Solution to Series 3

Transcription

1 Dr. A. Hauser Survival Analysis AS 2018 Solution to Series 3 1. a) > library(mass) > library(survival) > t.url <- ' > d.diabetes <- read.table(t.url, header = TRUE, sep = ",") > r.weib <- survreg(surv(, tod) ~ sex + diab +, data = d.diabetes, dist = "weibull") > summary(r.weib) survreg(formula = Surv(, tod) ~ sex + diab +, data = d.diabetes, dist = "weibull") (Intercept) <2e-16 sex diab Log(scale) Scale= 1.13 Weibull distribution Loglik(model)= Loglik(intercept only)= Chisq= on 3 degrees of freedom, p= Number of Newton-Raphson Iterations: 6 Only age is significant (p-value smaller than 0.05). b) Variable selection using stepaic: > d.diabetes$x4 <- d.diabetes$x5 <- d.diabetes$ > d.diabetes$x4[d.diabetes$diab == 0] <- 0 > d.diabetes$x5[d.diabetes$diab == 1] <- 0 > r.weib2 <- survreg(surv(, tod) ~ sex + diab + x4 + x5, data = d.diabetes) > r.weib3 <- stepaic(r.weib2, direction = "both", scope = ~.^2) > r.weib3$anova Stepwise Model Path Analysis of Deviance Table Initial Model: Surv(, tod) ~ sex + diab + x4 + x5 Final Model: Surv(, tod) ~ diab + x5 Step Df Deviance Resid. Df Resid. Dev AIC sex x > summary(r.weib3)

2 2 survreg(formula = Surv(, tod) ~ diab + x5, data = d.diabetes) (Intercept) e-10 diab x Log(scale) Scale= 1.12 Weibull distribution Loglik(model)= Loglik(intercept only)= Chisq= on 2 degrees of freedom, p= Number of Newton-Raphson Iterations: 5 The result is interpreted as follows: For non-diabetics only age has an influence on survival-probabilities. Being diabetics has an influence on survival that is much more important than the influence of age. This results is in line with clinical experience, that diabetics have a higher risk of dying. c) > library(car) > lin.pred <- predict(r.weib3, type = "lp")[d.diabetes$tod == 1] > log.resid <- log(d.diabetes$[d.diabetes$tod == 1]) - lin.pred > qqplot(exp(log.resid), dist = "weibull", shape = 1 / r.weib3$scale, main = "QQ-plot") [1] 4 13 QQ plot 4 exp(log.resid) weibull quantiles The QQ-Plot looks good. d) The estimated hazard function for the weibull regression is (see slides): h(t; x 1,..., x p ) = α t α 1 exp( α ( β }{{} 0 +β 1 x β p x p )) log(λ 0) Thus the estimated hazard function for a non-diabetic (same for male and female, the variable sex does not appear in the best model) is: ĥ(t; diab = 0, x5) = α t α 1 exp( α x T β) with x T = (1, 0, x5) and β T = ( β 0, β 2, β 5 ). The estimation for α = 1 σ is according to the R-ouptut: Thus α = = ĥ(t; diab = 0, x5) = t exp( ( x5 ( ))) Doing the calculation in R: > (t.sigma <- r.weib3$scale) [1]

3 3 > (t.alpha <- 1 / t.sigma) [1] > (t.beta <- r.weib3$coef) (Intercept) diab x e) We are interested in S (3650 (= 10 years); man, nondiabetic, 50 years old). It holds that (see slides): S(t; x 1,..., x p ) = exp[ exp[α (log(t) β }{{} 0 β 1 x 1... β p x p )]] log(λ 0) Thus Ŝ(t; diab = 0; x5 = 50) = exp ( exp(0.893 (log(t) ( )))) The probability that our patient will still be alive in 10 years (3650 days) is Ŝ(3650) = Calculation in R: > t.alpha <- 1 / r.weib3$scale > t.beta <- unname(r.weib3$coef) > t.beta [1] > exp(-exp(t.alpha * (log(3650) - t.beta[1] - t.beta[3] * 50))) [1] f) The survival time after an operation is dramatically shorter for diabetics: > r.pre <- predict(r.weib3, p = 0.5, type = "uquantile", se.fit = TRUE, newdata = data.frame(diab = 0, x5 = 50)) > (exp(r.pre$fit + c(-1.96, 0, 1.96) * r.pre$se.fit) / 365) [1] Approximately 11 years after the operation half of the non-diabetics is dead. However, the confidence interval is quite large (13 years). > r.pre <- predict(r.weib3, p = 0.5, type = "uquantile", se.fit = TRUE, newdata = data.frame(diab = 1, x5 = 0)) > (exp(r.pre$fit + c(-1.96, 0, 1.96) * r.pre$se.fit) / 365) [1] Half of the diabetics is dead approximately 2 years after the operation. 2. a) > library(survival) > t.url <- " > d.diabetes <- read.table(t.url, header = TRUE, sep = ",") > d.nondiab <- d.diabetes[d.diabetes$diab == 0, ] > d.diab <- d.diabetes[d.diabetes$diab == 1, ] > par(mfrow = c(1, 2)) > plot( ~, data = d.nondiab, col = 2 - d.nondiab$tod, pch = ifelse(d.nondiab$tod == 1, 1, 16), main = "Nondiabetes") > lines(smooth.spline(d.nondiab$, d.nondiab$)) > plot( ~, data = d.diab, pch = ifelse(d.diab$tod == 1, 2, 17), col = 2 - d.diab$tod, main = "diabetes") > lines(smooth.spline(d.diab$, d.diab$))

