1. Proportional Hazard Models
Bo Gallant

2. Presentation Topics Introduction Model Structure Partial Likelihoods
Background information
Main assumptions
Model Structure
PH function
Survival function
Partial Likelihoods
Without and with tied observations explanation
Without tied observations equations
With tied observations equations
Newton-Raphson equation

3. Introduction ℎ 𝑡 = 𝛽 0 + 𝛽 1 𝑋 = exp⁡(𝛽 0 + 𝛽 1 𝑋)
Model Structure
Partial Likelihood
Introduction
Background Information
Non-parametric: don’t need to know distribution
Constant hazard function over time: similar to Exponential
Hazard function of different individuals are proportional and independent of time
Main Assumptions
Hazard function of different individuals are proportional and independent of time
ℎ(𝑡| 𝒙 1 ) ℎ(𝑡| 𝒙 2 ) =𝐶
ℎ(𝑡| 𝒙 1 ) proportional to ℎ 𝑡 𝒙 2
PH Model Background Equation
Fix Negative
ℎ 𝑡 = 𝛽 0 + 𝛽 1 𝑋
= exp⁡(𝛽 0 + 𝛽 1 𝑋)
= exp⁡(𝛽 0 )exp⁡( 𝛽 1 𝑋)
= ℎ 0 (𝑡)exp⁡( 𝛽 1 𝑋)
Proportional Hazard Function
ℎ(𝑡| 𝒙 1 ) ℎ(𝑡| 𝒙 2 ) = ℎ 0 𝑡 exp⁡ 𝒃 ′ 𝒙 1 ℎ 0 𝑡 exp⁡ 𝒃 ′ 𝒙 2 = ℎ 0 𝑡 𝑔( 𝒙 1 ) ℎ 0 𝑡 𝑔( 𝒙 2 ) = 𝑔( 𝒙 1 ) 𝑔( 𝒙 2 )
Constant and independent of time

4. Model Structure S(t) = 𝑆 0 (𝑡) exp⁡( 𝒃 ′ 𝒙)
Introduction
Model Structure
Partial Likelihood
Model Structure
Proportional Hazard Function
g 𝒙 = exp 𝒃 ′ 𝒙 = exp 𝑗=1 𝑝 𝑏 𝑗 𝑥 𝑗
ℎ(𝑡| 𝒙 1 ) ℎ(𝑡| 𝒙 2 ) = 𝑔( 𝒙 1 ) 𝑔( 𝒙 2 ) = exp⁡ 𝒃 ′ 𝒙 1 exp⁡ 𝒃 ′ 𝒙 2
Survival Function
General Survival Equation:
S(t) = exp − 0 𝑡 ℎ 𝜏 𝑑𝜏
S(t) = exp − 0 𝑡 ℎ 0 𝜏 exp⁡( 𝒃 ′ 𝒙)𝑑𝜏
PH Survival Equation:
= exp − 0 𝑡 ℎ 0 𝜏 𝑑𝜏∗exp⁡( 𝒃 ′ 𝒙)
S(t) = 𝑆 0 (𝑡) exp⁡( 𝒃 ′ 𝒙)

5. Partial Likelihoods Types of PL Example Without Tied Observations
Introduction
Model Structure
Partial Likelihood
Partial Likelihoods
Types of PL
2 Types of Partial Likelihoods Equations:
Equation for data sets Without Tied Observations
Equations for data sets With Tied Observations
Reasons?
PL equation depends on order of failures
For Tied observations, order is unknown
Example
Without Tied Observations
With Tied Observations
5 Individuals Without Tied Failures:
5 Individuals with 3 Tied Failures:
Individual (i)
Time (t) 1 5 2 10 3 15 4 20 25
Individual (i)
Time (t) 1 5 2 3 4 10 15

