Question

Let X₁ ,X₂ ,…, Xₙ constitute a random sample from a U[0,θ] distribution. (a) Calculate the MSE when the sample mean is used to estimate θ. (b) Suppose there is another estimator, ϕ(x)=2 xˉ. Find the MSE of this estimation (c) Which of these two estimators would you choose to estimate θ and why?

Answer 1

(a) The mean squared error (MSE) when using the sample mean
`(\(\overline{X}\))` to estimate
`\(\theta\) is \((\theta^2)/(12n)\)`.

(b) The MSE for the estimator
`\(\phi(x) = 2\overline{X}\) is \((\theta^2)/(3n)\)`.

(c) The estimator
`\(\phi(x) = 2\overline{X}\)` is preferred as it has a smaller MSE, providing a more efficient estimate of
`\(\theta\)`.

Step-by-step explanation:

(a) To calculate the MSE for the sample mean
`(\(\overline{X}\))`, we use the formula:

`\[MSE(\overline{X}) = E\left[(\overline{X} - \theta)^2\right]\]`

Given that
`\(\overline{X}\) follows a U[0, \(\theta\)]` distribution, the variance of
`\(\overline{X}\)` is
`\((\theta^2)/(12n)\)`. Therefore, the MSE is
`\((\theta^2)/(12n)\)`.

(b) For the estimator
`\(\phi(x) = 2\overline{X}\)`, we calculate its MSE:

`\[MSE(\phi) = E\left[(2\overline{X} - \theta)^2\right]\]`

Since the variance of
`\(2\overline{X}\) is \((\theta^2)/(3n)\), the MSE is \((\theta^2)/(3n)\)`.

(c) Comparing the MSE values, we find that the MSE for
`\(\phi(x) = 2\overline{X}\)` is smaller than that for
`\(\overline{X}\)`. A smaller MSE indicates a more efficient estimator. Therefore,
`\(\phi(x) = 2\overline{X}\)` is preferred for estimating
`\(\theta\)` due to its lower MSE, implying better precision and accuracy in estimation.