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

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


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


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

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.

Thanks in advance.

  • 2
    $\begingroup$ 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 $\bar{X}$. $\endgroup$ Commented May 15, 2017 at 22:01
  • 2
    $\begingroup$ By the central limit theorem, $$\sqrt{n}\left(\bar{X} - \dfrac{\theta}{2}\right)\overset{d}{\to}\mathcal{N}\left(0, \dfrac{\theta^2}{12} \right)\text{.}$$ $\endgroup$ Commented May 15, 2017 at 22:43

1 Answer 1


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., $$ \sqrt{n}(2\bar{X}_n - \theta) \xrightarrow{D} N(0, \theta^2/3). $$

  • $\begingroup$ In order to clarify in your answer you explain that the procedure that I follow with Maximum likelihood estimator is wrong because X is not independent of he (unknown) parameter of interest. On other hand I'm more interested the case of method of moments estimator, can you go more deeper, what is the difference with the comment of @clarinetist. $\endgroup$ Commented May 16, 2017 at 17:43
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    $\begingroup$ No difference - If you divide the LHS by $2$ then you'll get his expression. $\endgroup$
    – V. Vancak
    Commented May 16, 2017 at 19:06
  • $\begingroup$ 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 $I_{\theta}=-E_{\theta}[\frac{\partial^2 \ln f(y| \theta)}{\partial \theta^2 }]$ $\endgroup$ Commented May 18, 2017 at 17:52
  • $\begingroup$ Yes, but it should equal $var(\partial / \partial \theta \ln f(\theta;x))$ that is $0$ in this case (because Uniform dist. is "irregular"). $\endgroup$
    – V. Vancak
    Commented May 18, 2017 at 19:31
  • $\begingroup$ It means that $\sqrt{n}(\tilde{\theta}_n-\theta)\xrightarrow[n\rightarrow\infty]{(l)} 0$. Can you suggest any reference to go more deeper? $\endgroup$ Commented May 18, 2017 at 20:37

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