1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Ian Dec 5 '16 at 20:32
1
$\begingroup$

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 $$ \begin{align} \lim_{n\rightarrow \infty}\mbox{Var }W_2 = \lim_{n\rightarrow \infty} n \mbox{Var } \hat{\theta} = 0. \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.