Cox PH regression

A basic Cox model with 2 predictors and their interaction takes the form:

h(t|x, z) = h₀ ⋅ exp (β₁x + β₂z + β₃x ⋅ z)

After estimation of the model, we want to predict the HR for X over levels of Z. This is given by

$HR_{10}=\frac{h(t|x=1,z)}{h(t|x=0,z)}=\frac{h_0\cdot\exp(\beta_1x+\beta_2z+\beta_3x\cdot z)}{h_0\cdot\exp(\beta_1x+\beta_2z+\beta_3x\cdot z)}=\frac{\exp(\beta_1+\beta_2z+\beta_3z)}{\exp(\beta_2z)}$ ,

which will be plotted against Z

To estimate 95% confidence intervals SE(HR₁₀) is required. This is obtained by focusing on log (HR₁₀) and calculating lower and upper bounds for this quantity, which are then exponentiated.

$SE(\log(HR_{10}))=SE(\log(\frac{\exp(\beta_1+\beta_2z+\beta_3z)}{\exp(\beta_2z)}))=SE(\log(\exp(\beta_1+\beta_2z+\beta_3z)-\log(\exp(\beta_2z)))=SE(\beta_1+\beta_3z)$

This is calculated by using the delta method. Upper and lower bounds are then derived with exp (log (HR₁₀) ± 1.96 ⋅ SE(log(HR₁₀)))

Logistic regression

A logistic regression model with 2 predictors and their interaction takes the form:

logit(p|x, z) = β₀ + β₁x + β₂z + β₃x ⋅ z

After estimation of the model, we want to predict the OR for X over levels of Z. This is given by

$OR_{10}=\frac{odds(p|x=1,z)}{odds(p|x=0,z)}=\frac{\exp(\beta_0+\beta_1x+\beta_2z+\beta_3x\cdot z)}{\exp(\beta_0+\beta_1x+\beta_2z+\beta_3x\cdot z)}=\frac{\exp(\beta_0+\beta_1+\beta_2z+\beta_3z)}{\exp(\beta_0+\beta_2z)}$ ,

which will be plotted against Z

To estimate 95% confidence intervals SE(OR₁₀) is required. This is obtained by focusing on log (OR₁₀) and calculating lower and upper bounds for this quantity, which are then exponentiated.

$SE(\log(HR_{10}))=SE(\log(\frac{\exp(\beta_0+\beta_1+\beta_2z+\beta_3z)}{\exp(\beta_0+\beta_2z)}))=SE(\log(\exp(\beta_0+\beta_1+\beta_2z+\beta_3z)-\log(\exp(\beta_0+\beta_2z)))=SE(\beta_1+\beta_3z)$

This is calculated by using the delta method. Upper and lower bounds are then derived with exp (log (OR₁₀) ± 1.96 ⋅ SE(log(OR₁₀)))

Linear regression

A linear regression model with 2 predictors and their interaction takes the form:

E[Y|x, z] = β₀ + β₁x + β₂z + β₃x ⋅ z

After estimation of the model, we want to predict the effect for X over levels of Z. This is given by

E[Y|x = 1, z] − E[Y = 0|x, z] = (β₀ + β₁ + β₂z + β₃ ⋅ z) − (β₀ + β₂z) = β₁ + β₃ ⋅ z ,

which will be plotted against Z

To estimate 95% confidence intervals we need the basic linear combination of standard errors calculated by $SE(\beta_1+\beta_3z)=\sqrt{SE(\beta_1)^2+SE(\beta_3z)^2+2cov(\beta_1,\beta_3)z}$,

or by using the Delta method

Cox PH regression

The interaction model, in this setting, takes the form:

h(t|x, z, c) = h₀ ⋅ exp (β₁x + sp(z) + sp₂(z ⋅ x))

If 3 knots are specified:

$sp(z)=\alpha_1z+\alpha_2\cdot[\frac{(z-t_1)^3_+-(z-t_2)^3_+\cdot\frac{ t_3-t_1}{ t_3-t_2} +(z-t_3)^3_+\cdot\frac{t_2-t_1}{t_3-t_2}}{(t_3-t_1)^2}]$

and

$sp_2(z\cdot x)=\gamma_1x+\gamma_2x\cdot[\frac{(z-t_1)^3_+-(z-t_2)^3_+\cdot\frac{ t_3-t_1}{ t_3-t_2} +(z-t_3)^3_+\cdot\frac{t_2-t_1}{t_3-t_2}}{(t_3-t_1)^2}]$

After estimation of the model, we want to predict the HR for X over levels of Z. This is given by

$HR_{10}=\frac{h(t|X=x_1,Z)}{h(t|X=x_2,Z)}=\frac{h_0\cdot\exp(\beta_1x_1+sp(z)+sp_2(z)\cdot x_1)}{h_0\cdot\exp(\beta_1x_2+sp(z)+sp_2(z)\cdot x_2)}=\frac{h(t|X=x_1,Z)}{h(t|X=x_2,Z)}=\frac{\exp(\beta_1+sp(z)+sp_2(z))}{\exp(sp(z))}$ ,

which will be plotted against Z

Similarly to the previous situation, the SE can be derived by using the delta method to calculate SE(β₁ + sp₂(z))

