# Gaussian sufficient statistic calculation

Consider the Gaussian model $$Y_i = \beta + \epsilon_i,\, i = 1, \cdots, n,\; \mbox{where}\; \epsilon_i \stackrel{i.i.d.}{\sim} \mathcal{N}(0, \sigma^2),$$ parametrized by $$\beta$$, with known $$\sigma$$. The likelihood function admits the factorization \begin{align*} L(Y) &= \frac{1}{ (\sqrt{2\pi \sigma^2}) ^n } e^{-\frac{1}{2\sigma^2} (\sum_{i = 1}^n (Y_i - \beta)^2)} \\\\ &= \frac{1}{ (\sqrt{2\pi \sigma^2}) ^n } e^{-\frac{1}{2\sigma^2} ( \sum_{i = 1}^n (Y_i - \bar{Y})^2 + n ( \bar{Y} - \beta)^2)}. \end{align*}

Evidently, by the Halmos-Savage factorization criterion, the sample mean $$\bar{Y}$$ is a sufficient statistic. (In fact, it is a minimal sufficient statistic.)

The decomposition
\begin{align*} L(Y) &= \frac{1}{ (\sqrt{2\pi \sigma^2}) ^n } e^{-\frac{1}{2\sigma^2} \sum_{i = 1}^n (Y_i - \bar{Y})^2} e^{-\frac{1}{2 \frac{\sigma^2}{n} } (\bar{Y} - \beta)^2 }\\\\ &\equiv h(Y | \bar{Y}) \; g(\bar{Y}|\beta) \end{align*} suggests that $$\bar{Y}$$ is normally distributed with mean $$\beta$$ and variance $$\frac{\sigma^2}{n}$$, as it should be. So putting $$g(\bar{Y}|\beta) = \frac{1}{ (\sqrt{2\pi \frac{ \sigma^2}{n} }) } e^{-\frac{1}{2 \frac{\sigma^2}{n} } (\bar{Y} - \beta)^2}$$ leads to $$h(Y | \bar{Y}) = \frac{1}{ (\sqrt{ \color{red}{ n } \cdot 2\pi \sigma^2 }) ^{n-1} } e^{-\frac{1}{2\sigma^2} \sum_{i = 1}^n (Y_i - \bar{Y})^2}.$$

Question What is the explanation for the extra factor of $$n$$ in $$\sqrt{ \color{red}{ n } \cdot 2\pi \sigma^2 }$$? One would expect the conditional density $$h(Y | \bar{Y})$$ to be a degenerate Gaussian---the joint density of $$n-1$$ i.i.d. Gaussians with variance $$\sigma^2$$---supported on a $$(n-1)$$-dimensional affine subspace of $$\mathbb{R}^n$$. So the normalization factor ought to be $$(\sqrt{ 2\pi \sigma^2 })^{n-1}$$, at first glance.

$$\left(\sqrt{2\pi\sigma^2}\right)^n = \sqrt{2\pi\sigma^2/n} \left(\sqrt{2\pi\sigma^2}\right)^{n-1}\sqrt{n}$$
So, $$h(Y | \bar{Y}) = \frac{1}{ \sqrt{ \color{red}{ n }} \cdot (\sqrt{2\pi \sigma^2 }) ^{n-1} } e^{-\frac{1}{2\sigma^2} \sum_{i = 1}^n (Y_i - \bar{Y})^2}.$$
$$\displaystyle f(\mathbf {x} )=\left(\det \nolimits ^{*}(2\pi {\boldsymbol {\Sigma }})\right)^{-{\frac {1}{2}}}\,e^{-{\frac {1}{2}}(\mathbf {x} -{\boldsymbol {\mu }})'{\boldsymbol {\Sigma }}^{+}(\mathbf {x} -{\boldsymbol {\mu }})}$$ where $$\displaystyle {\boldsymbol {\Sigma }}^{+}$$ is the generalized inverse and det* is the pseudo-determinant.
• So $\Sigma$ here is the orthogonal projection onto the orthogonal complement of $(1,...,1)'$ (say $\sigma^2 = 1$). Then $\Sigma = \Sigma^+$. The pseudo-determinant of $2 \pi \Sigma$ is $(2 \pi)^{n-1}$. Where does $n$ come from? – Michael Jul 10 '19 at 12:04