Tag Archives: Renglish

Function basis and regression

In the first part of the course on linear models, we’ve seen how to construct a linear model when the vector of covariates \boldsymbol{x} is given, so that \mathbb{E}(Y|\boldsymbol{X}=\boldsymbol{x}) is either simply \boldsymbol{x}^\top\boldsymbol{\beta} (for standard linear models) or a functional of \boldsymbol{x}^\top\boldsymbol{\beta} (in GLMs). But more generally, we can consider transformations of the covariates, so that a linear model can be used. In a very general setting, consider \sum_{j=1}^m\beta_j h_j(\boldsymbol{x})with h_j:\mathbb{R}^p\rightarrow\mathbb{R}. The standard linear model is obtained when m=p and h_j(\boldsymbol{x})=x_j , but of course, much more general models can be obtained, for instance with h_k(\boldsymbol{x})=x_j^2 or h_k(\boldsymbol{x})=x_{j}x_{j'}, that could be used to achieve high-order Taylor expansions. In that case, we will obtain the polynomial regression, that we will discuss first. We might also think of piecewise constant functions, h_k(\boldsymbol{x})=\boldsymbol{1}(x_j\in [a,b]) , that could be related to regression trees (but that is not in the scope in the STT5100 course). And if we go on step futher, we might think of piecewise linear or piecewise polynomial function, possibly with additional continuity constraints, that will lead us to spline basis.

  • Polynomial regression

For pedagogical purpose, when I talk about polynomial regression, I always have in mind (in the univariate case) y=\beta_0+\beta_1x+\beta_2x^2+\cdots+\beta_kx^k+\varepsilonbut if we use

lm(y~poly(x,k))

in R, the output is not the \beta_j‘s.

As discussed in Kennedy & Gentle (1980) Statistical Computing,

Recall that orthogonal polynomials are defined with respect to the classical inner-product (on the finite interval (a,b)){\displaystyle \langle f,g\rangle =\int _{a}^{b}f(x)g(x)~\mathrm {d} x} And a sequence of orthogonal polynomials is (P_n) where P_n is a polynomial of degree n, for all n, and such that P_m\perp P_n for all m\neq n. Note that those polyomials are orthogonal with respect to the inner product defined above, i.e. given some finite interval (a,b). But if (a,b) changes, the polynomials will be different.

A popular family of orthogonal polynomial, on finite interval (-1,+1) is the family of Legendre polynomials, satisfying{\displaystyle \int _{-1}^{1}P_{m}(x)P_{n}(x)~\mathrm {d} x=0}as soon as m\neq n. Those polynomials satisfy Bonnet’s recursion formula{\displaystyle (n+1)P_{n+1}(x)=(2n+1)xP_{n}(x)-nP_{n-1}(x)} or Rodrigues’ formula {\displaystyle P_{n}(x)={\frac {1}{2^{n}n!}}{\frac {d^{n}}{dx^{n}}}(x^{2}-1)^{n}}The first values are here{\displaystyle P_{0}(x)=1} {\displaystyle P_{1}(x)=x}{\displaystyle P_{2}(x)={\frac {3x^{2}-1}{2}}}{\displaystyle P_{3}(x)={\frac {5x^{3}-3x}{2}}} {\displaystyle P_{4}(x)={\frac {35x^{4}-30x^{2}+3}{8}}}

Interestingly, we can get those polynomial functions using

library(orthopolynom)
(leg4coef = legendre.polynomials(n=4))
[[1]]
1 
 
[[2]]
x 
 
[[3]]
-0.5 + 1.5*x^2 
 
[[4]]
-1.5*x + 2.5*x^3 
 
[[5]]
0.375 - 3.75*x^2 + 4.375*x^4

Of course, there are many families of orthogonal polynomials (Jacobi polynomials, Laguerre polynomials, Hermite polynomials, etc). Now, in R, there is the standard poly function, that we use in polynomial regression.

x = seq(-1,1,length=101)
y = poly(x,4)
y
                   1            2             3            4
  [1,] -1.706475e-01  0.215984813 -2.480753e-01  0.270362873
  [2,] -1.672345e-01  0.203025724 -2.183063e-01  0.216290298
...
[100,]  1.672345e-01  0.203025724  2.183063e-01  0.216290298
[101,]  1.706475e-01  0.215984813  2.480753e-01  0.270362873
attr(,"coefs")
attr(,"coefs")$alpha
[1] 3.157229e-17 2.655145e-16 9.799244e-17 5.368224e-16
 
attr(,"coefs")$norm2
[1]   1.0000000 101.0000000  34.3400000   9.3377328   2.4472330   0.6330176
 
attr(,"degree")
[1] 1 2 3 4
attr(,"class")
[1] "poly"   "matrix"

But these are not Legendre polynomials… As explained in 李哲源‘s post on stackoverflow, the idea is to start with P_{-1}(x)=0, P_{0}(x)=1 and P_{1}(x)=x, and then define \ell_n=\langle P_n,P_n\rangle  as well as \alpha_n=\langle P_nP_1,P_1\rangle/\ell_n=\langle P_n^2,P_1\rangle/\ell_i= and \beta_n=\ell_n/\ell_{n-1}. Finally, define recursively{\displaystyle P_{n}(x)=(x-\alpha_{n-1})P_{n-1}(x)-\beta_{i-1}P_{i-2}(x)}and its normalized version, \tilde{P}_{n}=P_n/\sqrt{\ell_n}. That is what poly computes.

So, for pedagogical purpose, I said that I like to use y=\boldsymbol{x}^\top\boldsymbol{\beta}+\varepsilon where\boldsymbol{x}=(1,x,x^2,\cdots,xˆ{k-1},x^k)And actually, when using poly, we use the QR decomposition of that matrix. As discussed in in 李哲源‘s post, we can almost reproduce the poly function using

my_poly - function (x, degree = 1) {
    xbar = mean(x)
    x = x - xbar
    QR = qr(outer(x, 0:degree, "^"))
    X = qr.qy(QR, diag(diag(QR$qr), length(x), degree + 1))[, -1, drop = FALSE]
    X2 = X * X
    norm2 = colSums(X * X)   
    alpha = drop(crossprod(X2, x)) / norm2
    beta = norm2 / (c(length(x), norm2[-degree]))
    colnames(X) = 1:degree
    scale = sqrt(norm2)
    X = X * rep(1 / scale, each = length(x))
    X}

Nevertheless, the two models are equivalent. More precisely,

plot(cars)
reg1 = lm(dist~speed+I(speed^2)+I(speed^3),data=cars)
reg2 = lm(dist~poly(speed,3),data=cars)
u = seq(3,26,by=.1)
v1 = predict(reg1,newdata=data.frame(speed=u))
v2 = predict(reg2,newdata=data.frame(speed=u))
lines(u,v1,col="blue")
lines(u,v2,col="red",lty=2)

We have exactly the same prediction here

v1[u==15]
     121 
38.43919 
v2[u==15]
     121 
38.43919

And probably also quite interesting : the coefficients do not have the same interpretation (since we do not have the same basis), but the p-value for the highest degree is exactly the same here ! Here the two models reject, with the same confidence, the polynomial of degree three,

summary(reg1)
 
Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) -19.50505   28.40530  -0.687    0.496
speed         6.80111    6.80113   1.000    0.323
I(speed^2)   -0.34966    0.49988  -0.699    0.488
I(speed^3)    0.01025    0.01130   0.907    0.369
 
Residual standard error: 15.2 on 46 degrees of freedom
Multiple R-squared:  0.6732,	Adjusted R-squared:  0.6519 
F-statistic: 31.58 on 3 and 46 DF,  p-value: 3.074e-11
 
summary(reg2)
 
Coefficients:
                Estimate Std. Error t value Pr(>|t|)    
(Intercept)        42.98       2.15  19.988  < 2e-16 ***
poly(speed, 3)1   145.55      15.21   9.573  1.6e-12 ***
poly(speed, 3)2    23.00      15.21   1.512    0.137    
poly(speed, 3)3    13.80      15.21   0.907    0.369    
---
Signif. codes:  0***0.001**0.01*0.05 ‘.’ 0.1 ‘ ’ 1
 
Residual standard error: 15.2 on 46 degrees of freedom
Multiple R-squared:  0.6732,	Adjusted R-squared:  0.6519 
F-statistic: 31.58 on 3 and 46 DF,  p-value: 3.074e-11
  • B-splines regression (and GAMs)

