# Verifying the statistics are complete and sufficient for two parameter Pareto distribution

Let$$(X_1,...,X_{n})$$ be a random sample from the Pareto distribution with pdf density $$\theta a^{\theta} x^{-(\theta+1)}I_{(a,\infty)}(x),$$ where $$\theta>0$$ and $$a>0$$

$$\textbf{(i)}$$ Show that when $$\theta$$ is known, $$X_{(1)}$$ is complete and sufficient for $$a$$.

$$\textbf{(ii)}$$ Show that when both $$a$$ and $$\theta$$ are unknown, $$\left (Y,X_{(1)} \right )$$ is complete and sufficient for $$\left (a,\theta \right ),$$ where $$Y =\sum_{i}(\log X_i −\log X_{(1)}).$$

The joint pdf of $$X_1,...,X_{n}$$ is $$f(x_1,...,x_{n})=\theta^{n}a^{n\theta}\exp\left\{-(\theta+1)\sum_{i=1}^{n}\log x_{i} \right\}I_{(a,\infty)}(x_{(1)}).$$

Applying Factorization theorem, I know $$X_{(1)}$$ is sufficient for $$a.$$ But how to show that $$X_{(1)}$$ is complete and $$\left (Y,X_{(1)} \right )$$is complete and sufficient for $$\left (a,\theta \right ).$$

• You can easily show $(\prod X_i, X_{(1) })$ is jointly minimal sufficient for $(a, \theta).$ Can you proceed then? Feb 28, 2023 at 3:55
• @User1865345:Applying Lehmann-Scheffe Theorem ,I know that $\left ( \prod_{i=1}^{n}X_i,X_{(1)} \right )$ is minimal sufficient statistics for $(a,\theta).$ But I have no ideal about $\left (Y,X_{(1)} \right )$is complete and sufficient for $\left (a,\theta \right ).$ Feb 28, 2023 at 4:39
• Warning! The Lehmann-Scheffé Theorem does not establish minimal sufficiency: It requires a complete and sufficient statistic as a given (to then deduce a unique best unbiased estimator). Feb 28, 2023 at 8:36
• @Xi'an I mean Lehmann-Scheffé Theorem for Minimal Sufficient Statistics (LSM). Let $f(\mathbf{y}|\mathbf{\theta})$ be the pmf or pdf of an iid sample $\mathbf{Y}$. Let $c_\mathbf{x,y}$ be a constant. Suppose there exists a function $\mathbf{T (y)}$ such that for any two sample points $\mathbf{x}$ and $\mathbf{y}$, the ratio $R_\mathbf{x,y}(\mathbf{\theta})=f(\mathcal{x}|\mathbf{\theta})/ f (\mathcal{y}|\mathbf{\theta}) = c_\mathbf{x,y}$ for all$\mathbf{\theta}$ in $\mathbf{\Theta}$ iff $\mathbf{T (x) = T (y)}$. Then $\mathbf{T(Y)}$ is a minimal sufficient statistic for $\mathbf{\theta}$. Feb 28, 2023 at 9:43

Given the density is

$$f(x; a, \theta) := \theta a^{\theta} x^{-(\theta+1)}\boldsymbol 1_{(a,\infty)}(x).\tag 1\label 1$$

The pdf of the first order statistic $$X_{(1)}$$ can be easily shown to be $$g(x_1):= n\theta a^{n\theta}{x_1}^{-(n\theta+1)} \boldsymbol 1_{(a,\infty)}(x_1)\tag 2\label 2.$$

Observation $$1.$$ For known $$\theta,~\hat a:= X_{(1)}$$ is sufficient and complete for $$a.$$

Use $$\eqref 2$$ to express the joint pdf of the sample in the form $$f(\mathbf x) = g\left(\hat a; a\right)\times \left\{\frac1n \theta ^{n-1}\left(\prod x_i\right)^{-(\theta+1)}{\hat a}^{n\theta+1}\right\} .\tag 3$$ Sufficiency follows from Neyman factorization theorem.

