Gebruiker:Pjetter/klad

<Londenp> Hi. Question for someone who knows the Parser functions: Is there a way how I can find out with parser language if a certain article is in a category. 
Something like if AAA is in CAT:BBB then display a certain image?
<Splarka> probably need DPL for that (depending on what you wanna do exactly)
<Londenp> In fact it is for Wikibooks
<Londenp> We have categorized the books in fases 0 to 4
<Splarka> or in JS or CSS
<Londenp> I would like to show on the front page a bookname together with a symbol for the development states
<Londenp> I couldn't find a solution with magic words and a parser
<Splarka> Londenp: well, what categorizes it? a template? or the [[category]] added directly?
<Londenp> with a template
<Splarka> so, like {{devstate|3}} ?
<Splarka> you can use the template to categorize, and generate the image
<Splarka> say you had {{devstate|3}} for example
<Splarka> [[Category:Development state {{{1|0}}}]]
<Splarka> {{#switch:{{{1|none}}}
<Londenp> OK
<Duesentrieb> Splarka: but he wants the image not on the page that is categorized, but on a pages that links to the page that is categorized
<Splarka> |0=[[Image:Devstate_0.png|40px]]
<Splarka> |1=[[Image:Devstate_1.png|40px]]
<Splarka> etc
<Splarka> really? hmmmm'
<Londenp> Duesentrieb exactly
<Splarka> DPL or javascript
<Duesentrieb> don't know if something like that exists.in any case, it would be tricky with parser cache
<Splarka> javascript could do it in a quick query to the API
<Londenp> We have Dynamicpagelist installed
<Duesentrieb> disabling it is the obvious solution, that would suck for the front page though
<Splarka> (but on the front page that would be messy)
<Splarka> DPL should be able to do it then
<Londenp> not DPL alas
<Splarka> which wikibooks?
<Londenp> nl
<Londenp> Rereading the remark from Duesentrieb it is still a little different
<Splarka> useful how it doesn't say what version of dpl it has
* Splarka grumbles
<Londenp> it is the old old one
<Londenp> http://www.mediawiki.org/wiki/Extension:DynamicPageList/old This one
<Splarka> drat, don't think you can do what you want with it then
<Londenp> That is what I thought
<Londenp> It is very useful though, but the most modern version of DPL is much better
<Londenp> but I was not allowed to let that be installed on nl.wikibooks.org
<Londenp> might try again thoug on bugzilla
<Londenp> So my idea is this {{template|bookname}} for each book
<Londenp> this will show a link to the book and display a image for the status of development of this book
<Londenp> this will automatically be updated if the status is changed
<Splarka> only way I can think of is the API, but on the mainpage that would be a lot of queries to it (and it isn't cached)
<Splarka> or a custom extension
<Londenp> The information for the status of the book is with a DPL-thing in a hidden category and the template would look in which 
hidden category for the status it is and then display the according image
<Londenp> OK too bad then
<Splarka> OR.. you could create subpages for each book, that indicate the status
<Splarka> such as {{Somebook/status 1}}
<Splarka> and use #ifexist
<Splarka> but that'd be... messy
<Londenp> That is a good idea
<Londenp> Sparkla Thanks I will try that then
<Splarka> uhoh
<Splarka> don't you hate it when someone is threatening to jump off a bridge, and you say "go for it" sarcastically.. and then they do...
<Splarka> Londenp: you could also create one subpage, like [[Bookname/status]]
<Londenp> You might give me another choice :-=)
<Splarka> and have that /status contain the image that you want to transclude
<Splarka> and then always transclude [[{{{1}}}/status]] if it exists
<Splarka> that way you don't have to create and delete pages continually
<Londenp> yep
<Splarka> {{#ifexist:{{{1}}}/status|{{{{{1}}}/status}}|[[Image:Nostatus.gif]]}}
<Londenp> In fact we have something like an Infobox with the book that contains all information already
* Splarka nods
<Londenp> I just don't know how to substract that
<Splarka> heh
<Splarka> well, then...
<Splarka> what if /status only contained the number
<Splarka> 0 to 4
<Splarka> and on the infobox:
<Splarka> [[Category:Development status {{{{{FULLPAGENAME}}/status}}]]
<Splarka> and on the main page:
<Splarka> {{#ifexist:{{{1}}}/status|[[Image:Status_{{{{{1}}}/status}}.gif]]|[[Image:Nostatus.gif]]}}
<Splarka> then you just update that /status page, and both the main page and the infobox are changed
<Londenp> OK, looks good
<Londenp> but I need a little time to understand this (not being educated)
* Splarka nods
<Londenp> Sparkla thanks, I am going to study your proposals and work it out. When I have some more questions I will return.
<Splarka> thanks for the warning ^_^