Splines are also important in regression models, especially when we start talking about Generalized Additive Models. See Perperoglou, Sauerbrei, Abrahamowicz & Schmid (2019) for a review. In the univariate case, I introduce (linear) splines through positive parts, in the sense thaty=\beta_0+\beta_1x+\beta_2(x-s_1)_++\cdots+\beta_k(x-s_{k-1})_++\varepsilonwhere (x-s)_+ equals 0 if x<s and x-s if x>s. Those functions are nice since they are continuous, so the model is continuous (the weighted sum of continuous functions is continuous). And we can go one step further, with y=\beta_0+\beta_1x+\beta_2x^2+\beta_3(x-s_1)^2_++\cdots+\beta_k(x-s_{k-2})^2_++\varepsilonwith quadratic splines, or y=\beta_0+\beta_1x+\beta_2x^2+\beta_3x^3+\beta_4(x-s_1)^3_++\cdots+\beta_k(x-s_{k-3})^3_++\varepsilonfor cubic splines. Interestingly, quadratic splines are not only continuous, but their first derivative is also continuous (and the second one for cubic splines). So the knot discontinuity is s_1,s_2,\cdots is now invisible…

I like those models since they are easy to interprete. For example, the simple model \beta_1 x+\beta_2(x-s)_+ is the following piecewise linear function, continuous, with a “rupture” at knot s.

Observe also the following interpretation: for small values of x, there is a linear increase, with slope \beta_1, and for lager values of x, there is a linear decrease, with slope \beta_1+\beta_2. Hence, \beta_2 is interpreted as a change of the slope.

Unfortunately, it is now what R is using when using the bs function in R, which are the standard B-splines. Just to visualize (I will skip the maths here), with R, we have

library(splines)
clr6 = c("#1b9e77","#d95f02","#7570b3","#e7298a","#66a61e","#e6ab02")
x = seq(5,25,by=.25)
B = bs(x,knots=c(10,20),Boundary.knots=c(5,55),degre=1)
matplot(x,B,type="l",lty=1,lwd=2,col=clr6)
B=bs(x,knots=c(10,20),Boundary.knots=c(5,55),degre=2)
matplot(x,B,type="l",col=clr6,lty=1,lwd=2)

while the functions I mentioned were (more or less) the following

pos = function(x,s) (x-s)*(x&gt;s)
par(mfrow=c(1,2))
clr6 = c("#1b9e77","#d95f02","#7570b3","#e7298a","#66a61e","#e6ab02")
x = seq(5,25,by=.25)
B = cbind(pos(x,5),pos(x,10),pos(x,20))
matplot(x,B,type="l",lty=1,lwd=2,col=clr6)
pos2 = function(x,s) (x-s)^2*(x&gt;s)
B = cbind(pos(x,5)*20,pos2(x,5),pos2(x,10),pos2(x,20))
matplot(x,B,type="l",col=clr6,lty=1,lwd=2)

And as for the polynomial regression, the two models are equivalent. For example

plot(cars)
reg1 = lm(dist~speed+pos(speed,10)+pos(speed,20),data=cars)
reg2 = lm(dist~bs(speed,degree=1,knots=c(10,20)),data=cars)
v1 = predict(reg1,newdata=data.frame(speed=u))
v2 = predict(reg2,newdata=data.frame(speed=u))
lines(u,v1,col="blue")
lines(u,v2,col="red",lty=2)

or more specifically

v1[u==15]
     121 
39.35747 
v2[u==15]
     121 
39.35747

So one more time, the two models are equivalent, but I still find the approach with the positive part more intuitive, and easy to understand. As well as the interpretation of coefficients,

summary(reg1)
 
Coefficients:
               Estimate Std. Error t value Pr(&gt;|t|)  
(Intercept)     -7.6305    16.2941  -0.468   0.6418  
speed            3.0630     1.8238   1.679   0.0998 .
pos(speed, 10)   0.2087     2.2453   0.093   0.9263  
pos(speed, 20)   4.2812     2.2843   1.874   0.0673 .
---
Signif. codes:  0***0.001**0.01*0.05 ‘.’ 0.1 ‘ ’ 1
 
Residual standard error: 15 on 46 degrees of freedom
Multiple R-squared:  0.6821,	Adjusted R-squared:  0.6613 
F-statistic: 32.89 on 3 and 46 DF,  p-value: 1.643e-11
 
summary(reg2)
 
Coefficients:
                                          Estimate Std. Error t value Pr(&gt;|t|)    
(Intercept)                                  4.621      9.344   0.495   0.6233    
bs(speed, degree = 1, knots = c(10, 20))1   18.378     10.943   1.679   0.0998 .  
bs(speed, degree = 1, knots = c(10, 20))2   51.094     10.040   5.089 6.51e-06 ***
bs(speed, degree = 1, knots = c(10, 20))3   88.859     12.047   7.376 2.49e-09 ***
---
Signif. codes:  0***0.001**0.01*0.05 ‘.’ 0.1 ‘ ’ 1
 
Residual standard error: 15 on 46 degrees of freedom
Multiple R-squared:  0.6821,	Adjusted R-squared:  0.6613 
F-statistic: 32.89 on 3 and 46 DF,  p-value: 1.643e-11

Here we can see directly that the first knot was not interesting (the slope did not change significantly) while the second one was…

Testing for a causal effect (with 2 time series)

A few days ago, I came back on a sentence I found (in a French newspaper), where someone was claiming that

“… an old variable explains 85% of the change in a new variable. So we can talk about causality”

and I tried to explain that it was just stupid : if we consider the regression of the temperature on day t+1 against the number of cyclist on day t, the R^2 exceeds 80%… but it is hard to claim that the number of cyclists on specific day will actually cause the temperature on the next day…

Nevertheless, that was frustrating, and I was wondering if there was a clever way to test for causality in that case. A popular one is Granger causality (I can mention a paper we published a few years ago where we use such a test, Tents, Tweets, and Events: The Interplay Between Ongoing Protests and Social Media). To explain that test, consider a bivariate time series (just like the one we have here), \boldsymbol{z}_t=(x_t,y_t), and consider some bivariate autoregressive model
{\displaystyle {\begin{bmatrix}x_{t}\\y_{t}\end{bmatrix}}={\begin{bmatrix}c_{1}\\c_{2}\end{bmatrix}}+{\begin{bmatrix}a_{1,1}&\textcolor{red}{a_{1,2}}\\\textcolor{blue}{a_{2,1}}&a_{2,2}\end{bmatrix}}{\begin{bmatrix}x_{t-1}\\y_{t-1}\end{bmatrix}}+{\begin{bmatrix}u_{t}\\v_{t}\end{bmatrix}}}where \boldsymbol{\varepsilon}_t=(u_t,v_t) is some bivariate white noise, in the sense that (i) {\displaystyle \mathbb{E} (\boldsymbol{\varepsilon}_{t})=\boldsymbol{0}} (the noise is centered) (ii) {\displaystyle \mathbb{E} (\boldsymbol{\varepsilon}_{t}\boldsymbol{\varepsilon}_{t}^\top)=\Omega } , so the variance matrix is constant, but possibly non-diagonal (iii) {\displaystyle \mathbb{E} (\boldsymbol{\varepsilon}_{t}\boldsymbol{\varepsilon}_{t-h}^\top)=\boldsymbol{0} } for all h\neq 0. Note that we can use the simplified expression{\displaystyle {\boldsymbol{z}_t=\boldsymbol{c}+\boldsymbol{A}\boldsymbol{z}_{t-1}+\boldsymbol{\varepsilon}_t}}Now, Granger test is based on several quantities. With off-diagonal terms of matrix \Omega, we have a so-called instantaneous causality, and since \Omega is symmetry, we will write x\leftrightarrow y. With off-diagonal terms of matrix \boldsymbol{A}, we have a so-called lagged causality, with either \textcolor{blue}{x\rightarrow y} or \textcolor{red}{x\leftarrow y} (and possibly both, if both terms are significant).

