagonal rotunda would contain. Georgia guy (talk) 13:11, 17 December 2012 (UTC)[reply]

A hexagonal rotunda would have to have a dodecagonal base (speaking of it as though it were in the same orientation as the pentagonal rotunda picture there), with each side having either a pentagon or two vertex-connected triangles on it (alternating), and then a hexagonal top. It is almost certainly possible with irregular polygons (but I haven't checked and don't have a picture). Double sharp (talk) 09:45, 20 December 2012 (UTC)[reply]

BTW, shouldn't the plural form of rotunda be rotundae? :-) Double sharp (talk) 09:45, 20 December 2012 (UTC)[reply]

According to Wiktionary, the only plural is "rotundas". Maybe it should also include "rotundae" as a British plural -- I don't know -- but American English tends to avoid "-ae" plurals, preferring e.g. "hyperbolas" over "hyperbolae" etc. (although AmEng uses "alumnae" as the plural for "alumna"). Duoduoduo (talk) 14:34, 20 December 2012 (UTC)[reply]

Yes, the only plural is "rotundas" in British English too (since 1753 at least) because it comes from a Latin adjective (Sancta Maria Rotunda) or possibly from the Italian rotonda (plural rotonde). Why do you think that the British make up false Latin plurals? Dbfirs 08:29, 22 December 2012 (UTC)[reply]

Noisy Chaos[edit]

I am familiar with the theory of dynamical chaos, but am interested in slightly different problems.

Suppose a n-dimensional set of dynamical equations ${\dot {x}}=f(x)$ is known to exhibit chaotic dynamics, I am interested in the related systems:

${\dot {x}}=f(x)+n(t)$ where $n(t)$ is stochastic noise, and $x(t)$ is observed directly.

and ${\dot {x}}=f(x)$ , but only $y(t)=x(t)+n(t)$ can be observed and $n(t)$ is again stochastic noise.

I would be appreciative of any useful suggested reading that I could do to learn more about such cases. Particularly regarding how to distinguish the two cases, and hoe the dynamical chaos is still evident given the introduction of noise.

Many thanks. — Preceding unsigned comment added by 123.136.64.14 (talk) 03:23, 17 December 2012 (UTC)[reply]

That's quite an interesting and useful thing to do, some industrial systems already do that for quality control I believe and there's also more glamorous stuff like diagnosing heart conditions.If you just google on distinguishing noise|random|sochastic from chaos you should get quite a few hits.

Our article on Stochastic differential equations is relevant to your problem. Bo Jacoby (talk) 13:46, 17 December 2012 (UTC).[reply]

On the question of how to spot chaos even in the presence of noise, see Chaos theory#Distinguishing random from chaotic data, correlation dimension, Lyapunov exponent, and the references therein. Duoduoduo (talk) 14:22, 17 December 2012 (UTC)[reply]