Question sur la façon de normaliser le coefficient de régression

16

Je ne sais pas si normaliser est le mot correct à utiliser ici, mais je ferai de mon mieux pour illustrer ce que j'essaie de demander. L'estimateur utilisé ici est le moindre carré.

Supposons que vous ayez y=β0+β1x1 , vous pouvez le centrer autour de la moyenne par où et , de sorte que n'a plus aucune influence sur l'estimation de . β 0 = β 0 + β 1 ˉ x 1 x 1 = x - ˉ x β 0 β 1y=β0+β1x1β0=β0+β1x¯1x1=xx¯β0β1

J'entends par là en est équivalent à en y = β 0 + β 1 x 1 . Nous avons réduit l'équation pour un calcul des moindres carrés plus facile.y=β1x ' 1 β 1β^1y=β1x1β^1y=β0+β1x1

Comment appliquez-vous cette méthode en général? Maintenant que j'ai le modèle y=β1ex1t+β2ex2t , j'essaie de le réduire à y=β1x .

Sabre CN
la source
Quel type de données analysez-vous et pourquoi souhaitez-vous supprimer une covariable, , de votre modèle. De plus, y a-t-il une raison pour laquelle vous supprimez l'interception? Si vous voulez centrer les données, la pente sera la même dans le modèle avec / sans interception, mais le modèle avec interception s'adaptera mieux à vos données. ex1t
caburke
@caburke Je ne suis pas préoccupé par l'ajustement du modèle, car après avoir calculé et β 2, je peux les remettre dans le modèle. Le but de cet exercice est d'estimer β 1 . En réduisant l'équation d'origine à seulement y = β 1 x , le calcul du moindre carré sera plus facile (x 'fait partie de ce que j'essaie de découvrir, il peut inclure e x 1 t ). J'essaie d'apprendre les mécanismes, c'est une question d'un livre de Tukey. β1β2β1y=β1xex1t
Sabre CN
@ca L'observation à la fin de votre commentaire est déroutante. Cela ne peut pas s'appliquer aux expressions non linéaires - elles ne contiennent rien qui puisse raisonnablement être considéré comme une "pente" - mais ce n'est pas correct dans le paramètre OLS: l'ajustement pour les données centrées sur la moyenne est précisément aussi bon que le s'adapter à une interception. Sabre, votre modèle est ambigu: quelles sont les variables et les paramètres? Quelle est la structure d'erreur prévue? (Et de quel livre de Tukey est la question?)β1,β2,x1,x2,t
whuber
1
@whuber Ceci est tiré du livre de Tukey "Analyse et régression des données: un deuxième cours de statistiques", chapitre 14A. sont les paramètres que nous essayons d'estimer, x 1 , x 2 sont les variables contenant chacune n observations, t je suppose que c'est la variable temporelle associée aux observations, mais elle ne l'a pas précisé. L'erreur doit être normale et peut être ignorée pour cette question. β1,β2x1,x2t
Sabre CN
1
@whuber Je faisais surtout référence à la première partie du message, mais ce n'était pas clair dans mon commentaire. Ce que je voulais dire, c'est que si vous ne voulez dire que le centre-centre , et non y , comme cela semblait être suggéré dans l'OP, puis supprimez l'ordonnée à l'origine, l'ajustement sera pire, car ce n'est pas nécessairement le cas que ˉ y = 0 . La pente n'est évidemment pas un bon terme pour le coefficient dans le modèle mentionné dans la dernière ligne du PO. xyy¯=0
caburke

Réponses:

38

Bien que je ne puisse pas rendre justice à la question ici - qui nécessiterait une petite monographie - il peut être utile de récapituler certaines idées clés.

La question

Commençons par reformuler la question et utiliser une terminologie sans ambiguïté. Les données consistent en une liste de paires ordonnées . Les constantes connues α 1 et α 2 déterminent les valeurs x 1 , i = exp ( α 1 t i ) et x 2 , i = exp ( α 2 t i ) . Nous posons un modèle dans lequel(ti,yi) α1α2x1,i=exp(α1ti)x2,i=exp(α2ti)

yi=β1x1,i+β2x2,i+εi

pour que les constantes et β 2 soient estimées, ε i sont aléatoires et - de toute façon à une bonne approximation - indépendantes et ayant une variance commune (dont l'estimation est également intéressante).β1β2εi

Contexte: "appariement" linéaire

Mosteller and Tukey refer to the variables x1 = (x1,1,x1,2,) and x2 as "matchers." They will be used to "match" the values of y=(y1,y2,) in a specific way, which I will illustrate. More generally, let y and x be any two vectors in the same Euclidean vector space, with y playing the role of "target" and x that of "matcher". We contemplate systematically varying a coefficient λ in order to approximate y by the multiple λx. The best approximation is obtained when λx is as close to y as possible. Equivalently, the squared length of yλx is minimized.

