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.


1 Answer 1


$$\left(\sqrt{2\pi\sigma^2}\right)^n = \sqrt{2\pi\sigma^2/n} \left(\sqrt{2\pi\sigma^2}\right)^{n-1}\sqrt{n}$$

So, $$h(Y | \bar{Y}) = \frac{1}{ \sqrt{ \color{red}{ n }} \cdot (\sqrt{2\pi \sigma^2 }) ^{n-1} } e^{-\frac{1}{2\sigma^2} \sum_{i = 1}^n (Y_i - \bar{Y})^2}.$$

The common pdf of degenerated normal distribution is

$$\displaystyle f(\mathbf {x} )=\left(\det \nolimits ^{*}(2\pi {\boldsymbol {\Sigma }})\right)^{-{\frac {1}{2}}}\,e^{-{\frac {1}{2}}(\mathbf {x} -{\boldsymbol {\mu }})'{\boldsymbol {\Sigma }}^{+}(\mathbf {x} -{\boldsymbol {\mu }})}$$ where $\displaystyle {\boldsymbol {\Sigma }}^{+}$ is the generalized inverse and det* is the pseudo-determinant.

  • $\begingroup$ 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? $\endgroup$
    – Michael
    Jul 10, 2019 at 12:04

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