ST3242

Soya beans Test Confidence intervals CI for β MVNs, Var, Cov 

Cox proportional hazards model 

ST3242: Introduction to Survival Analysis 

Alex Cook 

September 2008 

ST3242 : Cox proportional hazards model 1/44


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Plan for today 

Go over mock mid-term test 

3pm feedback 

Lecture: confidence intervals for the Cox phm 



Exam advice 

Read all questions at start of exam 

Read all instructions 

Ensure do everything asked and nothing more 

Write as neatly as you can 

Don’t “do a Ravi” 

Avoid rounding too much too soon 



Exam advice 

INSTRUCTIONS TO CANDIDATES: 

1 This test contains EIGHT questions and comprises 

FOUR printed pages. 

2 Answer ALL questions for a total of 100 marks. 

3 This is a closed-book test; only non-programmable 

calculators are allowed. 



Q1: [3] 

If the survival function is S(t), what is the hazard function 

h(t)? 



Q2: [6] 

If the hazard function is h(t) = θt for a parameter θ > 0, 

what are the survival and density functions? 



Q3: [10] 

If survival times in the absence of censoring are distributed 

according to a log-logistic distribution with parameters κ and 

λ, the hazard and survival functions are 

h(t) = λκtκ−1 

1 + λtκ S(t) = 

1 

1 + λtκ respectively. If we have observed data of the form (ti, δi), 

where δi = 1 if individual i fails at time ti and δi = 0 if i is 

right-censored at ti, for i = 1, . . . , m, what is the 

log-likelihood function? 



Q4: [6] 

Consider a study into the length of time people can survive 

without purchasing or receiving plastic goods. The study 

starts at time Tstart and ends at time Tend. Three nus 

students are among the participants. 

Student A is recruited to the study at time Tstart. She 

buys no plastic until time tA < Tend, when she buys a 

bottle of Pocari Sweat. 

For each student, say if the datum is uncensored, 

right-censored, left-censored or interval-censored. 



Q4: [6] 





Student B is recruited to the study at time Tstart. He 

reports to the organisers at time tB < Tend that he has 

bought no plastics, but they never hear from him again. 





Q4: [6] 





Student C is recruited to the study at time 

tC ∈ (Tstart, Tend). She buys no plastic for the remainder 

of the study. 





Q5: [12] 

Use the delta method to approximate the mean and variance 

of √ X , where X is a random variable with mean µ and 

variance σ 2 . When do you expect these will provide a good 

approximation to the true mean and variance of √ X ? 



Q6(i): [36] 

Calculate and sketch the Kaplan–Meier estimate of S(t) for 

the following data, where δi = 1 if individual i died at time ti 

and δi = 0 if i was censored at that time. 

i ti δi 

1 0.6 1 

2 0.9 1 

3 1.1 0 

4 1.5 1 

5 2.0 0 

6 2.0 0 



Q6(ii): [36] 

Recall that 

 

ˆV{ˆS(t)} = ˆS(t) 

2 

d(i) 

n(i 

t (i)≤t 

− )(n(i − ) − d(i)) 

where t(i) is the ith ordered event time, d(i) is the number of 

deaths at that time and n(i − ) is the number at risk an instant 

before that time. 

Construct the “naïve” 95% confidence interval described in 

the lecture notes for S(1), the probability of surviving at least 

to one time unit. What is the probability the true value of 

S(1) lies within your confidence interval? 



Q7: [12] 

The following are data on the times to recurrence of breast 

cancer following treatment in German women. These data 

have been grouped according to whether the patients have 

undergone the menopause (g = 2) or not (g = 1); t(i) is the 

time in years until the ith unique recurrence time, ng,(i − ) is the 

total number at risk in group g an instant before time t(i), 

dg,(i) is the number in group g that had a recurrence at time 

t(i), while n(i − ) = n1,(i − ) + n2,(i − ) and d(i) = d1,(i) + d2,(i). The 

columns marked êg,(i) and ˆvg,(i) are the expected value and 

variance of dg,(i) under the hypothesis that menopause is 

independent of recurrence, respectively. Note that for your 

convenience, the data have been grouped into years. [To find 

out more about these data, see Hosmer et al., 2008.] 



