Finding MLE and MSE of $\theta$ where $f_X(x\mid\theta)=\theta x^{−2} I_{x\geq\theta}(x)$

Consider i.i.d random variables $X_1$, $X_2$, . . . , $X_n$ having pdf

$$f_X(x\mid\theta) = \begin{cases} \theta x^{−2} & x\geq\theta \\ 0 &
x\lt\theta \end{cases}$$

where $\theta \gt 0$ and $n\geq 3$

(a) Give the likelihood function, expressing it clearly as a function
of $\theta$

(b) Give the MSE of the MLE of $\theta$

My Attempt:

(a) $$L(\theta\mid\vec{x}) = \begin{cases} \theta^n\left(\prod_{i=1}^n x_i\right)^{−2} & x_{(1)}\geq\theta \\ 0 &
x_{(1)}\lt\theta \end{cases}$$

(b) Clearly the MLE of $\theta$ is $X_{(1)}$. We have

$$\begin{align*}
F_{X_{(1)}}(x)
&=\mathsf P(\text{min}{\{X_1,…,X_n}\}\leq x)\\\\
&=1-\mathsf P(\text{min}{\{X_1,…,X_n}\}\gt x)\\\\
&=1-\left(1-F_X(x)\right)^n\\\\
&=1-\left(\frac{\theta}{x}\right)^n
\end{align*}$$

so

$$f_{X_{(1)}}(x)=\left(1-\left(\frac{\theta}{x}\right)^n\right)’=\frac{n\theta^n}{x^{n+1}}I_{[\theta,\infty)}(x)$$

It follows that

$$\begin{align*}
\mathsf E\left(X_{(1)}^2\right)
&=\int_{\theta}^{\infty}\frac{n\theta^n}{x^{n-1}}dx\\\\
&=n\theta^n\left(\frac{x^{-n+2}}{-n+2}\Biggr{|}_{\theta}^{\infty}\right)\\\\
&=\frac{n\theta^2}{n-2}
\end{align*}$$

and

$$\begin{align*}
\mathsf E\left(X_{(1)}\right)
&=\int_{\theta}^{\infty}\frac{n\theta^n}{x^{n}}dx\\\\
&=n\theta^n\left(\frac{x^{-n+1}}{-n+1}\Biggr{|}_{\theta}^{\infty}\right)\\\\
&=\frac{n\theta}{n-1}
\end{align*}$$

so

$$\begin{align*}
\mathsf{Var}\left(X_{(1)}\right)
&=\frac{n\theta^2}{n-2}-\left(\frac{n\theta}{n-1}\right)^2\\\\
&=\theta^2\left(\frac{n}{n-2}-\frac{n^2}{(n-1)^2}\right)
\end{align*}$$

We also have

$$\begin{align*}
\text{bias}^2\left(\hat{\theta}\right)
&=\left(\mathsf E\left(\hat{\theta}\right)-\theta\right)^2\\\\
&=\left(\frac{n\theta}{n-1}-\theta\right)^2\\\\
&=\left(\theta\left(\frac{n}{n-1}-1\right)\right)^2
\end{align*}$$

Finally the MSE is given by

$$\begin{align*}
\mathsf{Var}\left(X_{(1)}\right)+\text{bias}^2\left(\hat{\theta}\right)
&=\theta^2\left(\frac{n}{n-2}-\frac{n^2}{(n-1)^2}\right)+\left(\theta\left(\frac{n}{n-1}-1\right)\right)^2\\\\
&=\theta^2\left(\frac{n}{n-2}-\frac{n^2}{(n-1)^2}+\left(\frac{n}{n-1}-1\right)^2\right)\\\\
&=\frac{2\theta^2}{(n-1)(n-2)}
\end{align*}$$

Are these valid solutions?

