Let $X_1,...,X_n$ a i.i.d from a population with distribution $U[0,\theta]$, i.e.,

$f_{X_i}(x)=\frac{1}{\theta}g_{[0,\theta]}(x)$, for $i=1, \ldots, n$


\begin{align} g_{[0,\theta]}(x) = \begin{cases} 1 & x \in [0, \theta], \\ 0 & x \notin [0, \theta]. \end{cases} \end{align}

How to prove that $\hat\theta = \max\{X_1, \ldots, X_n\}$ is a root mean square consistent estimator for $\theta$?

I know that an estimator $\hat\theta_n$ is consistent in root mean square if it fulfills that

$$\lim\limits_{n \to \infty}MSE(\hat\theta_n)=0.$$

with $MSE(\hat\theta_n)=E[(\hat\theta_n-\theta)^2]=\operatorname{Var}(\hat\theta_n)+(E(\hat\theta_n)-\theta)^2$.

  • $\begingroup$ What have you tried? $\endgroup$
    – Glen_b
    Commented Apr 4, 2023 at 0:03
  • $\begingroup$ I tried to show consistency proving that $E(\hat\theta)=\theta$ and $Var(\hat\theta)=0$ but I'm not sure if that's enough to prove root mean square consistency $\endgroup$ Commented Apr 4, 2023 at 0:08
  • 1
    $\begingroup$ @WillowDouglas What you "tried to show" is impossible to hold (which would imply $\hat{\theta} \equiv \theta$). The correct procedure obviously relies on "$\lim_n$". $\endgroup$
    – Zhanxiong
    Commented Apr 4, 2023 at 0:41
  • $\begingroup$ @Zhanxiong, my guess is to find the bias from the estimator (by finding $E(\hat\theta_n)$) and $Var(\hat\theta_n)$. But first you have to find the CDF and the density function of the estimator. $\endgroup$ Commented Apr 4, 2023 at 1:41
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    $\begingroup$ @WillowDouglas Exactly, so can you find the CDF of $X_{(n)} := \max(X_1, \ldots, X_n)$? It is a pretty standard result. $\endgroup$
    – Zhanxiong
    Commented Apr 4, 2023 at 1:45

1 Answer 1


The standard way is to find the CDF of $\hat{\theta}_n$ first, which is \begin{align*} F(x) := P(\max(X_1, \ldots, X_n) \leq x) = \prod_{i = 1}^n \frac{x}{\theta} = \frac{x^n}{\theta^n}, \; 0 \leq x \leq \theta. \end{align*} This implies the PDF of $\hat{\theta}_n$ is $f(x) = nx^{n - 1}/\theta^n$ for $0 \leq x \leq \theta$ and $0$ otherwise, whence \begin{align*} E[(\hat{\theta}_n - \theta)^2] = \int_0^\theta(t - \theta)^2f(t)dt = \frac{n}{\theta^n}\int_0^\theta (t - \theta)^2t^{n - 1}dt. \tag{1} \end{align*} To evaluate the integral $(1)$, apply the variable substitution $u = t/\theta$, which transfers $(1)$ to \begin{align*} n\int_0^1\theta^2(1 - u)^2 u^{n - 1} du = \theta^2nB(3, n), \tag{2} \end{align*} where $B(a, b)$ is the Beta function. Since $B(a, b) = (a - 1)!(b - 1)!/(a + b - 1)!$ when $a, b$ are positive integers, the right-hand side of $(2)$ equals to \begin{align} \theta^2 \frac{2n(n - 1)! }{(n + 2)!} = \frac{2\theta^2}{(n + 2)(n + 1)}, \end{align} which converges to $0$ as $n \to \infty$. This shows that $\hat{\theta}_n \to \theta$ in quadratic mean.

The above method is straightforward. However, it involves pretty heavy calculations (you may circumvent summoning the Beta function by expanding $(1 - u)^2$ then applying linearity of the integral, which requires less machinery). This motivates me to consider the following alternative solution.

First, show that $\hat{\theta}_n$ converges to $\theta$ in probability. I will leave this to you as an exercise. It is similar to finding the CDF of $\hat{\theta}_n$.

Next, for any $\epsilon \in (0, \theta)$, we have
\begin{align*} & E[(\hat{\theta}_n - \theta)^2] = E\left[(\hat{\theta}_n - \theta)^2I_{[|\hat{\theta}_n - \theta| \leq \epsilon]}(\omega)\right] + E\left[(\hat{\theta}_n - \theta)^2I_{[|\hat{\theta}_n - \theta| > \epsilon]}(\omega)\right] \\ \leq & \epsilon^2 + 4\theta^2P[|\hat{\theta}_n - \theta| > \epsilon]. \tag{3} \end{align*} Since $\hat{\theta}_n$ converges to $\theta$ in probability, the second term in $(3)$ converges to $0$ as $n \to \infty$. Since $\epsilon$ is arbitrary, this implies that $E[(\hat{\theta}_n - \theta)^2] \to 0$ as $n \to \infty$, completing the proof.

  • $\begingroup$ Well if you show convergence in probability then by Vitali’s convergence theorem you have $L^p$ convergence since the sequence of random variables $\hat{\theta}_n$ is uniformly integrable (in fact, its bounded). I guess your last step is encapsulating part of the proof of that theorem. $\endgroup$ Commented Apr 4, 2023 at 22:14

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