Sufficiency of $|X|$ when $X\sim N(0,\sigma^2)$ without using Factorization theorem

Question

Question:

Given, $X\sim N(0,\sigma^2)$. By means of conditional approach show that $|X|$ is a sufficient estimator for $\sigma^2$.

My Attempt:

This problem is very easy if we use Fisher–Neyman Factorisation theorem. But in this problem we have to use the basic definition of Sufficient Statistics, i.e,

A statistic $t = T(X)$ is sufficient for underlying parameter $\theta$ precisely if the conditional probability distribution of the data $X$, given the statistic $t = T(X)$, does not depend on the parameter $θ$.

$X\sim N(0,\sigma^2)$, here parameter, $\theta=\sigma^2$. So we have to show that $P_{\sigma^2}(X\le\ x\mid T(X)\le\ t)$ is independent of $\sigma^2$.

For $x\in\ (-t,t]$,

\begin{align} P_{\sigma^2}(X\le\ x \mid T(X)\le\ t)&= \frac{P_{\sigma^2}(X\le\ x,T(X)\le\ t)}{P_{\sigma^2}(T(X)\le\ t)} \\&=\frac{P_{\sigma^2}(X\le\ x,|X|\le\ t)}{P_{\sigma^2}(|X|\le\ t)} \\&=\frac{P_{\sigma^2}(-t\le\ X\le\ x)}{P_{\sigma^2}(-t\le\ X\le\ t)} \end{align}

But how to show that this expression is free of $\sigma^2$?

Answer 1

The problem of your derivation is that you misunderstood the concept of conditional distribution. It is not $P_\sigma(X \leq x |T(X) \leq t)$ -- it should be $P_\sigma(X \leq x | T(X) = t)$. For a thorough discussion of the latter notation, see this answer.

To derive the correct conditional distribution, intuitively, given $|X| = t > 0$, then $X$ can only take value $t$ or $-t$. Therefore, for any $x \in \mathbb{R}$, the event $[X \leq x] \cap [|X| = t]$ is: \begin{align} \begin{cases} \varnothing & x < -t, \\ [X = -t] & -t \leq x < t, \\ [|X| = t] & x \geq t. \end{cases} \end{align} By symmetry of $N(0, \sigma^2)$, this implies the conditional distribution of $X$ given $|X| = t$ is \begin{align} P[X \leq x | |X| = t] = \begin{cases} \frac{1}{2}I_{[-x, \infty)}(t) & x \leq 0, \\ I_{(0, x]}(t) + \frac{1}{2}I_{(x, \infty)}(t) & x > 0. \end{cases} \tag{1} \end{align} Therefore the conditional distribution of $X$ given $|X|$ does not depend on $\sigma^2$, hence $|X|$ is sufficient for the distribution family $\{N(0, \sigma^2): \sigma > 0\}$.

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To prove $(1)$ rigorously, first rewrite $(1)$ as \begin{align} P[X \leq x | |X|] = \begin{cases} \frac{1}{2}I_{[|X| \geq -x]}(\omega) & x \leq 0, \\ I_{[|X| \leq x]}(\omega) + \frac{1}{2}I_{[|X| > x]}(\omega) & x > 0. \end{cases} \tag{2} \end{align} Since the right-hand side of $(2)$ is obviously $\sigma(|X|)$-measurable, it suffices to show for any generic $\sigma(|X|)$-set $[|X| \leq t]$, where $t > 0$, it holds that (these are two defining relations of the measure-theoretic conditional probability. For more details, refer to, for example, Equation (33.8) in Probability and Measure by Patrick Billingsley): \begin{align} P[[X \leq x]\cap [|X| \leq t]] = \int_{[|X| \leq t]}P[X \leq x ||X|]dP. \tag{3} \end{align}

When $x \leq 0$, the left-hand side of $(3)$ is $(\Phi_\sigma(x) - \Phi_\sigma(t))I_{[-t, 0]}(x)$, while the right-hand side of $(3)$ is \begin{align} \frac{1}{2}P[|X| \leq t, |X| \geq -x] = (\Phi_\sigma(x) - \Phi_\sigma(t))I_{[-t, 0]}(x). \end{align} Hence $(3)$ holds.

When $x > 0$, the left-hand side of $(3)$ is \begin{align} (\Phi_\sigma(x) - \Phi_\sigma(-t))I_{(0, t)}(x) + (\Phi_\sigma(t) - \Phi_\sigma(-t))I_{[t, \infty)}(x), \end{align} while the right-hand side of $(3)$ is \begin{align} & P[|X| \leq t, |X| \leq x] + \frac{1}{2}P[|X| \leq t, |X| > x] \\ =& \left[P[|X| \leq x] + \frac{1}{2}P[x < |X| \leq t]\right]I_{(0, t)}(x) + P[|X| \leq t]I_{[t, \infty)}(x) \\ =& (\Phi_\sigma(t) - \Phi_\sigma(-x))I_{(0, t)}(x) + (\Phi_\sigma(t) - \Phi_\sigma(-t))I_{[t, \infty)}(x)\\ =& (\Phi_\sigma(x) - \Phi_\sigma(-t))I_{(0, t)}(x) + (\Phi_\sigma(t) - \Phi_\sigma(-t))I_{[t, \infty)}(x). \end{align} Hence $(3)$ holds. This completes the proof.

Answer 2

$[\rm I]$ notes for $T(\mathbf X) $ to be sufficient for $\theta, $

$$ \mathbb P_\theta(\mathbf X=\mathbf x\mid T(\mathbf X) =T(\mathbf x))=\frac{p(\mathbf x\mid\theta)}{q(T(\mathbf x)\mid \theta)} \tag 1\label 1$$ should be constant as a function of $\theta.$

It is well-known that if $X\sim\mathrm N(\mu, \sigma^2), $ then $|X|\sim\mathrm{FoldedN}(\mu, \sigma^2). $

Here $X\sim\mathrm N(0, \sigma^2), ~n=1.$ So, evaluating $\eqref 1$

\begin{align}\frac{p(x\mid\sigma^2)}{q(|x| \mid \sigma^2)}&=\frac{(2\pi\sigma^2)^{-1/2}\exp\left[-\frac{x^2}{2\sigma^2}\right]}{(2\pi\sigma^2)^{-1/2}\left(\exp\left[-\frac{x^2}{2\sigma^2}\right]+\exp\left[-\frac{x^2}{2\sigma^2}\right]\right)}\\&=\frac12\end{align}

shows $|X|$ is sufficient for $\sigma^2.$

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Reference:

$\rm [I]$ Statistical Inference, George Casella, Roger L. Berger, Wadsworth, $2002, $ sec. $6.2.1, $ p. $273.$