MCMC and data augmentation

Question

I have been looking at an MCMC data augmentation question; the general form of the question is as follows:

Suppose data gathered on a process suggests $X_{i} \sim \text{Pois}(\lambda)$ and a prior for the rate parameter is suggested as $\lambda \sim \text{Exp}(\lambda_{0})$. The data is recorded and presented in a typical form (i.e. the number of occurrences of each value for $X_{i}$ from $0$ to $n$), however, the data gathered does not discriminate in instances where $X_{i} \leq 1$ (i.e. all occurrences where $X_{i} = 0$ and $X_{i} = 1$ are grouped into one category).

Given the data, the likelihood and the prior described above, the question asks for:

• The posterior form of $\lambda$,

• The number of occurrences where $X_{i} = 0$.

I'm not really sure how to answer this question, but I am aware that Gibbs Sampling can be used in data augmentation. Does anybody have any information on how this could be done?

EDIT:

I should specify that it's primarily the second part (the number of occurrences where $X_{i} = 0$) that I'm unsure about. For the first part (the posterior form of $\lambda$), given the likelihood and the prior suggested, I have reasoned (although I'm happy to be corrected):

Given:

$$ \pi(\lambda|\vec{x}) \propto p(\vec{x}|\lambda) \times p(\lambda) $$

So, for the model given above:

$$ \pi(\lambda|\vec{x}) = \frac{\lambda^{\sum_{i=1}^{n}x_{i}}}{\sum_{i=1}^{n}x_{i}!}e^{-n\lambda} \times \lambda_{0}e^{-\lambda \lambda_{0}} $$

Simplifying yields:

$$ \pi(\lambda|\vec{x}) = \frac{\lambda^{\sum_{i=1}^{n}x_{i}}}{\sum_{i=1}^{n}x_{i}!}e^{-\lambda(n+\lambda_{0})}\lambda_{0} $$

which is proportional to (and hence the posterior form is given by):

$$ \pi(\lambda|\vec{x}) \propto \lambda^{\sum_{i=1}^{n}x_{i}}e^{-\lambda(n+\lambda_{0})}\lambda_{0} $$

Answer 1

Your answer does not account for the fact that the observations equal to zero and to one are merged together: what you computed is the posterior for the complete Poisson data, $(X_1,\ldots,X_n)$, rather than for the aggregated or merged data, $(X_1^*,\ldots,X^*_n)$.

If we take the convention that cases when the observation $X_i^*=1$ correspond to $X_i=1$ or $X_i=0$ and the observation $X_i^*=k>1$ to $X_i=k$, the density of the observed vector $(X_1^*,\ldots,X^*_n)$ is (after some algebra and factorisation) $$ \pi(\lambda|x_1^*,\ldots,x^*_n) \propto \lambda^{\sum_{i=1}^n x_i^*\mathbb{I}(x_i^*>1)} \exp\{-\lambda(\lambda_0+n)\} \times \{1+\lambda\}^{n_1} $$ where $n_1$ is the number of times the $x_i^*$'s are equal to one. The last term between brackets above is the probability to get 0 or 1 in a Poisson draw.

So this is your true/observed posterior. From there, you can implement a Gibbs sampler by

1. Generating the "missing observations" given $\lambda$ and the observations, namely simulating $p(x_i|\lambda,x_i^*=1)$, which is given by $$ \mathbb{P}(x_i=0|\lambda,x_i^*=1)=1-\mathbb{P}(x_i=1|\lambda,x_i^*=1)=\dfrac{1}{1+\lambda}\,. $$
2. Generating $\lambda$ given the "completed data", which amounts to $$ \lambda|x_1,\ldots,x_n \sim \mathcal{G}(\sum_i x_i + 1,n+\lambda_0) $$ as you already computed.

(If you want more details, Example 9.7, p.346, in my Monte Carlo Statistical Methods book with George Casella covers exactly this setting.)