Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n\) from a distribution with pdf \(f(x ; \theta)=\theta x^{\theta-1}, 00\). (a) Show that the geometric mean \(\left(X_{1} X_{2} \cdots X_{n}\right)^{1 / n}\) of the sample is a complete sufficient statistic for \(\theta\). (b) Find the maximum likelihood estimator of \(\theta\), and observe that it is a function of this geometric mean.

Short answer

The geometric mean \((X_{1} X_{2} \cdots X_{n})^{1/n}\) is a complete sufficient statistic for \(\theta\). The maximum likelihood estimator of \(\theta\) is given by \(\theta = \frac{-n}{{\sum_{i=1}^{n}\log(\frac{1}{x_i})}}.\) The maximum likelihood estimator of \(\theta\) is a function of the geometric mean because the expression of MLE contains \(\log(x_i)\) which corresponds to the logarithm of the geometric mean.

Step-by-step solution

Compute the likelihood function

The likelihood function is given by\[ L(\theta; x) = (x_1\cdot x_2 \cdot...\cdot x_n)^{\theta-1}\cdot\theta^n \]since \( f(x ; \theta)=\theta x^{\theta-1}\) is the pdf.

Apply a logarithm to the likelihood function

By applying a logarithm, we get \[ \log L(\theta; x) = (\theta - 1) \cdot \sum_{i=1}^{n}\log(x_i) + n\log(\theta) \] The reason for taking a log is that it simplifies the algebraic manipulations.

Simplify the result

For \( \frac{\partial}{\partial\theta}\log L(\theta; x) = 0 \), we get \[ \theta = \frac{-n}{\sum_{i=1}^{n}\log(\frac{1}{x_i})} This gives us the maximum likelihood estimator of \theta.

Show the statistic is sufficient

The factorization theorem states that a statistic T(X) is sufficient for θ if and only if nonzero constants g and h exist such that the joint pdf or pmf of the sample observations can be expressed as\[ g(t(x), \theta) h(x) \]This can be obtained from our likelihood function. So, the geometric mean \((X_{1} X_{2} \cdots X_{n})^{1/n}\) is a complete sufficient statistic for \(\theta\) as it satisfies the factorization theorem.