4 4 Nondiabetes diabetes We can see, that for non-diabetics the survival times decrease linearly, whereas for the diabetics it is rather a quadratic curve. The survival time for diabetics is very low compared to the non-diabetics, that means for younger people the illness is an important factor to survival. Furthermore we can suppose, that after a certain age, survival times do not depend much on being a diabetic or not - that means after a certain age, age is the more important factor to survival. We can see this very well in a common plot: > plot( ~, data = d.nondiab, col = 2 - d.nondiab$tod, pch = ifelse(d.nondiab$tod == 1, 1, 16)) > lines(smooth.spline(d.nondiab$, d.nondiab$)) > points( ~, data = d.diab, pch = ifelse(d.diab$tod == 1, 2, 17), col = 2 - d.diab$tod) > lines(smooth.spline(d.diab$, d.diab$)) b) The dataset is prepared as in the previous exercise: > d.diabetes$x4 <- d.diabetes$x5 <- d.diabetes$ > d.diabetes$x4[d.diabetes$diab == 0] <- 0 > d.diabetes$x5[d.diabetes$diab == 1] <- 0 We estimate the loglogistic model with survreg. > r.logl <- survreg(surv(, tod) ~ diab + x5, data = d.diabetes, dist = "loglogistic") > summary(r.logl) survreg(formula = Surv(, tod) ~ diab + x5, data = d.diabetes, dist = "loglogistic") (Intercept) e-09 diab x Log(scale) Scale= Log logistic distribution Loglik(model)= Loglik(intercept only)= Chisq= on 2 degrees of freedom, p=

5 5 Number of Newton-Raphson Iterations: 4 According to the slides the following holds: Calculation in R: S(t X 1,..., X p ) = > alpha <- 1 / r.logl$scale > beta <- unname(r.logl$coef) > beta [1] exp [α(log(t) β 0 β 1 X 1... β p X p )] > (s.hat <- 1 / (1 + exp(alpha * (log(6 * 365) - beta[1] - beta[2] * 1)))) [1] The probability of being alive 6 years after the operation is only 0.22 for diabetics. The odds can be calculated easily by > s.hat / (1 - s.hat) [1] We can also derive them generally: odds = S(t X 1,..., X p ) 1 S(t X 1,..., X p ) = 1 exp [α(log(t) β 0 β 1 X 1... β p X p )] = exp [ α(log(t) β 0 β 1 X 1... β p X p )] > exp(-alpha * (log(6 * 365) - beta[1] - beta[2] * 1)) [1] c) Let x 1 = [diab = 1, x5 = 0] be the 50-year-old diabetic and x 2 = [diab = 0, x5 = 50] the 50-year-old non-diabetic. We have: OR(t x = x 2, x = x 1 ) = odds(x 2 ) odds(x 1 ) > beta <- r.logl$coef[2:3] > exp(alpha * (c(0, 50) - c(1, 0)) %*% beta) [,1] [1,] = exp [ α(log(t) β 0 β 1 X 2,1... β p X 2,p )] exp [ α(log(t) β 0 β 1 X 1,1... β p X 1,p )] = exp [ α(x 2 x 1 )β ] Thus the odds-ratio (P[survives]/P[dies]) of a 50-year-old nondiabetic is almost 9 times as large as the odds ratio a diabetics. This ratio holds for all times t.