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This morning during the lecture on nonlinear regression, I mentioned (very) briefly the case of convex regression. Since I forgot to mention the codes in R, I will publish them here. Assume that $y_i=m(\mathbf{x}_i)+\varepsilon_i$ where $m:\mathbb{R}^d\rightarrow \mathbb{R}$ is some convex function.

Then $m$ is convex if and only if $\forall\mathbf{x}_1,\mathbf{x}_2\in\mathbb{R}^d$, $\forall t\in[0,1]$, [latex display=”true”]m(t\mathbf{x}_1+[1-t]\mathbf{x}_2) \leq tm(\mathbf{x}_1)+[1-t]m(\mathbf{x}_2)[/latex]Hidreth (1954) proved that if[latex display=”true”] m^\star=\underset{m \text{ convex}}{\text{argmin}}\left\lbrace\sum_{i=1}^n \big(y_i-m(\mathbf{x_i})\big)^2\right\rbrace[/latex]then $\mathbf{\theta}^\star=(m^\star(\mathbf{x_1}),\cdots,m^\star(\mathbf{x_n}))$ is unique.

Let $\mathbf{y}=\mathbf{\theta}+\mathbf{\varepsilon}$, then [latex display=”true”]\mathbf{\theta}^\star=\underset{\mathbf{\theta}\in \mathcal{K}}{\text{argmin}}\left\lbrace\sum_{i=1}^n \big(y_i-\theta_i)\big)^2\right\rbrace[/latex]where[latex display=”true”]\mathcal{K}=\{\mathbf{\theta}\in\mathbb{R}^n:\exists m\text{ convex },m(\mathbf{x}_i)=\theta_i\}[/latex]. I.e. $\mathbf{\theta}^\star$ is the projection of $\mathbf{y}$ onto the (closed) convex cone $\mathcal{K}$. The projection theorem gives existence and unicity.

For convenience, in the application, we will consider the real-valued case, $m:\mathbb{R}\rightarrow \mathbb{R}$, i.e. $y_i=m(x_i)+\varepsilon_i$. Assume that observations are ordered $x_1\leq x_2\leq\cdots \leq x_n$. Here [latex display=”true”]\mathcal{K}=\left\lbrace\mathbf{\theta}\in\mathbb{R}^n:\frac{\theta_2-\theta_1}{x_2-x_1}\leq \frac{\theta_3-\theta_2}{x_3-x_2}\leq \cdots \leq \frac{\theta_n-\theta_{n-1}}{x_n-x_{n-1}}\right\rbrace[/latex]

Hence, quadratic program with $n-2$ linear constraints.

$m^\star$ is a piecewise linear function (interpolation of consecutive pairs $(x_i,\theta_i^\star)$).

If $m$ is differentiable, $m$ is convex if [latex display=”true”]m(\mathbf{x})+ \nabla m(\mathbf{x})^{\text{T}}\cdot[\mathbf{y}-\mathbf{x}] \leq m(\mathbf{y})[/latex]

More generally, if $m$ is convex, then there exists $\xi_{\mathbf{x}}\in\mathbb{R}^n$ such that [latex display=”true”]m(\mathbf{x})+ \xi_{\mathbf{x}}^{\text{ T}}\cdot[\mathbf{y}-\mathbf{x}] \leq m(\mathbf{y})[/latex]
$\xi_{\mathbf{x}}$ is a subgradient of $m$ at ${\mathbf{x}}$. And then [latex display=”true”]\partial m(\mathbf{x})=\big\lbrace m(\mathbf{x})+ \xi^{\text{ T}}\cdot[\mathbf{y}-\mathbf{x}] \leq m(\mathbf{y}),\forall \mathbf{y}\in\mathbb{R}^n\big\rbrace[/latex]

Hence, $\mathbf{\theta}^\star$ is solution of [latex display=”true”]\text{argmin}\big\lbrace\|\mathbf{y}-\mathbf{\theta}\|^2\big\rbrace[/latex][latex display=”true”]\text{subject to }\theta_i+\xi_i^{\text{ T}}[\mathbf{x}_j-\mathbf{x}_i]\leq\mathbf{\theta}_j,~\forall i,j[/latex] and $\xi_1,\cdots,\xi_n\in\mathbb{R}^n$. Now, to do it for real, use cobs package for constrained (b)splines regression,

library(cobs)

To get a convex regression, use

plot(cars)
x = cars$speed y = cars$dist
rc = conreg(x,y,convex=TRUE)
lines(rc, col = 2)

Here we can get the values of the knots

rc

Call:  conreg(x = x, y = y, convex = TRUE)
Convex regression: From 19 separated x-values, using 5 inner knots,
7,    8,    9,   20,   23.
RSS =  1356; R^2 = 0.8766;
needed (5,0) iterations

and actually, if we use them in a linear-spline regression, we get the same output here

reg = lm(dist~bs(speed,degree=1,knots=c(4,7,8,9,,20,23,25)),data=cars)
u = seq(4,25,by=.1)
v = predict(reg,newdata=data.frame(speed=u))
lines(u,v,col="green")

Let us add vertical lines for the knots

abline(v=c(4,7,8,9,20,23,25),col="grey",lty=2)