test 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890 1234567890

Citaten Papier

• The paper burns, but the words fly away - Ben Joseph Akiba
• What the world really needs is more love and less paper work - Pearl Bailey
• Anything on paper is obsolete - Craig Bruce
• This paper will no doubt be found interesting by those who take an interest in it - John Dalton
• When the weight of the paper equals the weight of the airplane, only then you can go flying - Donald Douglas
• When you sell a man a book you don't sell him just 12 ounces of paper and ink and glue-you sell him a whole new life - Christopher Morley
• A specification that will not fit on one page of 8.5x11 inch paper cannot be understood - Mark Ardis
• A child's life is like a piece of paper on which every person leaves a mark - Chinese Proverb

Version	Release Date	Comments
1.7	13/03/2007	Ignore non-list lines when generating blacklist
1.6	15/01/2007	Support multiple-language translations
1.5	16/12/2006	Don't block all usernames when the blacklist contains blank lines Use Unicode-friendly regular expressions Don't show errors when the blacklist contains only comments (above fixes from Brion Vibber)
1.4	19/06/2006	Fix fatal error
1.3	06/07/2006	Cache blacklist in memcached or similar, if available
1.2	25/04/2006	Performance rewrite Allow users with the uboverride permission to pass the blacklist
1.1	08/03/2006	Make compatible with MediaWiki 1.5.8 Allow commenting out lines with #
1.0	09/01/2006	Initial version

Boekenplank en boekenkast

Links:

• de:Wikibooks:Bücherregale - interessant is de opdeling vooral in werkgroepen per vakgebied en de organisatiestructuur
• en:Wikibooks:Departments - optisch niet bijzonder interessant
• fr:Wikilivres:Tous les livres - eigenlijk geen boekenplank en geen organisatie naar groepen, maar gebruikmakend van de categorie-structuur

[[{{{Vorighoofdstuk}}}|Vorig hoofdstuk]]

[[{{{Vorigepagina}}}|Vorige pagina]]

[[{{{Volgendepagina}}}|Volgende pagina]]

[[{{{Volgend hoofdstuk}}}|Volgend hoofdstuk]]

[[{{{Inhoudsopgave}}}|Inhoudsopgave]]

[[{{{boek}}}|Papier]]

visible text godsdienstige beweging in de 16e eeuw onder leiding van Maarten Luther en Johannes Calvijn

math

$\operatorname{MSE}(\hat{\theta})=\operatorname{E}[(\hat{\theta}-\theta)^2]$

Normal multiplication turns this into:

$\operatorname{MSE}(\hat{\theta})=\operatorname{E}[\hat{\theta}^2 - 2\hat{\theta}{\theta} +\theta^2]$

The expectation of theta is exactly theta, so you can write it like that:

$\operatorname{MSE}(\hat{\theta})=\operatorname{E}[\hat{\theta}^2] - 2{\theta}\operatorname{E}[\hat{\theta}] +\theta^2$

Add two parameters: one with a negative sign and the other with a positive sign (the sum of both is 0, but you can use that in additional calculations)

$\operatorname{MSE}(\hat{\theta})=\operatorname{E}[\hat{\theta}^2] - \operatorname{E}[\hat{\theta}]^2 + \operatorname{E}[\hat{\theta}]^2 - 2{\theta}\operatorname{E}[\hat{\theta}] +\theta^2$

Now the second part is the usual (a-b)^2, which can be written as a^2 - 2ab + b^2. You just work the other way around.

$\operatorname{MSE}(\hat{\theta})=\operatorname{E}[\hat{\theta}^2] - \operatorname{E}[\hat{\theta}]^2 + (\operatorname{E}[\hat{\theta}] -\theta)^2$

And the first part is the variance of theta-hat and the second part is the bias. This can be written as:

$\operatorname{MSE}(\hat{\theta})=\operatorname{V}(\hat{\theta}) + (bias(\hat{\theta}))^2$

Now when theta-hat is unbiassed this means that the expectation of theta-hat is equal to theta and therefore the bias is zero.
And that means that the MSE of theta-hat is exactly the variance of theta-hat.

