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

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.

| cite | improve this answer | |
  • $\begingroup$ 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 '19 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.