# Weak Convergence with Uniform Distribution $U[0; \theta]$ and Method of Moments

We have $(X_1, ... , X_n)$ an n-sample of a uniform distribution $U[0; \theta]$, where $\theta > 0$ and moment estimate $\widehat{\theta}_n$ of $\theta$. \begin{eqnarray*} \widehat{\theta}_n &=& 2 \overline{X} \end{eqnarray*}

What can I say about weak convergence (in law) of $\sqrt{n}(\widehat{\theta}_n-\theta)$

In the case of Maximum likelihood estimator gives $\tilde{\theta}_n$ if regular dominated model

$$\sqrt{n}(\tilde{\theta}_n-\theta)\xrightarrow[n\rightarrow\infty]{(l)} > N(0,I_{\theta}^{-1})$$

Thus

$$\sqrt{n}I_{\theta}^{1/2}(\tilde{\theta}_n-\theta)\sim N(0,I_{\theta})$$

and

$$n(\tilde{\theta}_n-\theta)^T > I_{\tilde{\theta}}(\tilde{\theta}_n-\theta)\sim \chi^2(k)$$

So we have an approximate $1-\alpha$ confidence region for

$$\big\lbrace \theta: (\theta-\tilde{\theta}_n)^T > I_{\tilde{\theta}}(\theta-\tilde{\theta}_n) \leq \frac{\chi^2_{k,\alpha}}{n} \big\rbrace$$

• So in summary my question is what is the analysis in the case of estimator of the method of moments.

• Regardless of the distribution, as long as the 2nd moment is finite (which it is in this case), you know the weak convergence of $\bar{X}$ from Central Limit Theorem. Then use the fact that $\hat{\theta}$ is a linear transformation of $\bar{X}$. Commented May 15, 2017 at 22:01
• By the central limit theorem, $$\sqrt{n}\left(\bar{X} - \dfrac{\theta}{2}\right)\overset{d}{\to}\mathcal{N}\left(0, \dfrac{\theta^2}{12} \right)\text{.}$$ Commented May 15, 2017 at 22:43

Note that the asymptotic normality holds only for regular cases, where one of the regularity conditions is that the support of the r.v $X$ is independent of the (unknown) parameter of interest. In the uniform case it isn't true and Fisher's information $I_{\theta}$ is basically irrelevant.
Recall that $\hat{\theta}_n = X_{(n)}$, thus $$F_{X_{(n)}}(y) = P( X_{(n)} \le y) = (F_X(y) )^n = (y/\theta)^n,$$
thus $$P(n(\theta-X_{(n)})\le y)= 1 - P(X_{(n)} < \theta - y/n)=1-(1-y/(n\theta))^n,$$ that is $n(\theta-X_{(n)})\xrightarrow{D} EXP(1)$. Where for any $0\le p<1$, $n^p (\theta - X_{(n)}) \xrightarrow{D} 0$.
For the method of moments estimator, just use the CLT, i.e., $$\sqrt{n}(2\bar{X}_n - \theta) \xrightarrow{D} N(0, \theta^2/3).$$
• No difference - If you divide the LHS by $2$ then you'll get his expression. Commented May 16, 2017 at 19:06
• But you can say that $\frac{\partial F(y)}{\partial y}=f(y)=(\frac{n}{\theta})(\frac{y}{\theta})^{n-1}$ so you can get Fisher's information $I_{\theta}=-E_{\theta}[\frac{\partial^2 \ln f(y| \theta)}{\partial \theta^2 }]$ Commented May 18, 2017 at 17:52
• Yes, but it should equal $var(\partial / \partial \theta \ln f(\theta;x))$ that is $0$ in this case (because Uniform dist. is "irregular"). Commented May 18, 2017 at 19:31
• It means that $\sqrt{n}(\tilde{\theta}_n-\theta)\xrightarrow[n\rightarrow\infty]{(l)} 0$. Can you suggest any reference to go more deeper? Commented May 18, 2017 at 20:37