Was passiert hier, wenn ich bei der Einstellung der logistischen Regression den quadratischen Verlust verwende?

Ich versuche, einen quadratischen Verlust zu verwenden, um eine binäre Klassifizierung für einen Spielzeugdatensatz durchzuführen.

Ich verwende einen mtcarsDatensatz, verwende Meile pro Gallone und Gewicht, um die Übertragungsart vorherzusagen. Das folgende Diagramm zeigt die zwei Arten von Übertragungstypdaten in verschiedenen Farben und die Entscheidungsgrenze, die durch verschiedene Verlustfunktionen erzeugt werden. Der quadrierte Verlust ist $\sum_i (y_i-p_i)^2$ wobei $y_i$ ist die Grundwahrheits Etikett (0 oder 1) und ist die vorhergesagte Wahrscheinlichkeit . Mit anderen Worten, ich ersetze den logistischen Verlust durch den quadratischen Verlust in der Klassifizierungseinstellung, andere Teile sind die gleichen.p i = Logit - 1 ( β T x i$p_i$$p_i=\text{Logit}^{-1}(\beta^Tx_i)$

Für ein Spielzeugbeispiel mit mtcarsDaten habe ich in vielen Fällen ein Modell erhalten, das der logistischen Regression "ähnlich" ist (siehe folgende Abbildung mit zufälligem Startwert 0).

Aber in einigen Fällen (wenn wir dies tun set.seed(1)) scheint der Quadratverlust nicht gut zu funktionieren. Was passiert hier? Die Optimierung konvergiert nicht? Ein logistischer Verlust ist einfacher zu optimieren als ein quadratischer Verlust? Jede Hilfe wäre dankbar.

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Code

d=mtcars[,c("am","mpg","wt")]
plot(d$mpg,d$wt,col=factor(d$am))
lg_fit=glm(am~.,d, family = binomial())
abline(-lg_fit$coefficients[1]/lg_fit$coefficients[3],
       -lg_fit$coefficients[2]/lg_fit$coefficients[3])
grid()

# sq loss
lossSqOnBinary<-function(x,y,w){
  p=plogis(x %*% w)
  return(sum((y-p)^2))
}

# ----------------------------------------------------------------
# note, this random seed is important for squared loss work
# ----------------------------------------------------------------
set.seed(0)

x0=runif(3)
x=as.matrix(cbind(1,d[,2:3]))
y=d$am
opt=optim(x0, lossSqOnBinary, method="BFGS", x=x,y=y)

abline(-opt$par[1]/opt$par[3],
       -opt$par[2]/opt$par[3], lty=2)
legend(25,5,c("logisitc loss","squared loss"), lty=c(1,2))

Anscheinend haben Sie das Problem in Ihrem Beispiel behoben, aber ich denke, es lohnt sich immer noch, den Unterschied zwischen der logistischen Regression der kleinsten Quadrate und der maximalen Wahrscheinlichkeit genauer zu untersuchen.

Lassen Sie uns etwas Notation bekommen. Let $L_S(y_i, \hat y_i) = \frac 12(y_i - \hat y_i)^2$und$L_L(y_i, \hat y_i) = y_i \log \hat y_i + (1 - y_i) \log(1 - \hat y_i)$. Wenn wir MaximumLikelihood tun (oder minimalen negativen LogLikelihood wie ich hier tue), haben wir β L:=arg minb∈ R

$$\hat \beta_L := \text{argmin}_{b \in \mathbb R^p} -\sum_{i=1}^n y_i \log g^{-1}(x_i^T b) + (1-y_i)\log(1 - g^{-1}(x_i^T b))$$ mit$g$unserer LinkFunktion ist.

