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 $(\mu_1,\mu_2,\sigma)$?.
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
The probability density function for the data is
\begin{align} f_\theta\left(\{X\}_i,\{Y\}_j\right)&=\prod_i\exp\left(\frac12\left(\frac{X_i-\mu_1}\sigma\right)^2\right)\prod_j\exp\left(\frac12\left(\frac{Y_j-\mu_2}\sigma\right)^2\right)\\ &=\exp\left(\frac1{2\sigma^2}\left(\sum_iX_i^2+\sum_jY_j^2-2\mu_1\sum_iX_i-2\mu_2\sum_jY_j+m\mu_1^2+n\mu_2^2\right)\right)\;, \end{align}
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.)
-
$\begingroup$ Thanks a lot for the answer! Also how can we check whether the obtained minimal sufficient statistic is complete or not? In case they have same mean and variance then they belong to the exponential family and the answers follows. But here they have different probability density functions and, therefore, do not belong to the exponential family. $\endgroup$– SambafCommented Feb 27, 2016 at 3:25
-
$\begingroup$ @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? $\endgroup$– jorikiCommented Feb 27, 2016 at 9:35