So I wanted to try on my two-variable problem.

df = read.csv("cyclistsTempHKI.csv")
dfts = cbind(C=ts(df$cyclists,start = c(2014, 1,2), frequency = 365),
             T=ts(df$meanTemp,start = c(2014, 1,2), frequency = 365))
library(vars)

I now have “time series” objects, and we can fit a VAR model,

var2 = VAR(dfts, p = 1, type = "const")
coefficients(var2)
$C
         Estimate   Std. Error   t value      Pr(&gt;|t|)
C.l1    0.8684009   0.02889424 30.054460 8.080226e-107
T.l1   70.3042012  20.07247411  3.502518  5.102094e-04
const 807.6394001 187.75472482  4.301566  2.110412e-05
 
$T
           Estimate   Std. Error   t value     Pr(&gt;|t|)
C.l1   0.0003865391 6.257596e-05  6.177118 1.540467e-09
T.l1   0.6611135594 4.347074e-02 15.208241 6.086394e-42
const -1.6413074565 4.066184e-01 -4.036481 6.446018e-05

For instant, we can run a causality, to test if the number of cyclists can cause the temperature (on the next day)

causality(var2, cause = "C")
$Granger
 
	Granger causality H0: C do not Granger-cause T
 
data:  VAR object var2
F-Test = 38.157, df1 = 1, df2 = 842, p-value = 1.015e-09

Here, we should clearly reject H_0, which is that there is no causal effect. Which is the way statisticians say that there should be some causal effect between the number of cyclist and the temperature…

So clearly, something is wrong here. Either it is some sort of superpower that cyclists are not aware of. Or this test that was used for forty years (Clive Granger even got a Nobel price for it) is not working. Or we missed something. Actually… I think we missed something here. Possibly because the series are not stationary. We can almost see it with

Phi = matrix(c(coefficients(var2)$C[1:2,1],coefficients(var2)$T[1:2,1]),2,2)
eigen(Phi)
eigen() decomposition
$values
[1] 0.9594810 0.5700335

where the highest eigenvalue is very close to one. But actually, we look here at the temperature…

plot(dfts)

so, at least, we should expect some seasonal unit root here. So let us use two techniques. The first one is a classical one-year difference, \Delta_{365}\boldsymbol{z}_t=\boldsymbol{z}_t-\boldsymbol{z}_{t-365}

var2 = VAR(diff(dfts,365), p = 1, type = "const")
coefficients(var2)
$C
          Estimate   Std. Error   t value     Pr(&gt;|t|)
C.l1     0.8376424   0.07259969 11.537823 1.993355e-16
T.l1    42.2638410  28.58783276  1.478386 1.449076e-01
const -507.5514795 219.40240747 -2.313336 2.440042e-02
 
$T
         Estimate   Std. Error   t value     Pr(&gt;|t|)
C.l1  0.000518209 0.0003277295 1.5812096 1.194623e-01
T.l1  0.598425288 0.1290511945 4.6371154 2.162476e-05
const 0.547828079 0.9904263469 0.5531235 5.823804e-01

The test on the fited VAR model yields

causality(var2, cause = "C") 
$Granger
 
	Granger causality H0: C do not Granger-cause T
 
data:  VAR object var2
F-Test = 2.5002, df1 = 1, df2 = 112, p-value = 0.1167

i.e., with a 11% p-value, we should reject the assumption that the number of cyclists cause the temperature (on the next day), and actually, we should also reject the other way

causality(var2, cause = "T") 
$Granger
 
	Granger causality H0: T do not Granger-cause C
 
data:  VAR object var2
F-Test = 2.1856, df1 = 1, df2 = 112, p-value = 0.1421

Nevertheless, if we look at the instantaneous causality, this one makes more sense

$Instant
 
	H0: No instantaneous causality between: T and C
 
data:  VAR object var2
Chi-squared = 13.081, df = 1, p-value = 0.0002982

The second idea would be to use a one day difference, \Delta_{1}\boldsymbol{z}_t=\boldsymbol{z}_t-\boldsymbol{z}_{t-1} and to fit a VAR model on that one

VARselect(diff(dfts,1), lag.max = 4, type="const")
$selection
AIC(n)  HQ(n)  SC(n) FPE(n) 
     3      3      2      3

but on that one, a VAR(1) model – with only one lag – might not be sufficient. It might be better to consider a VAR(3)

var2 = VAR(diff(dfts,1), p = 3, type = "const")

and on that one, one more time, we should reject the causal effect of the number of cyclists on the temperature (on the next day)

causality(var2, cause = "C")  
$Granger
 
	Granger causality H0: C do not Granger-cause T
 
data:  VAR object var2
F-Test = 0.67644, df1 = 3, df2 = 828, p-value = 0.5666

and this time, there could be a (lagged) causal effect of the temperature on the number of cyclists

causality(var2, cause = "T")  
$Granger
 
	Granger causality H0: T do not Granger-cause C
 
data:  VAR object var2
F-Test = 7.7981, df1 = 3, df2 = 828, p-value = 3.879e-05
 
$Instant
 
	H0: No instantaneous causality between: T and C
 
data:  VAR object var2
Chi-squared = 55.83, df = 1, p-value = 7.905e-14

but nothing instantaneously… So it looks like Granger causality performs well on that one !

Lasso Regression (home made)

Again, this post is related to my MAT7381 course, where we will see that it is actually possible to write our own code to compute Lasso regression, \min\left\lbrace\frac{1}{2}\|\mathbf{y}-\mathbf{X}\mathbf{\beta}\|_{\ell_2}^2+\lambda\|\mathbf{\beta}\|_{\ell_1}\right\rbraceWe have to define the soft-thresholding functionS(z,\gamma)=\text{sign}(z)\cdot(|z|-\gamma)_+=\begin{cases}z-\gamma&\text{ if }\gamma>|z|\text{ and }z<0\\z+\gamma&\text{ if }\gamma<|z|\text{ and }z<0 \\0&\text{ if }\gamma\geq|z|\end{cases}The R function would be

soft_thresholding = function(x,a){
sign(x) * pmax(abs(x)-a,0)
}

To solve our optimization problem, set\mathbf{r}_j=\mathbf{y} - \left(\beta_0\mathbf{1}+\sum_{k\neq j}\beta_k\mathbf{x}_k\right)=\mathbf{y}-\widehat{\mathbf{y}}^{(j)}
so that the optimization problem can be written, equivalently
\min\left\lbrace\frac{1}{2n}\sum_{j=1}^p [\mathbf{r}_j-\beta_j\mathbf{x}_j]^2+\lambda |\beta_j|\right\rbrace
hence\min\left\lbrace\frac{1}{2n}\sum_{j=1}^p \beta_j^2\|\mathbf{x}_j\|-2\beta_j\mathbf{r}_j^T\mathbf{x}_j+\lambda |\beta_j|\right\rbrace
and one gets
\beta_{j,\lambda} = \frac{1}{\|\mathbf{x}_j\|^2}S(\mathbf{r}_j^T\mathbf{x}_j,n\lambda)
or, if we develop
\beta_{j,\lambda} = \frac{1}{\sum_i x_{ij}^2}S\left(\sum_ix_{i,j}[y_i-\widehat{y}_i^{(j)}],n\lambda\right)
Again, if there are weights \mathbf{\omega}=(\omega_i), the coordinate-wise update becomes
\beta_{j,\lambda,{\color{red}{\omega}}} = \frac{1}{\sum_i {\color{red}{\omega_i}}x_{ij}^2}S\left(\sum_i{\color{red}{\omega_i}}x_{i,j}[y_i-\widehat{y}_i^{(j)}],n\lambda\right)
The code to compute this componentwise descent is

lasso_coord_desc = function(X,y,beta,lambda,tol=1e-6,maxiter=1000){
  beta = as.matrix(beta)
  X = as.matrix(X)
  omega = rep(1/length(y),length(y))
  obj = numeric(length=(maxiter+1))
  betalist = list(length(maxiter+1))
  betalist[[1]] = beta
  beta0list = numeric(length(maxiter+1))
  beta0 = sum(y-X%*%beta)/(length(y))
  beta0list[1] = beta0
  for (j in 1:maxiter){
    for (k in 1:length(beta)){
      r = y - X[,-k]%*%beta[-k] - beta0*rep(1,length(y))
      beta[k] = (1/sum(omega*X[,k]^2))*
        soft_thresholding(t(omega*r)%*%X[,k],length(y)*lambda)
    }
    beta0 = sum(y-X%*%beta)/(length(y))
    beta0list[j+1] = beta0
    betalist[[j+1]] = beta
    obj[j] = (1/2)*(1/length(y))*norm(omega*(y - X%*%beta - 
           beta0*rep(1,length(y))),'F')^2 + lambda*sum(abs(beta))
    if (norm(rbind(beta0list[j],betalist[[j]]) - 
             rbind(beta0,beta),'F') &lt; tol) { break } 
  } 
  return(list(obj=obj[1:j],beta=beta,intercept=beta0)) }

