# 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, $$\min\left\lbrace\frac{1}{2}\|\mathbf{y}-\mathbf{X}\mathbf{\beta}\|_{\ell_2}^2+\lambda\|\mathbf{\beta}\|_{\ell_1}\right\rbrace$$We have to define the soft-thresholding function$$S(z,\gamma)=\text{sign}(z)\cdot(|z|-\gamma)_+=\begin{cases}z-\gamma&\text{ if }\gamma>|z|\text{ and }z<0\\z+\gamma&\text{ if }\gamma<|z|\text{ and }z<0 \\0&\text{ if }\gamma\geq|z|\end{cases}$$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$$\min_\mu \left\lbrace\sum_{i=1}^n|y_i-\mu|\right\rbrace$$which can be solved using linear programming techniques. More precisely, this problem is equivalent to$$\min_{\mu,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^na_i+b_i\right\rbrace$$with $a_i,b_i\geq 0$ 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. $$a_i=(\varepsilon_i)_+=\max\lbrace0,\varepsilon_i\rbrace=|\varepsilon|\cdot\boldsymbol{1}_{\varepsilon_i>0}$$and$$b_i=(-\varepsilon_i)_+=\max\lbrace0,-\varepsilon_i\rbrace=|\varepsilon|\cdot\boldsymbol{1}_{\varepsilon_i<0}$$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,

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$$\min_{\boldsymbol{\beta}^+,\boldsymbol{\beta}^-,\mathbf{a},\mathbf{b}}\left\lbrace\sum_{i=1}^n\tau a_i+(1-\tau)b_i\right\rbrace$$with $a_i,b_i\geq 0$ 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 that$$A^2=P\Delta P^\top=P\Delta P^\top=P\Delta^2 P^\top$$where $\Delta$ 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 $$\boldsymbol{Z}^\top \Delta\boldsymbol{Z}=\sum_{i:\Delta_{i,i}-1}Z_i^2\sim\chi^2_r$$Nice, 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 $$f^{\star}(\boldsymbol{y})=\max_{\boldsymbol{x}}\lbrace\boldsymbol{x}^\top\boldsymbol{y}-f(\boldsymbol{x})\rbrace$$so, 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 $$f^{\star}(y)=\max_{x}\lbrace xy-f(x)\rbrace=\max_{x}\lbrace xy-x^2/2\rbrace$$The first order condition is here $x^{\star}=y$ and thus$$f^{\star}(y)=\max_{x}\lbrace xy-x^2/2\rbrace=\lbrace x^{\star}y-(x^{\star})^2/2\rbrace=\lbrace y^2-y^2/2\rbrace=y^2/2$$And 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$$\frac{1}{p}+\frac{1}{q}=1$$The 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})=p$$where classically $p=k+1$, where $k$ is the number of explanatory variables. Further, since $\boldsymbol{H}$ is idempotent, we can write (from $\boldsymbol{H}=\boldsymbol{H}^2$) that$$\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}^2$$One 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$$\boldsymbol{H}_{i_1,i_1}=2\boldsymbol{H}_{i_1,i_1}^2+\sum_{j\notin\{i_1,i_2\}}\boldsymbol{H}_{i_1,j}^2$$thus, we now have$$\boldsymbol{H}_{i_1,i_1}\big(1-2\boldsymbol{H}_{i_1,i_1}\big)>0$$i.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 : Networks At the Summer School of the Swiss Association of Actuaries, in Lausanne, I will start talking about networks and insurance this Friday. Slides are available online # Insurance data science : Text At the Summer School of the Swiss Association of Actuaries, in Lausanne, I will start talking about text based data and NLP this Thursday. Slides are available online Ewen Gallic (AMSE) will present a tutorial on tweets. I can upload a few additional slides on LSTM (recurrent neural nets) # 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) 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. # Pareto Models for Top Incomes With Emmanuel Flachaire, we uploaded on hal a paper on Pareto Models for Top Incomes, Top incomes are often related to Pareto distribution. To date, economists have mostly used Pareto Type I distribution to model the upper tail of income and wealth distribution. It is a parametric distribution, with an attractive property, that can be easily linked to economic theory. In this paper, we first show that modelling top incomes with Pareto Type I distribution can lead to severe over-estimation of inequality, even with millions of observations. Then, we show that the Generalized Pareto distribution and, even more, the Extended Pareto distribution, are much less sensitive to the choice of the threshold. Thus, they provide more reliable results. We discuss different types of bias that could be encountered in empirical studies and, we provide some guidance for practice. To illustrate, two applications are investigated, on the distribution of income in South Africa in 2012 and on the distribution of wealth in the United States in 2013. This paper was presented at and UCSB and in several workshops this spring, and this Summer, Emmanuel will present it at ECINEQ. Note that a R package is also available on github, TopIncomes. # 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$estimate) abline(a=0,b=1,lty=2,col="gray") abline(v=0,lty=2) abline(h=0,lty=2) } which are here, almost on the diagonal, meaning that the intercept on the two samples is (more or less) the same. We can then look at other parameters (which is actually more interesting). On that variable, it seems that it is significant on the training dataset (somehow, it is consistent with the fact that it is remaining in the model after the stepwise procedure) but not on the validation sample (or hardly significant). Others are much more consistent (with some possible outliers) On the next one, we have again significance on the training sample, but not on the validation sample, and probably more interesting where the two are very consistent. # What is the interpretation of the diagonal for a ROC curve Last Friday, we discussed the use of ROC curves to describe the goodness of a classifier. I did say that I will post a brief paragraph on the interpretation of the diagonal. If you look around some say that it describes the “strategy of randomly guessing a class“, that it is obtained with “a diagnostic test that is no better than chance level“, even obtained by “making a prediction by tossing of an unbiased coin“. Let us get back to ROC curves to illustrate those points. Consider a very simple dataset with 10 observations (that is not linearly separable) x1 = c(.4,.55,.65,.9,.1,.35,.5,.15,.2,.85) x2 = c(.85,.95,.8,.87,.5,.55,.5,.2,.1,.3) y = c(1,1,1,1,1,0,0,1,0,0) df = data.frame(x1=x1,x2=x2,y=as.factor(y)) here we can check that, indeed, it is not separable plot(x1,x2,col=c("red","blue")[1+y],pch=19) Consider a logistic regression (the course is on linear models) reg = glm(y~x1+x2,data=df,family=binomial(link = "logit")) but any model here can be used… We can use our own function Y=df$y S=predict(reg) roc.curve=function(s,print=FALSE){ Ps=(S&gt;=s)*1  FP=sum((Ps==1)*(Y==0))/sum(Y==0)  TP=sum((Ps==1)*(Y==1))/sum(Y==1)  if(print==TRUE){  print(table(Observed=Y,Predicted=Ps))  }  vect=c(FP,TP)  names(vect)=c("FPR","TPR")  return(vect) }

