CODEWITHSUNDEEP
rlinear-regression
IceCreamToucan
IceCreamToucan

Reputation: 28675

Predict function for lm object in R

Why are prediction_me and prediction_R not equal? I'm attempting to follow the formula given by Lemma 5 here. Does the predict function use a different formula, have I made a mistake in my computation somewhere, or is it just rounding error? (the two are pretty close)

set.seed(100)
# genrate data
x    <- rnorm(100, 10)
y    <- 3 + x + rnorm(100, 5)
data <- data.frame(x = x, y = y)
# fit model
mod  <- lm(y ~ x, data = data)

# new observation
data2 <- data.frame(x = rnorm(5, 10))

# prediction for new observation
d    <- as.matrix(cbind(1, data[,-2]))
d2   <- as.matrix(cbind(1, data2))
fit  <- d2 %*% mod$coefficients 
t    <- qt(1 - .025, mod$df.residual)
s    <- summary(mod)$sigma
half <- as.vector(t*s*sqrt(1 + d2%*%solve(t(d)%*%d, t(d2))))

prediction_me <- cbind(fit, fit - half, fit + half)

prediction_R <- predict(mod, newdata = data2, interval = 'prediction')


prediction_me
prediction_R

Upvotes: 1

Views: 224

Answers (1)

Julius Vainora
Julius Vainora

Reputation: 48191

Your current code is almost fine. Just note that the formula in Lemma 5 is for a single newly observed x. For this reason, half contains not only relevant variances but also covariances, while you only need the former ones. Thus, as.vector should be replaced with diag:

half <- diag(t * s * sqrt(1 + d2 %*% solve(t(d) %*%d , t(d2))))
prediction_me <- cbind(fit, fit - half, fit + half)
prediction_R <- predict(mod, newdata = data2, interval = 'prediction')

range(prediction_me - prediction_R)
# [1] 0 0

Upvotes: 2

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