For instance, consider the following (simple) dataset, with three covariates

chicago = read.table("http://freakonometrics.free.fr/chicago.txt",header=TRUE,sep=";")

that we can “normalize” (or “standardize“)

X = model.matrix(lm(Fire~.,data=chicago))[,2:4]
for(j in 1:3) X[,j] = (X[,j]-mean(X[,j]))/sd(X[,j])
y = chicago$Fire
y = (y-mean(y))/sd(y)

To initialize the algorithm, use the OLS estimate

beta_init = lm(Fire~0+.,data=chicago)$coef

For instance

lasso_coord_desc(X,y,beta_init,lambda=.001)
$obj
[1] 0.001014426 0.001008009 0.001009558 0.001011094 0.001011119 0.001011119
 
$beta
          [,1]
X_1  0.0000000
X_2  0.3836087
X_3 -0.5026137
 
$intercept
[1] 2.060999e-16

and we can get the standard Lasso plot by looping,

Quantile Regression (home made, part 2)

A few months ago, I posted a note with some home made codes for quantile regression… there was something odd on the output, but it was because there was a (small) mathematical problem in my equation. So since I should teach those tomorrow, let me fix them.

Median

Consider a sample \{y_1,\cdots,y_n\}. To compute the median, solve\min_\mu \left\lbrace\sum_{i=1}^n|y_i-\mu|\right\rbracewhich can be solved using linear programming techniques. More precisely, this problem is equivalent to\min_{\mu,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^na_i+b_i\right\rbracewith a_i,b_i\geq 0 and y_i-\mu=a_i-b_i, \forall i=1,\cdots,n. Heuristically, the idea is to write y_i=\mu+\varepsilon_i, and then define a_i‘s and b_i‘s so that \varepsilon_i=a_i-b_i and |\varepsilon_i|=a_i+b_i, i.e. a_i=(\varepsilon_i)_+=\max\lbrace0,\varepsilon_i\rbrace=|\varepsilon|\cdot\boldsymbol{1}_{\varepsilon_i>0}andb_i=(-\varepsilon_i)_+=\max\lbrace0,-\varepsilon_i\rbrace=|\varepsilon|\cdot\boldsymbol{1}_{\varepsilon_i<0}denote respectively the positive and the negative parts.

Unfortunately (that was the error in my previous post), the expression of linear programs is\min_{\mathbf{z}}\left\lbrace\boldsymbol{c}^\top\mathbf{z}\right\rbrace\text{ s.t. }\boldsymbol{A}\mathbf{z}=\boldsymbol{b},\mathbf{z}\geq\boldsymbol{0}In the equation above, with the a_i‘s and b_i‘s, we’re not far away. Except that we have \mu\in\mathbb{R}, while it should be positive. So similarly, set \mu=\mu^+-\mu^- where \mu^+=(\mu)_+ and \mu^-=(-\mu)_+.

Thus, let\mathbf{z}=\big(\mu^+;\mu^-;\boldsymbol{a},\boldsymbol{b}\big)^\top\in\mathbb{R}_+^{2n+2}and then write the constraint as \boldsymbol{A}\mathbf{z}=\boldsymbol{b} with \boldsymbol{b}=\boldsymbol{y} and \boldsymbol{A}=\big[\boldsymbol{1}_n;-\boldsymbol{1}_n;\mathbb{I}_n;-\mathbb{I}_n\big]And for the objective function\boldsymbol{c}=\big(\boldsymbol{0},\boldsymbol{1}_n,-\boldsymbol{1}_n\big)^\top\in\mathbb{R}_+^{2n+2}

To illustrate, consider a sample from a lognormal distribution,

n = 101 
set.seed(1)
y = rlnorm(n)
median(y)
[1] 1.077415

For the optimization problem, use the matrix form, with 3n constraints, and 2n+1 parameters,

library(lpSolve) 
X = rep(1,n) 
A = cbind(X, -X, diag(n), -diag(n))
b = y
c = c(rep(0,2), rep(1,n),rep(1,n))
equal_type = rep("=", n) 
r = lp("min", c,A,equal_type,b)
head(r$solution,1)
[1] 1.077415

It looks like it’s working well…

Quantile

Of course, we can adapt our previous code for quantiles

tau = .3
quantile(y,tau)
      30% 
0.6741586

The linear program is now\min_{q^+,q^-,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^n\tau a_i+(1-\tau)b_i\right\rbracewith a_i,b_i,q^+,q^-\geq 0 and y_i=q^+-q^-+a_i-b_i, \forall i=1,\cdots,n. The R code is now

c = c(rep(0,2), tau*rep(1,n),(1-tau)*rep(1,n))
r = lp("min", c,A,equal_type,b)
head(r$solution,1)
[1] 0.6741586

So far so good…

Quantile Regression

Consider the following dataset, with rents of flat, in a major German city, as function of the surface, the year of construction, etc.

base=read.table("http://freakonometrics.free.fr/rent98_00.txt",header=TRUE)

The linear program for the quantile regression is now\min_{\boldsymbol{\beta}^+,\boldsymbol{\beta}^-,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^n\tau a_i+(1-\tau)b_i\right\rbracewith a_i,b_i\geq 0 and y_i=\boldsymbol{x}^\top[\boldsymbol{\beta}^+-\boldsymbol{\beta}^-]+a_i-b_i\forall i=1,\cdots,n and \beta_j^+,\beta_j^-\geq 0 \forall j=0,\cdots,k. So use here

require(lpSolve) 
tau = .3
n=nrow(base)
X = cbind( 1, base$area)
y = base$rent_euro
K = ncol(X)
N = nrow(X)
A = cbind(X,-X,diag(N),-diag(N))
c = c(rep(0,2*ncol(X)),tau*rep(1,N),(1-tau)*rep(1,N))
b = base$rent_euro
const_type = rep("=",N)
r = lp("min",c,A,const_type,b)
beta = r$sol[1:K] -  r$sol[(1:K+K)]
beta
[1] 148.946864   3.289674

Of course, we can use R function to fit that model

library(quantreg)
rq(rent_euro~area, tau=tau, data=base)
Coefficients:
(Intercept)        area 
 148.946864    3.289674

Here again, it seems to work quite well. We can use a different probability level, of course, and get a plot

plot(base$area,base$rent_euro,xlab=expression(paste("surface (",m^2,")")),
     ylab="rent (euros/month)",col=rgb(0,0,1,.4),cex=.5)
sf=0:250
yr=r$solution[2*n+1]+r$solution[2*n+2]*sf
lines(sf,yr,lwd=2,col="blue")
tau = .9
r = lp("min",c,A,const_type,b)
tail(r$solution,2)
[1] 121.815505   7.865536
yr=r$solution[2*n+1]+r$solution[2*n+2]*sf
lines(sf,yr,lwd=2,col="blue")

And we can adapt the later to multiple regressions, of course,

X = cbind(1,base$area,base$yearc)
K = ncol(X)
N = nrow(X)
A = cbind(X,-X,diag(N),-diag(N))
c = c(rep(0,2*ncol(X)),tau*rep(1,N),(1-tau)*rep(1,N))
b = base$rent_euro
const_type = rep("=",N)
r = lp("min",c,A,const_type,b)
beta = r$sol[1:K] -  r$sol[(1:K+K)]
beta
[1] -5542.503252     3.978135     2.887234

to be compared with

library(quantreg)
rq(rent_euro~ area + yearc, tau=tau, data=base)
 
Coefficients:
 (Intercept)         area        yearc 
-5542.503252     3.978135     2.887234 
 
Degrees of freedom: 4571 total; 4568 residual

On Cochran Theorem (and Orthogonal Projections)