Logistic regression

The interaction model, in this setting, takes the form:

logit(p|x, z) = β₀ + β₁x + sp(z) + sp₂(z ⋅ x)

After estimation of the model, we want to predict the HR for X over levels of Z. This is given by

$OR_{10}=\frac{\exp(\beta_0+\beta_1+sp(z)+sp_2(z))}{\exp(\beta_0+sp(z))}=\exp(\beta_1+sp_2(z))$ ,

which will be plotted against Z

Similarly to the previous situation, the SE can be derived by using the delta method to calculate SE(β₁ + sp₂(z))

Linear regression

The interaction model, in this setting, takes the form:

E[Y|x, z] = β₀ + β₁x + sp(z) + sp₂(z ⋅ x)

After estimation of the model, we want to predict

E[Y|x = 1, z] − E[Y = 0|x, z] = (β₀ + β₁ + sp(z) + sp₂(z)) − (β₀ + sp(z)) = β₁ + sp₂(z) ,

which will be plotted against Z

The SE can be derived by using the delta method to calculate SE(β₁ + sp₂(z))

More than 3 knots

Based on Harrell’s book (chapter 2-23), we derive equations for more than 3 knots

The general splines function for k knots is:

f(X) = β₀ + β₁ ⋅ X₁ + β₂ ⋅ X₂ + … + β_k − 1 ⋅ X_k − 1

where X₁ = X and for j = 1, …, k − 2,

$X_{j+1}=\frac{(z-t_j)^3_+-(z-t_{k-1})^3_+\cdot\frac{ t_k-t_j}{t_k-t_{k-1}} +(z-t_k)^3_+\cdot\frac{t_{k-1}-t_j}{t_k-t_{k-1}}}{(t_k-t_1)^2}$

The Cox model with k = 4 will therefore be:

h(t|x, z, c) = h₀ ⋅ exp (β₁x + sp(z) + sp₂(z ⋅ x))

where

$sp(z)=\alpha_1z+\alpha_2\cdot[\frac{(z-t_1)^3_+-(z-t_3)^3_+\cdot\frac{ t_4-t_1}{ t_4-t_3} +(z-t_3)^3_+\cdot\frac{t_3-t_1}{t_4-t_3}}{(t_4-t_1)^2}]+\alpha_3\cdot[\frac{(z-t_2)^3_+-(z-t_3)^3_+\cdot\frac{ t_4-t_2}{ t_4-t_3} +(z-t_4)^3_+\cdot\frac{t_3-t_2}{t_4-t_3}}{(t_4-t_1)^2}]$

and

$sp_2(z\cdot x)=\gamma_1x+\gamma_2x\cdot[\frac{(z-t_1)^3_+-(z-t_3)^3_+\cdot\frac{ t_4-t_1}{ t_4-t_3} +(z-t_4)^3_+\cdot\frac{t_3-t_2}{t_4-t_3}}{(t_3-t_1)^2}]+\gamma_3x\cdot[\frac{(z-t_2)^3_+-(z-t_3)^3_+\cdot\frac{ t_4-t_2}{ t_4-t_3} +(z-t_4)^3_+\cdot\frac{t_3-t_2}{t_4-t_3}}{(t_4-t_1)^2}]$

The generic Cox model will be:

h(t|x, z, c) = h₀ ⋅ exp (β₁x + sp(z) + sp₂(z ⋅ x))

where

$sp(z)=\alpha_1z+\alpha_2\cdot[\frac{(z-t_j)^3_+-(z-t_{k-1})^3_+\cdot\frac{ t_k-t_j}{t_k-t_{k-1}} +(z-t_k)^3_+\cdot\frac{t_{k-1}-t_j}{t_k-t_{k-1}}}{(t_k-t_1)^2}]+\alpha_3\cdot[\frac{(z-t_j)^3_+-(z-t_{k-1})^3_+\cdot\frac{ t_k-t_j}{t_k-t_{k-1}} +(z-t_k)^3_+\cdot\frac{t_{k-1}-t_j}{t_k-t_{k-1}}}{(t_k-t_1)^2}]$

and

$sp_2(z\cdot x)=\gamma_1x+\gamma_2x\cdot[\frac{(z-t_j)^3_+-(z-t_{k-1})^3_+\cdot\frac{ t_k-t_j}{t_k-t_{k-1}} +(z-t_k)^3_+\cdot\frac{t_{k-1}-t_j}{t_k-t_{k-1}}}{(t_k-t_1)^2}]+\gamma_3x\cdot[\frac{(z-t_j)^3_+-(z-t_{k-1})^3_+\cdot\frac{ t_k-t_j}{t_k-t_{k-1}} +(z-t_k)^3_+\cdot\frac{t_{k-1}-t_j}{t_k-t_{k-1}}}{(t_k-t_1)^2}]$