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Let $X_1, . . . , X_m, Y_1, . . . , Y_n$ be independent with $X_i ∼ N(ξ, σ^2)$ and $ Y_j ∼ N(η, τ^2).$

What is the minimal sufficient statistic for $(ξ,η,σ^2)$ where $σ^2 = τ^2$?

I've seen MSS derived from a single distribution by using the result shown here. Not sure how to approach the above question however.

Edit:

Let $Z=(X^T,Y^T)^T$, f(z)/f(w) becomes

$$\prod_{i=1}^m\exp\left(\frac{(w_{i}-\mu)^2-(z_{i}-\mu)^2}{2\sigma^2}\right) \prod_{j=m+1}^{m+n}\exp\left(\frac{(w_{j}-\eta)^2-(z_{j}-\eta)^2}{2\sigma^2}\right)$$

expanding it further: $$\exp\left(\frac{\sum_{i=1}^m w_{i}^2-\sum_{i=1}^m z_{i}^2+2\mu\left(\sum_{i=1}^m z_{i}-\sum_{i=1}^m w_{i}\right)}{2\sigma^2}\right) \exp\left(\frac{\sum_{j=1+m}^{n+m} w_{i}^2-\sum_{j=1+m}^{n+m} z_{i}^2+2\eta\left(\sum_{j=1+m}^{n+m} z_{i}-\sum_{j=1+m}^{n+m} w_{i}\right)}{2\sigma^2}\right) $$

so MSS for $\mu$ and $\eta$ are obviously just the sums from 1 to m and m+1 to n+m respectively. while for $\sigma^2$ we have $\left(\sum_{i=1}^{m}z_i,\sum_{i=1}^{m}z_i^2,\sum_{j=1+m}^{n+m}z_j,\sum_{j=1+m}^{n+m}z_j^2\right)$.

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    $\begingroup$ Do you know of some ways to characterize or identify minimal sufficient statistics, in addition to the definition? Which methods have you attempted to apply to this case? $\endgroup$
    – whuber
    May 29 '17 at 17:18
  • $\begingroup$ The "direct consequence from Fisher's factorization theorem" part in the link above basically. $\endgroup$
    – ChuckP
    May 29 '17 at 17:23
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    $\begingroup$ Okay. So begin by writing down the probability density function and examining how it depends on the parameters. What function(s) of $X_i$ and $Y_j$ appear to be involved in the expression? $\endgroup$
    – whuber
    May 29 '17 at 18:13
  • $\begingroup$ I just gave it a go. Felt like I may have overcomplicated things. Am I on the right path? Is there an easier way to obtain MSS for this tho? $\endgroup$
    – ChuckP
    May 29 '17 at 19:17
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    $\begingroup$ You did fine, but you're right: you have overcomplicated things. You know that the sum of the $X_i$ has a Normal$(n\xi, n\sigma^2)$ distribution (and a similar fact holds for the sum of the $Y_j$). Use this information to greatly simplify the expression. $\endgroup$
    – whuber
    May 29 '17 at 19:19

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