or any R package actually

library(ROCR)  perf=performance(prediction(S,Y),"tpr","fpr")

We can plot the two simultaneously here

plot(performance(prediction(S,Y),"tpr","fpr")) V=Vectorize(roc.curve)(seq(-5,5,length=251))  points(V[1,],V[2,])  segments(0,0,1,1,col="light blue")

So our code works just fine, here. Let us consider various strategies that should lead us to the diagonal.

The first one is : everyone has the same probability (say 50%)

S=rep(.5,10)  plot(performance(prediction(S,Y),"tpr","fpr"))  V=Vectorize(roc.curve)(seq(0,1,length=251))  points(V[1,],V[2,])

Indeed, we have the diagonal. But to be honest, we have only two points here : $(0,0)$ and $(1,1)$. Claiming that we have a straight line is not very satisfying… Actually, note that we have this situation whatever the probability we choose

S=rep(.2,10)  plot(performance(prediction(S,Y),"tpr","fpr"))  V=Vectorize(roc.curve)(seq(0,1,length=251))  points(V[1,],V[2,])

We can try another strategy, like “making a prediction by tossing of an unbiased coin“. This is what we obtain

set.seed(1)  S=sample(0:1,size=10,replace=TRUE)  plot(performance(prediction(S,Y),"tpr","fpr"))  V=Vectorize(roc.curve)(seq(0,1,length=251))  points(V[1,],V[2,])  segments(0,0,1,1,col="light blue")

We can also try some sort of “random classifier”, where we choose the score randomly, say uniform on the unit interval

set.seed(1)  S=runif(10)  plot(performance(prediction(S,Y),"tpr","fpr"))  V=Vectorize(roc.curve)(seq(0,1,length=251))  points(V[1,],V[2,])  segments(0,0,1,1,col="light blue")

