Minimal sufficient statistics for normal distributions

Question

I have the following variables $X_1,..,X_n,Y_1,..,Y_m$ with distribution $X_i\sim N(\mu,\sigma_1^2), Y_j\sim N(\mu,\sigma_2^2)$ that are all iid. How do I find the minimal sufficient statistic for $(\mu,\sigma_1^2,\sigma_2^2)$?

Answer 1

Use the Neyman-Pearson factorization criteria, i.e., \begin{align} & L(\mu, \sigma_1^2, \sigma_2^2; Z=(X_1,..,X_n, Y_1,...,Y_m)) & \\ &=\frac{1}{\sigma_1^{n} \sigma_2^m (2\pi)^{(n+m)/2}}\exp\{\left( \sigma_2^2\sum_{i=1}^n(Z_i-\mu)^2 + \sigma_1^2\sum_{j=n+1}^{n+m}(Z_j-\mu)^2 \right)/(2\sigma_1^2\sigma_2^2) \}, \end{align}

so the MSS for the vector $\theta = (\mu, \sigma_1^2, \sigma_2^2)$, is $$ T(X) = (\frac{1}{m+n}\sum_{i=1}^{m+n}Z_i,S_n^2, S_m^2), $$ where $S_i^2 = \frac{1}{n_i}\sum(X_i - \bar{X})^2$, which is the MLE of the $i$th group, $i=n,m$.