Sufficient statistic for normal distribution by a bijective map

Question

For $X_i$~$\mathcal{N}(\mu,\sigma^2)$, I know that since $T=(\Sigma^{n}_{i=1}x_i,\Sigma^{n}_{i=1}x_i^2)$ is a sufficient statistic for $(\mu,\sigma^2)$, $\bar{T}=(\bar{X}_n,S_n^2)$ ($\bar{X}_n$ is the sample mean and $S_n^2$ is the sample variance) is also a sufficient statistic for $(\mu,\sigma^2)$ since there is a bijective map from $T$ to $\bar{T}$. However, is $\dot{T}=(\bar{X}_n,S_n^2,X_1^2)$ also a sufficient statistic? And can I find a bijective map between one of the sufficient statistics to $\dot{T}$?

Answer 1

Such a bijection exists only for minimally sufficient statistics. It follows from the factorization theorem that if $S$ is sufficient, and $S'$ is minimally sufficient, then $S'=f(S)$ for some $f$. If $S_1$ and $S_2$ are minimally sufficient, then there must exist an $f_1$ so that $S_1=f_1(S_2)$, and an $f_2$ so that $S_2=f_2(S_1)$. $f_1$ and $f_2$ are hence bijective (and are eachother's inverse).

We may show that $\dot{T}$ is not minimally sufficient. Notice that there cannot be a function that return $\cdot{T}$ from $T$, since we may change $X_1$ and $X_2$ so that $T$ remains the same but $\cdot{T}$ changes. As such no bijection exists between it and any other minimally sufficient statistic.