6. Partial Likelihoods – Without Tied Observations
Introduction
Model Structure
Partial Likelihood
Partial Likelihoods – Without Tied Observations
Probability Failure of an Individual at time t(i)
R(t(i)): Risk set which is everyone at Risk at time t(i)
Partial Likelihood
𝐿 𝒃 = 𝒊=𝟏 𝒌 𝑒𝑥𝑝( 𝒋=𝟏 𝒑 𝒃 𝒋 𝒙 𝒋(𝒊) ) 𝑙∈𝑹( 𝒕 𝒊 ) 𝑒𝑥𝑝( 𝒋=𝟏 𝒑 𝒃 𝒋 𝒙 𝒋𝒍 ) = 𝒊=𝟏 𝒌 𝑒𝑥𝑝( 𝒃′𝒙 (𝒊) ) 𝑙∈𝑹( 𝒕 𝒊 ) 𝑒𝑥𝑝(𝒃′ 𝒙 𝒍 )

7. Partial Likelihoods – Without Tied Observations
Introduction
Model Structure
Partial Likelihood
Partial Likelihoods – Without Tied Observations
5 Individuals Without Tied Failures:
Hazard functions:
Individual (i)
Time (t) 1 5 2 10 3 15 4 20 25
ℎ(𝑡|𝑥 1 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟏 )
ℎ(𝑡|𝑥 2 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟐 )
ℎ(𝑡|𝑥 3 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟑 )
ℎ(𝑡|𝑥 4 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟒 )
ℎ(𝑡|𝑥 5 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟓 )
Partial Likelihood
𝐿 𝒃 = 𝒊=𝟏 𝒌 𝑒𝑥𝑝( 𝒋=𝟏 𝒑 𝒃 𝒋 𝒙 𝒋(𝒊) ) 𝑙∈𝑹( 𝒕 𝒊 ) 𝑒𝑥𝑝( 𝒋=𝟏 𝒑 𝒃 𝒋 𝒙 𝒋𝒍 ) = 𝒊=𝟏 𝒌 𝑒𝑥𝑝( 𝒃′𝒙 (𝒊) ) 𝑙∈𝑹( 𝒕 𝒊 ) 𝑒𝑥𝑝(𝒃′ 𝒙 𝒍 )
L 𝒃 = ℎ(5|𝑥 1 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 ) ℎ(10|𝑥 2 ) ℎ(10|𝑥 2 )+ ℎ(10|𝑥 3 )+ ℎ(10|𝑥 4 )+ ℎ(10|𝑥 5 )
* ℎ(15|𝑥 3 ) ℎ(15|𝑥 3 )+ ℎ(15|𝑥 4 )+ ℎ(15|𝑥 5 ) ℎ(20|𝑥 4 ) ℎ(20|𝑥 4 )+ ℎ(20|𝑥 5 ) ℎ(25|𝑥 5 ) ℎ(25|𝑥 5 )

8. Partial Likelihood - Tied Observations
Introduction
Model Structure
Partial Likelihood
Partial Likelihood - Tied Observations
5 Individuals with 3 Tied Failures:
Hazard functions:
Individual (i)
Time (t) 1 5 2 3 4 10 15
ℎ(𝑡|𝑥 1 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟏 )
ℎ(𝑡|𝑥 2 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟐 )
ℎ(𝑡|𝑥 3 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟑 )
ℎ(𝑡|𝑥 4 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟒 )
ℎ(𝑡|𝑥 5 )= ℎ 0 (𝑡)𝑒𝑥𝑝( 𝒃 ′ 𝒙 𝟓 )
1) Breslow Estimation (1974)
𝐿 𝑏 𝒃 = 𝒊=𝟏 𝒌 𝑒𝑥𝑝( 𝒙′ 𝒖∗(𝒊) 𝒃) 𝑙∈𝑹( 𝒕 𝒊 ) 𝑒𝑥𝑝( 𝒙 ′ 𝒍 𝒃)
R(t(i)): Risk set: everyone at Risk at time t(i)
m(i) : # individual fail at t(i)
𝒖 ∗ 𝒊 : set of m(i) individuals failed at time t(i)
𝐿 𝐵 𝒃 = ℎ(5|𝑥 1 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 ) ℎ(5|𝑥 2 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 )
* ℎ(5|𝑥 3 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 ) ℎ(10|𝑥 4 ) ℎ(10|𝑥 4 )+ ℎ(20|𝑥 5 ) ℎ(15|𝑥 5 ) ℎ(15|𝑥 5 )

