ch3l4 survival analysis

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

Cox proportional hazards modelST3242: 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 comprisesFOUR printed pages.

2 Answer ALL questions for a total of 100 marks.

3 This is a closed-book test; only non-programmablecalculators are allowed.



Q1: [3]

If the survival function is S(t), what is the hazard functionh(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 distributedaccording to a log-logistic distribution with parameters κ andλ, 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 isright-censored at ti , for i = 1, . . . ,m, what is thelog-likelihood function?



Q4: [6]

Consider a study into the length of time people can survivewithout purchasing or receiving plastic goods. The studystarts at time Tstart and ends at time Tend. Three nusstudents are among the participants.

Student A is recruited to the study at time Tstart. Shebuys no plastic until time tA < Tend, when she buys abottle 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. Hereports to the organisers at time tB < Tend that he hasbought no plastics, but they never hear from him again.




Q4: [6]


Student C is recruited to the study at timetC ∈ (Tstart,Tend). She buys no plastic for the remainderof the study.




Q5: [12]

Use the delta method to approximate the mean and varianceof√

X , where X is a random variable with mean µ andvariance σ2. When do you expect these will provide a goodapproximation to the true mean and variance of

√X?



Q6(i): [36]

Calculate and sketch the Kaplan–Meier estimate of S(t) forthe following data, where δi = 1 if individual i died at time tiand δi = 0 if i was censored at that time.

i ti δi1 0.6 12 0.9 13 1.1 04 1.5 15 2.0 06 2.0 0



Q6(ii): [36]

Recall that

V{S(t)} = S(t)2∑t(i)≤t

d(i)

n(i−)(n(i−) − d(i))

where t(i) is the ith ordered event time, d(i) is the number ofdeaths at that time and n(i−) is the number at risk an instantbefore that time.Construct the “naıve” 95% confidence interval described inthe lecture notes for S(1), the probability of surviving at leastto one time unit. What is the probability the true value ofS(1) lies within your confidence interval?



Q7: [12]

The following are data on the times to recurrence of breastcancer following treatment in German women. These datahave been grouped according to whether the patients haveundergone the menopause (g = 2) or not (g = 1); t(i) is thetime in years until the ith unique recurrence time, ng ,(i−) is thetotal 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 timet(i), while n(i−) = n1,(i−) + n2,(i−) and d(i) = d1,(i) + d2,(i). Thecolumns marked eg ,(i) and vg ,(i) are the expected value andvariance of dg ,(i) under the hypothesis that menopause isindependent of recurrence, respectively. Note that for yourconvenience, the data have been grouped into years. [To findout more about these data, see Hosmer et al., 2008.]



Q7: [12]

i t(i) n1,(i−) n2,(i−) n(i−) d1,(i) d2,(i) d(i) e1,(i) e2,(i) v1,(i) v2,(i)

1 1 290 396 686 29 27 56 23.7 32.3 12.6 12.62 2 245 357 602 44 65 109 44.4 64.6 21.6 21.63 3 183 275 458 20 39 59 23.6 35.4 12.4 12.44 4 140 191 331 16 23 39 16.5 22.5 8.4 8.45 5 98 130 228 4 18 22 9.5 12.5 4.9 4.96 6 56 65 121 6 5 11 5.1 5.9 2.5 2.57 7 14 22 36 0 3 3 1.2 1.8 0.7 0.7

Test the hypothesis that the menopause is independent ofbreast cancer recurrence.



Q8(i): [15]

Given that the Cox proportional hazards model for a singlecontinuous covariate xi is

h(t, xi) = h0(t,α) exp{βxi}

for individual i , what is the corresponding survival functionincorporating this covariate?



Q8(ii): [15]

What is the hazard ratio comparing an individual withcovariate x1 to one with covariate x2 at time t?



CIs for β, hazard ratios

So far. . .

Saw an equation for asymptotic distribution of β:

β ∼ N(β,E{I(β)}−1)

Can replace by observed information:

β ∼ N(β, I(β)−1)

R gives us standard errors for the individual βs




So far. . .


β ∼ N(β,E{I(β)}−1)


β ∼ N(β, I(β)−1)





So far. . .


β ∼ N(β,E{I(β)}−1)


β ∼ N(β, I(β)−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 theobserved information

β ∼ N(β, I(β)−1)

If β is vector of length p, I(β) is p × p matrixi , j element is

− d2lp(β)

dβi dβj



Example

β1

β2

β3

∼ N

β1

β2

β3

,

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

where

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

=

− d2lp(β)

dβ1 dβ1− d2lp(β)

dβ1 dβ2− d2lp(β)

dβ1 dβ3

− d2lp(β)dβ2 dβ1






−1



Example

If β1

β2

β3

∼ N

β1

β2

β3

,

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

the marginal distribution of β1 is

β1 ∼ N(β1, σ11)

R

gives√σii as standard output!



