具有二次多项式和在断点处平滑连接的直线的分段回归 [英] Piecewise regression with a quadratic polynomial and a straight line joining smoothly at a break point
问题描述
我想用一个断点xt拟合分段线性回归，这样对于x < xt我们有一个二次多项式，对于x >= xt我们有一条直线.两段应该平滑连接，连续性最高为xt的一阶导数.这是它的外观图片:
I want to fit a piecewise linear regression with one break point xt, such that for x < xt we have a quadratic polynomial and for x >= xt we have a straight line. Two pieces should join smoothly, with continuity up to 1st derivative at xt. Here's picture of what it may look like:
我将分段回归函数参数化为:
I have parametrize my piecewise regression function as:
其中a，b，c和xt是要估计的参数.
where a, b, c and xt are parameters to be estimated.
我想将此模型与调整后R平方的整个范围内的二次多项式回归进行比较.
I want to compare this model with a quadratic polynomial regression over the whole range in terms of adjusted R-squared.
这是我的数据:
y <- c(1, 0.59, 0.15, 0.078, 0.02, 0.0047, 0.0019, 1, 0.56, 0.13,
0.025, 0.0051, 0.0016, 0.00091, 1, 0.61, 0.12, 0.026, 0.0067,
0.00085, 4e-04)
x <- c(0, 5.53, 12.92, 16.61, 20.3, 23.07, 24.92, 0, 5.53, 12.92,
16.61, 20.3, 23.07, 24.92, 0, 5.53, 12.92, 16.61, 20.3, 23.07,
24.92)
对于一个已知的xt，我的尝试如下:
My attempt goes as follows, for a known xt:
z <- pmax(0, x - xt)
x1 <- pmin(x, xt)
fit <- lm(y ~ x1 + I(x1 ^ 2) + z - 1)
但是直线似乎与xt处的二次多项式不相切.我在哪里做错了?
But the straight line does not appear to be tangent to the quadratic polynomial at xt. Where am I doing wrong?
类似的问题:
- 以直线和水平线在断点处连接的分段分割
- 在我的数据中拟合V形曲线(经过交叉验证)
- Piecewise regression with a straight line and a horizontal line joining at a break point
- Fitting a V-shape curve to my data (on Cross Validated)
推荐答案
在本节中，我将演示一个可重现的示例.请确保您已在其他答案中定义了源函数.
In this section I will be demonstrating a reproducible example. Please make sure you have sourced functions defined in the other answer.
## we first generate a true model
set.seed(0)
x <- runif(100) ## sample points on [0, 1]
beta <- c(0.1, -0.2, 2) ## true coefficients
X <- getX(x, 0.6) ## model matrix with true break point at 0.6
y <- X %*% beta + rnorm(100, 0, 0.08) ## observations with Gaussian noise
plot(x, y)
现在，假设我们不知道c，我们想在一个均匀分布的网格上进行搜索:
Now, assume we don't know c, and we would like to search on a evenly spaced grid:
c.grid <- seq(0.1, 0.9, 0.05)
fit <- choose.c(x, y, c.grid)
fit$c
RSS选择了0.55.这与真实值0.6略有不同，但是从图中可以看出，RSS曲线在[0.5, 0.6]之间变化不大，所以我对此感到满意.
RSS has chosen 0.55. This is slightly different from the true value 0.6, but from the plot we see that RSS curve does not vary much between [0.5, 0.6] so I am happy with this.
生成的模型fit包含丰富的信息:
The resulting model fit contains rich information:
#List of 12
# $ coefficients : num [1:3] 0.114 -0.246 2.366
# $ residuals : num [1:100] 0.03279 -0.01515 0.21188 -0.06542 0.00763 ...
# $ fitted.values: num [1:100] 0.0292 0.3757 0.2329 0.1087 0.0263 ...
# $ R : num [1:3, 1:3] -10 0.1 0.1 0.292 2.688 ...
# $ sig2 : num 0.00507
# $ coef.table : num [1:3, 1:4] 0.1143 -0.2456 2.3661 0.0096 0.0454 ...
# ..- attr(*, "dimnames")=List of 2
# .. ..$ : chr [1:3] "beta0" "beta1" "beta2"
# .. ..$ : chr [1:4] "Estimate" "Std. Error" "t value" "Pr(>|t|)"
# $ aic : num -240
# $ bic : num -243
# $ c : num 0.55
# $ RSS : num 0.492
# $ r.squared : num 0.913
# $ adj.r.squared: num 0.911
我们可以提取系数汇总表:
We can extract the summary table for coefficients:
fit$coef.table
# Estimate Std. Error t value Pr(>|t|)
#beta0 0.1143132 0.009602697 11.904286 1.120059e-20
#beta1 -0.2455986 0.045409356 -5.408546 4.568506e-07
#beta2 2.3661097 0.169308226 13.975161 5.730682e-25
最后，我们希望看到一些预测图.
Finally, we want to see some prediction plot.
x.new <- seq(0, 1, 0.05)
p <- pred(fit, x.new)
head(p)
# fit se.fit lwr upr
#[1,] 0.9651406 0.02903484 0.9075145 1.0227668
#[2,] 0.8286400 0.02263111 0.7837235 0.8735564
#[3,] 0.7039698 0.01739193 0.6694516 0.7384880
#[4,] 0.5911302 0.01350837 0.5643199 0.6179406
#[5,] 0.4901212 0.01117924 0.4679335 0.5123089
#[6,] 0.4009427 0.01034868 0.3804034 0.4214819
我们可以绘制一个图:
plot(x, y, cex = 0.5)
matlines(x.new, p[,-2], col = c(1,2,2), lty = c(1,2,2), lwd = 2)
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