Sufficient statistic for $N(\mu,1)$

Let $$X_1,\ldots,X_n$$ be a random sample from $$N(\mu,1)$$, where $$\mu$$ is an unknown parameter. Show that $$(\overline{x}/{S^2}, S^2)$$ is a sufficient statistic for $$\mu$$, where $$S^2$$ is the sample variance.

My Approach

$$\sum_{i=1}^n (x_i-\mu)^2 = n(\bar x-\mu)^2 + \sum_{i=1}^n (x_i-\bar x)^2.$$

By Neyman Fisher Factorisation, \begin{align} f_{X_1,\ldots,X_n}(x_1,\ldots,x_n) \propto {} & \prod_{i=1}^n \frac 1 {(\sqrt2\pi)^n} \exp\left( \frac{-1} 2 \left( {x_i-\mu} \right)^2 \right) = \frac 1 {(\sqrt2\pi)^n} \exp\left( \frac{-1}{2} \sum_{i=1}^n (x_i-\mu)^2 \right) \\[10pt] = {} & (\sqrt2\pi)^{-n} \exp\left( \frac{-1}{2} \left( n(\bar x - \mu)^2 + \sum_{i=1}^n (x_i - \bar x)^2 \right) \right) \\[10pt] = {} & (\sqrt2\pi)^{-n} \exp\left( - \frac n {2} ((\bar x - \mu)^2 + s^2) \right). \end{align}

How do I proceed from here?

• If $T$ is sufficient for $\mu$, then so is $(T,T')$ for any other statistic $T'$. – StubbornAtom Jan 31 at 7:07

Since the variance in this problem is a known constant, it is actually the case that $$\bar{x}$$ is sufficient for $$\mu$$, and since $$\bar{x}$$ can be obtained from the stated statistic (as the product of the two elements), that statistic is sufficient also sufficient for $$\mu$$. To see that $$\bar{x}$$ is sufficient for $$\mu$$ we can apply the Fisher-Neyman factorisation theorem just as you have attempted to do. In this problem your likelihood function is given by:$$^\dagger$$

\begin{aligned} L_\mathbf{x}(\mu) &= (2 \pi)^{-n/2} \cdot \exp \Big( - \frac{1}{2} (x_i-\mu)^2 \Big) \\[6pt] &= (2 \pi)^{-n/2} \cdot \exp \Big( - \frac{n}{2} \cdot (\bar{x}-\mu)^2 - \frac{n-1}{2} \cdot s^2 \Big) \\[6pt] &= \underbrace{(2 \pi)^{-n/2} \cdot \exp \Big( - \frac{n-1}{2} \cdot s^2 \Big)}_{h(\mathbf{x})} \cdot \underbrace{\exp \Big( - \frac{n}{2} \cdot (\bar{x}-\mu)^2 \Big)}_{g_\mu(\bar{x},n)}. \\[6pt] \end{aligned}

It follows that $$\bar{x}$$ is sufficient for $$\mu$$. With some more work it is possible to show that it is minimal sufficient, but that is unnecessary for the result you want in this case.

$$^\dagger$$ I have used the standard notation where the sample variance uses Bessel's correction) so my result is slightly different to yours.

• how do I express $(\overline{x}/{S^2}, S^2)$ as a sufficient statistic for $\mu$? – Lady Jan 31 at 11:08
• It is already sufficient. Since $\mu$ is sufficient, then $\mu / S^2 \times S^2$ is sufficient, so the vector of both elements is sufficient (but not minimal sufficient). – Ben Feb 1 at 23:34