Sufficient statistics, how compute it. Guys can you help me with this? I’ve tried to apply the definition of sufficient statistics but the professor only gave us simple examples so I don’t know how to compute it.

$$X\sim f_θ$$

$$f_θ(x)=\frac{θ}{2}(θx)^2\exp\{−θx\}\text{ for }x&gt;0$$

Moreover, let $X_1,…,X_n$ be a sample from $X$. a) Find a sufficient statistic for the model b) Find the MLE of $θ$ and the MLE of $E(X)=\frac{3}{θ}$
\begin{align}
\prod_{i=1}^n f_\theta(x_i) &amp; = \left(\frac \theta 2\right)^n ((\theta x_1) \cdots (\theta x_n))^2 \exp\{−\theta (x_1+\cdots+x_n)\} \\[10pt]
&amp; = \Big(\, \overbrace{\theta^{3n} \exp(-\theta(x_1+\cdots + x_n))}^\text{first factor}\,\Big) \cdot \Big(\, \overbrace{ \frac{x_1\cdots x_n}{2^n}}^\text{second factor} \, \Big) \tag 1
\end{align}

Fisher's factorization theorem says that if this can be written as a product of two factors in which


one factor depends on $x_1,\ldots,x_n$ only through some specified function of $x_1,\ldots,x_n$ and
the other factor does not depend on $\theta,$


then that specified function, evaluated at the sample, is a sufficient statistic.

Therefore $X_1+\cdots+X_n$ is a sufficient statistic.

If $L(\theta) = $ the expression on line $(1)$ above, then
$$
\ell(\theta) = \log L(\theta) = 3n\log\theta - \theta(x_1+\cdots+x_n) + \text{constant}
$$
so
$$
\ell\,'(\theta) = \frac{3n}\theta - (x_1+\cdots+x_n) \quad \begin{cases} &gt;0 &amp; \text{if } 0 \le \theta &lt; 3/\overline{x}, \\[6pt] &lt;0 &amp; \text{if } \theta&gt;3/\overline{x}. \end{cases}
$$
Therefore $\widehat \theta = 3/\overline x.$

By equivariance of MLEs, the MLE of $3/\theta$ is $\overline x.$
\begin{align}
\prod_{i=1}^n f_\theta(x_i) &amp; = \left(\frac \theta 2\right)^n ((\theta x_1) \cdots (\theta x_n))^2 \exp\{−\theta (x_1+\cdots+x_n)\} \\[10pt]
&amp; = \Big(\, \overbrace{\theta^{3n} \exp(-\theta(x_1+\cdots + x_n))}^\text{first factor}\,\Big) \cdot \Big(\, \overbrace{ \frac{x_1\cdots x_n}{2^n}}^\text{second factor} \, \Big) \tag 1
\end{align}

Fisher's factorization theorem says that if this can be written as a product of two factors in which


one factor depends on $x_1,\ldots,x_n$ only through some specified function of $x_1,\ldots,x_n$ and
the other factor does not depend on $\theta,$


then that specified function, evaluated at the sample, is a sufficient statistic.

Therefore $X_1+\cdots+X_n$ is a sufficient statistic.

If $L(\theta) = $ the expression on line $(1)$ above, then
$$
\ell(\theta) = \log L(\theta) = 3n\log\theta - \theta(x_1+\cdots+x_n) + \text{constant}
$$
so
$$
\ell\,'(\theta) = \frac{3n}\theta - (x_1+\cdots+x_n) \quad \begin{cases} &gt;0 &amp; \text{if } 0 \le \theta &lt; 3/\overline{x}, \\[6pt] &lt;0 &amp; \text{if } \theta&gt;3/\overline{x}. \end{cases}
$$
Therefore $\widehat \theta = 3/\overline x.$

By equivariance of MLEs, the MLE of $3/\theta$ is $\overline x.$