If we take for theta the mu and for theta-hat the X-bar, you will use the normal variance for a n-sample from a population (X) with mean μ and variance σ2 you will get:

$\operatorname{MSE}=\left(\frac{\sigma}{\sqrt{n}}\right)^2$

Another calculation will be as follows: We start again with:

$\operatorname{MSE}(\hat{\theta})=\operatorname{E}[(\hat{\theta}-\theta)^2]$

Now we replace theta-hat with X-bar and theta with mu and you get:

$\operatorname{MSE}(\bar{X})=\operatorname{E}[(\bar{X}-\mu)^2]$

We will replace X-bar in this equation with the following:

$\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i = \frac{1}{n} (X_1+\cdots+X_n).$

So you get:

$\operatorname{MSE}(\bar{X})=\operatorname{E}[\frac{1}{n} (X_1+\cdots+X_n)-\mu)^2]$

This you can write differently according to:

$\operatorname{MSE}(\bar{X})=\operatorname{E}[\frac{X_1+\cdots+X_n}{n}-\mu)^2]$

and

$\operatorname{MSE}(\bar{X})=\operatorname{E}[\frac{X_1+\cdots+X_n}{n}-\frac{n}{n}\mu)^2]$

which will lead to:

$\operatorname{MSE}(\bar{X})=\operatorname{E}[\frac{X_1+\cdots+X_n-{n}\mu)}{n}^2]$

We take the n in the denominator out and this becomes squarred:

$\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\operatorname{E}[X_1+\cdots+X_n-{n}\mu)^2]$

which turns into:

$\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\operatorname{E}[(X_1-\mu) +\cdots+(X_n-\mu))^2]$

Now the expectation is redistributed in the equation:

$\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\operatorname{E}[(X_1-\mu)^2] +\cdots + \frac{1}{n^2}\operatorname{E}[(X_n-\mu)^2]$

We know that the definition for sigma is

$\operatorname{E}[(X_1-\mu)^2] = \sigma_1^2$

so you get:

$\operatorname{MSE}(\bar{X})=\frac{1}{n^2}\sigma_1^2 +\cdots + \frac{1}{n^2}\sigma_n^2=\frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}$

This is the variance of the unbiassed sample distribution of the mean

regression

$\hat{y} = 0.0027 + 0.7916{x}$

$\hat{y} = 49.48 + 1.96{x}$

$y_i = \beta x_i + \varepsilon_i \,$

The sum of squares to be minimized is

$S = \sum_{i=1}^{n} \left(Y_i - \hat\beta x_i\right)^2$

so

$S = \sum_{i=1}^{n} \left(Y_i^2 -2 Y_i \hat\beta x_i\ + \hat\beta x_i^2\right)$

So we do some derivation and you get for minimization the derivative is set to 0:

$\frac {\delta} {\delta {\beta}} \sum_{i=1}^{n} \left(Y_i^2 -2 Y_i \hat\beta x_i\ + \hat\beta x_i^2\right) = \sum_{i=1}^{n} -2 Y_i x_i + 2 \hat\beta x_i^2 = 0$

Divide both sides by -2 and redistribute the summation any will get:

$\sum_{i=1}^{n} \hat\beta x_i^2 = \sum_{i=1}^{n} Y_i x_i$

Take beta out and divide by Sum xi2:

$\hat\beta = \frac {\sum_{i=1}^{n} Y_i x_i} {\sum_{i=1}^{n} x_i^2}$

And therefore the least squares estimator for β, is given by

$\hat \beta=\frac{\sum_i^{n} x_i Y_i}{\sum_i^{n} x_i^2}$

$\hat \beta=\frac{\sum_i^{n} x_i y_i}{\sum_i^{n} x_i^2}$

Biasedness

$B(\widehat{\theta}) = \operatorname{E}(\widehat{\theta}) - \theta$

$B(\widehat{\theta}) = 0$

So check if

$\operatorname{E}(\widehat{\theta}) = \theta$

Let us use this on the estimator in a):

$\widehat{\beta} = \frac{\sum_i^{n} x_i Y_i}{\sum_i^{n} x_i^2}$

Now let us fill in the proposed population model into this equation:

$\operatorname{Y}_i = \beta x_i + \epsilon_i$

and you will get:

$\widehat{\beta} = \frac{\sum_i^{n} x_i (\beta x_i + \epsilon_i) }{\sum_i^{n} x_i^2}$

Which is:

$\widehat{\beta} = \frac{\sum_i^{n} (\beta x_i^2 + \epsilon_i x_i) }{\sum_i^{n} x_i^2}$

And this can be written as:

