Categories

## Maximum Likelihood Estimation | ISI MStat 2017 PSB Problem 8

This problem based on Maximum Likelihood Estimation, gives a detailed solution to ISI M.Stat 2017 PSB Problem 8, with a tinge of simulation and code.

## Problem

Let $\theta>0$ be an unknown parameter, and $X_{1}, X_{2}, \ldots, X_{n}$ be a random sample from the distribution with density.

$f(x) = \begin{cases} \frac{2x}{\theta^{2}} & , 0 \leq x \leq \theta \\ 0 & , \text { otherwise } \end{cases}$

Find the maximum likelihood estimator of $\theta$ and its mean squared error.

### Prerequisites

• Proof algorithm to find the MLE of $\theta$ for U$(0, \theta)$
• Order Statistics $X_{(1)}, X_{(2)}, \ldots, X_{(n)}$
• Mean Square Error

## Solution

Do you remember the method of finding the MLE of $\theta$ for U$(0, \theta)$ ? Just proceed along a similar line.

$L(\theta) = f\left(x_{1}, \cdots, x_{n} | \theta\right) \overset{ X_{1}, X_{2}, \ldots, X_{n} \text{are iid}}{=}f\left(x_{1} | \theta\right) \cdots f\left(x_{n} | \theta\right) \\ = \begin{cases} \frac{ 2^n \prod_{i=1}^{\infty} X_{i}}{ \theta^{2n}} & , 0 \leq X_{(1)} \leq X_{(2)} \leq \ldots \leq X_{(n)} \leq \theta \\ 0 & , \text { otherwise } \end{cases}$

Let’s draw the diagram.

Thus, you can see that $L(\theta)$ is maximized at $\theta = X_{(n)}$.

Hence, $\hat{\theta}_{mle} = X_{(n)}$.

#### MSE

Now, we need to find the distribution of $X_{(n)}$.

For, that we need to find the distribution function of $X_i$.

Observe $F_{X_i}(x) = \begin{cases} \frac{x^2}{\theta^{2}} & , 0 \leq x \leq \theta \\ 0 & , \text { otherwise } \end{cases}$

$F_{X_{(n)}}(x) \overset{\text{Order Statistics}}{=} \begin{cases} 0 &, x \leq 0 \\ \frac{x^{2n}}{\theta^{2n}} & , 0 \leq x \leq \theta \\ 1 & , \text { otherwise } \end{cases}$

$f_{X_{(n)}}(x) = \begin{cases} \frac{2n.x^{2n-1}}{\theta^{2n}} & , 0 \leq x \leq \theta \\ 0 & , \text { otherwise } \end{cases}$

MSE($X_{(n)}$) = E$((X_{(n)} – \theta)^2)$

= $\int_{0}^{\theta} (x-\theta)^2 f_{X_{(n)}}(x) dx$

= $\int_{0}^{\theta} (x-\theta)^2 \frac{2n.x^{2n-1}}{\theta^{2n}} dx$

= $\int_{0}^{\theta} (x^2 + {\theta}^2 – 2x\theta) \frac{2n.x^{2n-1}}{\theta^{2n}} dx$

= $\int_{0}^{\theta} \frac{2n.x^{2n+1}}{\theta^{2n}} dx$ + $\int_{0}^{\theta} \frac{2n.x^{2n-1}}{\theta^{2n-2}} dx$ – $\int_{0}^{\theta} \frac{4n.x^{2n}}{\theta^{2n-1}} dx$

= ${\theta}^2(\frac{2n}{2n+2} + 1 – \frac{4n}{2n+1}) = \frac{1}{(2n+1)(n+1)}$

Observe that $\lim_{ n \to \infty} { MSE( X_{(n)})} = 0$.

### Let’s add a computing dimension to it and verify it by simulation.

Let’s take $\theta = 1, n = 5$. MSE is expected to be around 0.002. You can change the $\theta$ and n and play around.

v = NULL
n = 15
theta = 1
for (i in 1:1000) {
r = runif(n, 0, theta)
s = theta*sqrt(r) #We use Inverse Transformation Method to generate the random variables from the distribution.
m = max(s)
v = c(v,m)
}
hist(v, freq = FALSE)
k = replicate(1000,1)
mse(v,k) =  0.001959095

You should also check out this link: Triangle Inequality Problems and Solutions

I hope that helps you. Stay tuned.