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July 4

Estimated Date of 2,000,000th Article

First, we all know that the 1,000,000th article was created on March 2, 2006 at 5:09 PM Central time. As of 2:20 AM Central on July 4, there was 1,863,700 articles.

The duration since the creation of the 1,000,000th article was 488 days, 9 hours, 11 minutes and 0 seconds as of the above date.

At the rate the articles are being created, which is:

863,700 / ((488 + (9*60+11) = 551 / 1440) = 488.38263888888888888888888888888889) = 1768.4903827969587826030079443063 articles per day.

We should have the 2,000,000th article created on:

1,000,000 / 1768.4903827969587826030079443063 = 565.45402210129546010060077444585 days after the creation of the 1,000,000th article.

To convert to hours: .45402210129546010060077444585 * 24 = 10.8965304310910424144185867004 hours

To minutes: .8965304310910424144185867004 * 60 = 53.791825865462544865115202024 minutes

To seconds: .791825865462544865115202024 * 60 = 47.50955192775269190691212144 seconds

Which calculates that: The 2,000,000th article will estimatably be created 565 days, 10 hours, 53 minutes, and ~48 seconds after the 1,000,000th article.

Therefore, the estimated date and time of the 2,000,000th article creation is:

Wednesday, September 19, 2007

4:02:48 AM (CST)

Is this an accurate estimate, or did I miss another formula? If I did, can you recalculate it? --70.133.218.43 07:58, 4 July 2007 (UTC)[reply]

The thing is, Wikipedia article creation is more exponential than linear. Titoxd^{(?!? - cool stuff)} 08:05, 4 July 2007 (UTC)[reply]

... and an exponential growth model based on these two data points predicts 543.75 days to go from 1m articles to 2m articles. Gandalf61 10:15, 4 July 2007 (UTC)[reply]

Perhaps Wikipedia's growth has looked exponential so far, but I doubt it will remain this way for long. There aren't that many things to write about, and I doubt they grow any faster than quadratic. -- Meni Rosenfeld (talk) 10:56, 4 July 2007 (UTC)[reply]

You may be interested in Wikipedia:Modelling Wikipedia's growth, a somewhat old article that suggests an exponential model, and Wikipedia:Size of Wikipedia, which suggests that in the last few years things have slowed down to something more linear. Personally I'd like to pass the whole thing through X-12-ARIMA to see if there's a decent (non-deterministic) model you can fit to it. Confusing Manifestation 11:40, 4 July 2007 (UTC)[reply]

See WP:2MP for estimates of other people. – b_jonas 14:41, 4 July 2007 (UTC)[reply]

Well now, we're just 2 months away from the tentative 2,000,000th article date! Someone ought to start a countdown. --70.179.170.119 05:05, 19 July 2007 (UTC)[reply]

According to Gandalf61, the estimated time & date of the 2,000,000th article creation is:

Tuesday, August 28, 2007

11:09 AM (CST)

Oh, I appear to have been 10 days and ~44 minutes off, while Gandalf was off by a few days more. If only we could find the most accurate formula to calculate when the next # of articles will arrive... --70.179.175.240 06:51, 11 September 2007 (UTC)[reply]

What is the upperbound of E(exp{-alpha*x^2})

I am having trouble to quantify the upperbound of $E(exp{-\alpha x^{2}})$ , where alpha is non-negative coefficient, and E(.) is the expected value of the argument. The lower bound can be easily found based on Jensen's inequality, i.e., E(exp{-alpha*x^2}) \geq exp(-alpha*[E{x}]^2), but the upperbound is more difficult to find. Does anyone have idea about what the upperbound is? If the general upperbound is hard to be solved, the upperbound for the case of a large alpha is of my interest. 141.83.61.65 13:37, 4 July 2007 (UTC)[reply]

If

E(X)

and

Var(X)

are given, I have a hunch that

E(e^{-\alpha X^{2}})

might be maximal when:

X={\begin{cases}0&{\mbox{with probability }}1-{\frac {E(X)^{2}}{E(X^{2})}}\\{\frac {E(X^{2})}{E(X)}}&{\mbox{with probability }}{\frac {E(X)^{2}}{E(X^{2})}}\end{cases}}

In this case,

E(e^{-\alpha X^{2}})

is, of course,

1+{\frac {E(X)^{2}}{E(X^{2})}}\left(e^{-\alpha \left({\frac {E(X^{2})}{E(X)}}\right)^{2}}-1\right)

, which reduces to

1-{\frac {E(X)^{2}}{E(X^{2})}}

if

\alpha

is very large. -- Meni Rosenfeld (talk) 14:05, 4 July 2007 (UTC)[reply]

Actually, that's not true. A higher value (perhaps this one is the maximum?) can be obtained with

X={\begin{cases}a&{\mbox{with probability }}p_{1}\\b&{\mbox{with probability }}p_{2}\end{cases}}

where

p_{1}+p_{2}=1

,

p_{1}a+p_{2}b=E(X)

,

p_{1}a^{2}+p_{2}b^{2}=E(X^{2})

and

a

is chosen to be optimal. I don't think the final solution can be represented algebraically, but it can be approximted one way or another. -- Meni Rosenfeld (talk) 14:26, 4 July 2007 (UTC)[reply]

Further results: I have run simulations and realized that $E(e^{-\alpha x^{2}})$ is upperbounded by ${\frac {1}{\alpha }}e^{-(E{(x)})^{2}}$ when $\alpha$ is large enough. However, I cannot prove it mathematically as well as quantify how large $\alpha$ should be, depending on $E(X)$ and $Var(X)$ . Does anybody have idea about this? 141.83.61.65 07:46, 5 July 2007 (UTC)[reply]

What information is given here, really? We have

\operatorname {E} (X)

and

\operatorname {Var} (X)

it seems. Is that all, or is perhaps the distribution of

X

fully known? —Bromskloss 12:53, 5 July 2007 (UTC)[reply]

X can be normally distributed or log-normally distributed random variables. The later case is of my interest. Thus $E(x)=e^{\mu +\sigma ^{2}/2}$ and the k-th moment of x is $e^{k\mu +k^{2}\sigma ^{2}/2}$ . Therefore, Var(x), E(x), moments, and distribution are known. 141.83.61.65 14:08, 5 July 2007 (UTC)[reply]