$\widehat{\beta} = \frac{\sum_i^{n} \beta x_i^2} {\sum_i^{n} x_i^2} + \frac{\sum_i^{n} \epsilon_i x_i}{\sum_i^{n} x_i^2}$

$\widehat{\beta} = \beta + \frac{\sum_i^{n} \epsilon_i x_i}{\sum_i^{n} x_i^2}$

According to biasedness equation above we can take the expectation from both sides according to:

$\operatorname{E} (\widehat{\beta}) = \beta + \operatorname{E} (\frac{\sum_i^{n} \epsilon_i x_i}{\sum_i^{n} x_i^2})$

and as

$\operatorname{E} (\epsilon_i) = 0$

you get:

$\operatorname{E} (\widehat{\beta}) = \beta$

Therefore unbiasedness applies.

Sigma

First we start with the equation:

$y_i = \beta x_i + \varepsilon_i$

rearrange:

$\widehat \varepsilon_i = y_i - \widehat \beta x_i$

SSR

${SSR}=\sum_i^{n} ( \hat y_i - \bar y)^2 = \beta^2 \sum_i^{n} ( \hat x_i - \bar x)^2$

$\hat{\beta}=\frac{\sum(x_i-\bar{x})(y_i-\bar{y})}{\sum(x_i-\bar{x})^2}$

$\hat{\alpha}=\bar{y}-\hat{\beta}\bar{x}$

$\hat{\sigma}^2=\frac{SSR}{n-2}$

${t}=\frac{\hat \beta - \beta}{s_{\hat \beta}}$

sigma

$\hat{\epsilon}_i = y_i - \hat \beta x_i$

$\hat{\epsilon}_i = \left( \beta x_i + \epsilon _i \right) - \hat \beta x_i$

$\hat{\epsilon}_i = \left( \beta - \hat \beta\right) x_i + \epsilon _i$

$\hat{\sigma}^2 = \frac{1} { n-1} \sum \hat{\epsilon}_i^2$

Take the variance of beta-hat:

$V \left( \widehat{\beta} \right) = V \left( \frac{\sum_i^{n} x_i Y_i}{\sum_i^{n} x_i^2} \right)$

Fill in the estimator of a):

$V \left( \widehat{\beta} \right) = V \left( \frac{\sum_i^{n} x_i Y_i}{\sum_i^{n} x_i^2} \right)$

Now use the equation for Y_i:

$y_i = \beta x_i + \varepsilon_i$

and fill that in the previous equation and you get:

$V \left( \widehat{\beta} \right) = V \left( \frac{\sum_i^{n} x_i ( \beta x_i + \varepsilon_i )}{\sum_i^{n} x_i^2} \right)$

Two typical properties of variance when X and Y are independent are the following:

${Var} \left( X + Y \right) = Var (X) + Var (Y)$

and

${Var} \left( aX + b \right) = a^2 Var (X)$

this will be used:

$V \left( \widehat{\beta} \right) = V \left( \frac{\sum_i^{n} ( \beta x_i^2 + \varepsilon_i x_i)}{\sum_i^{n} x_i^2} \right)$

$V \left( \widehat{\beta} \right) = V \left( \frac{\sum_i^{n} \beta x_i^2}{\sum_i^{n} x_i^2} + \frac{\sum_i^{n} \varepsilon_i x_i}{\sum_i^{n} x_i^2} \right)$

$V \left( \widehat{\beta} \right) = V \left( \frac{\sum_i^{n} \beta x_i^2}{\sum_i^{n} x_i^2} + \frac{\sum_i^{n} \varepsilon_i x_i}{\sum_i^{n} x_i^2} \right)$

$V \left( \widehat{\beta} \right) = V \left( \beta + \frac{\sum_i^{n} \varepsilon_i x_i}{\sum_i^{n} x_i^2} \right)$

$V \left( \widehat{\beta} \right) = x_i^2 V \left( \frac{\sum_i^{n} \varepsilon_i}{\sum_i^{n} x_i^2} \right)$

Using the definition given for Variance of epsilon:

$Var \left( {\varepsilon_i} \right) = \sigma^2$

$V \left( \widehat{\beta} \right) = \frac{\sigma^2} {\sum_i^{n} x_i^2}$

Nav

Vorig hoofdstuk ___ Papier

Vorige pagina ___ Gebruiker:Londenp

Volgende pagina ___ Hoofdpagina

Volgend hoofdstuk ___ Wikijunior

Inhoudsopgave ___ Kookboek

HTML ___ HTML

Sjabloon

Afbeelding:Crystal Clear action player play.png

Question 1

Question 2

Question 3

Question 4