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Let $X_1,\ldots,X_n$ identically indepedent and distributed like $N(b,1)$ .

I'm supposed to find a sufficient statistic for $a=P[X_1<1]$.

Michael Hardy
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Itried
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You have $$ a=\Pr(X_1<1) = \Pr(X_1-b < 1-b) = \Phi(1-b) $$ so $a$ is a one-to-one function of $b$. Since the conditional distribution of $X_1,\ldots,X_n$ given $X_1+\cdots+X_n$ is the same for all values of $b$, and $a$ and $b$ determine each other completely, you can say that the conditional distribution of $X_1,\ldots,X_n$ given $X_1+\cdots+X_n$ is the same for all values of $a.$

Michael Hardy
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  • So if I understand corretly the explicit form of my sufficient statistic would be $ X_1+\ldots,+X_n$? actually to be more accurate it would be the same sufficient statistic as the sufficient statistic of b ? – Itried Jun 24 '17 at 16:19
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    Correct. A sufficient statistic is for a family of probability distributions, and the two parameters, $a$ and $b,$ since they completely determine each other, both parametrize the same family of probability distributions. $\qquad$ – Michael Hardy Jun 24 '17 at 16:25