Consider an arbitrary Borel measurable function $$\varphi$$ such that $$\mathbb E\left[\varphi\left(\hat a\right)\right] = 0, ~\forall a\in (0,\infty).\tag 4\label 4$$

$$\eqref 4$$ implies $$\int_a^\infty\varphi\left(\hat a\right){\hat a}^{-(n\theta+1)}~\mathrm d\hat a = 0, ~\forall a\in (0,\infty).\tag 5\label 5$$ Define (denoting $$\varphi:= \varphi^+-\varphi^-$$) $$\nu^{\pm}(A) :=\int_A\varphi^{\pm}\left(\hat a\right){\hat a}^{-(n\theta+1)}~\mathrm d\hat a; ~~A:= [a, \infty);$$ from $$\eqref 5, ~\nu^+(A) = \nu^-(A)$$ whence $$\varphi^+ = \varphi^- ~\textrm{a.s.} ~[\lambda].$$ The result follows.

$$\blacksquare$$

Observation $$2.$$ Define $$Z:= \sum_{i=1}^n\ln\frac{Y_i}{Y_1}$$ where $$Y_i$$ are the order statistics. Then $$\hat \theta:= Z$$ and $$\hat a = Y_1$$ are stochastically independent.

The approach would be to show $$M_Z(t),$$ the moment generating function of $$Z$$ doesn't depend on $$a,$$ whence by Observation $$1.$$ and Basu's theorem, $$Z$$ would be independent of $$Y_1.$$

$$M_Z(t) = n!\int_a^\infty\int_a^{y_n}\cdots\int_a^{y_2}\exp\left(t\sum_{i=1}^n\ln\frac{Y_i}{Y_1}\right)\theta^na^{n\theta}\prod_{i=1}^n y_1^{-(\theta+1)}~\mathrm dy_i ;$$

substitute $$y_i\mapsto \frac a{y_i}~\forall i\in\{1,2,\ldots, n\}.$$ It is easy to see it is one-to-one and $$|\mathcal J| = a^n.$$ Therefore, $$M_Z(t)$$ reduces to a form that doesn't depend on $$a.$$

$$\blacksquare$$

Now, we would concentrate on the distribution of $$Z.$$ Note that $$\sum_{i=1}^n\ln\frac{X_i}{X_1}$$ doesn't depend on the ordering of $$x_2, x_3, \ldots, x_n.$$ So assuming $$x_1< x_2, \ldots, x_n, ~~\sum_{i=1}^n\ln\frac{X_i}{X_1} = \sum_{i=1}^n\ln\frac{Y_i}{Y_1}.$$ Also

$$g(x_2, \ldots, x_n\mid x_1) = \frac{\prod_{i=2}^nf(x_i)}{[1- F(x_1)]^{n-1}} .\tag 6$$ Therefore, the characteristic function of $$Z$$ given $$X_1 = x_1$$

\begin{align}\phi(t) &= \mathbb E\left[\exp\left(it\sum_{i=1}^n\ln\frac{X_i}{X_1}\right)\bigg \vert ~x_1\right]\\ &=\left[\int_{x_2}^\infty\frac{\exp\left(it\ln\frac{x_2}{x_1}\right)f(x_2)}{1- F(x_1)}~\mathrm dx_2\right]^{n-1} \tag 7.\end{align}

From this, the pdf of $$Z$$ is (by Observation $$2.$$)

\begin{align}f(z) &= \frac1{2\pi}\int_{-\infty}^{\infty}\exp{(-itz)}\phi(t)~\mathrm dt.\end{align}

Simplify it to deduce (cf. $$\rm [II]$$)

$$f(z) = \frac{\theta}{\Gamma{(n-1)}}z^{n-2}e^{-\theta z}.\tag 8 \label 8$$

Observation $$3.$$$$\hat \theta$$ is complete.

