**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 $\theta$.


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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 that **"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 contional distrbution 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? 


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

Since

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

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


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 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.