Example

If β1

β2

β3

∼ N

β1

β2

β3

,

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

the marginal distribution of β1 is

β1 ∼ N(β1, σ11)

R

gives√σii as standard output!



Confidence interval for β

p(β1 − 1.96

√σ11 < β1 < β1 + 1.96

√σ11

)= 0.95



Confidence interval for β

p(β1 − 1.96

√σ11 < β1 < 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,

(ea, eb) is a confidence interval for eβ

(eax , ebx) is a confidence interval for eβx

What about a CI for eβ1−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 ifnot. 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 ifnot. 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 aNormal distribution with corresponding rows of µ and rowsand columns of Σ.

For example

β1

β2

β3

∼ N

β1

β2

β3

,

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

⇒(β1

β3

)∼ N

((β1

β3

),

(σ11 σ13

σ31 σ33

))



Multivariate normal distribution

If x ∼ N(µ,Σ) then any subset of the rows of x also has aNormal distribution with corresponding rows of µ and rowsand columns of Σ.

For example

β1

β2

β3

∼ N

β1

β2

β3

,

σ11 σ12 σ13

σ21 σ22 σ23

σ31 σ32 σ33

⇒(β1

β3

)∼ N

((β1

β3

),

(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 )?



Properties of Cov



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




Properties of Cov



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




Properties of Cov



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




Properties of Cov

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 )



Properties of Cov


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

C(X ,X ) = V(X )



A:

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

= V(X ) + V(Y )− 2C(X ,Y )



Properties of Cov


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

C(X ,X ) = V(X )



A:

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

= V(X ) + V(Y )− 2C(X ,Y )



Properties of Cov


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

C(X ,X ) = V(X )



A:

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

= V(X ) + V(Y )− 2C(X ,Y )



Properties of Cov


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

C(X ,X ) = V(X )



A:

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

= V(X ) + V(Y )− 2C(X ,Y )



Properties of Cov


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

C(X ,X ) = V(X )


Q: what is V(βi − βj)?

A:

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

= V(βi) + V(βj)− 2C(βi , βj)



CIs for βi − βj

A 95% CI for βi − βj is

βi − βj ± 1.96√σii + σjj − 2σij

R does it for us!

The coxph functions silently outputs var, thevariance–covariance matrixphm.cell=coxph(Surv(t,delta)∼factor(Cell))phm.cell$var

phm.cell$coefficients



CIs for βi − βj

A 95% CI for βi − βj is

βi − βj ± 1.96√σii + σjj − 2σij

R does it for us!

The coxph functions silently outputs var, thevariance–covariance matrixphm.cell=coxph(Surv(t,delta)∼factor(Cell))phm.cell$var

phm.cell$coefficients



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 largecells is

(e−0.31, e0.77) = (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 ratiosSuppose we want a hazard ratio comparing individuals with xa

and xb

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

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

Same approach

obtain β and covariance matrix via coxph

evaluate βTxa and β

Txb

evaluate V(βTxa − β

Txb)

95% CI is βTxa − β

Txb ±

√V(β

Txa − β

Txb)




and xb



Same approach



Txb


Txb)


Txb ±

√V(β

Txa − β

Txb)




and xb



Same approach



Txb


Txb)


Txb ±

√V(β

Txa − β

Txb)




and xb



Same approach



Txb


Txb)


Txb ±

√V(β

Txa − β

Txb)




and xb



Same approach



Txb


Txb)


Txb ±

√V(β

Txa − β

Txb)




and xb



Same approach



Txb


Txb)


Txb ±

√V(β

Txa − β

Txb)



Example

xa = (xa1 , x

a2 ) and xb = (xb

1 , xb2 ) Variance of βxa − βxb is

V(βxa − βxb) = V{β1(xa1 − xb

1 ) + β2(xa2 − xb

2 )}= (xa

1 − xb1 )2V(β1) + (xa

2 − xb2 )2V(β2)

+2C{β1(xa1 − xb

1 ), β2(xa2 − xb

2 )}= (xa

1 − xb1 )2V(β1) + (xa

2 − xb2 )2V(β2)

+2(xa1 − xb

1 )(xa2 − xb

2 )C{β1, β2}


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