One way to visualize this matching process is to make a scatterplot of x and y on which is drawn the graph of xλx. The vertical distances between the scatterplot points and this graph are the components of the residual vector yλx; the sum of their squares is to be made as small as possible. Up to a constant of proportionality, these squares are the areas of circles centered at the points (xi,yi) with radii equal to the residuals: we wish to minimize the sum of areas of all these circles.

Here is an example showing the optimal value of λ in the middle panel:

Panel

The points in the scatterplot are blue; the graph of xλx is a red line. This illustration emphasizes that the red line is constrained to pass through the origin (0,0): it is a very special case of line fitting.

Multiple regression can be obtained by sequential matching

Returning to the setting of the question, we have one target y and two matchers x1 and x2. We seek numbers b1 and b2 for which y is approximated as closely as possible by b1x1+b2x2, again in the least-distance sense. Arbitrarily beginning with x1, Mosteller & Tukey match the remaining variables x2 and y to x1. Write the residuals for these matches as x21 and y1, respectively: the 1 indicates that x1 has been "taken out of" the variable.

We can write

y=λ1x1+y1 and x2=λ2x1+x21.

Having taken x1 out of x2 and y, we proceed to match the target residuals y1 to the matcher residuals x21. The final residuals are y12. Algebraically, we have written

y1=λ3x21+y12; whencey=λ1x1+y1=λ1x1+λ3x21+y12=λ1x1+λ3(x2λ2x1)+y12=(λ1λ3λ2)x1+λ3x2+y12.

This shows that the λ3 in the last step is the coefficient of x2 in a matching of x1 and x2 to y.

We could just as well have proceeded by first taking x2 out of x1 and y, producing x12 and y2, and then taking x12 out of y2, yielding a different set of residuals y21. This time, the coefficient of x1 found in the last step--let's call it μ3--is the coefficient of x1 in a matching of x1 and x2 to y.

Finally, for comparison, we might run a multiple (ordinary least squares regression) of y against x1 and x2. Let those residuals be ylm. It turns out that the coefficients in this multiple regression are precisely the coefficients μ3 and λ3 found previously and that all three sets of residuals, y12, y21, and ylm, are identical.

Depicting the process

None of this is new: it's all in the text. I would like to offer a pictorial analysis, using a scatterplot matrix of everything we have obtained so far.

Scatterplot

Because these data are simulated, we have the luxury of showing the underlying "true" values of y on the last row and column: these are the values β1x1+β2x2 without the error added in.

The scatterplots below the diagonal have been decorated with the graphs of the matchers, exactly as in the first figure. Graphs with zero slopes are drawn in red: these indicate situations where the matcher gives us nothing new; the residuals are the same as the target. Also, for reference, the origin (wherever it appears within a plot) is shown as an open red circle: recall that all possible matching lines have to pass through this point.

Much can be learned about regression through studying this plot. Some of the highlights are:

  • The matching of x2 to x1 (row 2, column 1) is poor. This is a good thing: it indicates that x1 and x2 are providing very different information; using both together will likely be a much better fit to y than using either one alone.

  • Once a variable has been taken out of a target, it does no good to try to take that variable out again: the best matching line will be zero. See the scatterplots for x21 versus x1 or y1 versus x1, for instance.

  • The values x1, x2, x12, and x21 have all been taken out of ylm.

  • Multiple regression of y against x1 and x2 can be achieved first by computing y1 and x21. These scatterplots appear at (row, column) = (8,1) and (2,1), respectively. With these residuals in hand, we look at their scatterplot at (4,3). These three one-variable regressions do the trick. As Mosteller & Tukey explain, the standard errors of the coefficients can be obtained almost as easily from these regressions, too--but that's not the topic of this question, so I will stop here.

Code

These data were (reproducibly) created in R with a simulation. The analyses, checks, and plots were also produced with R. This is the code.

#
# Simulate the data.
#
set.seed(17)
t.var <- 1:50                                    # The "times" t[i]
x <- exp(t.var %o% c(x1=-0.1, x2=0.025) )        # The two "matchers" x[1,] and x[2,]
beta <- c(5, -1)                                 # The (unknown) coefficients
sigma <- 1/2                                     # Standard deviation of the errors
error <- sigma * rnorm(length(t.var))            # Simulated errors
y <- (y.true <- as.vector(x %*% beta)) + error   # True and simulated y values
data <- data.frame(t.var, x, y, y.true)

par(col="Black", bty="o", lty=0, pch=1)
pairs(data)                                      # Get a close look at the data
#
# Take out the various matchers.
#
take.out <- function(y, x) {fit <- lm(y ~ x - 1); resid(fit)}
data <- transform(transform(data, 
  x2.1 = take.out(x2, x1),
  y.1 = take.out(y, x1),
  x1.2 = take.out(x1, x2),
  y.2 = take.out(y, x2)
), 
  y.21 = take.out(y.2, x1.2),
  y.12 = take.out(y.1, x2.1)
)
data$y.lm <- resid(lm(y ~ x - 1))               # Multiple regression for comparison
#
# Analysis.
#
# Reorder the dataframe (for presentation):
data <- data[c(1:3, 5:12, 4)]