Cochran Theorem – from The distribution of quadratic forms in a normal system, with applications to the analysis of covariance published in 1934 – is probably the most import one in a regression course. It is an application of a nice result on quadratic forms of Gaussian vectors. More precisely, we can prove that if \boldsymbol{Y}\sim\mathcal{N}(\boldsymbol{0},\mathbb{I}_d) is a random vector with d \mathcal{N}(0,1) variable then (i) if A is a (squared) idempotent matrix \boldsymbol{Y}^\top A\boldsymbol{Y}\sim\chi^2_r where r is the rank of matrix A, and (ii) conversely, if \boldsymbol{Y}^\top A\boldsymbol{Y}\sim\chi^2_r then A is an idempotent matrix of rank r. And just in case, A is an idempotent matrix means that A^2=A, and a lot of results can be derived (for instance on the eigenvalues). The prof of that result (at least the (i) part) is nice: we diagonlize matrix A, so that A=P\Delta P^\top, with P orthonormal. Since A is an idempotent matrix observe thatA^2=P\Delta P^\top=P\Delta P^\top=P\Delta^2 P^\topwhere \Delta is some diagonal matrix such that \Delta^2=\Delta, so terms on the diagonal of \Delta are either 0 or 1‘s. And because the rank of A (and \Delta) is r then there should be r 1‘s and d-r 1‘s. Now write\boldsymbol{Y}^\top A\boldsymbol{Y}=\boldsymbol{Y}^\top P\Delta P^\top\boldsymbol{Y}=\boldsymbol{Z}^\top \Delta\boldsymbol{Z}where \boldsymbol{Z}=P^\top\boldsymbol{Y} that satisfies\boldsymbol{Z}\sim\mathcal{N}(\boldsymbol{0},PP^\top) i.e. \boldsymbol{Z}\sim\mathcal{N}(\boldsymbol{0},\mathbb{I}_d). Thus \boldsymbol{Z}^\top \Delta\boldsymbol{Z}=\sum_{i:\Delta_{i,i}-1}Z_i^2\sim\chi^2_rNice, isn’t it. And there is more (that will be strongly connected actually to Cochran theorem). Let A=A_1+\dots+A_k, then the two following statements are equivalent (i) A is idempotent and \text{rank}(A)=\text{rank}(A_1)+\dots+\text{rank}(A_k) (ii) A_i‘s are idempotents, A_iA_j=0 for all i\neq j.

Now, let us talk about projections. Let \boldsymbol{y} be a vector in \mathbb{R}^n. Its projection on the space \mathcal V(\boldsymbol{v}_1,\dots,\boldsymbol{v}_p) (generated by those p vectors) is the vector \hat{\boldsymbol{y}}=\boldsymbol{V} \hat{\boldsymbol{a}} that minimizes \|\boldsymbol{y} -\boldsymbol{V} \boldsymbol{a}\| (in \boldsymbol{a}). The solution is\hat{\boldsymbol{a}}=( \boldsymbol{V}^\top \boldsymbol{V})^{-1} \boldsymbol{V}^\top \boldsymbol{y} \text{ and } \hat{\boldsymbol{y}} = \boldsymbol{V} \hat{\boldsymbol{a}}
Matrix P=\boldsymbol{V} ( \boldsymbol{V}^\top \boldsymbol{V})^{-1} \boldsymbol{V}^\top is the orthogonal projection on \{\boldsymbol{v}_1,\dots,\boldsymbol{v}_p\} and \hat{\boldsymbol{y}} = P\boldsymbol{y}.

Now we can recall Cochran theorem. Let \boldsymbol{Y}\sim\mathcal{N}(\boldsymbol{\mu},\sigma^2\mathbb{I}_d) for some \sigma>0 and \boldsymbol{\mu}. Consider sub-vector orthogonal spaces F_1,\dots,F_m, with dimension d_i. Let P_{F_i} be the orthogonal projection matrix on F_i, then (i) vectors P_{F_1}\boldsymbol{X},\dots,P_{F_m}\boldsymbol{X} are independent, with respective distribution \mathcal{N}(P_{F_i}\boldsymbol{\mu},\sigma^2\mathbb{I}_{d_i}) and (ii) random variables \|P_{F_i}(\boldsymbol{X}-\boldsymbol{\mu})\|^2/\sigma^2 are independent and \chi^2_{d_i} distributed.

We can try to visualize those results. For instance, the orthogonal projection of a random vector has a Gaussian distribution. Consider a two-dimensional Gaussian vector

library(mnormt)
r = .7
s1 = 1
s2 = 1
Sig = matrix(c(s1^2,r*s1*s2,r*s1*s2,s2^2),2,2)
Sig
Y = rmnorm(n = 1000,mean=c(0,0),varcov = Sig)
plot(Y,cex=.6)
vu = seq(-4,4,length=101)
vz = outer(vu,vu,function (x,y) dmnorm(cbind(x,y),
mean=c(0,0), varcov = Sig))
contour(vu,vu,vz,add=TRUE,col='blue')
abline(a=0,b=2,col="red")

Consider now the projection of points \boldsymbol{y}=(y_1,y_2) on the straight linear with directional vector \overrightarrow{\boldsymbol{u}} with slope a (say a=2). To get the projected point \boldsymbol{x}=(x_1,x_2) recall that x_2=ay_1 and \overrightarrow{\boldsymbol{x},\boldsymbol{y}}\perp\overrightarrow{\boldsymbol{u}}. Hence, the following code will give us the orthogonal projections

p = function(a){
x0=(Y[,1]+a*Y[,2])/(1+a^2)
y0=a*x0
cbind(x0,y0)
}

with

P = p(2)
for(i in 1:20) segments(Y[i,1],Y[i,2],P[i,1],P[i,2],lwd=4,col="red")
points(P[,1],P[,2],col="red",cex=.7)

Now, if we look at the distribution of points on that line, we get… a Gaussian distribution, as expected,

z = sqrt(P[,1]^2+P[,2]^2)*c(-1,+1)[(P[,1]>0)*1+1]
vu = seq(-6,6,length=601)
vv = dnorm(vu,mean(z),sd(z))
hist(z,probability = TRUE,breaks = seq(-4,4,by=.25))
lines(vu,vv,col="red")

Or course, we can use the matrix representation to get the projection on \overrightarrow{\boldsymbol{u}}, or a normalized version of that vector actually

a=2
U = c(1,a)/sqrt(a^2+1)
U
[1] 0.4472136 0.8944272
matP = U %*% solve(t(U) %*% U) %*% t(U)
matP %*% Y[1,]
[,1]
[1,] -0.1120555
[2,] -0.2241110
P[1,]
x0 y0
-0.1120555 -0.2241110

(which is consistent with our manual computation). Now, in Cochran theorem, we start with independent random variables,

Y = rmnorm(n = 1000,mean=c(0,0),varcov = diag(c(1,1)))

Then we consider the projection on \overrightarrow{\boldsymbol{u}} and \overrightarrow{\boldsymbol{v}}=\overrightarrow{\boldsymbol{u}}^\perp

U = c(1,a)/sqrt(a^2+1)
matP1 = U %*% solve(t(U) %*% U) %*% t(U)
P1 = Y %*% matP1
z1 = sqrt(P1[,1]^2+P1[,2]^2)*c(-1,+1)[(P1[,1]>0)*1+1]
V = c(a,-1)/sqrt(a^2+1)
matP2 = V %*% solve(t(V) %*% V) %*% t(V)
P2 = Y %*% matP2
z2 = sqrt(P2[,1]^2+P2[,2]^2)*c(-1,+1)[(P2[,1]>0)*1+1]

We can plot those two projections

plot(z1,z2)

and observe that the two are indeed, independent Gaussian variables. And (of course) there squared norms are \chi^2_{1} distributed.

On the conjugate function

In the MAT7381 course (graduate course on regression models), we will talk about optimization, and a classical tool is the so-called conjugate. Given a function f:\mathbb{R}^p\to\mathbb{R} its conjugate is function f^{\star}:\mathbb{R}^p\to\mathbb{R} such that f^{\star}(\boldsymbol{y})=\max_{\boldsymbol{x}}\lbrace\boldsymbol{x}^\top\boldsymbol{y}-f(\boldsymbol{x})\rbraceso, long story short, f^{\star}(\boldsymbol{y}) is the maximum gap between the linear function \boldsymbol{x}^\top\boldsymbol{y} and f(\boldsymbol{x}).

