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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?

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

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

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.

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  • $\begingroup$ how do I express $(\overline{x}/{S^2}, S^2)$ as a sufficient statistic for $\mu$? $\endgroup$ – Lady Jan 31 at 11:08
  • $\begingroup$ 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). $\endgroup$ – Ben Feb 1 at 23:34

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