Alternativ haben wir β S : = arg min b ∈ R p 1

$$\hat \beta_S := \text{argmin}_{b \in \mathbb R^p} \frac 12 \sum_{i=1}^n (y_i - g^{-1}(x_i^T b))^2$$ als Lösung der kleinsten Quadrate. So β SminimiertLSundähnlicher Weise fürLL.$\hat \beta_S$$L_S$$L_L$

Lassen $f_S$ und $f_L$ sein , die Zielfunktionen zur Minimierung der entsprechenden $L_S$ und $L_L$ jeweils wie für erfolgt β$\hat \beta_S$ und β L . Schließlich sei h = g - 1 so y i = h ( x T i b ) . Beachten Sie, dass bei Verwendung des kanonischen Links h ( z ) = 1 ist$\hat \beta_L$$h = g^{-1}$$\hat y_i = h(x_i^T b)$

$$h(z) = \frac{1}{1+e^{-z}} \implies h'(z) = h(z) (1 - h(z)).$$

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Für eine regelmäßige logistische Regression haben wir

$$\frac{\partial f_L}{\partial b_j} = -\sum_{i=1}^n h'(x_i^T b)x_{ij} \left( \frac{y_i}{h(x_i^T b)} - \frac{1-y_i}{1 - h(x_i^T b)}\right).$$ Mit$h' = h \cdot (1 - h)$wir dies zu∂fLvereinfachen $$\frac{\partial f_L}{\partial b_j} = -\sum_{i=1}^n x_{ij} \left( y_i(1 - \hat y_i) - (1-y_i)\hat y_i\right) = -\sum_{i=1}^n x_{ij}(y_i - \hat y_i)$$ also $$\nabla f_L(b) = -X^T (Y - \hat Y).$$

Als nächstes machen wir zweite Ableitungen. Der Hessische

$$H_L:= \frac{\partial^2 f_L}{\partial b_j \partial b_k} = \sum_{i=1}^n x_{ij} x_{ik} \hat y_i (1 - \hat y_i).$$ $H_L = X^T A X$$A = \text{diag} \left(\hat Y (1 - \hat Y)\right)$$H_L$$\hat Y$$Y$$H_L$$b$

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Vergleichen wir dies mit den kleinsten Quadraten.

$$\frac{\partial f_S}{\partial b_j} = - \sum_{i=1}^n (y_i - \hat y_i) h'(x^T_i b)x_{ij}.$$

This means we have

$$\nabla f_S(b) = -X^T A (Y - \hat Y).$$ This is a vital point: the gradient is almost the same except for all $i$ $\hat y_i (1 - \hat y_i) \in (0,1)$ so basically we're flattening the gradient relative to $\nabla f_L$. This'll make convergence slower.

For the Hessian we can first write

$$\frac{\partial f_S}{\partial b_j} = - \sum_{i=1}^n x_{ij}(y_i - \hat y_i) \hat y_i (1 - \hat y_i) = - \sum_{i=1}^n x_{ij}\left( y_i \hat y_i - (1+y_i)\hat y_i^2 + \hat y_i^3\right).$$

This leads us to

$$H_S:=\frac{\partial^2 f_S}{\partial b_j \partial b_k} = - \sum_{i=1}^n x_{ij} x_{ik} h'(x_i^T b) \left( y_i - 2(1+y_i)\hat y_i + 3 \hat y_i^2 \right).$$

Let $B = \text{diag} \left( y_i - 2(1+y_i)\hat y_i + 3 \hat y_i ^2 \right)$. We now have

$$H_S = -X^T A B X.$$

Unfortunately for us, the weights in $B$ are not guaranteed to be non-negative: if $y_i = 0$ then $y_i - 2(1+y_i)\hat y_i + 3 \hat y_i ^2 = \hat y_i (3 \hat y_i - 2)$ which is positive iff $\hat y_i > \frac 23$. Similarly, if $y_i = 1$ then $y_i - 2(1+y_i)\hat y_i + 3 \hat y_i ^2 = 1-4 \hat y_i + 3 \hat y_i^2$ which is positive when $\hat y_i < \frac 13$ (it's also positive for $\hat y_i > 1$ but that's not possible). This means that $H_S$ is not necessarily PSD, so not only are we squashing our gradients which will make learning harder, but we've also messed up the convexity of our problem.

