Log-linear model: Difference between revisions

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{{unreferenced|date=July 2010}}
{{sections|date=October 2012}}
{{tone|date=July 2012}}
{{expert-subject|mathematics|date=August 2011}}
}}


{{merge to|Coupon collector's problem|date=August 2011}}
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The '''[[coupon collector's problem]]''' can be solved in several different ways. The [[generating function]] approach  is a [[combinatorial]] technique that allows to obtain precise results.
 
This technique uses the [[probability generating function]] (PGF) <math>G(z)</math> where <math>[z^q] G(z)</math> is the probability that ''q'' steps are taken to collect the ''n'' coupons i.e. <math>T=q</math>
 
and the expectation is given by
:<math>\operatorname{E}(T) = \left. \frac{\mathrm{d}}{\mathrm{d}z} G(z) \right|_{z=1}.</math>
 
<math>G(z)</math> can be calculated explicitly as follows :
:<math> G(z) =
\frac{n}{n} z
\frac{1}{1-\frac{1}{n} z}
\frac{n-1}{n} z
\frac{1}{1-\frac{2}{n} z}
\frac{n-2}{n} z
\frac{1}{1-\frac{3}{n} z}
\frac{n-3}{n} z
\cdots
\frac{1}{1-\frac{n-1}{n} z}
\frac{n-(n-1)}{n} z.</math>
 
In this equation, It can be found some features :
:<math> \frac{1}{1 - p z} = 1 + p z + p^2 z^2 + p^3 z^3 + \cdots,</math>
This is the PGF of an event that has probability ''p'' occurring zero or more times, with the exponent of ''z'' counting the number of times. We split the sequence of coupons into segments. A new segment begins every time a new coupon is retrieved for the first time. The PGF is the product of the PGFs of the individual segments.
 
Applying this to <math>G(z)</math>, It represents the following sequence of events:
* retrieve the first coupon (no restrictions at this time)
* retrieve the first coupon some number of times
* retrieve the second coupon (probability <math>(n-1)/n</math>)
* retrieve a mix of the first and second coupons some number of times
* retrieve the third coupon (probability <math>(n-2)/n</math>)
* retrieve a mix of the first, second, and third coupons some number of times
* retrieve the fourth coupon (probability <math>(n-3)/n</math>)
* <math>\ldots</math>
* retrieve the last coupon (probability <math>(n-(n-1))/n</math>).
 
In the following, <math>H_n</math> and <math>H_n^{(2)}</math> denote [[harmonic number]]s.
 
The function <math>G(z)</math>, first simplified before deriving the expectation, can be expressed as follows :
:<math> G(z) = z^n
\frac{n-1}{n-z}
\frac{n-2}{n-2z}
\frac{n-3}{n-3z}
\cdots
\frac{n-(n-1)}{n-(n-1)z}
</math>.
 
In this function, the following formula can be used to obtain the derivative of <math>G(z)</math> :
:<math> \frac{\mathrm{d}}{\mathrm{d}z} \frac{n-k}{n-kz} = \frac{k(n-k)}{(n-kz)^2}</math>
 
so thus
:<math>\frac{\mathrm{d}}{\mathrm{d}z} G(z) =
G(z)
\left(
\frac{n}{z}
+ \frac{1}{n-z}
+ \frac{2}{n-2z}
+ \frac{3}{n-3z}
\cdots
+ \frac{n-1}{n-(n-1)z}
\right)
</math>.
 
From this equation, it can be found <math>\operatorname{E}(T)</math> as following :
:<math>
\operatorname{E}(T) = \left. \frac{\mathrm{d}}{\mathrm{d}z} G(z) \right|_{z=1}
= G(1)
\left(
n
+ \frac{1}{n-1}
+ \frac{2}{n-2}
+ \frac{3}{n-3}
\cdots
+ \frac{n-1}{n-(n-1)}
\right)
</math>
 
or
 
:<math>\operatorname{E}(T) = n + \sum_{k=1}^{n-1} \frac{k}{n-k}.</math>
 
The next formula is useful for simplifying the <math>\operatorname{E}(T)</math>:
:<math> \sum_{k=1}^{n-1} \frac{k}{n-k} =
\sum_{k=1}^{n-1} \left( \frac{k}{n-k} - \frac{n}{n-k} \right) + n H_{n-1} =
n H_{n-1} - (n-1)</math>
 
so that
:<math> \operatorname{E}(T) = n + n H_{n-1} - (n-1) = n H_{n-1} + 1 = n H_n,
\quad \mbox{QED.}</math>
 
The PGF <math>G(z)</math> makes it possible to obtain an exact value for the [[variance]].
:<math>\operatorname{Var}(T) =
\operatorname{E}(T(T-1)) + \operatorname{E}(T) - \operatorname{E}(T)^2,</math>
This formula consists entirely of [[factorial moment]]s that can be calculated from the PGF. The value of <math>\operatorname{E}(T)</math> is already found.
 
