Proportional Hazard Models Bo Gallant

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

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

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)

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

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 )

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 )

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 )

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 )

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 Check Every Time