# MCMC and data augmentation

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):

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

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.)

• (2) Any MCMC algorithm can start with arbitrary values because the Markov chain is recurrent, this is the core idea behind Markov chain Monte Carlo methods. Note that $\lambda_0$ is a parameter of the prior: it is chosen a priori and does not change once the data is observed. Commented Nov 14, 2012 at 9:42
• (3) When sampling from the Gamma distribution in step 2 of the Gibbs sampler, note that I condition upon the complete data, generated on step 1 of the Gibbs sampler. I thus "know" every value of the $x_i$'s, even those for which $x_i^*=1$. Please try to understand the distinction between the $x_i$'s and the $x_i^*$'s, this is the fundamental idea behind the data augmentation principle. Commented Nov 14, 2012 at 9:45
• (1) The $[\{\lambda+1\}\exp(-\lambda)]^{n_1}$ part corresponds to the grouped observations. Commented Nov 16, 2012 at 20:33
• (2) This is conditional probability (please try to do the math by yourself): $\mathbb{P}(x_i=0|\lambda,x^∗_i=1)=\mathbb{P}(x_i=0,x^∗_i=1|\lambda)/\mathbb{P}(x^∗_i=1|\lambda)=\mathbb{P}(x_i=0|\lambda)/\mathbb{P}(x^∗_i=1|\lambda)$ Commented Nov 16, 2012 at 20:37
• (3) Gibbs sampling works by conditionals. So on step 2, we condition on the $x_i$'s we simulated in step 1 (and in step 1 on the $\lambda$ we simulated in step 2). This means those $x_i$'s are known (even though they will change at the next iteration) and so is the sum. You definitely need to read some introduction to Gibbs if this fundamental point remains unclear to you... Commented Nov 16, 2012 at 20:40