Just to visualize, consider a simple parabolic function (in dimension 1) f(x)=x^2/2, then f^{\star}(\color{blue}{2}) is the maximum gap between the line x\mapsto\color{blue}{2}x and function f(x).

x = seq(-100,100,length=6001)
f = function(x) x^2/2
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)

We can see it on the figure below.

viz = function(x0=1,YL=NA){
idx=which(abs(x)<=3) par(mfrow=c(1,2)) plot(x[idx],vf[idx],type="l",xlab="",ylab="",col="blue",lwd=2) abline(h=0,col="grey") abline(v=0,col="grey") idx2=which(x0*x>=vf)
polygon(c(x[idx2],rev(x[idx2])),c(vf[idx2],rev(x0*x[idx2])),col=rgb(0,1,0,.3),border=NA)
abline(a=0,b=x0,col="red")
i=which.max(x0*x-vf)
segments(x[i],x0*x[i],x[i],f(x[i]),lwd=3,col="red")
if(is.na(YL)) YL=range(vfstar[idx])
plot(x[idx],vfstar[idx],type="l",xlab="",ylab="",col="red",lwd=1,ylim=YL)
abline(h=0,col="grey")
abline(v=0,col="grey")
segments(x0,0,x0,fstar(x0),lwd=3,col="red")
points(x0,fstar(x0),pch=19,col="red")
}
viz(1)

or

viz(1.5)

In that case, we can actually compute f^{\star}, since f^{\star}(y)=\max_{x}\lbrace xy-f(x)\rbrace=\max_{x}\lbrace xy-x^2/2\rbraceThe first order condition is here x^{\star}=y and thusf^{\star}(y)=\max_{x}\lbrace xy-x^2/2\rbrace=\lbrace x^{\star}y-(x^{\star})^2/2\rbrace=\lbrace y^2-y^2/2\rbrace=y^2/2And actually, that can be related to two results. The first one is to observe that f(\boldsymbol{x})=\|\boldsymbol{x}\|_2^2/2 and in that case f^{\star}(\boldsymbol{y})=\|\boldsymbol{y}\|_2^2/2 from the following general result : if f(\boldsymbol{x})=\|\boldsymbol{x}\|_p^p/p with p>1, where \|\cdot\|_p denotes the standard \ell_p norm, then f^{\star}(\boldsymbol{y})=\|\boldsymbol{y}\|_q^q/q where\frac{1}{p}+\frac{1}{q}=1The second one is the conjugate of a quadratic function. More specifically if f(\boldsymbol{x})=\boldsymbol{x}^{\top}\boldsymbol{Q}\boldsymbol{x}/2 for some definite positive matrix \boldsymbol{Q}f^{\star}(\boldsymbol{y})=\boldsymbol{y}^{\top}\boldsymbol{Q}^{-1}\boldsymbol{y}/2. In our case, it was a univariate problem with \boldsymbol{Q}=1.

For the conjugate of the \ell_p norm, we can use the following code to visualize it

p = 3
f = function(x) abs(x)^p/p
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)
viz(1.5)

or

p = 1.1
f = function(x) abs(x)^p/p
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)
viz(1, YL=c(0,10))

Actually, in that case, we almost visualize that if f(x)=|x| then\displaystyle{f^{\star}\left(y\right)={\begin{cases}0,&\left|y\right|\leq 1\\\infty ,&\left|y\right|>1.\end{cases}}}

To conclude, another popular case, f(x)=\exp(x) then{\displaystyle f^{\star}\left(y\right)={\begin{cases}y\log(y)-y,&y>0\\0,&y=0\\\infty ,&y<0.\end{cases}}}We can visualize that case below

f = function(x) exp(x)
vf = Vectorize(f)(x)
fstar = function(y) max(y*x-vf)
vfstar = Vectorize(fstar)(x)
viz(1,YL=c(-3,3))

Combining automatically factor levels with trees

Last year, in a post, I discussed how to merge levels of factor variables, using combinatorial techniques (it was for my STT5100 cours, and trees are not in the syllabus), with an extension on trees at the end of the post.

consider the following (simulated dataset)

n=200
set.seed(1)
x1=runif(n)
x2=runif(n)
y=1+2*x1-x2+rnorm(n,0,.2)
LB=sample(LETTERS[1:10])
b=data.frame(y=y,x1=x1,
  x2=cut(x2,breaks=
  c(-1,.05,.1,.2,.35,.4,.55,.65,.8,.9,2),
  labels=LB))
str(b)
'data.frame':	200 obs. of  3 variables:
 $ y : num  1.345 1.863 1.946 2.481 0.765 ...
 $ x1: num  0.266 0.372 0.573 0.908 0.202 ...
 $ x2: Factor w/ 10 levels "I","A","H","F",..: 4 4 6 4 3 6 7 3 4 8 ...
table(b$x2)[LETTERS[1:10]]
 
 A  B  C  D  E  F  G  H  I  J 
11 12 23 34 23 36 12 32  3 14

Just by looking at the data (see the previous post), we could easily get the feeling that 10 levels was too much.

Following my post, Przemyslaw sent a comment suggesting to use

library(factorMerger)

It is indeed a nice package (unless you have really really big datasets with a lot of categories in your factor variables – as I experienced recently), and you can get great graphs

MF = mergeFactors(response = b$y, 
             factor = b$x2, 
             family = "gaussian")
plot(MF)

Here is suggests to create three categories. Recall that with student t-tests (changing the reference), we got

Another interesting package, by Piro Polo, is

library(tree.bins)

To use it, we simply call the following function, and we transform automatically our dataset : the continuous variables remain unchanged, and (possibly) categories of categorical variables are merged

b.bins = tree.bins(data=b, y=y)
str(b.bins)
Classes ‘data.table’ and 'data.frame':	200 obs. of  3 variables:
 $ y : num  1.345 1.863 1.946 2.481 0.765 ...
 $ x1: num  0.266 0.372 0.573 0.908 0.202 ...
 $ x2: chr  "Group.4" "Group.4" "Group.4" "Group.4" ...
 - attr(*, ".internal.selfref")= 
table(b.bins$x2)

Group.1 Group.2 Group.3 Group.4 
     23      35      26     116

here in four groups. To get the correspondance, use

tree.bins(data=b, y=y, return = "lkup.list")
[[1]]
   x2 Categories
1   E    Group.1
2   G    Group.2
3   C    Group.2
4   B    Group.3
5   J    Group.3
6   I    Group.4
7   A    Group.4
8   H    Group.4
9   F    Group.4
10  D    Group.4

(we have a list with one element, one dataframe, since there is only one factor variable). Cool, isn’t it ? I miss Przemyslaw’s plot, but this is rather quick, and efficient..

 

On leverage

Last week, in our STT5100 (applied linear models) class, I’ve introduce the hat matrix, and the notion of leverage. In a classical regression model, \boldsymbol{y}=\boldsymbol{X}\boldsymbol{\beta} (in a matrix form), the ordinary least square estimator of parameter \boldsymbol{\beta} is \widehat{\boldsymbol{\beta}}=(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top\boldsymbol{y}The prediction can then be written\widehat{\boldsymbol{y}}=\boldsymbol{X}\widehat{\boldsymbol{\beta}}=\underbrace{\color{blue}{\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top}}_{\color{blue}{\boldsymbol{H}}}\boldsymbol{y}where \color{blue}{\boldsymbol{H}} is called the hat matrix.

The matrix is idempotent, i.e. \boldsymbol{H}^2={\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\textcolor{grey}{\boldsymbol{X}^\top{\boldsymbol{X}}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}}\boldsymbol{X}^\top}={\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top}=\boldsymbol{H}so it can be interpreted as a projection matrix. Furthermore, since\boldsymbol{H}\boldsymbol{X}=\boldsymbol{X} (just do the maths), the projection is on a subspace that contains all the linear combinations of columns of \boldsymbol{X}. One can also observe that \mathbb{I}-\boldsymbol{H} is also a projection matrix. And we can write\boldsymbol{y}=\underbrace{\boldsymbol{H}\boldsymbol{y}}_{\widehat{\boldsymbol{y}}}+\underbrace{(\mathbb{I}-\boldsymbol{H})\boldsymbol{y}}_{\widehat{\boldsymbol{\varepsilon}}}where \widehat{\boldsymbol{y}} is the orthogonal projection of \boldsymbol{y} on the (linear) space of linear combinations of columns of \boldsymbol{X}, and \widehat{\boldsymbol{y}}\perp\widehat{\boldsymbol{\varepsilon}}, which gives the classical interpretation of residuals, being unpredictible (at least with a linear model using variables \boldsymbol{X}).