For arbitrary Borel measurable function $$\varphi,$$

\begin{align}\mathbb E\left[\varphi\left(\hat \theta\right)\right] &= 0, ~\forall \theta\in (0, \infty)\\\implies \int_0^\infty \varphi\left(\hat \theta\right){\hat\theta}^{n-2}\exp{\left(-\theta\hat\theta\right)}~\mathrm d\hat\theta &= 0; \tag 9\label 9\end{align}

$$\eqref 9$$ resembles the kernel of a one-dimensional exponential family. So, $$\hat \theta$$ is complete for $$\theta.$$

$$\blacksquare$$

Theorem $$1.$$ The family $$\{F\left(\hat\theta, \hat a;\theta, a\right): (\theta, a)\in \mathbb R_+\times \mathbb R_+\}$$ is complete.

From $$\eqref 2, \eqref 8$$ and by Observation $$2.,$$ write down the density $$f\left(\hat\theta, \hat a;\theta, a\right).$$

For an arbitrary Borel measurable function $$\varphi\left(\hat\theta, \hat a\right)$$

\begin{align}\mathbb E\left[\varphi\left(\hat\theta, \hat a\right)\right] &= 0, ~\forall (\theta, a)\in \mathbb R_+\times \mathbb R_+\\ \implies \int_0^\infty\int_k^\infty f\left(\hat\theta, \hat a;\theta, a\right)\varphi\left(\hat\theta, \hat a\right)~\mathrm d\hat\theta\mathrm d\hat a & = 0\\ \overset{\textrm{Fubini}}{\implies} \int_k^\infty n\theta a^{n\theta}{\hat a}^{-(n\theta+1)}\left[\underbrace{\int_0^\infty\varphi\left(\hat\theta, \hat a\right) \frac{\theta}{\Gamma{(n-1)}}{\hat\theta}^{n-2}e^{-\theta \hat\theta}~\mathrm d\hat\theta}_{:= h\left(\hat a, \theta\right)}\right] ~\mathrm d\hat a & =0 ;\tag{ 10}\label{ 10}\end{align}

by Observation $$1.$$ and $$\eqref{ 10}, ~\lambda \{\hat a: h\left(\hat a, \theta\right) \ne 0\} = 0, ~\forall \theta > 0.$$ This means $$\lambda\otimes\lambda\{\left(\theta,\hat a\right): h\left(\hat a, \theta\right) \ne 0\} = 0$$ whence $$\lambda\{\left(\theta,\hat a\right): h\left(\hat a, \theta\right) \ne 0\} = 0, ~\forall \hat a~\textrm{a.s.}.$$ By continuity of $$h\left(\hat a, \cdot\right), ~h\left(\hat a, \cdot\right) = 0,~\forall \hat a~\textrm{a.s.}\overset{\textrm{Obs.}~3.}\implies \lambda\left\{\hat a: \varphi\left(\hat\theta,\hat a\right) \ne 0\right\} = 0, ~\forall \hat a~\textrm{a.s.}\implies \lambda\otimes\lambda\left\{\left(\hat\theta,\hat a\right): \varphi\left(\hat\theta,\hat a\right) \ne 0\right\} = 0.$$

$$\blacksquare$$

## References:

$$\rm [I]$$ Best Unbiased Estimators for the Parameters of a Two-Parameter Pareto Distribution, S.K. Saksena, A.M. Johnson, Metrika, Volume $$31,~ 1984,$$ page $$77-83.$$

$$\rm[II]$$ Estimation of the Parameters of the Pareto Distribution, H. J. Malik, Skandinavisk Aktuarietidskrift $$49, ~1966, ~144-157.$$

• To the downvoter, I would appreciate if they care to explain why they downvoted. It is a well-researched answer, so if there is anything missing, it is better to let me know. Feb 28, 2023 at 21:23