Question

The method of moment estimator of parameter '$\alpha$' in the following probability density function $f_{X}(x)=\begin{cases}\frac{\alpha}{x^{\alpha+1}}, & x>1, \alpha>1 \\0 & \text{otherwise}\end{cases}$ is

• A. $\frac{1}{\overline{x}-1}$
• B. $\frac{\overline{x}}{\overline{x}-1}$
• C. $\frac{\overline{x}-1}{\overline{x}}$
• D. $\overline{x}-1$

Hint:

For Pareto-like distributions, the mean is $\frac{\alpha}{\alpha-1}$. If the mean is $\mu$, then $\alpha = \frac{\mu}{\mu-1}$. Just replace $\mu$ with the sample mean $\overline{x}$ to get the MoM estimator instantly.

Answer 1

The Correct Option is B

We use the Method of Moments (MoM) by equating the population mean to the sample mean.

Step 1: \color{redCalculate the Population Mean $E(X)$
$E(X) = \int_{1}^{\infty} x \cdot f(x) dx = \int_{1}^{\infty} x \cdot \frac{\alpha}{x^{\alpha+1}} dx$
$E(X) = \alpha \int_{1}^{\infty} x^{-\alpha} dx$
Evaluating the integral:
$E(X) = \alpha \left[ \frac{x^{-\alpha+1}}{-\alpha+1} \right]_{1}^{\infty}$
Since $\alpha > 1$, as $x \to \infty$, $x^{1-\alpha} \to 0$.
$E(X) = \alpha \left[ 0 - \frac{1}{1-\alpha} \right] = \frac{\alpha}{\alpha-1}$.

Step 2: \color{redEquate to Sample Mean
Set $E(X) = \overline{x}$:
$\frac{\alpha}{\alpha-1} = \overline{x}$.

Step 3: \color{redSolve for $\alpha$
$\alpha = \overline{x}(\alpha - 1)$
$\alpha = \overline{x}\alpha - \overline{x}$
$\overline{x} = \overline{x}\alpha - \alpha$
$\overline{x} = \alpha(\overline{x} - 1)$
$\hat{\alpha} = \frac{\overline{x}}{\overline{x}-1}$.
Thus, the Method of Moment estimator is $\frac{\overline{x}}{\overline{x}-1}$.