Q7: [12] 

i t (i) n 1,(i − ) n 2,(i − ) n (i − ) d 1,(i) d 2,(i) d (i) ê 1,(i) ê 2,(i) ˆv 1,(i) 

1 1 290 396 686 29 27 56 23.7 32.3 12.6 

2 2 245 357 602 44 65 109 44.4 64.6 21.6 

3 3 183 275 458 20 39 59 23.6 35.4 12.4 

4 4 140 191 331 16 23 39 16.5 22.5 8.4 

5 5 98 130 228 4 18 22 9.5 12.5 4.9 

6 6 56 65 121 6 5 11 5.1 5.9 2.5 

7 7 14 22 36 0 3 3 1.2 1.8 0.7 

Test the hypothesis that the menopause is independent of 

breast cancer recurrence. 



Q8(i): [15] 

Given that the Cox proportional hazards model for a single 

continuous covariate xi is 

h(t, xi) = h0(t, α) exp{βxi} 

for individual i, what is the corresponding survival function 

incorporating this covariate? 



Q8(ii): [15] 

What is the hazard ratio comparing an individual with 

covariate x1 to one with covariate x2 at time t? 



CIs for β, hazard ratios 

So far. . . 

Saw an equation for asymptotic distribution of ˆβ: 

ˆβ ∼ N(β, E{I(β)} −1 ) 

Can replace by observed information: 

ˆβ ∼ N(β, I(ˆβ) −1 ) 

R gives us standard errors for the individual βs 




So far. . . 


ˆβ ∼ N(β, E{I(β)} −1 ) 


ˆβ ∼ N(β, I(ˆβ) −1 ) 





So far. . . 


ˆβ ∼ N(β, E{I(β)} −1 ) 


ˆβ ∼ N(β, I(ˆβ) −1 ) 








We want more! 

CIs for β 

CIs for hazard ratio between two categories 

CIS for hazard ratio between two individuals 



CIs for β 

The asymptotic distribution of β can be estimated using the 

observed information 

ˆβ ∼ N(β, I( ˆ β) −1 ) 

If β is vector of length p, I( ˆ β) is p × p matrix 

i, j element is 

− d2 lp(ˆβ) 

dβi dβj 



where 

⎛ 

⎝ 

Example 

⎛ 

⎝ 

ˆβ1 

ˆβ2 

ˆβ3 

⎞ 

σ11 σ12 σ13 

σ21 σ22 σ23 

σ31 σ32 σ33 

⎛⎛ 

⎠ ∼ N ⎝⎝ 

⎞ 

⎠ = 

⎛ 

⎜ 

⎝ 

β1 

β2 

β3 

⎞ 

⎛ 

⎠ , ⎝ 

σ11 σ12 σ13 

σ21 σ22 σ23 

σ31 σ32 σ33 

⎞⎞ 

⎠⎠ 


dβ1 dβ1 − d2 lp( ˆ β) 


dβ1 dβ3 


dβ2 dβ1 − d2 lp(ˆβ) 

dβ2 dβ2 − d2 lp(ˆβ) 

dβ2 dβ3 

− d2 lp( ˆ β) 



dβ3 dβ3 

ST3242 : Cox proportional hazards model 28/44 

⎞ 

⎟ 

⎠ 

−1


Example 

If ⎛ 

⎝ 

ˆβ1 

ˆβ2 

ˆβ3 

⎞ 

⎛⎛ 

⎠ ∼ N ⎝⎝ 

β1 

β2 

β3 

⎞ 

the marginal distribution of ˆβ1 is 

R 

gives √ σii as standard output! 

⎛ 

⎠ , ⎝ 

ˆβ1 ∼ N(β1, σ11) 

σ11 σ12 σ13 

σ21 σ22 σ23 

σ31 σ32 σ33 

⎞⎞ 

⎠⎠ 



Example 

If ⎛ 

⎝ 

ˆβ1 

ˆβ2 

ˆβ3 

⎞ 

⎛⎛ 

⎠ ∼ N ⎝⎝ 

β1 

β2 

β3 

⎞ 

the marginal distribution of ˆβ1 is 

R 

gives √ σii as standard output! 

