Uniform distribution parameter estimate

Question

Consider parameter estimate $\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{X_{(n)}}{2}$, where $X_1, X_2, ..., X_n$ independent identically distributed random variables with uniform distribution on the interval $[0, \theta], \theta>0$.

I want to check if the given parameter estimate is unbiased, consistent, strongly consistent.

So, i need to check that:

$$E\left(\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{X_{(n)}}{2}\right) = \theta\\ \lim_{n\to\infty}P\left(\left|\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{X_{(n)}}{2} - \theta \right| \geq \epsilon\right) = 0, \forall\epsilon>0\\ P\left(\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{X_{(n)}}{2} = \theta\right) = 1 $$

To test for unbias, I can use the linearity of the expectation, so $$E(\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{X_{(n)}}{2}) = E(\frac{1}{n}\sum_{i=1}^{n}X_i) + E(\frac{X_{(n)}}{2}) = \theta/2 + \frac{n\theta}{2(n+1)} = \frac{(2n+1)\theta}{2(n+1)} \neq \theta$$ (Just want to check if the solution is correct).

To prove consistency I need to check the second statement. My idea was to use Chebyshev's inequality: $$ \lim_{n\to\infty}P(|\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{X_{(n)}}{2} - \theta| \geq \epsilon) =\\ = \lim_{n\to\infty}P(|\frac{2n + 1}{2(n+1)}\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{2n + 1}{2(n+1)}\frac{X_{(n)}}{2} - \frac{2n + 1}{2(n+1)}\theta| \geq \frac{2n + 1}{2(n+1)}\epsilon) \leq \frac{(\frac{2n + 1}{2(n+1)})^2Var(\frac{1}{n}\sum_{i=1}^{n}X_i + \frac{X_{(n)}}{2})}{\epsilon^2} $$ But I find it difficult to calculate the variance of this estimate. And for the proof of the third statement, I do not have any ideas. I will be extremely grateful for your help or hints.

Answer 1

To show strong consistency, it is enough to argue that

$$\frac1n\sum_{i=1}^n X_i \stackrel{\text{a.s.}}\longrightarrow \frac{\theta}{2} \tag{1}$$

and $$X_{(n)} \stackrel{\text{a.s.}}\longrightarrow \theta \tag{2}$$

This would imply

$$\frac1n\sum_{i=1}^n X_i+\frac12X_{(n)} \stackrel{\text{a.s.}}\longrightarrow \frac{\theta}2+\frac{\theta}2=\theta$$

Strong consistency implies weak consistency.

$(1)$ is of course true by strong law of large numbers.

You only need to prove $(2)$, and check that $Y_n \to Y$ a.s and $Z_n \to Z$ a.s implies $Y_n+Z_n \to Y+Z$ a.s.

For $(2)$, you can skip a direct proof if you want by first showing that $X_{(n)}$ converges in probability to $\theta$ and then use this result.