– R-Blogger

Mein Schüler hat heute gefragt, wie er die AIC-Statistik (Akaike ‚ s InformationCriteria) für die Modellauswahl interpretieren soll. Am Ende haben wir einen Rcode ausgeblendet, um zu demonstrieren, wie die AIC für ein einfaches GLM (Generallinear model) berechnet wird. Ich denke immer, wenn Sie die Ableitung von astatistic verstehen können, ist es viel einfacher, sich daran zu erinnern, wie man es benutzt.,

Wenn Sie nun die Ableitung des AIC googeln, werden Sie wahrscheinlich auf eine Menge Mathematik stoßen. Aber die Prinzipien sind wirklich nicht so komplex. Also hierWir passen einige einfache GLMs, dann leiten Sie ein Mittel ab, um das „Beste“ zu wählen.

Überspringen Sie bis zum Ende, wenn Sie nur die Grundprinzipien durchgehen möchten.

Bevor wir jedoch die AIC verstehen können, müssen wir diestatistische Methodik der Lebensgrundlagen verstehen.,

Wahrscheinlichkeiten erklären

Angenommen, Sie haben einige Daten, die normalerweise mit einem Mittelwert von 5und einer sd von 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")

Es ist nur eine normale Verteilung.
Denken Sie dazu darüber nach, wie Sie die Wahrscheinlichkeit vonmultiple (unabhängige) Ereignisse. Sagen wir, die Chance, dass ich an einem bestimmten Tag mit dem Fahrrad zur Arbeit fahre, ist 3/5 und die Chance, dass es regnet, ist 161/365 (likevancing!,), dann ist die chance, ich werde Reiten in der Regen 3/5 *161/365 = 1/4, so dass ich am besten tragen Sie einen Mantel, wenn Reiten in Vancouver.
Wir können dasselbe für den Lebensunterhalt tun, multiplizieren Sie einfach die Wahrscheinlichkeit ofeach einzelnen y Wert und wir haben die Gesamtwahrscheinlichkeit. Dies wird eine sehr kleine Zahl sein, weil wir viele kleine Zahlen miteinander multiplizieren., Ein Trick, den wir verwenden, besteht also darin, das Protokoll der Wahrscheinlichkeiten zu summieren, anstatt sie zu multiplizieren:
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))

Die predict(m1) gibt die Linie der besten Passform an, dh den Mittelwert von ygiven für jeden x1-Wert. Wir verwenden dann predict, um die Wahrscheinlichkeiten für jedes Modell zu ermitteln:
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
Für m1 gibt es drei Parameter, einen Intercept, eine Steigung und eine Standardabweichung. Berechnen wir nun den AIC für alle drei Modelle:
-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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