Let us try to go further on that one. For convenience, let us consider another function to plot the ROC curve

V=Vectorize(roc.curve)(seq(0,1,length=251))  roc_curve=Vectorize(function(x) max(V[2,which(V[1,]&lt;=x)]))

We have the same line as previously

x=seq(0,1,by=.025)  y=roc_curve(x)   lines(x,y,type="s",col="red")

But now, consider many scoring strategies, all randomly chosen

MY=matrix(NA,500,length(y))  for(i in 1:500){  S=runif(10)  V=Vectorize(roc.curve)(seq(0,1,length=251))  MY[i,]=roc_curve(x)  }  plot(performance(prediction(S,df$y),"tpr","fpr"),col="white") for(i in 1:500){ lines(x,MY[i,],col=rgb(0,0,1,.3),type="s") } lines(c(0,x),c(0,apply(MY,2,mean)),col="red",type="s",lwd=3) segments(0,0,1,1,col="light blue") The red line is the average of all random classifiers. It is not a straight line, be we observe oscillations around the diagonal. Consider a dataset with more observations  myocarde = read.table("http://freakonometrics.free.fr/myocarde.csv",head=TRUE, sep=";") myocarde$PRONO = (myocarde$PRONO=="SURVIE")*1 reg = glm(PRONO~.,data=myocarde,family=binomial(link = "logit")) Y=myocarde$PRONO  S=predict(reg)  plot(performance(prediction(S,Y),"tpr","fpr"))  V=Vectorize(roc.curve)(seq(-5,5,length=251))  points(V[1,],V[2,])  segments(0,0,1,1,col="light blue")

Here is a “random classifier” where we draw scores randomly on the unit interval

S=runif(nrow(myocarde)  plot(performance(prediction(S,Y),"tpr","fpr"))  V=Vectorize(roc.curve)(seq(-5,5,length=251))  points(V[1,],V[2,])  segments(0,0,1,1,col="light blue")

And if we do that 500 times, we obtain, on average

MY=matrix(NA,500,length(y))  for(i in 1:500){  S=runif(length(Y))  V=Vectorize(roc.curve)(seq(0,1,length=251))  MY[i,]=roc_curve(x) }  plot(performance(prediction(S,Y),"tpr","fpr"),col="white")  for(i in 1:500){  lines(x,MY[i,],col=rgb(0,0,1,.3),type="s")  }  lines(c(0,x),c(0,apply(MY,2,mean)),col="red",type="s",lwd=3) segments(0,0,1,1,col="light blue")

So, it looks like me might say that the diagonal is what we have, on average, when drawing randomly scores on the unit interval…

I did mention that an interesting visual tool could be related to the use of the Kolmogorov Smirnov statistic on classifiers. We can plot the two empirical cumulative distribution functions of the scores, given the response $Y$

score=data.frame(yobs=Y,  ypred=predict(reg,type="response"))  f0=c(0,sort(score$ypred[score$yobs==0]),1)  f1=c(0,sort(score$ypred[score$yobs==1]),1)  plot(f0,(0:(length(f0)-1))/(length(f0)-1),col="red",type="s",lwd=2,xlim=0:1)  lines(f1,(0:(length(f1)-1))/(length(f1)-1),col="blue",type="s",lwd=2)

we can also look at the distribution of the score, with the histogram (or density estimates)

S=score\$ypred  hist(S[Y==0],col=rgb(1,0,0,.2),  probability=TRUE,breaks=(0:10)/10,border="white")  hist(S[Y==1],col=rgb(0,0,1,.2),  probability=TRUE,breaks=(0:10)/10,border="white",add=TRUE)  lines(density(S[Y==0]),col="red",lwd=2,xlim=c(0,1))  lines(density(S[Y==1]),col="blue",lwd=2)

The underlying idea is the following : we do have a “perfect classifier” (top left corner)

is the supports of the scores do not overlap

otherwise, we should have errors. That the case below

we in 10% of the cases, we might have misclassification

or even more missclassification, with overlapping supports

Now, we have the diagonal

when the two conditional distributions of the scores are identical

Of course, that only valid when $n$ is very large, otherwise, it is only what we observe on average….