## The Annals of Mathematical Statistics

### A Note on Mutual Singularity of Priors

David A. Freedman

#### Abstract

Let $0 < r < 1$. Let $\mu$ be a probability on $S$, the closed unit square, which assigns probability 1 to the vertical line with abscissa $r$. Let $\Delta$ be the set of distribution functions on the closed unit interval $I$. Endow $\Delta$ with its weak $\ast$ Borel $\sigma$-field. For any closed sub-interval $H$ of $I$, let $\langle H\rangle$ be the linear function which maps $I$ onto $H$ and sends 0 to the left endpoint of $H$; and $H_0$ (respectively, $H_1$) be the image of $\lbrack 0, r\rbrack$ (respectively, $\lbrack r, 1\rbrack$) under $\langle H\rangle$. Write $B$ for the set of all finite sequences of 0's and 1's (including the empty sequence $\varnothing$). For $b \varepsilon B$ and $\epsilon = 0$ or 1, $I_{b\epsilon} = (I_b)_\epsilon$, where $I_\varnothing = I$. For $G \varepsilon \Delta$, let $|G|$ be the unique probability on $I$ with $G(x) = |G|\lbrack 0, x\rbrack$ for all $x \varepsilon I$. Let $Y_b :b \varepsilon B$ be independent random variables, such that $(r, Y_b)$ has distribution $\mu$. Let $P_\mu$, a probability on $\Delta$, be the distribution of the (random) $F \varepsilon \Delta$ satisfying: $|F|(I_\varnothing) = 1$ and $|F|(I_{b0}) = Y_b |F|(I_b)$ for all $b \varepsilon B$. For a more detailed description, see [2], or Sections 1 and 2 of [3]. Say $F \varepsilon \Delta$ is strictly singular with respect to $G \varepsilon \Delta$ if there is no $x$ for which the ratio of $F(x + h) - F(x)$ to $G(x + h) - G(x)$ converges to a finite, positive limit as $h$ tends to 0. The object of this note is to prove the Theorem. Let $0 < r < 1$. Let $\mu$ and $\nu$ be distinct probabilities on $S$, both assigning measure 1 to the vertical line with abscissa $r$. Then there are weak $\ast$ Borel subsets $C$ and $D$ of $\Delta$, with $P_\mu(C) = P_\nu(D) = 1$, and each distribution function in $C$ strictly singular with respect to each distribution function in $D$. Let $\mu^\ast = \mu\{(r,0)\} + \mu\{(r,1)\}$, and $\nu^\ast = \nu\{(r,0)\} + \nu\{(r,1)\}$. The easy case (either $\mu^\ast = 1$ or $\nu^\ast = 1$) of the Theorem is proved in Section 5. The harder case ($\mu^\ast < 1$ and $\nu^\ast < 1$) is proved in Section 4. Section 3 contains preliminary material on branching processes, which may be of general interest.

#### Article information

Source
Ann. Math. Statist., Volume 37, Number 2 (1966), 375-381.

Dates
First available in Project Euclid: 27 April 2007

https://projecteuclid.org/euclid.aoms/1177699518

Digital Object Identifier
doi:10.1214/aoms/1177699518

Mathematical Reviews number (MathSciNet)
MR189081

Zentralblatt MATH identifier
0214.16303

JSTOR