Question

4. (10pts) Let X 1 ​ ,X 2 ​ ,…,X n ​ denote a random sample of size n>1 from a distribution with pdf f(x;θ)=6θe −6θx ,00. (1) (5pts) Show that Y=∑ i=1 n ​ X i ​ is a complete and sufficient statistic for θ. (2) (5pts) Find the MVUE of θ. Justify your answer.

Answer 1

(1) The random variable
`\(Y = \sum_(i=1)^(n)X_i\)`is a complete and sufficient statistic for the parameter \(\theta\) in the given distribution. (2) The Minimum Variance Unbiased Estimator (MVUE) of
`\(\theta\) is \(\frac{1}{6\bar{X}}\`), where \(\bar{X}\) is the sample mean.

Step-by-step explanation:

(1) To show that Y is a complete and sufficient statistic, we need to demonstrate that the likelihood function
`\(f(\mathbf{X}|\theta)\)`factors into a product that is a function of the data only through the statistic
`\(Y = \sum_(i=1)^(n)X_i\),` and that
`\(E[g(Y)] = 0\)` implies
`\(P[g(Y) = 0] = 1\)` for all functions g. Given the distribution
`\(f(x;\theta) = 6\theta e^(-6\theta x)\)`, the joint probability density function for the sample is
`\(f(\mathbf{X}|\theta) = \prod_(i=1)^(n)6\theta e^(-6\theta x_i)\)`. Simplifying this expression and applying the factorization theorem confirms that Y is a complete and sufficient statistic.

(2) To find the MVUE of
`\(\theta\)`, we can use the Lehmann-Scheffé Theorem, which states that if Tis a complete and sufficient statistic for
`\(\theta\)` and there exists an unbiased estimator
`\(U(\mathbf{X})\)` for
`\(\tau(\theta)\)`, then
`\(U(\mathbf{X})\)`is the unique MVUE of
`\(\tau(\theta)\)`. In this case,
`\(\bar{X}\)`is the unbiased estimator for \(\frac{1}{\theta}\), so the MVUE for
`\(\theta\) is \(\frac{1}{6\bar{X}}\).`