# Ancillary function of a random vector, which is independent of change of origin and scale

Let $$(X_1,\ldots,X_n)$$ be a random vector, whose distribution involves unknown: location parameter $$\mu$$ and a scale parameter $$\sigma>0$$. It follows, that any measurable function $$f(X_1,\ldots,X_n)$$, satisfying condition: $$f(\frac{X_1-a}{b},\ldots,\frac{X_n-a}{b})=f(X_1,\ldots,X_n) \quad \forall a\in \mathbb{R}\quad\land b>0,$$ is ancillary statistic for the unknown parameter $$(\mu,\sigma)$$. How to prove the above implication? This statement appears in: "On statistics independent of a complete sufficient statistics", D. Basu.

You say that $$X$$ has some parametric distribution, which is a location-scale family. Then aspect of the distribution which is not influenced by the location or scale is said to measure some aspect of the distributions shape. By transforming $$X_i \mapsto \frac{X_i-a}{b}$$ you changes $$X$$ to have some distribution from the same family, only with location and scale changed. Your requirement is that this should not change the value of $$f$$. Then compare to the definition of ancillary statistic.
• Ok, so starting from $X$, which has some distribution $\mathbb{P_{\mu,\sigma}}$, or starting with $Y=\frac{X-a}{b}$, which has some distribution $\mathbb{P_{\mu',\sigma'}}$, does not change value of function $f$. But i further can't see, that the distribution of this function will be the same. Could you give me some hint about how i can show it analytically? – Mentossinho Jul 5 at 21:17
• @Mentossinho: If its value does not change by this transformation, and that is true for all $x$, how can its distribution change? – kjetil b halvorsen Jul 21 at 20:27
• i think i've got it: if $X$ has distribution $\mathbb{P_{\mu,\sigma}}$, and $Y=b^{-1}X-a \mathbb{1}$ has distribution $\mathbb{P_{\mu',\sigma'}}$, then $\mathbb{P_{\mu',\sigma'}}(\{f(Y)\in B\})=\mathbb{P_{\mu,\sigma}}(\{f(b^{-1}X-a)\}\in B)=\mathbb{P_{\mu,\sigma}}(\{f(X)\in B\})$ for every measurable $B$ from range of $f$. – Mentossinho Jul 21 at 22:49