9. Partial Likelihood - Tied Observations
Introduction
Model Structure
Partial Likelihood
Partial Likelihood - Tied Observations
2) Efron Estimation (1977)
𝐿 𝐸 𝒃 = 𝒊=𝟏 𝒌 𝑒𝑥𝑝( 𝒙′ 𝒖∗(𝒊) 𝒃) 𝒋=𝟏 𝒎 (𝒊) 𝑙∈𝑹( 𝒕 𝒊 ) 𝑒𝑥𝑝( 𝒙 ′ 𝒍 𝒃) − 𝒋−𝟏 𝒎 (𝒊) 𝑙∈ 𝒖 (𝒊) ∗ 𝑒𝑥𝑝( 𝒙 ′ 𝒍 𝒃)
R(t(i)): Risk set: everyone at Risk at time t(i)
m(i) : # individual fail at t(i)
𝒖 ∗ 𝒊 : set of m(i) individuals failed at time t(i)
𝐿 𝐸 𝒃 = ℎ(5|𝑥 1 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 )− 𝟏−𝟏 𝟑 [ ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )]
* ℎ(5|𝑥 2 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 )− 𝟐−𝟏 𝟑 [ ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )]
* ℎ(5|𝑥 3 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 )− 𝟑−𝟏 𝟑 [ ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )] ℎ(10|𝑥 4 ) ℎ(10|𝑥 4 )+ ℎ(10|𝑥 5 ) ℎ(15|𝑥 5 ) ℎ(15|𝑥 5 )
𝐿 𝐸 𝒃 = ℎ(5|𝑥 1 ) ℎ(5|𝑥 1 )+ ℎ(5|𝑥 2 )+ ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 )
* ℎ(5|𝑥 2 ) 𝟐 𝟑 ℎ(5|𝑥 1 )+ 𝟐 𝟑 ℎ(5|𝑥 2 )+ 𝟐 𝟑 ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 )
* ℎ(5|𝑥 3 ) 𝟏 𝟑 ℎ(5|𝑥 1 )+ 𝟏 𝟑 ℎ(5|𝑥 2 )+ 𝟏 𝟑 ℎ(5|𝑥 3 )+ ℎ(5|𝑥 4 )+ ℎ(5|𝑥 5 ) ℎ(10|𝑥 4 ) ℎ(10|𝑥 4 )+ ℎ(10|𝑥 5 ) ℎ(15|𝑥 5 ) ℎ(15|𝑥 5 )

10. Newton-Raphson Algorithm
Introduction
Model Structure
Partial Likelihood
Newton-Raphson Algorithm
Follow these Steps to find Estimates:
Step 1) Let the initial values of 𝑏 1 ,…, 𝑏 𝑝 be the 𝑏 (0) .
For example, we can let 𝑏 (0) = 0 or some other value.
Step 2) The changes ∆ 𝑗 for b is ∆ 𝑗 = − 𝜕 2 𝑙( 𝒃 𝑗−1 ) 𝜕𝒃𝜕 𝒃 ′ −1 𝜕𝑙( 𝒃 𝑗−1 ) 𝜕𝒃
Step 3) The new value of 𝒃 𝑗 = 𝒃 (𝑗−1) +∆ 𝑗
Step 4) Repeat steps 2) and 3) until ||∆ 𝑚 ||<𝛿 where 𝛿 is a specified small
value such as 10 −4 or 10 −5 . Once the algorithm is stopped, the estimated value
will be 𝐛= 𝒃 (𝑚−1)
Only 1 Time
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