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studentul meu a întrebat astăzi cum să interpreteze statistica AIC (Akaike ‘ s InformationCriteria) pentru selectarea modelului. Am ajuns să batem niște coduri Rcode pentru a demonstra cum se calculează AIC pentru un simplu GLM (model generalliniar). Întotdeauna cred că dacă puteți înțelege derivarea astatistică, este mult mai ușor să vă amintiți cum să o utilizați.,

acum, dacă Google derivare a AIC, vă sunt susceptibile de a rula în o mulțime de matematică. Dar principiile nu sunt chiar atât de complexe. Deci, aici vom potrivi câteva GLM-uri simple, apoi vom obține un mijloc de a alege „cel mai bun” unul.

Treci la final dacă vrei doar să treci peste principiile de bază.

înainte de a putea înțelege AIC, trebuie să înțelegemmetodologia statistică a probabilităților.,

Explicând likelihoods

Spune că ai niște date care nu sunt normal distribuite, cu o medie de 5si un sd de 3:

set.seed(126)n 
Now we want to estimate some parameters for the population that 
 wassampled from, like its mean and standard devaiation (which we know hereto be 5 and 3, but in the real world you won’t know that).
We are going to use frequentist statistics to estimate those parameters.Philosophically this means we believe that there is ‘one true value’ foreach parameter, and the data we observed are generated by this truevalue.
m1 
The estimate of the mean is stored here 
 =4.38, the estimatedvariance here 
= 5.91, or the SD 
=2.43. Just to be totally clear, we also specified that we believe thedata follow a normal (AKA "Gaussian”) distribution.
We just fit a GLM asking R to estimate an intercept parameter (
),which is simply the mean of 
. We also get out an estimate of the SD(= $\sqrt variance$) You might think its overkill to use a GLM toestimate the mean and SD, when we could just calculate them directly.
Well notice now that R also estimated some other quantities, like theresidual deviance and the AIC statistic.
summary(m1)#### Call:## glm(formula = y ~ 1, family = "gaussian")#### Deviance Residuals:## Min 1Q Median 3Q Max## -5.7557 -0.9795 0.2853 1.7288 3.9583#### Coefficients:## Estimate Std. Error t value Pr(>|t|)## (Intercept) 4.3837 0.3438 12.75 
You might also be aware that the deviance is a measure of model fit,much like the sums-of-squares. Note also that the value of the AIC issuspiciously close to the deviance. Despite its odd name, the conceptsunderlying the deviance are quite simple.
As I said above, we are observing data that are generated from apopulation with one true mean and one true SD. Given we know haveestimates of these quantities that define a probability distribution, wecould also estimate the likelihood of measuring a new value of 
 thatsay = 7.
To do this, we simply plug the estimated values into the equation forthe normal distribution and ask for the relative likelihood of 
. Wedo this with the R function 
sdest 
Formally, this is the relative likelihood of the value 7 given thevalues of the mean and the SD that we estimated (=4.8 and 2.39respectively if you are using the same random seed as me).
You might ask why the likelihood is greater than 1, surely, as it comesfrom a probability distribution, it should be y values, so the probability of any given value will be zero.The relative likelihood on the other hand can be used to calculate theprobability of a range ofvalues.
So you might realise that calculating the likelihood of all the datawould be a sensible way to measure how well our ‘model’ (just a mean andSD here) fits the data.
Here’s what the likelihood looks like:
plot(y, dnorm(y, mean = coef(m1), sd = sdest), ylab = "Likelihood")

