Let's imagine I have a machine giving me an independent random number from a normal distribution $N(\mu,1)$ whenever I push a button. I have a stopping rule to decide how many samples to collect - I will collect samples until the sum of observations exceeding 1.

Mathematically, my stopping rule is to collect $N$ samples where $N$ is a stopping time defined as $N = \inf\{n \geq 1 : \sum_{i=1}^n X_i > 1\}$ where $X_1, X_2, \dots \sim N(\mu, 1)$. Let $D_N := \{X_1, \dots, X_N\}$ be the observed data set.

My question is that if I put a standard normal prior on $\mu$, that is if $\mu \sim N(0,1)$, what is the posterior mean of $\mu$ given $D_N$?

If I simply collected a fixed $n$ numbers of samples, the answer is $$ \mathbb{E}[\mu |D_n] = \frac{n}{n+1}\bar{X}_n, $$ where $\bar{X}_n$ is the sample mean based on $n$ observations.

Intuitively, I think for a randomly stopped $D_N$ the posterior mean should be similar which is given as $$ \mathbb{E}[\mu |D_N] = \frac{N}{N+1}\bar{X}_N. $$ However, I cannot rigorously justify it since it is unclear how to define the likelihood function of $\mu$ given the randomly stopped data set $D_N$.

Thanks in advance for your helpful answers!


The system you are describing is an asymmetrical Gaussian random walk with unknown $\mu$. Try checking out some related questions like this one, although your problem is significantly harder.

I think the key is to figure out $P(N=n|\mu)$ (probability of stopping after $n$ samples) and $P(D_N|N=n,\mu)$ (probability of your observing your sampled values $D_N$). Then you can compute the conditional likelihood of $\mu$ as:

$$ P(\mu | D_N) = \frac{P(D_N | N=n, \mu) \times P(\mu)}{P(D_N)} $$

| cite | improve this answer | |
  • 1
    $\begingroup$ Thanks! Yes, the most tricky part is $P(D_N | N = n, \mu)$ has no simple form in general since condition on $N = n$, all observations are dependent to each other. $\endgroup$ – JaeHyeok Shin Jul 11 '19 at 21:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.