⎛ 

⎠ , ⎝ 

ˆβ1 ∼ N(β1, σ11) 

σ11 σ12 σ13 

σ21 σ22 σ23 

σ31 σ32 σ33 

⎞⎞ 

⎠⎠ 



Confidence interval for β 

 

p β1 − 1.96 √ σ11 < ˆ β1 < β1 + 1.96 √ 

σ11 = 0.95 



Confidence interval for β 

 

p ˆβ1 − 1.96 √ σ11 < β1 < ˆ β1 + 1.96 √ 

σ11 = 0.95 



Example 

A 95% confidence interval for βbmi is 

(−0.0985 − 1.96 × 0.0148, −0.0985 + 1.96 × 0.0148) 

= (−0.13, −0.07) 



Confidence interval for g(β) 

If (a, b) is a confidence interval for β, 

(e a , e b ) is a confidence interval for e β 

(e ax , e bx ) is a confidence interval for e βx 

What about a CI for e β1−β2 ? 
























Example 

h(t, xla, xad, xsm, xsq) = h0(t) exp(βlaxla+βadxad+βsmxsm+βsqxsq) 

where xc = 1 if the individual has cell type c and xc = 0 if 

not. Output relative to one category (adeno) as baseline 



Example 

h(t, xla, xad, xsm, xsq) = h0(t) exp(βlaxla+βadxad+βsmxsm+βsqxsq) 

where xc = 1 if the individual has cell type c and xc = 0 if 

not. Output relative to one category (adeno) as baseline 



Multivariate normal distribution 

If x ∼ N(µ, Σ) then any subset of the rows of x also has a 

Normal distribution with corresponding rows of µ and rows 

and columns of Σ. 

For example 

⎛ ⎞ ⎛⎛ 

ˆβ1 

⎝ ˆβ2 

⎠ ∼ N ⎝⎝ 

ˆβ3 

 

ˆβ1 

⇒ ∼ N 

ˆβ3 

β1 

β2 

β3 

β1 

β3 

⎞ 

⎛ 

⎠ , ⎝ 

 

, 

σ11 σ12 σ13 

σ21 σ22 σ23 

σ31 σ32 σ33 

 

σ11 σ13 

σ31 σ33 

 

⎞⎞ 

⎠⎠ 



Multivariate normal distribution 

If x ∼ N(µ, Σ) then any subset of the rows of x also has a 

Normal distribution with corresponding rows of µ and rows 

and columns of Σ. 

For example 

⎛ ⎞ ⎛⎛ 

ˆβ1 

⎝ ˆβ2 

⎠ ∼ N ⎝⎝ 

ˆβ3 

 

ˆβ1 

⇒ ∼ N 

ˆβ3 

β1 

β2 

β3 

β1 

β3 

⎞ 

⎛ 

⎠ , ⎝ 

 

, 

σ11 σ12 σ13 

σ21 σ22 σ23 

σ31 σ32 σ33 

 

σ11 σ13 

σ31 σ33 

 

⎞⎞ 

⎠⎠ 



Properties of Cov 

Q: what is V(X + Y )? 

A: V(X + Y ) = V(X ) + V(Y ) if X & Y independent 

A: V(X + Y ) = V(X ) + V(Y ) + 2C(X , Y ) 

Q: what is V(X − Y )? 






A: V(X + Y ) = V(X ) + V(Y ) + 2C(X , Y ) 







A: V(X + Y ) = V(X ) + V(Y ) + 2C(X , Y ) 







A: V(X + Y ) = V(X ) + V(Y ) + 2C(X , Y ) 





Properties of the covariance 

C(X , Y ) = C(Y , X ) 

C(X , X ) = V(X ) 

C(aX + b, cY + d) = acC(X , Y ) 


A: 

V(X − Y ) = V(X ) + V(−Y ) + 2C(X , −Y ) 

= V(X ) + V(Y ) − 2C(X , Y ) 





C(X , Y ) = C(Y , X ) 

C(X , X ) = V(X ) 

C(aX + b, cY + d) = acC(X , Y ) 


A: 

V(X − Y ) = V(X ) + V(−Y ) + 2C(X , −Y ) 

= V(X ) + V(Y ) − 2C(X , Y ) 





C(X , Y ) = C(Y , X ) 

C(X , X ) = V(X ) 

C(aX + b, cY + d) = acC(X , Y ) 


A: 

V(X − Y ) = V(X ) + V(−Y ) + 2C(X , −Y ) 

= V(X ) + V(Y ) − 2C(X , Y ) 





C(X , Y ) = C(Y , X ) 