este doar o distribuție normală.pentru a face acest lucru, gândiți-vă cum ați calcula probabilitateamultiple (independente) evenimente. Spuneți că șansa de a merge cu bicicleta pentru a lucra laorice zi dată este de 3/5, iar șansa de a ploua este de 161/365 (likeVancouver!,), atunci șansa pe care o voi călări în ploaie este de 3/5 *161/365 = aproximativ 1/4, așa că cel mai bine port o haină dacă călătoresc în Vancouver.
putem face același lucru pentru likelihoods, pur și simplu multiplica probabilitatea fiecarei individuale y valoare și avem totală probabilitatea. Acesta va fiun număr foarte mic, pentru că înmulțim o mulțime de numere mici între ele., Deci, un truc pe care o folosim este suma jurnalul de likelihoods loc multiplicarea lor:
y_lik 
The larger (the less negative) the likelihood of our data given themodel’s estimates, the ‘better’ the model fits the data. The deviance iscalculated from the likelihood and for the deviance smaller valuesindicate a closer fit of the model to the data.
The parameter values that give us the smallest value of the-log-likelihood are termed the maximum likelihood estimates.
Comparing alternate hypotheses with likelihoods
Now say we have measurements and two covariates, 
 and 
, eitherof which we think might affect y:
a 
So x1 is a cause of y, but x2 does not affect y. How would we choosewhich hypothesis is most likely? Well one way would be to compare modelswith different combinations of covariates:
m1 
Now we are fitting a line to y, so our estimate of the mean is now theline of best fit, it varies with the value of x1. To visualise this:
plot(x1, y)lines(x1, predict(m1))

predict(m1) dă linia cea mai potrivită, adică valoarea medie a ygiven fiecare x1 valoare. Apoi folosim anticipa pentru a obține likelihoods pentru eachmodel:
sm1 
The likelihood of 
 is larger than 
, which makes sense because
 has the ‘fake’ covariate in it. The likelihood for 
 (which hasboth x1 and x2 in it) is fractionally larger than the likelihood 
,so should we judge that model as giving nearly as good a representationof the data?
Because the likelihood is only a tiny bit larger, the addition of 
has only explained a tiny amount of the variance in the data. But wheredo you draw the line between including and excluding x2? You run into asimilar problem if you use R^2 for model selection.
So what if we penalize the likelihood by the number of paramaters wehave to estimate to fit the model? Then if we include more covariates(and we estimate more slope parameters) only those that account for alot of the variation will overcome the penalty.
What we want a statistic that helps us select the most parsimoniousmodel.
The AIC as a measure of parsimony
One way we could penalize the likelihood by the number of parameters isto add an amount to it that is proportional to the number of parameters.First, let’s multiply the log-likelihood by -2, so that it is positiveand smaller values indicate a closer fit.
LLm1 
Why its -2 not -1, I can’t quite remember, but I think just historicalreasons.
Then add 2*k, where k is the number of estimated parameters.
-2*LLm1 + 2*3## 257.2428
Pentru m1 există trei parametri, unul de interceptare, pe o pantă și onestandard abatere. Acum, să calculăm AIC pentru toate cele trei modele:
-2*LLm1 + 2*3## 257.2428LLm2 
We see that model 1 has the lowest AIC and therefore has the mostparsimonious fit. Model 1 now outperforms model 3 which had a slightlyhigher likelihood, but because of the extra covariate has a higherpenalty too.
AIC basic principles
So to summarize, the basic principles that guide the use of the AIC are:
 
 
Lower indicates a more parsimonious model, relative to a model fitwith a higher AIC.
 
 
 
It is a relative measure of model parsimony, so it only hasmeaning if we compare the AIC for alternate hypotheses (= differentmodels of the data).
 
 
 
We can compare non-nested models. For instance, we could compare alinear to a non-linear model.
 
 
 
The comparisons are only valid for models that are fit to the same responsedata (ie values of y).
 
 
 
Model selection conducted with the AIC will choose the same model asleave-one-out cross validation (where we leave out one data pointand fit the model, then evaluate its fit to that point) for largesample sizes.
 
 
 
You shouldn’t compare too many models with the AIC. You will runinto the same problems with multiple model comparison as you wouldwith p-values, in that you might by chance find a model with thelowest AIC, that isn’t truly the most appropriate model.
 
 
 
When using the AIC you might end up with multiple models thatperform similarly to each other. So you have similar evidenceweights for different alternate hypotheses. In the example above m3is actually about as good as m1.
 
 
 
You should correct for small sample sizes if you use the AIC withsmall sample sizes, by using the AICc statistic.
 
 
 Assuming it rains all day, which is reasonable for Vancouver.
 
 
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