Let’s move a bit faster now (we’ve seen many other properties last week), and consider elements on the diagonal of matrix \boldsymbol{H}. Recall that we have

so entry \boldsymbol{H}_{i,i} is a measure of the influence of entry \boldsymbol{y}_i on its prediction latex]\widehat{\boldsymbol{y}}_i[/latex].

We have seen that\sum_{i=1}^n\boldsymbol{H}_{i,i}=\text{trace}(\boldsymbol{H})=\text{trace}(\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}^\top)which can be written\sum_{i=1}^n\boldsymbol{H}_{i,i}=\text{trace}\boldsymbol{X}^\top(\boldsymbol{X}(\boldsymbol{X}^\top\boldsymbol{X})^{-1})=\text{trace}(\mathbb{I})=pwhere classically p=k+1, where k is the number of explanatory variables. Further, since \boldsymbol{H} is idempotent, we can write (from \boldsymbol{H}=\boldsymbol{H}^2) that\boldsymbol{H}_{i,i}=\boldsymbol{H}_{i,i}^2 + \sum_{j\neq i}\boldsymbol{H}_{i,j}\boldsymbol{H}_{j,i}=\boldsymbol{H}_{i,i}^2 + \sum_{j\neq i}\boldsymbol{H}_{i,j}^2One the one hand, since the second term is positive \boldsymbol{H}_{i,i}\geq\boldsymbol{H}_{i,i}^2, i.e. 1\geq\boldsymbol{H}_{i,i}. And since both terms are positive, then \boldsymbol{H}_{i,i}\in[0,1]. And there was a question in the course on the sharpeness of the bounds.

Using Anscombe’s dataset, we’ve seen that it was possible to get a leverage of 1. Using something rather similar

df = data.frame(x = c(rep(1,10),6), y = c(1:10,8))
plot(df)

we obtain

model = lm(y~x,data=df)
abline(model,col="red",lwd=2)
H = lm.influence(model)$hat
plot(1:11,H,type="h")

The very last observation, the one one the right, is here extremely influencial : if we remove it, the model is completely different ! And here, we reach the upper bound, \boldsymbol{H}_{11,11}=1. Observe that all other points are equally influencial, and because on the constraint on the trace of the matrix, \boldsymbol{H}_{i,i}=1/10 when i\in\{1,2,\cdots,10\}.

Now, what about the lower bound ? In order to have some sort of “non-influencial” observations, consider the two following case.

  • the case where one observation (below the first one) is such that \widehat{\boldsymbol{y}}_{i}=\boldsymbol{y}_{i} (perfect prediction)
  • the case where one observation (below the tenth one) is such that \boldsymbol{x}_{i}=\overline{\boldsymbol{x}} and \boldsymbol{y}_{i}=\overline{\boldsymbol{y}} (from the first order condition – or normal equation), the fitted regression line always go through point (\overline{\boldsymbol{x}},\overline{\boldsymbol{y}})

Let us move two observations from our dataset,

mean(c(4,rep(1,8),6))
[1] 1.8
df = data.frame(x = c(4,rep(1,8),6,1.8),
y = c(predict(model,newdata=data.frame(x=4)),
2:9,8,
predict(model,newdata=data.frame(x=1.8))))

We now have

If we compute the leverages, we obtain

model = lm(y~x,data=df)
H = lm.influence(model)$hat
plot(1:11,H,type="h")

so, for the first observation, its leverage actually increased (the blue part), and for the tenth one, we have the lowest influence, but it is not zero. Is it possible to reach zero ?

Here, observe that for the tenth observation, \boldsymbol{H}_{i,i}=1/n. And actually, that’s the best we can do… We can prove that, in the case of a simple regression (as above)\boldsymbol{H}_{i,i}=\frac{1}{n}+\frac{(x_i-\overline{x})^2}{n\text{Var}(x)}which is minimum when x_i=\overline{x}, and then \boldsymbol{H}_{i,i}=1/n, otherwise \boldsymbol{H}_{i,i}>1/n. And this property is also valid in a multiple regression (as soon as an intercept is included in the regression – which should always be the case). To prove that result, let \tilde{\boldsymbol{X}} denote the matrix of centered variables \boldsymbol{X}, then we can prove that \boldsymbol{H}_{i,i}=\frac{1}{n}+\big[\tilde{\boldsymbol{X}}(\tilde{\boldsymbol{X}}^\top\tilde{\boldsymbol{X}})^{-1}\tilde{\boldsymbol{X}}^\top\big]_{i,i}(which is basically a matrix version of the previous equation).

I can maybe add another comment on Anscombe’s data. We’ve seen that on the right that we did reach 1. But I did not prove it. One way to prove it is actually to focus on the remaining n-1 points, on the left. Those have all the same x values. We can prove that if \boldsymbol{X}_{i_1}=\boldsymbol{X}_{i_2}, then \boldsymbol{H}_{i_1,i_2}=\boldsymbol{X}_{i_1}^\top(\boldsymbol{X}^\top\boldsymbol{X})^{-1}\boldsymbol{X}_{i_2}=\boldsymbol{H}_{i_1,i_1}hence, using the relationship obtained since the hat matrix is idempotent\boldsymbol{H}_{i_1,i_1}=2\boldsymbol{H}_{i_1,i_1}^2+\sum_{j\notin\{i_1,i_2\}}\boldsymbol{H}_{i_1,j}^2thus, we now have\boldsymbol{H}_{i_1,i_1}\big(1-2\boldsymbol{H}_{i_1,i_1}\big)>0i.e. \boldsymbol{H}_{i_1,i_1}\in[0,1/2], where the upper bound becomes 1/(n-1) “duplicates”. So for n-1 \boldsymbol{H}_{i,i}‘s, we have values below 1/(n-1), the last one should be below 1 and the sum has to be k=2 . So we have the value of the n \boldsymbol{H}_{i,i}‘s.

 

Insurance data science : Pictures

At the Summer School of the Swiss Association of Actuaries, in Lausanne, following the part of Jean-Philippe Boucher (UQAM) on telematic data, I will start talking about pictures this Wednesday. Slides are available online

Ewen Gallic (AMSE) will present a tutorial on satellite pictures, and a simple classification problem, related to Alzeimher detection.

We will try to identify what is on the following pictures, starting with the car

(we will see that the car is indeed identified)

a skier,

and a fire,

We will also discuss previous pictures from the summer school

Insurance data science : use and value of unusual data #1

Next week, with , I will be at the Summer School of the Swiss Association of Actuaries, in Lausanne, with Jean-Philippe Boucher (UQAM) and Ewen Gallic (AMSE).

I will give an introductionary talk on Monday morning, and the slides are now available

There will be some hands-on applications, on R. I will share some codes in the slides.

Optimal transport on large networks

With Alfred Galichon and Lucas Vernet, we recently uploaded a paper entitled optimal transport on large networks on arxiv.

This article presents a set of tools for the modeling of a spatial allocation problem in a large geographic market and gives examples of applications. In our settings, the market is described by a network that maps the cost of travel between each pair of adjacent locations. Two types of agents are located at the nodes of this network. The buyers choose the most competitive sellers depending on their prices and the cost to reach them. Their utility is assumed additive in both these quantities. Each seller, taking as given other sellers prices, sets her own price to have a demand equal to the one we observed. We give a linear programming formulation for the equilibrium conditions. After formally introducing our model we apply it on two examples: prices offered by petrol stations and quality of services provided by maternity wards (only the later is described here for privacy issues). These examples illustrate the applicability of our model to aggregate demand, rank prices and estimate cost structure over the network. We insist on the possibility of applications to large scale data sets using modern linear programming solvers such as Gurobi.

Demand for gas in gas stations in Britanny, and demand for maternity in France (with border correction)

In addition to this paper we released a R toolbox to implement our results and an online tutorial, optimalnetwork.github.io.

Estimates on training vs. validation samples

Before moving to cross-validation, it was natural to say “I will burn 50% (say) of my data to train a model, and then use the remaining to fit the model”. For instance, we can use training data for variable selection (e.g. using some stepwise procedure in a logistic regression), and then, once variable have been selected, fit the model on the remaining set of observations. A natural question is usually “does it really matter ?”.

