The posterior predictive

Idea: Forecast new data, not θ

The future has $2$ layers of doubt: $\theta$ is unknown, and even a known $\theta$ gives a random outcome.

• The posterior predictive carries both
• Draw $\theta$ from the posterior, then new data from the model

Formula: The posterior predictive

$p(y\mid D)=\int p(y\mid \theta )p(\theta \mid D)d\theta$

Weight every possible θ by its posterior and average what it forecasts.

Example: Both orders fill

The fill rate has posterior $\mathrm{Beta}(6,4)$. What is the chance that the next $2$ orders both fill?

1. First fills with $\frac{6}{10}$, update to $\mathrm{Beta}(7,4)$
2. Second fills with $\frac{7}{11}$
3. Both: $\frac{6}{10}\times \frac{7}{11}=\frac{21}{55}$

Common mistake: Plugging in the estimate

Plug-in gives $\frac{3}{5}\times \frac{3}{5}=\frac{9}{25}$ for $2$ fills: it keeps the noise and drops the doubt.

Instead Average over the posterior. Fills cluster, and plug-in ranges come out too narrow, worst at small $n$.

Quick check: Exactly 1 fill

Same posterior $\mathrm{Beta}(6,4)$. What is the chance of exactly $1$ fill in the next $2$ orders?

Formula: Normal: noise plus doubt

$y\sim N(m,{\sigma }^2+V)$

Variances add. As data grow V falls toward zero, but σ² stays.

Idea: The gap grows with the horizon

Over $h$ days the noise is independent, but the mean’s error is shared: variance $h{\sigma }^2+h^{2}V$.

• $V=2$, ${\sigma }^2=12$: a day has $12+2=14$
• A week of $5$ days: $5\times 12+25\times 2=110$
• The plug-in says $5\times 12=60$

Quick check: No event at all

An event rate has posterior $\mathrm{Gamma}(2,3)$. What is the chance of no event in the next period?

Idea: Checking a model

Simulate fresh data sets from the predictive and compare a feature you care about, such as the worst day.

• Real figure far outside the simulated range: the model misses something