带有“lm”的线性模型:如何得到预测值之和的预测方差

时间:2016-09-05 21:25:39

标签: r regression linear-regression lm predict

我将具有多个预测变量的线性模型的预测值相加,如下例所示,并希望计算此总和的组合方差,标准误差和可能的置信区间。

lm.tree <- lm(Volume ~ poly(Girth,2), data = trees)

假设我有一组Girths

newdat <- list(Girth = c(10,12,14,16)

我想要预测总Volume

pr <- predict(lm.tree, newdat, se.fit = TRUE)
total <- sum(pr$fit)
# [1] 111.512

如何获取total的差异?

类似的问题是here (for GAMs),但我不确定如何继续vcov(lm.trees)。我很感激该方法的参考。

1 个答案:

答案 0 :(得分:5)

您需要获得完整的方差 - 协方差矩阵,然后对其所有元素求和。以下是小证明:

enter image description here

这里的证据是使用另一个定理,您可以从Covariance-wikipedia找到:

enter image description here

具体来说,我们采用的线性变换是所有1的列矩阵。生成的二次形式计算为as following,所有x_ix_j均为1。

enter image description here

设置

## your model
lm.tree <- lm(Volume ~ poly(Girth, 2), data = trees)

## newdata (a data frame)
newdat <- data.frame(Girth = c(10, 12, 14, 16))

重新实施predict.lm以计算方差 - 协方差矩阵

有关predict.lm的工作原理,请参阅How does predict.lm() compute confidence interval and prediction interval?。以下小函数lm_predict模仿它的作用,除了

  • 它不构建置信区间或预测区间(但是Q&amp; A中所解释的构造非常简单);
  • 如果diag = FALSE;
  • ,它可以计算预测值的完全方差 - 协方差矩阵
  • 它返回方差(对于预测值和残差),而不是标准误差;
  • 它不能type = "terms";它只预测响应变量。

lm_predict <- function (lmObject, newdata, diag = TRUE) {
  ## input checking
  if (!inherits(lmObject, "lm")) stop("'lmObject' is not a valid 'lm' object!")
  ## extract "terms" object from the fitted model, but delete response variable
  tm <- delete.response(terms(lmObject))      
  ## linear predictor matrix
  Xp <- model.matrix(tm, newdata)
  ## predicted values by direct matrix-vector multiplication
  pred <- c(Xp %*% coef(lmObject))
  ## efficiently form the complete variance-covariance matrix
  QR <- lmObject$qr   ## qr object of fitted model
  piv <- QR$pivot     ## pivoting index
  r <- QR$rank        ## model rank / numeric rank
  if (is.unsorted(piv)) {
    ## pivoting has been done
    B <- forwardsolve(t(QR$qr), t(Xp[, piv]), r)
    } else {
    ## no pivoting is done
    B <- forwardsolve(t(QR$qr), t(Xp), r)
    }
  ## residual variance
  sig2 <- c(crossprod(residuals(lmObject))) / df.residual(lmObject)
  if (diag) {
    ## return point-wise prediction variance
    VCOV <- colSums(B ^ 2) * sig2
    } else {
    ## return full variance-covariance matrix of predicted values
    VCOV <- crossprod(B) * sig2
    }
  list(fit = pred, var.fit = VCOV, df = lmObject$df.residual, residual.var = sig2)
  }

我们可以将其输出与predict.lm的输出进行比较:

predict.lm(lm.tree, newdat, se.fit = TRUE)
#$fit
#       1        2        3        4 
#15.31863 22.33400 31.38568 42.47365 
#
#$se.fit
#        1         2         3         4 
#0.9435197 0.7327569 0.8550646 0.8852284 
#
#$df
#[1] 28
#
#$residual.scale
#[1] 3.334785

lm_predict(lm.tree, newdat)
#$fit
#[1] 15.31863 22.33400 31.38568 42.47365
#
#$var.fit    ## the square of `se.fit`
#[1] 0.8902294 0.5369327 0.7311355 0.7836294
#
#$df
#[1] 28
#
#$residual.var   ## the square of `residual.scale`
#[1] 11.12079

特别是:

oo <- lm_predict(lm.tree, newdat, FALSE)
oo
#$fit
#[1] 15.31863 22.33400 31.38568 42.47365
#
#$var.fit
#            [,1]      [,2]       [,3]       [,4]
#[1,]  0.89022938 0.3846809 0.04967582 -0.1147858
#[2,]  0.38468089 0.5369327 0.52828797  0.3587467
#[3,]  0.04967582 0.5282880 0.73113553  0.6582185
#[4,] -0.11478583 0.3587467 0.65821848  0.7836294
#
#$df
#[1] 28
#
#$residual.var
#[1] 11.12079

请注意,方差 - 协方差矩阵不是以天真的方式计算的:Xp %*% vcov(lmObject) % t(Xp),这很慢。

聚合(总和)

在您的情况下,聚合操作是oo$fit中所有值的总和。此聚合的均值和方差为

sum_mean <- sum(oo$fit)  ## mean of the sum
# 111.512

sum_variance <- sum(oo$var.fit)  ## variance of the sum
# 6.671575

您可以通过在模型中使用t分布和剩余自由度来进一步构建此聚合值的置信区间(CI)。

alpha <- 0.95
Qt <- c(-1, 1) * qt((1 - alpha) / 2, lm.tree$df.residual, lower.tail = FALSE)
#[1] -2.048407  2.048407

## %95 CI
sum_mean + Qt * sqrt(sum_variance)
#[1] 106.2210 116.8029

构建预测区间(PI)需要进一步考虑残差方差。

## adjusted variance-covariance matrix
VCOV_adj <- with(oo, var.fit + diag(residual.var, nrow(var.fit)))

## adjusted variance for the aggregation
sum_variance_adj <- sum(VCOV_adj)  ## adjusted variance of the sum

## 95% PI
sum_mean + Qt * sqrt(sum_variance_adj)
#[1]  96.86122 126.16268

聚合(一般而言)

一般聚合操作可以是oo$fit

的线性组合
w[1] * fit[1] + w[2] * fit[2] + w[3] * fit[3] + ...