In order to visualize this problem, consider my (simple) dataset

MYOCARDE=read.table(
  "http://freakonometrics.free.fr/saporta.csv",
  head=TRUE,sep=";")

Let us generate 100 training samples (where we keep about 50% of the observations). On each of them, we use a stepwise procedure, and we keep the estimates of the remaining variables (and their standard deviation actually)

n=nrow(MYOCARDE)
M=matrix(NA,100,ncol(MYOCARDE))
colnames(M)=c("(Intercept)",names(MYOCARDE)[1:7])
S1=S2=M1=M2=M
for(i in 1:100){
idx = which(sample(0:1,size=n, replace=TRUE)==1)
reg=step(glm(PRONO=="DECES"~.,data=MYOCARDE[idx,]))
nm=names(reg$coefficients)
M1[i,nm]=reg$coefficients
S1[i,nm]=summary(reg)$coefficients[,2]
f=paste("PRONO=='DECES'~",paste(nm[-1],collapse="+"),sep="")
reg=glm(f,data=MYOCARDE[-idx,])
M2[i,nm]=reg$coefficients
S2[i,nm]=summary(reg)$coefficients[,2]
}

Then, for the 7 covariates (and the constant) we can look at the value of the coefficient in the model fitted on the training sample, and the value on the model fitted on the validation sample (of course, only when they were remaining)

for(j in 1:8){
idx=which(!is.na(M1[,j]))
plot(M1[idx,j],M2[idx,j])
abline(a=0,b=1,lty=2,col="gray")
segments(M1[idx,j]-2*S1[idx,j],M2[idx,j],M1[idx,j]+2*S1[idx,j],M2[idx,j])  
segments(M1[idx,j],M2[idx,j]-2*S2[idx,j],M1[idx,j],M2[idx,j]+2*S2[idx,j])  
}

For instance, with the intercept, we have the following

 

where horizontal segments are confidence intervals of the parameter on the model fitted on the training sample, the vertical on the validation sample. The green part means some sort of consistency, while the red one means that actually, the coefficient was negative with one model, positive with the other one. Which is odd (but in that case, observe that coefficients are rarely significant).

We can also visualize the joint distribution of the two estimators,

for(j in 1:8){
library(ks)
idx = which(!is.na(M1[,j]))
Z = cbind(M1[idx,j],M2[idx,j])
H = Hpi(x=Z)
fhat = kde(x=Z, H=H)
image(fhat$eval.points[[1]],
fhat$eval.points[[2]],fhat$estimate)
abline(a=0,b=1,lty=2,col="gray")
abline(v=0,lty=2)
abline(h=0,lty=2)
}

which are here, almost on the diagonal,

meaning that the intercept on the two samples is (more or less) the same. We can then look at other parameters (which is actually more interesting).

On that variable, it seems that it is significant on the training dataset (somehow, it is consistent with the fact that it is remaining in the model after the stepwise procedure) but not on the validation sample (or hardly significant).

Others are much more consistent (with some possible outliers)

 

 

On the next one, we have again significance on the training sample, but not on the validation sample,

 

 

and probably more interesting

where the two are very consistent.

Exotic link functions for GLMs

In my previous post on GLMs, I discussed power link functions. But there are much more links that can be used :

  • The square root link (for the Poisson model)

Consider some random variable Y with mean \mu and variance \sigma^2. Using Taylor’s expansion,g(Y)\sim g(\mu)+(Y-\mu)g'(\mu)+\frac{1}{2}(Y-\mu)^2g''(\mu)we can write\mathbb{E}[g(Y)]\sim g(\mu)+\frac{\sigma^2}{2}g''(\mu) \text{Var}[g(Y)]\sim [g'(\mu)]^2\sigma^2

Assume that Y\sim\mathcal{P}(\lambda), a consider a square root transformation, g(y)=\sqrt{y}, then the second equality becomes \text{Var}[\sqrt{Y}]\sim \left[\frac{1}{2\sqrt{\mathbb{E}[Y]}}\right]^2\text{Var}[Y]=\frac{1}{4}

So, somehow, with a square-root transformation, we have variance stability, which might be interpreted as some homoscedasticity.

  • The complementary log-log function for the Bernoulli model

Assume that the true variable of interest is a Poisson one, N|\mathbf{X}=\mathbf{x}\sim\mathcal{P}(\lambda_{\mathbf{x}}) where \lambda_{\mathbf{x}}=\exp[\mathbf{x}^T\mathbf{\beta}]Thus,\mathbb{P}[N=0|\mathbf{X}=\mathbf{x}]=\exp[-\lambda_{\mathbf{x}}]=\exp[-(\exp[\mathbf{x}^T\mathbf{\beta}])]while\mathbb{P}[N>0|\mathbf{X}=\mathbf{x}]=1-\exp[-(\exp[\mathbf{x}^T\mathbf{\beta}])]=H(\mathbf{x}^T\mathbf{\beta})where H(s)=1-\exp[-\exp(s)]. Let Y=\mathbf{1}(N>0). The previous model seems like a Bernoulli regression with H as link function,\mathbb{P}[Y=1|\mathbf{X}=\mathbf{x}]=H(\mathbf{x}^T\mathbf{\beta})

So, assume now that instead of observing N we observe Y=\boldsymbol{1}(N>0). In that case, running a Bernoulli regression with a complementary log-log link function would be the same (?) as running first a Poisson regression on the original data, and then use it on our binary variable, zero vs. non-zero. Let us generate some data, and see what’s going on. Let us compare e^{\lambda_{\mathbf{x}}} and p_{\mathbf{x}} obtained from a standard logistic regression

n=563
set.seed(1)
base=data.frame(X1=rnorm(n),X2=rnorm(n))
lambda=base$X1+base$X2
base$Y=rpois(n,exp(lambda))
regPois = glm(Y~.,data=base,family=poisson(link="log"))
lambda = predict(regPois,type="response")
regBinom = glm((Y==0)~.,data=base,family=binomial(link="probit"))
prob = predict(regBinom, type="response")
plot(prob,exp(-lambda),xlim=0:1,ylim=0:1)
abline(a=0,b=1,lty=2,col="red")

What if p_{\mathbf{x}} was obtained from a Bernoulli regression, with a cloglog link function ?

regBinom = glm((Y&gt;0)~.,data=base,family=binomial(link="cloglog"))
prob = predict(regBinom, type="response")
plot(prob,1-exp(-lambda),xlim=0:1,ylim=0:1)
abline(a=0,b=1,lty=2,col="red")

It looks like the fit is very good here ! Now, what if we have real data, like the dataset from A Theory of Extramarital Affairs, by Ray Fair, published in 1978 in the Journal of Political Economy (with 563 observations, and nine variables)

base = read.table("http://freakonometrics.free.fr/baseaffairs.txt",header=TRUE)
str(base)
x=base$SEX
base$SEX="M"
base$SEX[x=="0"]="F"
x=base$CHILDREN
base$CHILDREN="YES"
base$CHILDREN[x==0]="NO"
regPois = glm(Y~.,data=base,family=poisson(link="log"))
lambda = predict(regPois,type="response")
regBinom = glm((Y==0)~.,data=base,family=binomial(link="probit"))
prob = predict(regBinom, type="response")
plot(prob,exp(-lambda),xlim=0:1,ylim=0:1)
abline(a=0,b=1,lty=2,col="red")

In that case the two models are very different. But actually, so is the second one

regBinom = glm((Y&gt;0)~.,data=base,family=binomial(link="cloglog"))
prob = predict(regBinom, type="response")
plot(prob,1-exp(-lambda),xlim=0:1,ylim=0:1)
abline(a=0,b=1,lty=2,col="red")

How can we interpret that ? Could it be because the Poisson model is not good ? Actually, if we run a zero-inflated model here,

library(pscl)
regZIP = zeroinfl(Y ~ . | ., data = base)
summary(regZIP)
 
Count model coefficients (poisson with log link):
             Estimate Std. Error z value Pr(&gt;|z|)    
(Intercept) -0.002274   0.048413  -0.047    0.963    
X1           1.019814   0.026186  38.945   &lt;2e-16 ***
X2           1.004814   0.024172  41.570   &lt;2e-16 *** 
Zero-inflation model coefficients (binomial with logit link): 
            Estimate Std. Error z value Pr(&gt;|z|)  
(Intercept) -4.90190    2.07846  -2.358   0.0184 *
X1          -2.00227    0.86897  -2.304   0.0212 *
X2          -0.01545    0.96121  -0.016   0.9872  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Hence, we reject here the Poisson distribution assumption, because of the inflation of zeros… It looks like the cloglog link can be used to check if the Poisson distribution is a good model, or not…