# Confirm that the three ways to obtain the fit are the same:
pairs(subset(data, select=c(y.12, y.21, y.lm)))

# Explore what happened:
panel.lm <- function (x, y, col=par("col"), bg=NA, pch=par("pch"),
   cex=1, col.smooth="red",  ...) {
  box(col="Gray", bty="o")
  ok <- is.finite(x) & is.finite(y)
  if (any(ok))  {
    b <- coef(lm(y[ok] ~ x[ok] - 1))
    col0 <- ifelse(abs(b) < 10^-8, "Red", "Blue")
    lwd0 <- ifelse(abs(b) < 10^-8, 3, 2)
    abline(c(0, b), col=col0, lwd=lwd0)
  }
  points(x, y, pch = pch, col="Black", bg = bg, cex = cex)    
  points(matrix(c(0,0), nrow=1), col="Red", pch=1)
}
panel.hist <- function(x, ...) {
  usr <- par("usr"); on.exit(par(usr))
  par(usr = c(usr[1:2], 0, 1.5) )
  h <- hist(x, plot = FALSE)
  breaks <- h$breaks; nB <- length(breaks)
  y <- h$counts; y <- y/max(y)
  rect(breaks[-nB], 0, breaks[-1], y,  ...)
}
par(lty=1, pch=19, col="Gray")
pairs(subset(data, select=c(-t.var, -y.12, -y.21)), col="Gray", cex=0.8, 
   lower.panel=panel.lm, diag.panel=panel.hist)

# Additional interesting plots:
par(col="Black", pch=1)
#pairs(subset(data, select=c(-t.var, -x1.2, -y.2, -y.21)))
#pairs(subset(data, select=c(-t.var, -x1, -x2)))
#pairs(subset(data, select=c(x2.1, y.1, y.12)))

# Details of the variances, showing how to obtain multiple regression
# standard errors from the OLS matches.
norm <- function(x) sqrt(sum(x * x))
lapply(data, norm)
s <- summary(lm(y ~ x1 + x2 - 1, data=data))
c(s$sigma, s$coefficients["x1", "Std. Error"] * norm(data$x1.2)) # Equal
c(s$sigma, s$coefficients["x2", "Std. Error"] * norm(data$x2.1)) # Equal
c(s$sigma, norm(data$y.12) / sqrt(length(data$y.12) - 2))        # Equal
whuber
la source
1
Could multiple regression of y against x1 and x2 still be achieved by first computing y.1 and x2.1 if x1 and x2 were correlated? Wouldn't it then make a big difference whether we sequentially regressed y on x1 and x2.1 or on x2 and x1.2 ? How does this relate to one regression equation with multiple explanatory variables?
miura
1
@miura, One of the leitmotifs of that chapter in Mosteller & Tukey is that when the xi are correlated, the partials xij have low variances; because their variances appear in the denominator of a formula for the estimation variance of their coefficients, this implies the corresponding coefficients will have relatively uncertain estimates. That's a fact of the data, M&T say, and you need to recognize that. It makes no difference whether you start the regression with x1 or x2: compare y.21 to y.12 in my code.
whuber
1
I came across this today, here is what I think on the question by @miura, Think of a 2 dimensional space where Y is to be projected as a combination of two vectors. y = ax1 + bx2 + res (=0). Now think of y as a combination of 3 variables, y = ax1 + bx2 + cx3. and x3 = mx1 + nx2. so certainly, the order in which you choose your variables is going to effect the coefficients. The reason for this is: the minimum error here can be obtained by various combinations. However, in few examples, the minimum error can be obtained by only one combination and that is where the order will not matter.
Gaurav Singhal
@whuber Can you elaborate on how this equation might be used for a multivariate regression that also has a constant term ? ie y = B1 * x1 + B2 * x2 + c ? It is not clear to me how the constant term can be derived. Also I understand in general what was done for the 2 variables, enough at least to replicate it in Excel. How can that be expanded to 3 variables ? x1, x2, x3. It seems clear that we would need to remove x3 first from y, x1, and x2. then remove x2 from x1 and y. But it is not clear to me how to then get the B3 term.
Fairly Nerdy
I have answered some of my questions I have in the comment above. For a 3 variable regression, we would have 6 steps. Remove x1 from x2, from x3, and from y. Then remove x2,1 from x3,1 and from y1. Then remove x3,21 from y21. That results in 6 equations, each of which is of the form variable = lamda * different variable + residual. One of those equations has a y as the first variable, and if you just keep substituting the other variables in, you get the equation you need
Fairly Nerdy