Exercise. Let $X_1,\cdots,X_{n}$ be i.i.d.r.v.'s from $N(\theta,1),$ where $\theta$ is unknown.Show the statistic $T(\mathbf{X})=\sum_{i=1}^{n}X_{i}/n=\bar{X} $ is sufficient for $\theta$.

The answer as following:

Suppose that we transform variables $X_1,\cdots,X_{n}$ to $Y_1,\cdots,Y_{n}$ with the help of an orthogonal transformation so that $Y_{1}$ is $N(\sqrt{n}\theta,1), Y_{2},\cdots,Y_{n}$ are i.i.d. $N(\theta,1)$ and $Y_{1},\cdots,Y_{n}$ are independent. We take $$y_{1}=\sqrt{n}\bar{x},$$ and ,for $$k=2,\cdots,n,\quad y_{k}=\left [(k-1)x_{k}-(x_1+\cdot+x_{k-1}) \right ]/\sqrt{k(k-1)}.$$

And then, the author said

the conditional distribution of $\mathbf{X}=(X_{1},\cdots,X_{n})$ given $\bar{X}$, does not depend on $\theta$ ${\color{Red}{\text{is equivalent to}}}$ the conditional distribution of $(Y_1,\cdots,Y_{n})$,given $Y_{1},$ is independent of $\theta.$

I don't understand why the equivalence between them holds. How to prove it rigorously?

I know that the conditional distribution of $(Y_1,\cdots,Y_{n})$,given $Y_{1}$,is independent of $\theta.$


$$f_{\theta}(y_1,\cdots,y_{n})=(2\pi)^{-n/2}\exp \left [-\frac{1}{2}\sum_{i=2}^{n}y^{2}_{i}-\frac{1}{2}(y_{1}-\sqrt{n}\theta)^{2} \right ];$$ $$f_{Y_{1}}(y_{1})=\frac{1}{\sqrt{2\pi}}\left [ -\frac{1}{2}(y_{1}-\sqrt{n}\theta)^{2} \right ];$$

$$f(y_{1},\cdots,y_{n}|y_{1})=\frac{f_{\theta}(y_1,\cdots,y_{n})}{f_{Y_{1}}(y_{1})}=(2\pi)^{-\frac{n-1}{n}}\exp \left [ -\frac{1}{2}\sum_{i=2}^{n}y^{2}_{i}\right ]$$ is independent of $\theta.$

Applying Fisher–Neyman factorization theorem and change of variables formula,the equivalence is obvious.But the author gave this exercise before introducing the factorization theorem,there must be some other ways to get the equivalence.

Every (measurable) set $\mathcal{S}$ in $\mathbf{R}^{n}, \mathbb{P}_{\theta}[(X_1,\cdots, X_n)^{'}\in \mathcal{S}|\bar{X}=t]$ is independent of $\theta$ a.s. $\mathbf{y}=Q_{n}\mathbf{x},\mathbf{x}=Q_{n}^{T}\mathbf{y}$ and $Q_{n}^{T}Q_{n}=I_{n},y_{1}=\sqrt{n}\bar{x}.$ Then $$\mathbb{P}_{\theta}[(X_1,\cdots, X_n)^{'}\in\mathcal{S}|\bar{X}=t]=\mathbb{P}_{\theta}[(Y_1,\cdots, Y_n)^{'}\in Q_{n}^{T}(\mathcal{S})|Y_{1}=\sqrt{n}t].$$


1 Answer 1


The transformation you have written can be expressed succinctly in the form:

$$\mathbf{y} = H(\mathbf{x},\bar{x}) \quad \quad \quad \quad \quad \mathbf{x} = G(\mathbf{y},\bar{x}).$$

The equivalence mentioned in your question holds because, once we condition on $\bar{x}$, there is a one-to-one relationship between $\mathbf{x}$ and $\mathbf{y}$. For example, if we have $\mathbf{x} \ \bot \ \theta|\bar{x}$ then we get:

$$\begin{align} \mathbb{P}(\mathbf{y} \in \mathscr{Y}|\bar{x},\theta) &= \mathbb{P}(H(\mathbf{x},\bar{x}) \in \mathscr{Y}|\bar{x},\theta) \\[6pt] &= \mathbb{P}(H(\mathbf{x},\bar{x}) \in \mathscr{Y}|\bar{x}) \\[6pt] &= \mathbb{P}(\mathbf{y} \in \mathscr{Y}|\bar{x}) \\[6pt] \end{align}$$

which means that $\mathbf{y} \ \bot \ \theta|\bar{x}$. Similiarly, if we have $\mathbf{y} \ \bot \ \theta|\bar{x}$ then we get:

$$\begin{align} \mathbb{P}(\mathbf{x} \in \mathscr{X}|\bar{x},\theta) &= \mathbb{P}(G(\mathbf{y},\bar{x}) \in \mathscr{X}|\bar{x},\theta) \\[6pt] &= \mathbb{P}(G(\mathbf{y},\bar{x}) \in \mathscr{X}|\bar{x}) \\[6pt] &= \mathbb{P}(\mathbf{x} \in \mathscr{X}|\bar{x}) \\[6pt] \end{align}$$

which means that $\mathbf{x} \ \bot \ \theta|\bar{x}$.

  • $\begingroup$ This clearly answers OP's query. +1. $\endgroup$ Dec 8, 2022 at 7:08

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