# Method of moments estimate of Pareto Distribution

The Pareto distribution has the following $$cumulative \ distribution \ function$$ : $$F(x;\alpha ,\Theta ) = \left\{\begin{matrix} 1 - (\frac{\alpha}{x})^{\theta}\ \ if \ \alpha \leq x\ & \\ 0 \ \ \ \ \ \ \ \ otherwise & \end{matrix}\right.$$

where $$0< \alpha<\infty$$ and $$1< \theta<\infty$$ are the parameters. Using moments method, find the estimates of $$\alpha$$ and $$\theta$$ based on a sample of size $$5$$ for value $$3,5,2,7 \ and \ 8.$$

To solve this question I began by determining the expectation using the CDF.

I found that you can compute the expectation directly from CDF when integrating as follows link

$$E(X) = \int \left( 1 - F_X(x) \right) \,\mathrm{d}x$$

$$= \int_\alpha^\infty \left( (\frac{\alpha}{x})^{\theta} \right) \,\mathrm{d}x$$

$$= \frac{\alpha}{\theta -1}$$

$$= \overline{X} = 5$$

How do I proceed from here?

• Please add the self-study tag and read its wiki. Commented Sep 10, 2021 at 10:33

The formula you use for the expectation holds for any non-negative valued random variable. However, if $$X$$ has the Pareto distribution in your post, then it is $$X-\alpha$$ and not $$X$$ that is non-negative.

The general formula changes as $$E(X-\alpha)I_{X\ge\alpha}=\int_{\alpha}^\infty (1-F(x))\,dx$$

Therefore, in this problem,

$$E(X)=E(XI_{X\ge \alpha})=\alpha+\int_{\alpha}^\infty (1-F(x))\,dx=\frac{\theta\alpha}{\theta-1} \qquad\left[\because\, \theta>1 \right]$$

Suppose $$X_1,X_2,\ldots,X_n$$ are i.i.d with the given cdf $$F$$.

The first moment equation is $$\frac1n\sum\limits_{i=1}^n X_i=\frac{\theta\alpha}{\theta-1} \tag{1}$$

You need another equation to solve for the two unknown parameters. This is provided by the second moment $$E(X^2)$$, but the second moment is finite provided $$\theta>2$$. As you do not have this assumption, one option is to find $$E\left(\frac1{X}\right)$$ using the pdf of $$X$$:

$$E\left(\frac1{X}\right)=\int_{\alpha}^\infty \frac1x\cdot\frac{\theta \alpha^\theta}{x^{\theta+1}}\,dx=\frac{\theta}{(\theta+1)\alpha}\qquad\left[\because\, \theta+1>0 \right]$$

The corresponding moment equation is

$$\frac1n\sum\limits_{i=1}^n \frac1{X_i}=\frac{\theta}{(\theta+1)\alpha} \tag{2}$$

Now you can eliminate $$\alpha$$ using $$(1)$$ and $$(2)$$ and solve for the method of moment estimator $$\hat\theta$$ of $$\theta$$. The corresponding estimator of $$\alpha$$ can then be obtained in terms of $$\hat\theta$$ from $$(1)$$. It is easy to verify that both estimators are consistent. If it was given that $$\theta>2$$, then a similar approach would have worked using the second moment.

Note that this is just one possible solution (see How to use method of moment to find Pareto distribution estimator?). In Johnson-Kotz-Balakrishnan's Continuous Univariate Distributions (Volume 1, 2nd ed.), they discuss a solution due to Quandt (1966) where consistent method of moment estimators are obtained using the first moment and expectation of the first order statistic $$X_{(1)}$$, under the only assumption $$\theta>1$$.