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> https://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

$S=\sum _{i=1}^{n}\left(Y_{i}^{2}-2Y_{i}{\hat {\beta }}x_{i}\ +{\hat {\beta }}x_{i}^{2}\right)$

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

${\frac {\delta }{\delta {\beta }}}\sum _{i=1}^{n}\left(Y_{i}^{2}-2Y_{i}{\hat {\beta }}x_{i}\ +{\hat {\beta }}x_{i}^{2}\right)=\sum _{i=1}^{n}-2Y_{i}x_{i}+2{\hat {\beta }}x_{i}^{2}=0$

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:

$Var\left({\varepsilon _{i}}\right)=\sigma ^{2}$

$V\left({\widehat {\beta }}\right)={\frac {\sigma ^{2}}{\sum _{i}^{n}x_{i}^{2}}}$

Sjabloon

Afbeelding:Crystal Clear action player play.png

Question 1

Question 2

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Question 4