Dirichlet Distribution & Log-Logistic Survival

The Problem

You are building a 3-state Markov model for Chronic Kidney Disease (Stable -> Progressed -> Dead). From a cohort of 200 patients observed for 1 year starting in “Stable”:

• 150 remained Stable
• 35 progressed to CKD Stage 4
• 15 died

These three probabilities (0.75, 0.175, 0.075) must sum to 1.0 in every PSA iteration. If you sample them independently using three Beta distributions, they will almost never sum to 1 – breaking the model.

The Dirichlet Solution

The Dirichlet distribution is the multivariate generalisation of the Beta. Its parameters are the observed counts:

\[\boldsymbol{\alpha} = (150, 35, 15)\]

Each sample from a Dirichlet is a complete probability vector that sums to exactly 1.0.

Sampling via Gamma Decomposition

ParCC uses the standard algorithm:

1. Draw \(X_i \sim \text{Gamma}(\alpha_i, 1)\) for each state
2. Compute \(p_i = X_i / \sum_j X_j\)
3. The resulting \((p_1, p_2, p_3)\) is Dirichlet-distributed and sums to 1

In ParCC

1. Navigate to Uncertainty (PSA) and select Dirichlet (Multinomial).
2. Enter counts: 150, 35, 15.
3. Enter labels: Stable, Progressed, Dead.
4. Click Fit & Sample.

ParCC displays the Dirichlet parameters, mean proportions, a bar chart of sampled proportions, and a ready-to-use R code snippet for your PSA loop.

When to Use Dirichlet vs Independent Betas

The Problem

You are modelling recovery after hip replacement surgery. The hazard of revision is:

• Low immediately after surgery (close monitoring)
• Peaks around year 5-7 (implant loosening)
• Declines after year 10 (survivors have well-fixed implants)

Neither Exponential (constant hazard) nor Weibull (monotonic hazard) can capture this hump-shaped pattern.

The Log-Logistic Distribution

The survival function is:

\[S(t) = \frac{1}{1 + (t/\alpha)^\beta}\]

The hazard function is:

\[h(t) = \frac{(\beta/\alpha)(t/\alpha)^{\beta-1}}{1 + (t/\alpha)^\beta}\]

When \(\beta > 1\), the hazard rises to a peak then falls – exactly the hump shape needed.

In ParCC

From a published Kaplan-Meier curve, identify two time-survival points:

• Point 1: At Year 5, implant survival = 92%
• Point 2: At Year 15, implant survival = 78%

1. Navigate to Survival Curves > Fit Survival Curve.
2. Select method: Log-Logistic (From 2 Time Points).
3. Enter the values.
4. ParCC solves for alpha (scale) and beta (shape).
5. Verify beta > 1 in the output to confirm the expected hump-shaped hazard.

Calibration Method

ParCC uses the log-odds transformation. Since \(S(t) = 1/(1 + (t/\alpha)^\beta)\):

\[\ln\left(\frac{1 - S(t)}{S(t)}\right) = \beta \ln(t) - \beta \ln(\alpha)\]

Two points yield two equations, solved for alpha and beta.

Choosing the Right Survival Distribution

Extrapolation Warning

As with all parametric survival models, extrapolation beyond the observed data requires clinical justification. The Log-Logistic’s long tail means it predicts higher long-term survival than the Weibull – validate this against clinical expectations.