Sufficient statistic for $N(\theta,\theta^2)$

Question

Let $X_1, \ldots, X_n$ be a random sample of the normal distribution with parameters $(\theta, \theta^2)$. How can I find a sufficient statistic for $\theta$?

Answer 1

Let $\bar x = (x_1+\cdots+x_n)/n$ and recall from algebra that $$ \sum_{i=1}^n (x_i-\theta)^2 = n(\bar x-\theta)^2 + \sum_{i=1}^n (x_i-\bar x)^2. $$

Then deal with the density: \begin{align} & f_{X_1,\ldots,X_n}(x_1,\ldots,x_n) \\[10pt] \propto {} & \prod_{i=1}^n\frac 1 \theta \exp\left( \frac{-1} 2 \left( \frac{x_i-\theta}{\theta} \right)^2 \right) \\[10pt] = {} & \frac 1 {\theta^n} \exp\left( \frac{-1}{2\theta^2} \sum_{i=1}^n (x_i-\theta)^2 \right) \\[10pt] = {} & \theta^{-n} \exp\left( \frac{-1}{2\theta^2} \left( n(\bar x - \theta)^2 + \sum_{i=1}^n (x_i - \bar x)^2 \right) \right) \\[10pt] = {} & \theta^{-n} \exp\left( - \frac n {2\theta^2} ((\bar x - \theta)^2 + s^2) \right). \end{align} That the density depends on $(x_1,\ldots,x_n)$ only through $\bar x$ and $s^2$ is enough to conclude that that pair is sufficient. (If there had also been an additional factor depending on $(x_1,\ldots,x_n)$ through other functions but not depending on $\theta$, that would also be enough. That happens, e.g., with the Poisson distribution. This is Fisher's factorization criterion.)