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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 $\left (a,\theta \right ).$

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  • $\begingroup$ You can easily show $(\prod X_i, X_{(1) }) $ is jointly minimal sufficient for $(a, \theta). $ Can you proceed then? $\endgroup$ Feb 28, 2023 at 3:55
  • $\begingroup$ @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 $\left (a,\theta \right ).$ $\endgroup$ Feb 28, 2023 at 4:39
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    $\begingroup$ 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). $\endgroup$
    – Xi'an
    Feb 28, 2023 at 8:36
  • $\begingroup$ @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 $\mathbf{\theta}$. $\endgroup$ Feb 28, 2023 at 9:43

1 Answer 1

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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 $49, ~1966, ~144-157.$

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  • $\begingroup$ 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. $\endgroup$ Feb 28, 2023 at 21:23

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