Suppose I have data $\{X_i\}_{i=1}^{n}\sim \text{Uniform}[0,\theta_0]$. We know that the maximum likelihood estimator of $\theta_0$ is $\widehat{\theta}=\max\{X_1,...,X_n\}$. How to show that $\widehat{\theta}$ converges to $\theta_0$ at rate $n^{-1}$? Thanks!


1 Answer 1


When $(X_i)$ is a sequence of uniformly distributed variables $\mathcal{U}_{[0,\theta_0]}$ we can derive the distribution of the MLE $\hat \theta = \max X_i$.
We will show that $n( \theta_0 - \hat \theta)$ converge towards a non trivial (i.e. not $0$) distribution, which means that the rate of convergence of the MLE is $n^{-1}$.

First note that since $\max X_i \leq \theta_0$, $n( \theta_0 - \hat \theta) \geq 0$.

Then, for $t \geq 0$,

\begin{align*} \mathbb P\left( n(\theta_0 -\hat \theta) \leq t \right) &=1- \mathbb P\left( \hat \theta \leq \theta_0 - \frac{t}{n} \right) \\ &=1-\mathbb P\left( \max X_i \leq \theta_0 - \frac{t}{n} \right) \\ \end{align*}

Since $\max X_i \leq \theta_0 - \frac{t}{n} \iff \forall i, X_i \leq \theta_0 - \frac{t}{n}$ and since all $X_i$ are independent and identically distributed, \begin{align*} \mathbb P\left( \max X_i \leq \theta_0 - \frac{t}{n} \right) &=\prod_{i=1}^n\mathbb P\left( X_i \leq \theta_0 - \frac{t}{n} \right) \\ &=\mathbb P\left( X_1 \leq \theta_0 - \frac{t}{n} \right)^n \end{align*}

For $n$ large enough we have $\frac{t}{n} \leq \theta_0$ and then, \begin{align*} \mathbb P\left( X_1 \leq \theta_0 - \frac{t}{n} \right)^n &= \left(\frac{\theta_0 - \frac{t}{n} }{\theta_0} \right)^n \\ &=\left(1 - \frac{t}{\theta_0 n}\right)^n \end{align*}

As $n \to \infty$ the last line is equivalent to $\exp\left(-\frac{t}{\theta_0}\right).$

So for $t \geq 0$: $$ \mathbb P\left( n(\theta_0 -\hat \theta) \leq t \right) \xrightarrow[n \to \infty]{} 1- \exp\left(-\frac{t}{\theta_0}\right). $$

We can see that $1-\exp\left(-\frac{t}{\theta_0}\right)$ is the cumulative distribution function of an exponential random variable with rate $\frac{1}{\theta_0}$.

Thus, $n (\theta_0 - \hat \theta )$ is asymptotically exponentially distributed, and the rate of convergence of $\hat \theta$ is $n^{-1}$.

Here is a R code to illustrate this result (with $\theta_0 = 3$ and $n=50$):

#P_emp is an empirical cumulative distribution of n*(theta0 - MLE)
#P_emp should be close to the CDF of an exponential distribution

x<-seq(0,20,length.out = 1000)
y1<-sapply(x,function(t) P_emp(t))

legend(x=0,y=1,legend = c("Empirical CDF","Exponential 
  • $\begingroup$ Thanks! This is completely solved my problem! If I change the distribution to other distributions on $[0,\theta_0]$ and want to estimate the boundary ($\theta_0$) using $\underset{i}{max}\{X_i\}$, does the rate $n^{-1}$ still hold? $\endgroup$ Jul 11, 2020 at 6:40
  • 2
    $\begingroup$ @JTS365 No I don't think it works for any distribution over $[0, \theta_0]$. Take for example $\theta_0 =1$ and the distribution with density $f(x) = 2-2x$ over $[0,1]$, then $\forall t$, $\mathbb P( n(1- \max X_i) \leq t) \to 0$ which means that $n(1-\max X_i) \to \infty$. In that case the rate is $n^{-\frac{1}{2}}$ $\endgroup$
    – periwinkle
    Jul 11, 2020 at 14:34
  • $\begingroup$ Thanks! Does the result generalize to bivariate uniform distribution on a rectangle $[0,\theta_0]\times [0,\theta_1]$? $\endgroup$ Jul 16, 2020 at 23:41
  • $\begingroup$ @JTS365 Yes it does, if $(X_0,X_1) \sim U_{[0,\theta_0]\times [0,\theta1]}$ then $X_0 \perp X_1$ and $X_i \sim \mathcal U_{[0,\theta_i]}$ thus $\hat \theta_0 = \max X_{0,i}$ and $\hat \theta_1 = \max X_{1,i}$ and the rate for both estimators is $n^{-1}$. $\endgroup$
    – periwinkle
    Jul 17, 2020 at 22:05

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