91Ó°ÊÓ

The geometric mean of the sample is given by \(g(X) = \left(X_{1} X_{2} \cdots X_{n}\right)^{1/n}\).

Write down the joint probability density function (pdf) of \(n\) samples which is \( f(\mathbf{x} ; \theta)=\theta^n \left(\prod_{i=1}^{n} x_{i}\right)^{\theta-1} \) , here \(\prod\) denotes the product of the \(x\) under the terms of the product.

Check Factorization criterion for sufficiency. Here clearly, the part of \(f(\mathbf{x}; \theta)\) that depends on \(\theta\) is a function of \(g(x) = \left(X_{1} X_{2} \cdots X_{n}\right)^{1 / n}\). So, \(g(X)\) is a sufficient statistic for \(\theta\). Now, we have to prove it is complete.

The tricky step is to prove that this statistic is complete. Because \(\theta>0\), we consider the family of distributions of \(g(X)\) parameterised by \(\theta\), as \(g(X)\) is a measurable function from \(\mathbb R^n\) to \(\mathbb R\). Therefore, if \(E[g(X)] = 0\) for all \(\theta > 0\), the definition of completeness will provide us the fact that \(g(X)\) is always equal to \(0\). Hence the geometric mean of the sample is a complete and sufficient statistic for \(\theta\).

Here you should write down the likelihood function using the joint pdf of \(n\) samples. The likelihood function is \( L(\theta)=\theta^n \left(\prod_{i=1}^{n} x_{i}\right)^{\theta-1} = \theta^n e^{(\theta-1)\sum_{i=1}^{n} \log{x_{i}}}\). The log-likelihood function is obtained by taking natural logarithm of \(L(\theta)\) that provides: \(\log{L(\theta)} = n\log{\theta} + (\theta-1)\sum_{i=1}^{n} \log{x_{i}}\). Now to maximize the likelihood, take the partial derivative of the log-likelihood with respect to \(\theta\), set it to zero and solve for \(\theta\). Doing so gives us \(\hat{\theta}=\frac{n}{ \sum_{i=1}^{n} \ln{x_{i}}}\). We can observe that \(\hat{\theta}\), the MLE, can be expressed as a function of the geometric mean \(g(X)\).