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Michael Brenndoerfer

Cox Proportional Hazards › Example

Michael BrenndoerferOctober 26, 20265 min read

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Part 6 of 15Survival Analysis

Chapter 50Cox Proportional Hazards › Example

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We work through a minimal dataset of four subjects so every number can be checked by hand. Each subject has a single covariate $x \in \{0, 1\}$ , an observed time, and an event indicator.

Four-subject teaching dataset with a binary covariate, observed time, and event indicator.
Subject	$x$	$T$	$d$
1	0	2	1
2	1	4	1
3	0	6	1
4	1	8	0

There are three events, at times $2$ , $4$ , and $6$ . Subject 4 is censored at time 8, so they are at risk at every event time but never contribute a numerator term. The three risk sets are:

at $t=2$ : $\{1, 2, 3, 4\}$ (everyone is still at risk)
at $t=4$ : $\{2, 3, 4\}$ (subject 1 has already failed)
at $t=6$ : $\{3, 4\}$ (subject 2 has already failed)

The partial likelihood is

L(\beta) = \underbrace{\frac{e^{0}}{e^{0} + e^{\beta} + e^{0} + e^{\beta}}}_{t=2} \cdot \underbrace{\frac{e^{\beta}}{e^{\beta} + e^{0} + e^{\beta}}}_{t=4} \cdot \underbrace{\frac{e^{0}}{e^{0} + e^{\beta}}}_{t=6}.

Writing $q = e^{\beta}$ and simplifying,

L(\beta) = \frac{1}{2 + 2q} \cdot \frac{q}{1 + 2q} \cdot \frac{1}{1 + q}.

Taking logs,

\ell(\beta) = -\log(2 + 2q) + \beta - \log(1 + 2q) - \log(1 + q).

Let us evaluate this at three values of $\beta$ to see the shape of the surface. At $\beta = 0$ ( $q=1$ ):

\ell(0) = -\log 4 + 0 - \log 3 - \log 2 \approx -1.386 - 1.099 - 0.693 = -3.178.

At $\beta = -1$ ( $q \approx 0.368$ ):

\ell(-1) = -\log(2.736) - 1 - \log(1.736) - \log(1.368) \approx -1.006 - 1 - 0.552 - 0.313 = -2.871.

At $\beta = -1.5$ ( $q \approx 0.223$ ):

\ell(-1.5) = -\log(2.446) - 1.5 - \log(1.446) - \log(1.223) \approx -2.965.

The displayed evaluations bracket the maximum: $\ell(-1.5) \approx -2.965$ , $\ell(-1) \approx -2.871$ , and $\ell(0) \approx -3.178$ , so the largest value is at $\beta=-1$ and the optimum lies between $-1.5$ and $0$ . The score can be simplified to $1-2q/(1+q)-2q/(1+2q)$ . At $\beta=-1$ it is positive, so increasing $\beta$ locally increases the log partial likelihood. Multiplying the zero-score equation $1-2q/(1+q)-2q/(1+2q)=0$ by its positive denominators yields $1-q-4q^2=0$ , whose positive root $q=(\sqrt{17}-1)/8$ determines the finite optimum in this example. We compute it and an approximate Wald interval below; with only three events, the interval is a teaching calculation rather than a precise effect estimate.

PythonIn [3]
CollapseExpandhand_q = (np.sqrt(17.0) - 1.0) / 8.0
hand_score = 1 - 2 * hand_q / (1 + hand_q) - 2 * hand_q / (1 + 2 * hand_q)
hand_beta = np.log(hand_q)
hand_information = (
    2 * hand_q / (1 + hand_q) ** 2 + 2 * hand_q / (1 + 2 * hand_q) ** 2
)
hand_se = 1 / np.sqrt(hand_information)
hand_hr_interval = np.exp(hand_beta + np.array([-1.96, 1.96]) * hand_se)

ConsoleOut [4]
CollapseExpandhand optimum beta: -0.940614; hazard ratio: 0.390388
score at the hand optimum: 1.11e-16
approximate model-based 95% Wald hazard-ratio interval: [0.0343374  4.43839458]

The hand calculation also has a visual counterpart. Over a fixed grid of beta values, the log partial likelihood is a smooth concave curve with a single finite maximum at the value derived above.

PythonIn [5]
CollapseExpandbeta_grid = np.linspace(-2.5, 1.0, 400)
hand_q_grid = np.exp(beta_grid)
hand_loglike_grid = (
    -np.log(2.0 + 2.0 * hand_q_grid)
    + beta_grid
    - np.log(1.0 + 2.0 * hand_q_grid)
    - np.log(1.0 + hand_q_grid)
)
hand_loglike_opt = (
    -np.log(2.0 + 2.0 * hand_q)
    + hand_beta
    - np.log(1.0 + 2.0 * hand_q)
    - np.log(1.0 + hand_q)
)

VisualizationOut [6]
CollapseExpand

A concave log partial likelihood curve with a marked maximum near beta equals minus one. — Log partial likelihood for the four-subject hand example over a fixed beta grid. The curve is concave, with the marked point at the finite maximum derived from the score equation. This illustrates the optimization problem behind Cox fitting, not a general guarantee of finite estimability in every dataset.

The lesson from the hand calculation is not the specific coefficient. It is that the optimization operates on observed covariate values within risk sets. At an event where the failing individual has $x=1$ , increasing $\beta$ raises its numerator weight; when the failing individual has $x=0$ , that numerator stays at one. In either case, increasing $\beta$ raises the denominator weights of the subjects with $x=1$ who remain at risk. A positive $\beta$ helps only when the failing individual has high covariate values relative to those still at risk. Fitting the Cox model means finding the $\beta$ that balances those two forces across all events.

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