Suppose I have a Bernoulli process: $2n$ Bernoulli observations whose locations are distributed over an interval, say $(0,1]$. Conditional on the locations, $x_k\stackrel{iid}{\sim}Bern(\theta)$, $k=1,...,2n$. $n$ is known, but $\theta$ is not. The likelihood is then $$ \mathcal{L}(\theta|x_k's)= \prod_{x_k=\{0,1\},k=1,...,2n}\theta^{x_k}(1-\theta)^{1-x_k}=\theta^{\sum_k x_k}(1-\theta)^{2n-\sum_k x_k} $$ I have two partitioning schemes:

1) Partition interval $(0,1]$ into equal-length segments, $(0,1/2]$ and $(1/2,1]$ with potentially unequal number of $x_k$'s in each of the two equal-length segments.

2) Partition interval $(0,1]$ into segments of potentially unequal length, but $n$ $x_k$'s in each segment.

Let's say $X_1=\sum_{x_k\in\text{segment 1}}x_k$ and $X_2=\sum_{x_k\in\text{segment 2}}x_k$ are two random variables. Given my data generating process, is there any reason to think that my parameter $\theta$ should be different for $X_1$ and $X_2$ when I choose one partition scheme over the other?

In other words, for partition scheme 2, I expect $X_1$ and $X_2$ to follow a binomial distribution with size $n$, but can the parameter be the same because they came from $iid$ Bernoulli trials? For partition scheme 1, my gut tells me that the parameters won't be equal.


The probabilistic model you describe is such that the Bernoulli variates $X_1,\ldots,X_{2n}$ are iid conditional on the locations. Since you do not describe the distribution of those locations, it is impossible to know whether or not the distribution depends on the parameter $\theta$. If not, that is, if the locations are governed by an independent process that does not involve $\theta$, or if the locations are deterministic rather than random, the term "conditional" in your description is irrelevant. And we can assume we have $2n$ iid Bernoulli variates $X_1,\ldots,X_{2n}$.

Since the partition of the index set $\{1,2,\ldots,2n\}$ into two groups is unrelated with $\theta$ or the realisations of $X_1,X_2,\ldots,X_{2n}$, the distribution of the variables within the resulting groups of random variables remains the same, iid Bernoulli variates with parameter $\theta$. However, in the second case, the number of variables is fixed and you obtain for the sums, two iid Binomial $\text{B}(n,\theta)$ variables, while for the first case, the number of variables $N$ in the first group is random and hence $$X_1\sim\sum_{i=1}^N X_i\qquad X_2\sim\sum_{i=N+1}^{2n} X_i$$ implies that

  • $X_1$ and $X_2$ are not Binomial variates, but instead compound Binomial variates;
  • $X_1$ and $X_2$ are not independent.

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.