例如,总和操作的所有权重均为1;平均操作的所有权重均为0.25(在4个数据的情况下)。这是一个函数,它采用权重向量,显着性级别以及lm_predict返回的内容来生成聚合的统计信息。

agg_pred <- function (w, predObject, alpha = 0.95) {
  ## input checing
  if (length(w) != length(predObject$fit)) stop("'w' has wrong length!")
  if (!is.matrix(predObject$var.fit)) stop("'predObject' has no variance-covariance matrix!")
  ## mean of the aggregation
  agg_mean <- c(crossprod(predObject$fit, w))
  ## variance of the aggregation
  agg_variance <- c(crossprod(w, predObject$var.fit %*% w))
  ## adjusted variance-covariance matrix
  VCOV_adj <- with(predObject, var.fit + diag(residual.var, nrow(var.fit)))
  ## adjusted variance of the aggregation
  agg_variance_adj <- c(crossprod(w, VCOV_adj %*% w))
  ## t-distribution quantiles
  Qt <- c(-1, 1) * qt((1 - alpha) / 2, predObject$df, lower.tail = FALSE)
  ## names of CI and PI
  NAME <- c("lower", "upper")
  ## CI
  CI <- setNames(agg_mean + Qt * sqrt(agg_variance), NAME)
  ## PI
  PI <- setNames(agg_mean + Qt * sqrt(agg_variance_adj), NAME)
  ## return
  list(mean = agg_mean, var = agg_variance, CI = CI, PI = PI)
  }

对前一次总和操作的快速测试:

agg_pred(rep(1, length(oo$fit)), oo)
#$mean
#[1] 111.512
#
#$var
#[1] 6.671575
#
#$CI
#   lower    upper 
#106.2210 116.8029 
#
#$PI
#    lower     upper 
# 96.86122 126.16268 

快速测试平均操作:

agg_pred(rep(1, length(oo$fit)) / length(oo$fit), oo)
#$mean
#[1] 27.87799
#
#$var
#[1] 0.4169734
#
#$CI
#   lower    upper 
#26.55526 29.20072 
#
#$PI
#   lower    upper 
#24.21531 31.54067 

备注

此答案已得到改进,可为Linear regression with `lm()`: prediction interval for aggregated predicted values提供易于使用的功能。

升级(适用于大数据)

  

太好了!非常感谢!有一件事我忘了提到:在我的实际应用中,我需要总结约300,000个预测,这将产生一个大约为700GB大小的完全方差 - 协方差矩阵。你有没有想过是否有一种计算上更有效的方法来直接得到方差 - 协方差矩阵之和?

感谢Linear regression with `lm()`: prediction interval for aggregated predicted values的OP对此非常有用的评论。是的,这是可能的,而且(显着)计算上更便宜。目前,lm_predict形成了方差 - 协方差:

enter image description here

agg_pred将预测方差(用于构建CI)计算为二次形式:w'(B'B)w,并将预测方差(用于构建PI)计算为另一种二次形式w'(B'B + D)w,其中{{} 1}}是残差方差的对角矩阵。显然,如果我们融合这两个函数,我们有一个更好的计算策略:

enter image description here

避免了DB的计算;我们已经将所有矩阵 - 矩阵乘法替换为矩阵向量乘法。 B'BB没有内存存储空间;仅适用于B'B,它只是一个向量。这是融合的实现。

u

让我们快速测试一下。

## this function requires neither `lm_predict` nor `agg_pred`
fast_agg_pred <- function (w, lmObject, newdata, alpha = 0.95) {
  ## input checking
  if (!inherits(lmObject, "lm")) stop("'lmObject' is not a valid 'lm' object!")
  if (!is.data.frame(newdata)) newdata <- as.data.frame(newdata)
  if (length(w) != nrow(newdata)) stop("length(w) does not match nrow(newdata)")
  ## extract "terms" object from the fitted model, but delete response variable
  tm <- delete.response(terms(lmObject))      
  ## linear predictor matrix
  Xp <- model.matrix(tm, newdata)
  ## predicted values by direct matrix-vector multiplication
  pred <- c(Xp %*% coef(lmObject))
  ## mean of the aggregation
  agg_mean <- c(crossprod(pred, w))
  ## residual variance
  sig2 <- c(crossprod(residuals(lmObject))) / df.residual(lmObject)
  ## efficiently compute variance of the aggregation without matrix-matrix computations
  QR <- lmObject$qr   ## qr object of fitted model
  piv <- QR$pivot     ## pivoting index
  r <- QR$rank        ## model rank / numeric rank
  u <- forwardsolve(t(QR$qr), c(crossprod(Xp, w))[piv], r)
  agg_variance <- c(crossprod(u)) * sig2
  ## adjusted variance of the aggregation
  agg_variance_adj <- agg_variance + c(crossprod(w)) * sig2
  ## t-distribution quantiles
  Qt <- c(-1, 1) * qt((1 - alpha) / 2, lmObject$df.residual, lower.tail = FALSE)
  ## names of CI and PI
  NAME <- c("lower", "upper")
  ## CI
  CI <- setNames(agg_mean + Qt * sqrt(agg_variance), NAME)
  ## PI
  PI <- setNames(agg_mean + Qt * sqrt(agg_variance_adj), NAME)
  ## return
  list(mean = agg_mean, var = agg_variance, CI = CI, PI = PI)
  }

是的,答案是对的!