Answer

This question is now old enough to give a full succinct solution confirming your calculations. Using standard notation for order statistics, the likelihood function here is:

$$\begin{aligned}
L_\mathbf{x}(\theta)
&= \prod_{i=1}^n f_X(x_i|\theta) \\[6pt]
&= \prod_{i=1}^n \frac{\theta}{x_i^2} \cdot \mathbb{I}(x_i \geqslant) \\[6pt]
&\propto \prod_{i=1}^n \theta \cdot \mathbb{I}(x_i \geqslant \theta) \\[12pt]
&= \theta^n \cdot \mathbb{I}(0 < \theta \leqslant x_{(1)}). \\[6pt]
\end{aligned}$$

This function is strictly increasing over the range $0 < \theta \leqslant x_{(1)}$ so the MLE is:

$$\hat{\theta} = x_{(1)}.$$


Mean-squared-error of MLE: Rather than deriving the distribution of the estimator, it is quicker in this case to derive the distribution of the estimation error. Define the estimation error as $T \equiv \hat{\theta} – \theta$ and note that it has distribution function:

$$\begin{aligned}
F_T(t) \equiv \mathbb{P}(\hat{\theta} – \theta \leqslant t)
&= 1-\mathbb{P}(\hat{\theta} > \theta + t) \\[6pt]
&= 1-\prod_{i=1}^n \mathbb{P}(X_i > \theta + t) \\[6pt]
&= 1-(1-F_X(\theta + t))^n \\[6pt]
&= \begin{cases}
0 & & \text{for } t < 0, \\[6pt]
1 – \Big( \frac{\theta}{\theta + t} \Big)^n & & \text{for } t \geqslant 0. \\[6pt]
\end{cases}
\end{aligned}$$

Thus, the density has support over $t \geqslant 0$, where we have:

$$\begin{aligned}
f_T(t)
\equiv \frac{d F_T}{dt}(t)
&= – n \Big( – \frac{\theta}{(\theta + t)^2} \Big) \Big( \frac{\theta}{\theta + t} \Big)^{n-1} \\[6pt]
&= \frac{n \theta^n}{(\theta + t)^{n+1}}. \\[6pt]
\end{aligned}$$

Assuming that $n>2$, the mean-squared error of the estimator is therefore given by:

$$\begin{aligned}
\text{MSE}(\hat{\theta})
= \mathbb{E}(T^2)
&= \int \limits_0^\infty t^2 \frac{n \theta^n}{(\theta + t)^{n+1}} \ dt \\[6pt]
&= n \theta^n \int \limits_0^\infty \frac{t^2}{(\theta + t)^{n+1}} \ dt \\[6pt]
&= n \theta^n \int \limits_\theta^\infty \frac{(r-\theta)^2}{r^{n+1}} \ dr \\[6pt]
&= n \theta^n \int \limits_\theta^\infty \Big[ r^{-(n-1)} – 2 \theta r^{-n} + \theta^2 r^{-(n+1)} \Big] \ dr \\[6pt]
&= n \theta^n \Bigg[ -\frac{r^{-(n-2)}}{n-2} + \frac{2 \theta r^{-(n-1)}}{n-1} – \frac{\theta^2 r^{-n}}{n} \Bigg]_{r = \theta}^{r \rightarrow \infty} \\[6pt]
&= n \theta^n \Bigg[ \frac{\theta^{-(n-2)}}{n-2} – \frac{2 \theta^{-(n-2)}}{n-1} + \frac{\theta^{-(n-2)}}{n} \Bigg] \\[6pt]
&= n \theta^2 \Bigg[ \frac{1}{n-2} – \frac{2}{n-1} + \frac{1}{n} \Bigg] \\[6pt]
&= \theta^2 \cdot \frac{n(n-1) – 2n(n-2) + (n-1)(n-2)}{(n-1)(n-2)} \\[6pt]
&= \theta^2 \cdot \frac{n^2 – n – 2n^2 + 4n + n^2 – 3n + 2}{(n-1)(n-2)} \\[6pt]
&= \frac{2\theta^2}{(n-1)(n-2)}. \\[6pt]
\end{aligned}$$

Attribution
Source : Link , Question Author : Remy , Answer Author : Ben

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