C(X , X ) = V(X ) 

C(aX + b, cY + d) = acC(X , Y ) 


A: 

V(X − Y ) = V(X ) + V(−Y ) + 2C(X , −Y ) 

= V(X ) + V(Y ) − 2C(X , Y ) 





C(X , Y ) = C(Y , X ) 

C(X , X ) = V(X ) 

C(aX + b, cY + d) = acC(X , Y ) 


A: 

V(X − Y ) = V(X ) + V(−Y ) + 2C(X , −Y ) 

= V(X ) + V(Y ) − 2C(X , Y ) 





C(X , Y ) = C(Y , X ) 

C(X , X ) = V(X ) 

C(aX + b, cY + d) = acC(X , Y ) 

Q: what is V( ˆ βi − ˆ βj)? 

A: 

V( ˆ βi − ˆ βj) = V( ˆ βi) + V(− ˆ βj) + 2C( ˆ βi, − ˆ βj) 

= V( ˆ βi) + V( ˆ βj) − 2C( ˆ βi, ˆ βj) 



CIs for βi − βj 

A 95% CI for βi − βj is 

ˆβi − ˆ βj ± 1.96 σii + σjj − 2σij 

R does it for us! 

The coxph functions silently outputs var, the 

variance–covariance matrix 

phm.cell=coxph(Surv(t,delta)∼factor(Cell)) 

phm.cellvar 

phm.cellcoefficients 



CIs for βi − βj 

A 95% CI for βi − βj is 

ˆβi − ˆ βj ± 1.96 σii + σjj − 2σij 

R does it for us! 

The coxph functions silently outputs var, the 

variance–covariance matrix 

phm.cell=coxph(Surv(t,delta)∼factor(Cell)) 

phm.cellvar 

phm.cellcoefficients 



Example 



Example 

95%CI for βsq − βla is 

−0.77 − 1.00 ± 1.96 √ 0.0642 + 0.0638 − 2 × 0.0256 

i.e. (−0.31, 0.77) 

A 95% CI for the hazard ratio comparing squamous to large 

cells is 

(e −0.31 , e 0.77 ) = (0.73, 2.17) 




We want more! 

CIs for β 

CIs for hazard ratio between two categories 

CIS for hazard ratio between two individuals 



CIs for hazard ratios 

Suppose we want a hazard ratio comparing individuals with x a 

and x b 

squamous cell, performance of 60, aged 70, with no prior 

therapy 

large cell, performance of 20, aged 50, with no prior 

therapy 

Same approach 

obtain ˆ β and covariance matrix via coxph 

evaluate ˆ β T 

xa and ˆ β T 

xb evaluate V( ˆ β T 

x a − ˆ β T 

x b ) 

95% CI is ˆβ T 

x a − ˆβ T 

x b ± 

 

V(ˆβ T 


x b ) 





and x b 


therapy 


therapy 

Same approach 






x b ) 



x b ± 

 

V(ˆβ T 


x b ) 





and x b 


therapy 


therapy 

Same approach 






x b ) 



x b ± 

 

V(ˆβ T 


x b ) 





and x b 


therapy 


therapy 

Same approach 






x b ) 



x b ± 

 

V(ˆβ T 


x b ) 





and x b 


therapy 


therapy 

Same approach 






x b ) 



x b ± 

 

V(ˆβ T 


x b ) 





and x b 


therapy 


therapy 

Same approach 






x b ) 



x b ± 

 

V(ˆβ T 


x b ) 



Example 

x a = (x a 1 , x a 2 ) and x b = (x b 1 , x b 2 ) Variance of ˆ βx a − ˆ βx b is 

V( ˆ βx a − ˆ βx b ) = V{ ˆ β1(x a 1 − x b 1 ) + ˆ β2(x a 2 − x b 2 )} 

= (x a 1 − x b 1 ) 2 V( ˆ β1) + (x a 2 − x b 2 ) 2 V( ˆ β2) 

+2C{ ˆ β1(x a 1 − x b 1 ), ˆ β2(x a 2 − x b 2 )} 

= (x a 1 − x b 1 ) 2 V( ˆ β1) + (x a 2 − x b 2 ) 2 V( ˆ β2) 

+2(x a 1 − x b 1 )(x a 2 − x b 2 )C{ ˆβ1, ˆβ2}

ST3242

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