This is the first time I've come across this type of problem. I really appreciate any comments or suggestions.

In a Bayesian model implemented through Stan, with populations as the units of analysis (e.g., hospitals or institutions) , I have estimated a continuous parameter denoted as $\mu$. In this context:

$\mu_i \sim N(\theta, \tau^2)$

where $i$ denotes the mean for the $i^{th}$ population (for $i$ = 1,...,$k$) and $\tau^2$ represents the between-population variance. Thus, $\theta$ represents the average of true means across populations.

Based on $n$ samples from the posterior mean distribution, I want to calculate the probability that a person/individual (across all populations) has a value greather than 50.

In other words, assuming that the value for the $j^{th}$ individual in the $i^{th}$ population is represented by:

$\mu_{ji} \sim N(\mu_i, \sigma^2)$

I want to calculate Pr($\mu_{ji}>50$)

where $\sigma^2$ represents the population-specific variance, assumed to be the same for all populations for simplicity.

To compute Pr($\mu_{ji}>50$), I calculated for each sample from the posterior distribution Pr($\mu_{ji}>50$) = 1-$\Phi\left(\frac{50-\mu_{i}}{\sigma} \right)$

and obtained the median across all $k$ samples

Is this a reasonable approach?

Any suggestions are welcome.


1 Answer 1


This is a hierarchical Bayesian model, hence $\mu_{ij}$ only depends on $\mu_i$, and $$\mathbb P(\mu_{ji}>50|D)=\mathbb E\left[ \mathbb P(\mu_{ji}>50\big|\mu_i)|D\right]$$ where $D$ denotes the actual sample (to be distinguished from $\mu_{ij}$, which is distributed from the predictive distribution). Now, $$\mathbb P(\mu_{ji}>50\big|\mu_i)=1-\Phi(\sigma^{-1}\{50-\mu_i\})$$ Therefore $$\mathbb P(\mu_{ji}>50|D)=1-\mathbb E\left[ \Phi(\sigma^{-1}\{50-\mu_i\}) | D \right]$$ which can be approximated by averaging over the simulated posterior sample.

  • 1
    $\begingroup$ Thank so much you. Extremely clear. $\endgroup$
    – Jacob
    Aug 13, 2023 at 2:56

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