The remainder can be calculated as follows :
 
:<math>\operatorname{E}(T(T-1)) =
\left. \left( \frac{\mathrm{d}}{\mathrm{d}z} \right)^2 G(z) \right|_{z=1}.</math>
 
This derivative is
:<math>
\begin{align}
& G(z)
\left(
\frac{n}{z}
+ \frac{1}{n-z}
+ \frac{2}{n-2z}
+ \frac{3}{n-3z}
\cdots
+ \frac{n-1}{n-(n-1)z}
\right)^2 \\
+ & \;
G(z)
\left(
- \frac{n}{z^2}
+ \frac{1^2}{(n-z)^2}
+ \frac{2^2}{(n-2z)^2}
+ \frac{3^2}{(n-3z)^2}
\cdots
+ \frac{(n-1)^2}{(n-(n-1)z)^2}
\right)
\end{align}
</math>
 
Evaluation at <math>z=1</math> yields
:<math>
\begin{align} &
n^2 H_n^2 - n + \sum_{k=1}^{n-1} \frac{k^2}{(n-k)^2} =
n^2 H_n^2 - n + \sum_{k=1}^{n-1} \frac{(n-k)^2}{k^2} \\ = & \;
n^2 H_n^2 - n + n^2 H_{n-1}^{(2)} - 2 n H_{n-1} + (n-1).
\end{align}
</math>
 
As all value has been found, the conclusion can be reached at the following:
:<math>
\begin{align}
\operatorname{Var}(T) & = \;
n^2 H_n^2 - 1 + n^2 H_{n-1}^{(2)} - 2 n H_{n-1} + n H_{n-1} + 1 - n^2 H_n^2 \\ & = \;
n^2 H_{n-1}^{(2)} - n H_{n-1} < \frac{\pi^2}{6} n^2, \quad \text{QED.}
\end{align}</math>
 
{{DEFAULTSORT:Coupon Collector's Problem (Generating Function Approach)}}
[[Category:Named probability problems]]
[[Category:Games (probability)]]
[[Category:Combinatorics]]
[[Category:Articles containing proofs]]

Latest revision as of 12:43, 12 May 2014

Bonus: WP Twin and WP Twin Auto Backup: (link to ) While not a theme, I think this software is essential if you are maintaining your Wordpress blog or regularly create new blog sites. What I advise you do next is save the backup data file to a remote place like a CD-ROM, external disk drive if you have one or a provider such as Dropbox. One really cool features about this amazing and free wp plugin is that the code it generates is completely portable. If you're using Wordpress and want to make your blog a "dofollow" blog, meaning that links from your blog pass on the benefits of Google pagerank, you can install one of the many dofollow plugins available. provided by Word - Press Automatic Upgrade, so whenever you need to update the new version does not, it automatically creates no webmaster.

Thus, it is imperative that you must Hire Word - Press Developers who have the expertise and proficiency in delivering theme integration and customization services. Infertility can cause a major setback to the couples due to the inability to conceive. If you have any queries relating to where by and how to use wordpress backup, you can call us at our internet site. This plugin is a must have for anyone who is serious about using Word - Press. Now, I want to anxiety that not every single query will be answered. That's a total of 180$ for each Wordpress theme if you sell 3 links and keep the designer link for your own website, or 240$ if you sell all links.

Saying that, despite the launch of Wordpress Express many months ago, there has still been no sign of a Wordpress video tutorial on offer UNTIL NOW. After sending these details, your Word - Press blog will be setup within a few days. You've got invested a great cope of time developing and producing up the topic substance. These frequent updates have created menace in the task of optimization. Search engine optimization pleasant picture and solution links suggest you will have a much better adjust at gaining considerable natural site visitors.

Word - Press installation is very easy and hassle free. But the Joomla was created as the CMS over years of hard work. To do this, you should link your posts to other relevant posts that you've created. If you just want to share some picture and want to use it as a dairy, that you want to share with your friends and family members, then blogger would be an excellent choice. The popularity of Word - Press has increased the demand for Word - Press themes and these themes sells like hot cake on the internet.

Every single module contains published data and guidelines, usually a lot more than 1 video, and when pertinent, incentive links and PDF files to assist you out. An ease of use which pertains to both internet site back-end and front-end users alike. As a result, it is really crucial to just take aid of some experience when searching for superior quality totally free Word - Press themes, Word - Press Premium Themes for your web site. Thus, Word - Press is a good alternative if you are looking for free blogging software. Definitely when you wake up from the slumber, you can be sure that you will be lagging behind and getting on track would be a tall order.

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