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Consider the Gaussian model $y_i \sim_{\text{iid}} \mathcal{N}(\mu,\sigma^2)$, $i = 1, \ldots, n$, with unknown mean $\mu$ and unknown standard deviation $\sigma$.

The random variable $t = \tfrac{\overline{y}-\mu}{\text{sd}(y)}$ has a Student-$t$ distribution with degrees of freedom parameter $n-1$. This distribution is free of unknown parameters.

One has $$ \color{blue}{\boxed{\mu = \overline{y} - \text{sd}(y) \cdot t}} $$ The somehow esoteric fiducial argument consists in "switching the roles of the data and the parameters", and from the above boxed formula, the fiducial distribution of $\mu$ is the distribution of $\overline{y} - \text{sd}(y) \cdot t$ considering $\overline{y}$ and $\text{sd}(y)$ as fixed constants and $t \sim \text{Student}_{n-1}$ as the random variable.

With a similar approach, the fiducial distribution of $\sigma^2$ is the distribution of $(n-1)\tfrac{\text{sd}(y)}{\chi}$ where $\chi \sim \text{Chi}^2_{n-1}$ is the random variable.

These are, I think (not a master in this topic), the fiducial distributions considered by Fisher.

My question is: did Fisher consider a joint fiducial distribution of $(\mu,\sigma)$, and if yes, what is this distribution?

Stéphane Laurent
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1 Answers1

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I see. One has the bivariate pivot $$ (z, w) = \left(\frac{\sqrt{n}(\overline{y}-\mu)}{\sigma}, \frac{\text{sd}^2(y)}{\sigma^2} \right) \sim \mathcal{N}(0,1) \otimes \frac{\chi^2_{n-1}}{n-1}. $$ By "inverting", $$ \mu = \overline{y} - \frac{\text{sd}(y)\cdot z}{\sqrt{n w}} \qquad \sigma^2 = \frac{\text{sd}^2(y)}{w}. $$ This leads to the joint fiducial distribution of $(\mu,\sigma)$.

Stéphane Laurent
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