# Ignorability in Rubin's theory of missing data mechanisms

I am trying to understand Rubin's theory of bayesian inference with missing data, specifically how the missing data mechanism affects the inference on a superpopulation parameter. The theory is exposed for example in the chapter 7 of [1].

The complete $N$-dimensional data vector $y$ is splitted into an observed and unobserved components: $y = (y_{\mathrm{obs}}, y_{\mathrm{nob}})$ and $I$ is the $N$-vector of indicators whose components $I_{i}$ equal 0 when $y_{i}$ is unobserved and 1 when it is observed. The sample space is thus the product of the $y$ space and the $I$ space. For simplicity, I consider a missing data mechanism $p(I ~|~ y_{\mathrm{obs}})$ which does not depend on a fully observed covariate $x$ or an unknown parameter $\phi$.

The key result in Rubin is that when $p(I ~|~ y) = p(I ~|~ y_{\mathrm{obs}})$, the missing data mechanism is ignorable and the analysis can proceed as usual. This can be seen in the following development:

\begin{align*} p(\theta ~|~ y_{\mathrm{obs}}, I) &~=~ \frac{p(\theta) \int p(y~|~\theta)\, p(I ~|~ y)~d y_{\mathrm{nob}}~}{\int\int p(y~|~\theta)\,p(\theta) ~d\theta p(I ~|~ y)~d y_{\mathrm{nob}}~} \\ \end{align*}

If $p(I ~|~ y) = p(I ~|~ y_{\mathrm{obs}})$, it can be taken out of the integrals and cancelled out such that the posterior distribution $p(\theta ~|~ y_{\mathrm{obs}}, I) = p(\theta ~|~ y_{\mathrm{obs}})$.

So far so good. But what I don't understand is the following derivation that contradicts the preceding result:

\begin{align*} p(\theta ~|~ y_{\mathrm{obs}}, I) &~=~ \frac{p(y_{\mathrm{obs}}, I, \theta)}{p(y_{\mathrm{obs}}, I)} \\ &~=~ \frac{p(y_{\mathrm{obs}} ~|~ \theta) p(\theta) p(I ~|~ y_{\mathrm{obs}})}{\int p(y_{\mathrm{obs}} ~|~\theta)p(\theta) ~d\theta~ p(I ~|~ y_{\mathrm{obs}})} \\ &~=~ \frac{p(y_{\mathrm{obs}} ~|~ \theta)p(\theta)}{\int p(y_{\mathrm{obs}} ~|~\theta)p(\theta) ~d\theta} \\ &~=~ p(\theta ~|~ y_{\mathrm{obs}}). \end{align*}

Here the sampling mechanism cancels out no matter what. It does not matter if $p(I ~|~ y) = p(I ~|~ y_{\mathrm{obs}})$. I must be doing something wrong, but I can't see what.

[1] Gelman, A., Carlin, J. B., Stern, H. S., & Rubin, D. B. (2004). Bayesian Data Analysis (2nd ed.). Boca Raton: Chapman & Hall/CRC.

In the second row of your derivation, you seem to assume $$p(y_{obs},I,\theta) = p(y_{obs}|\theta)p(\theta)p(I|y_{obs}),$$ but this is not in general true - the correct factorization would be $$p(y_{obs},I,\theta) = p(y_{obs}|\theta)p(\theta)p(I|y_{obs},\theta).$$

Without the assumption $p(I|y) = p(I|y_{obs})$, $\theta$ 'impacts' $I$ via $y_{nob}$ even when conditioning on $y_{obs}$, and thus $p(I|y_{obs},\theta)\neq p(I|y_{obs})$.

• Thanks for your insight. I think I see what you mean. This may also be connected to the issue of distinct parameters (in his 1976 paper, Rubin shows that the missingness mechanism is ignorable when the missing data are MAR and when $\phi$ and $\theta$ are distinct). I'll have to think more about this. – Lionel Henry Apr 12 '14 at 8:02

Juho Kokkala provided the right answer and I will just add an example here, based on p. 121 of Little and Rubin, 2002, 2nd ed.

The correct factorization of the missingness mechanism is indeed $p(I ~|~ y_{\mathrm{obs}}, \theta)$. More specifically, it can be further factored into the components depending on $y_{\mathrm{obs}}$ and those which depend on $y_{\mathrm{nob}}$. Since the latter are unobserved data, we rely on their predictive distribution for a given $\theta$: \begin{align*} p(I ~|~ y_{\mathrm{obs}}, \theta) &~=~ p(I_{\mathrm{obs}} ~|~ y_{\mathrm{obs}}) p(I_{\mathrm{nob}} ~|~ \theta) \\ &~=~\prod_{i=1}^{n} p(I_{i} ~|~ y_{i}) \prod_{i=n+1}^{N} p(I_{i} ~|~ \theta) \end{align*} Note the $n$ first components of $y$ constitute $y_{\mathrm{obs}}$, and the complete data has $N$ observations.

Now the example. Imagine that $y \sim \mathrm{Exp}(\theta)$ and our missing data is due to the censoring mechanism \begin{align} p(I_{i} ~|~ y_{i}) ~=~ \begin{cases} 1, & \mathrm{if}~I_{i}= 1 \mathrm{~and~} y_{i} \ge c, \mathrm{~or~} I_{i} = 0 \mathrm{~and~} y_{i} < c \\ 0, & \mathrm{otherwise.} \end{cases} \end{align}

Then we have the likelihood \begin{align*} p(y_{\mathrm{obs}}, I ~|~ \theta) &~=~ p(y_{\mathrm{obs}} ~|~ \theta) p(I_{\mathrm{obs}} ~|~ y_{\mathrm{obs}}) p(I_{\mathrm{nob}} ~|~ \theta) \\ % &~=~ \prod_{i=1}^{n} p(y_{i}, I_{i} ~|~ \theta) \prod_{i=n+1}^{N} % p(I_{i} ~|~ \theta) \\ &~=~ \prod_{i=1}^{n} p(y_{i} ~|~ \theta) \mathrm{Pr}(y_{i} < c ~|~ y_{i}) \prod_{i=n+1}^{N} \mathrm{Pr}(y_{i} \ge c ~|~ \theta) \\ &~=~ \theta^{-n} \exp\left (-\sum_{i=1}^{n} \frac{y_{i}}{\theta} \right) \exp\left (-\frac{(N-n)c}{\theta}\right). \end{align*}

Using the hypothesized missing data mechanism and the hypothesized distribution of $y_{\mathrm{nob}}$, we compute the conditional density of the unobserved components of $I$.