# Some More Results on the Theory of Statistical Learning

Yesterday, I did mention a popular graph discussed when studying theoretical foundations of statistical learning. But there is usually another one, which is the following,

As previously, it is a graph with the risk on the $y$-axis, the red line being on the training sample, and the black line on the validation sample, as a function of something that can be related to the complexity of the model.

Let us get back to the underlying formulas. On the traning sample, we have some empirical risk, defined as

$R_n=\frac{1}{n}\sum_{i=1}^n L(y_i,\widehat{m}_n(\boldsymbol{x}))$

for some loss function $L$. From the law of large numbers,

$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n U_i = \mathbb{E}[U]$

when the $U_i$‘s are i.i.d., and $U_i\sim U$. But here we look for

$\lim_{n\rightarrow\infty} \underbrace{\frac{1}{n}\sum_{i=1}^n L(y_i,\widehat{m}_n(\boldsymbol{x})) }_{R_n}$

It is difficult to say something about the limit, since the $(y,\boldsymbol{x}_i)$‘s are independent, but not the $L(y_i,\widehat{m}_n(\boldsymbol{x}_i))$‘s, because of $\widehat{m}_n(\cdot)$ (which depends on the entire sample).

But if we look at the empirical risk on a validation sample

$\lim_{n\rightarrow\infty} \underbrace{\frac{1}{n}\sum_{i=1}^n L(\tilde{y}_i,\widehat{m}_n(\tilde{\boldsymbol{x}}))}_{\tilde{R}_i} =\mathbb{E}[L(Y,\boldsymbol{X})]$

One can prove that, with probability $\alpha$,

$\widehat{R}_n\leq R_n +\sqrt{\frac{{{VC}}[\log(2n/d)+1]-\log[\alpha/4]}{n}}$

which depends on $n$ (as discussed in the previous post), but also about that $VC$ parameter, the so-called Vapnik-Chervonenkis dimension. The part on the right is the blue curve on the graph, above.

I won’t spend hours on that dimension $VC$, but the idea is that this dimension is related to the model complexity. For instance, in dimension one (one covariate), if $m(\cdot)$ is a polynomial of degree $d$, then $VC=d+1$. In dimension two (two covariates), if $m(\cdot)$ is a (bivariate) polynomial of degree $d$, then $VC=(d+1)(d+2)/2$, while it would be $2(d+1)$ if $m(\cdot)$ is additive, with two polynomials of degree $d$.

Let us try to get a graph which looks like the one above, using the same idea as the one in our previous post.

MissClassU=rep(NA,25)
MissClassV=rep(NA,25)
n=200
U=data.frame(X1=runif(n),X2=runif(n))
p=(U[,1]+U[,2])/2
U$Y=rbinom(n,size=1,prob=p) V=data.frame(X1=runif(n),X2=runif(n)) p=(V[,1]+V[,2])/2 V$Y=rbinom(n,size=1,prob=p)
for(s in 1:25){
reg=glm(Y~poly(X1,s)+poly(X2,s),data=U,
family=binomial)
pd=function(x1,x2) predict(reg,newdata=data.frame(X1=x1,X2=x2),type="response")>.5
MissClassU[s]=mean(abs(pd(U$X1,U$X2)-U$Y)) MissClassV[s]=mean(abs(pd(V$X1,V$X2)-V$Y))
}

If we plot the missclassification rate, as a function of the polynomial degree, in purple on the validation sample, and in black on the training sample, we get

Again, it is on one sample, only. We can run it on hundreds, and see how the average risk of misclassification changes with complexity.

MCU=rep(NA,500)
MCV=rep(NA,500)

missclassification=function(s){
for(i in 1:500){
U=data.frame(X1=runif(n),X2=runif(n))
p=(U[,1]+U[,2])/2
U$Y=rbinom(n,size=1,prob=p) reg=glm(Y~bs(X1,s)+bs(X2,s),data=U, family=binomial) pd=function(x1,x2) predict(reg,newdata=data.frame(X1=x1,X2=x2),type="response")>.5 MCU[i]=mean(abs(pd(U$X1,U$X2)-U$Y))
V=data.frame(X1=runif(n),X2=runif(n))
p=(V[,1]+V[,2])/2
V$Y=rbinom(n,size=1,prob=p) MCV[i]=mean(abs(pd(V$X1,V$X2)-V$Y))
}
MissClassV=mean(MCU)
MissClassU=mean(MCV)
return(c(MissClassU,MissClassV))
}

Here, we cannot see the optimal dimension, because our risk on the validation samples keeps increasing. Which makes sence since our data are generated from a linear model, so the optimal transformation should be optained with linear transformation (and not polynomials with higher degrees).

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