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Suppose there are two random variables, $X_1$ and $X_2$, and we're trying to infer $\theta$.

If $X_1$ and $X_2$ are conditionally independent, then is $f(\theta|X_1)$ a sufficient statistic for $X_1$?

I.e., will this hold:

$f(\theta|X_1,X_2) = f(\theta|X_2,f(\theta|X_1))$

This seems to be the case for the Kalman filter (yesterday's posterior is sufficient statistic for all prior information). Is this a general rule?

I'm sure this is well-studied but I can't find the right terminology.

Tom Cunningham
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  • The function of $\theta$ $f(\theta|X_1)$ is not a statistic: a statistic is a transform of the data, i.e., a function from $\mathcal{X}$ onto $\mathbb{R}^p$. While $f(\theta|X_1)$ is a function of $\theta$ indexed by $X_1$. – Xi'an Apr 16 '15 at 08:58
  • Thanks, I see your point, but I think I can construct a random variable which represents the posterior. Suppose that $\theta\in\Theta$, then we can have a posterior as a distribution over values of $theta$, i.e. the statistic $P(X_1)$, where $P\in R^\Theta$, and $P$ has the same shape as $f(\theta|X_1)$. – Tom Cunningham Apr 16 '15 at 20:15
  • This is irrelevant for the definition of sufficiency. – Xi'an Apr 16 '15 at 20:19
  • OK I think I found the right terminology: the link below derives the result that I need: if you're adding new information, that is conditionally independent of the previous information, then you only need to know the posterior given the previous information, not the whole distribution. http://en.wikipedia.org/wiki/Recursive_Bayesian_estimation – Tom Cunningham Apr 16 '15 at 22:23

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