# MLE of $\theta$, where $\text{Unif}([0, \theta])$

Let $X_1,...,X_n$ be a sample from $\text{Unif}([0, \theta])$ where $\theta> 0$ is the unknown parametr.

(a) Find $\hat{\theta}_n$ the MLE of $\theta$.

I've received that $\hat{\theta}_n=\max\{X_1,...X_n\}$.

(b) Find limiting distribution of following sequences $n(\hat{\theta}_n-\theta)$ and $\sqrt{n}(\hat{\theta}_n-\theta)$.

Let $Y=\max\{X_1,...,X_n\}$, then:

$$g_{Y,\theta}(t)=\frac{n}{\theta^n}t^ {n-1}\mathbb{I}_{[0,\theta]}(t)$$ Let $W=n(\theta-Y)$, then:

$$g_{W,\theta}(t)=\frac{1}{n^{n-1}\theta^n}(n\theta-t)^{n-1}\mathbb{I}_{[0,n\theta]}(t)$$

$$\frac{1}{n^{n-1}\theta^n}(n\theta-t)^{n-1}\mathbb{I}_{[0,n\theta]}(t)=\frac{1}{\theta}(1-\frac{t}{n\theta})^{n-1}\mathbb{I}_{[0,n\theta]}(t)\longrightarrow_{n \rightarrow \infty}\frac{1}{\theta}\exp(-\frac{t}{\theta})$$.

So the first sequence has limiting distribution $\text{Exp}(\frac{1}{\theta})$.

But I don't know what I can do to find the limiting distribution of the second sequence. Could someone give some hint?

• Minor correction: $\hat{\theta}_n$ is always less than $\theta$, so the negative of the first thing would have an exponential distribution, by your argument. Anyway, can you at least write down the PDF in the second case? – Ian Dec 5 '16 at 20:32

## 1 Answer

As has been pointed out as it stands your statistic $n(\hat{\theta}-\theta)$ is negative but that is ok, let \begin{align} W_1 &=n(\hat{\theta}-\theta), \\ W_2 &= \sqrt{n}(\hat{\theta}-\theta). \end{align} You have already given a derivation in the first case using the probability mass function, but working with the cumulative distribution then let $T_{n}(X) = \hat{\theta} = \max(X_1,\ldots,X_n)$, \begin{align} \mathbb{P}\left[ T_n(X) \leq x\right] = \left( \mathbb{P}\left[ X_1 \leq x\right]\right)^n = \left( \frac{x}{\theta}\right)^n. \end{align} So let $w \leq 0$ and we consider \begin{align} \mathbb{P} \left[ W_1 \leq w \right] &= \mathbb{P}\left[ n(T_n(x)-\theta) \leq w\right] \\ &= \mathbb{P}\left[ T_n(x) \leq \theta + \frac{w}{n}\right] \\ &= \left(\frac{\theta + \frac{w}{n}}{\theta} \right)^n \\ &= \left( 1 + \frac{w}{\theta}\frac{1}{n} \right)^n \end{align} So taking the limit we have \begin{align} \mathbb{P}\left[ W_1 \leq w\right] = e^{\frac{w}{\theta}}, \; w <= 0. \end{align} which agrees with your approach using the mass function. Now applying the same methodology as before we get the limit for $x < 0$ we have \begin{align} \lim_{n\rightarrow \infty}\left(1 + \frac{x}{\sqrt{n}}\right)^n = 0. \end{align} Another way of thinking about this is that using your pmf you can show that \begin{align} \mbox{Var }\hat{\theta} = \frac{\theta^2 n}{(n+1)^2(n+2)} \approx \theta^2 \frac{1}{n^2} \end{align} So taking the second statistic we have \begin{align} \lim_{n\rightarrow \infty}\mbox{Var }W_2 = \lim_{n\rightarrow \infty} n \mbox{Var } \hat{\theta} = 0. \end{align}