# Probability density function and the minimal sufficient statistics for two samples from normal distribution

Suppose $X_1,\ldots, X_m$ is a random sample of size $m$ from the normal distribution $N(\mu_1,\sigma^2)$ with mean $\mu_1$ and standard deviation $\sigma$, and that $Y_1,\ldots, Y_n$ is a random sample of size $n$ from the normal distribution $N(\mu_2,\sigma^2)$ with mean $\mu_2$ and standard deviation $\sigma$. Also, suppose that the samples $X$ and $Y$ are independent. What are the probability density function and the minimal sufficient statistic for $(\mu_1,\mu_2,\sigma)$?.

so $\left(\sum_iX_i^2+\sum_jY_j^2,\sum_iX_i,\sum_jY_j\right)$ is a sufficient statistic. Since
$$\frac{f_\theta\left(\{X\}_i,\{Y\}_j\right)}{f_\theta\left(\{X'\}_i,\{Y'\}_j\right)}$$
is independent of $\theta$ if and only if this statistic is the same for the two sets of data, this is also a minimal sufficient statistic. (There's no such thing as "the" minimal sufficient statistic, since you can apply any bijective function to a minimal sufficient statistic to obtain another one.)
• @Sambaf: I don't understand why you think they don't form an exponential family. The distribution function has the form $$\exp\left(\sum_{i=1}^3\eta_i(\theta)T_i(x)+A(\theta)\right)$$ with \begin{align} \eta_1(\theta)=\frac1{2\sigma^2}&\quad&T_1(x)&=\sum_iX_i^2+\sum_jY_j^2\;,\\ \eta_2(\theta)=-\frac{\mu_1}{\sigma^2}&\quad&T_2(x)&=\sum_iX_i\;,\\ \eta_3(\theta)=-\frac{\mu_2}{\sigma^2}&\quad&T_3(x)&=\sum_iY_i\;,\\ A(\theta)=\frac{m\mu_1^2+n\mu_2^2}{2\sigma^2}\;. \end{align} Thus the $\eta_i$ form a complete statistic. Or a I missing something? – joriki Feb 27 '16 at 9:35