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All in all, it's no surprise that least squares logistic regression struggles sometimes, and in your example you've got enough fitted values close to $0$ or $1$ so that $\hat y_i (1 - \hat y_i)$ can be pretty small and thus the gradient is quite flattened.

Connecting this to neural networks, even though this is but a humble logistic regression I think with squared loss you're experiencing something like what Goodfellow, Bengio, and Courville are referring to in their Deep Learning book when they write the following:

One recurring theme throughout neural network design is that the gradient of the cost function must be large and predictable enough to serve as a good guide for the learning algorithm. Functions that saturate (become very flat) undermine this objective because they make the gradient become very small. In many cases this happens because the activation functions used to produce the output of the hidden units or the output units saturate. The negative log-likelihood helps to avoid this problem for many models. Many output units involve an exp function that can saturate when its argument is very negative. The log function in the negative log-likelihood cost function undoes the exp of some output units. We will discuss the interaction between the cost function and the choice of output unit in Sec. 6.2.2.

and, in 6.2.2,

Unfortunately, mean squared error and mean absolute error often lead to poor results when used with gradient-based optimization. Some output units that saturate produce very small gradients when combined with these cost functions. This is one reason that the cross-entropy cost function is more popular than mean squared error or mean absolute error, even when it is not necessary to estimate an entire distribution $p(y|x)$.

(both excerpts are from chapter 6).

I would thank to thank @whuber and @Chaconne for help. Especially @Chaconne, this derivation is what I wished to have for years.

The problem IS in the optimization part. If we set the random seed to 1, the default BFGS will not work. But if we change the algorithm and change the max iteration number it will work again.

As @Chaconne mentioned, the problem is squared loss for classification is non-convex and harder to optimize. To add on @Chaconne's math, I would like to present some visualizations on to logistic loss and squared loss.

We will change the demo data from mtcars, since the original toy example has $3$ coefficients including the intercept. We will use another toy data set generated from mlbench, in this data set, we set $2$ parameters, which is better for visualization.

Here is the demo

• The data is shown in the left figure: we have two classes in two colors. x,y are two features for the data. In addition, we use red line to represent the linear classifier from logistic loss, and the blue line represent the linear classifier from squared loss.

• The middle figure and right figure shows the contour for logistic loss (red) and squared loss (blue). x, y are two parameters we are fitting. The dot is the optimal point found by BFGS.

From the contour we can easily see how why optimizing squared loss is harder: as Chaconne mentioned, it is non-convex.

Here is one more view from persp3d.

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Code

set.seed(0)
d=mlbench::mlbench.2dnormals(50,2,r=1)
x=d$x
y=ifelse(d$classes==1,1,0)

lg_loss <- function(w){
  p=plogis(x %*% w)
  L=-y*log(p)-(1-y)*log(1-p)
  return(sum(L))
}
sq_loss <- function(w){
  p=plogis(x %*% w)
  L=sum((y-p)^2)
  return(L)
}

w_grid_v=seq(-15,15,0.1)
w_grid=expand.grid(w_grid_v,w_grid_v)

opt1=optimx::optimx(c(1,1),fn=lg_loss ,method="BFGS")
z1=matrix(apply(w_grid,1,lg_loss),ncol=length(w_grid_v))

opt2=optimx::optimx(c(1,1),fn=sq_loss ,method="BFGS")
z2=matrix(apply(w_grid,1,sq_loss),ncol=length(w_grid_v))

par(mfrow=c(1,3))
plot(d,xlim=c(-3,3),ylim=c(-3,3))
abline(0,-opt1$p2/opt1$p1,col='darkred',lwd=2)
abline(0,-opt2$p2/opt2$p1,col='blue',lwd=2)
grid()
contour(w_grid_v,w_grid_v,z1,col='darkred',lwd=2, nlevels = 8)
points(opt1$p1,opt1$p2,col='darkred',pch=19)
grid()
contour(w_grid_v,w_grid_v,z2,col='blue',lwd=2, nlevels = 8)
points(opt2$p1,opt2$p2,col='blue',pch=19)
grid()


# library(rgl)
# persp3d(w_grid_v,w_grid_v,z1,col='darkred')