Basu's theorem

Let ${\displaystyle P_{\theta }^{T}}$ and ${\displaystyle P_{\theta }^{A}}$ be the marginal distributions of ${\displaystyle T}$ and ${\displaystyle A}$ respectively.

Denote by ${\displaystyle A^{-1}(B)}$ the preimage of a set ${\displaystyle B}$ under the map ${\displaystyle A}$. For any measurable set ${\displaystyle B\in {\mathcal {B}}}$ we have

${\displaystyle P_{\theta }^{A}(B)=P_{\theta }(A^{-1}(B))=\int _{Y}P_{\theta }(A^{-1}(B)\mid T=t)\ P_{\theta }^{T}(dt).}$

The distribution ${\displaystyle P_{\theta }^{A}}$ does not depend on ${\displaystyle \theta }$ because ${\displaystyle A}$ is ancillary. Likewise, ${\displaystyle P_{\theta }(\cdot \mid T=t)}$ does not depend on ${\displaystyle \theta }$ because ${\displaystyle T}$ is sufficient. Therefore

${\displaystyle \int _{Y}{\big [}P(A^{-1}(B)\mid T=t)-P^{A}(B){\big ]}\ P_{\theta }^{T}(dt)=0.}$

Note the integrand (the function inside the integral) is a function of ${\displaystyle t}$ and not ${\displaystyle \theta }$. Therefore, since ${\displaystyle T}$ is boundedly complete the function

${\displaystyle g(t)=P(A^{-1}(B)\mid T=t)-P^{A}(B)}$

is zero for ${\displaystyle P_{\theta }^{T}}$ almost all values of ${\displaystyle t}$ and thus

${\displaystyle P(A^{-1}(B)\mid T=t)=P^{A}(B)}$

for almost all ${\displaystyle t}$. Therefore, ${\displaystyle A}$ is independent of ${\displaystyle T}$.

Let X₁, X₂, ..., X_n be independent, identically distributed normal random variables with mean μ and variance σ².

Then with respect to the parameter μ, one can show that

${\displaystyle {\widehat {\mu }}={\frac {\sum X_{i}}{n}},}$

the sample mean, is a complete sufficient statistic – it is all the information one can derive to estimate μ, and no more – and

${\displaystyle {\widehat {\sigma }}^{2}={\frac {\sum \left(X_{i}-{\bar {X}}\right)^{2}}{n-1}},}$

the sample variance, is an ancillary statistic – its distribution does not depend on μ.

Therefore, from Basu's theorem it follows that these statistics are independent.

This independence result can also be proven by Cochran's theorem.

Further, this property (that the sample mean and sample variance of the normal distribution are independent) characterizes the normal distribution – no other distribution has this property.[3]