Let $\{ P_{\gamma} \}$ be a parametric family of probability measures on $(\Omega, \mathcal{F})$, such that $P_{\gamma} \ll \mu$ for all $\gamma$, for some $\sigma$-finite $\mu$. Consider the Radon-Nikodym densities $\{ \frac{d P_{\gamma}}{d \mu} \}$. By the Halmos-Savage factorization criterion, a $\sigma$-subalgebra $\mathcal G$ is sufficient for $\{ \frac{d P_{\gamma}}{d \mu} \}$if and only if there exists a measurable $h$, and a $\mathcal G$-measurable family $\{ g_{\gamma} \}$ such that $$ \frac{d P_{\gamma}}{d \mu} = g_{\gamma} \cdot h $$ for each $\gamma$.

Question Given $\{ P_{\gamma} \}$, does the family of sufficient $\sigma$-subalgebras depend on the reference measure $\mu$? In other words, suppose $\mu \ll \mu'$ and $\mu' \ll \mu$, is it true that $\mathcal G$ is sufficient for $\{ \frac{d P_{\gamma}}{d \mu} \}$ if and only if it is sufficient for $\{ \frac{d P_{\gamma}}{d \mu'} \}$?

Informally, at first glance, the answer must be yes. Sufficiency should be invariant with respect to the measure used to compute the density. Since $$ \frac{d P_{\gamma}}{d \mu'} = \frac{d P_{\gamma}}{d \mu} \frac{d \mu }{d \mu'} = g_{\gamma} \cdot h \frac{d \mu }{d \mu'} = g_{\gamma} \cdot h' $$ where $h' = h \frac{d \mu }{d \mu'}$, the factorization criterion indeed says answer is yes.

On the other hand, there is an example below that, apparently, says otherwise. Why the apparent inconsistency here?

Example Consider the model is given by $$ y_t = \rho y_{t-1}, \, t = 1, \cdots, T, \; \epsilon_i \stackrel{i.i.d.}{\sim} \mathcal{N}(0, \sigma^2), $$ parametrized by $\gamma = \rho$. Take $\sigma^2$ and $y_0$ as fixed. Let $y = (y_1, \cdots, y_T)'$.

Take $\mu$ to be the Lebesgue on $\mathbb{R}^T$. Then the RN density is (see this question) \begin{align*} \frac{d P_{\rho}}{d \mu}(y) &= \frac{1}{ ( \sqrt{2 \pi \sigma^2} )^n } e^{-\frac{1}{2\sigma^2} ( \sum_{t = 1}^T (y_t - \rho y_{t-1})^2)} \\\\ &= \frac{1}{ ( \sqrt{2 \pi \sigma^2} )^n } e^{-\frac{1}{2\sigma^2} ((1 + \rho^2) \sum_{t = 1}^T y_{t-1}^2 + y_T^2 - 2 \rho \sum_{t = 1}^T y_{t-1} y_t)^2)}, \end{align*} which implies $$ \Phi(y) = (\sum_{t = 1}^T y_{t-1}^2,\, y_T^2,\, \sum_{t = 1}^T y_{t-1} y_t) $$ is a minimal sufficient statistic.

Now take $\mu'$ to be $P_1$, where

\begin{align*} \frac{d P_{1}}{d \mu}(y) &= \frac{1}{ ( \sqrt{2 \pi \sigma^2} )^n } e^{-\frac{1}{2\sigma^2} ( \sum_{t = 1}^T (y_t - y_{t-1})^2)}. \end{align*}

So \begin{align*} \frac{d P_{\rho}}{d \mu'}(y) &= \frac{d P_{\rho}}{d P_1}(y) \\\\ &= \frac{ \frac{d P_{\rho}}{d \mu} (y)}{ \frac{d P_{1}}{d \mu} (y)} \\\\ &= e^{-\frac{1}{2\sigma^2} ( \sum_{t = 1}^T (y_t - \rho y_{t-1})^2) - \sum_{t = 1}^T (y_t - y_{t-1})^2)} \\\\ &= e^{-\frac{1}{2\sigma^2} ( -2 (\rho -1) \sum_{t = 1}^T y_t y_{t-1} + (\rho^2 - 1) \sum_{t = 1}^T y_{t-1})^2}. \end{align*}

This suggests $$ \Phi'(y) = ( \sum_{t = 1}^T y_t y_{t-1}, \sum_{t = 1}^T y_{t-1}^2 ) $$ is minimal sufficient, apparently contradicting the minimality of $\Phi$.


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