Robustness of a truncated estimator for the smaller of two ordered means

Abstract

In this note, we consider the problem of estimating the smaller of two ordered means. Such problems frequently arise in applications where, for example, aggregated data are observed. In order to combine information from direct and indirect observations, we use the Stein-type truncated estimator. We show that it dominates the direct estimator for distributions with log-concave or log-convex densities.

Appendix

1.1 Proofs

Here we prove Propositions 2.2, 3.1, 3.2 and Lemma 2.2.

Proof of Proposition 2.2. We have

$$\begin{aligned} {2 {\widetilde{R}}' (v) \over E[ Z 1(X> (1 + v) Z) ]}&= 2 E \Big [ \Big \{ {X + (1 + v) Z \over 2} - \mu \Big \} Z 1(X> (1 + v) Z) \Big ] \nonumber \\&\quad / E[ Z 1(X > (1 + v) Z) ] \nonumber \\&= \frac{ \int _{0}^{\infty } x f(x) \big \{ \int _{0}^{x / (1 + v)} z f(z) dz \big \} dx }{ \int _{0}^{\infty } f(x) \big \{ \int _{0}^{x / (1 + v)} z f(z) dz \big \} dx } \nonumber \\&\quad + \frac{ \int _{0}^{\infty } (1 + v) z^2 f(z) \big \{ \int _{(1 + v) z}^{\infty } f(x) dx \big \} dz }{ \int _{0}^{\infty } z f(z) \big \{ \int _{(1 + v) z}^{\infty } f(x) dx \big \} dz } - 2 \mu \nonumber \\&= \frac{ \int _{0}^{\infty } x f(x) \big \{ \int _{0}^{x} z f(z / (1 + v)) dz \big \} dx }{ \int _{0}^{\infty } f(x) \big \{ \int _{0}^{x} z f(z / (1 + v)) dz \big \} dx } \nonumber \\&\quad + \frac{ \int _{0}^{\infty } z^2 f(z / (1 + v)) \big \{ \int _{z}^{\infty } f(x) dx \big \} dz }{ \int _{0}^{\infty } z f(z / (1 + v)) \big \{ \int _{z}^{\infty } f(x) dx \big \} dz } \nonumber \\&\quad - \frac{ \int _{0}^{\infty } x f(x) \big \{ \int _{0}^{x} f(z) dz \big \} dx }{ \int _{0}^{\infty } f(x) \big \{ \int _{0}^{x} f(z) dz \big \} dx } - \frac{ \int _{0}^{\infty } z f(z) \big \{ \int _{z}^{\infty } f(x) dx \big \} dz }{ \int _{0}^{\infty } f(z) \big \{ \int _{z}^{\infty } f(x) dx \big \} dz } \text {.} \end{aligned}$$

(5.1)

Now, by assumption,

$$\begin{aligned} z {\partial \over \partial z} \log {z f(z / (1 + v)) \over f(z)}&= 1 + {z \over 1 + v} {f' (z / (1 + v)) \over f(z / (1 + v))} - z {f' (z) \over f(z)} \ge 0 \text {.} \end{aligned}$$

Therefore, $z f(z / (1 + v)) / f(z)$ is a nondecreasing function of $z > 0$, which implies that $\int _{0}^{x} z f(z / (1 + v)) dz / \int _{0}^{x} f(z) dz$ is a nondecreasing function of $x > 0$. Thus, by the covariance inequality, the right-hand side of (5.1) is nonnegative. The result follows. $\square$

Proof of Lemma 2.2. We prove parts (i) and (ii) based on the proofs of Lemmas 3 and 4 of Bagnoli and Bergstrom (2005), which are for the continuous case. Without loss of generality, we can assume that $k = 1$ and $x_2 = x_1 + 1$. Let $F(- 1) = f(- 1) = 0$.

For part (i), note that $f(x - 1) / f(x) \ge f(z - 1) / f(z)$ for all $x, z \in \mathbb {N} _0$ with $x \ge z$ by the log-concavity of f. Then, for all $x \in \mathbb {N} _0$,

$$\begin{aligned} {f(x) - f(x - 1) \over f(x)} F(x)&= {f(x) - f(x - 1) \over f(x)} \sum _{z = 0}^{x} f(z) \\&\le \sum _{z = 0}^{x} {f(z) - f(z - 1) \over f(z)} f(z) = f(x) \text {,} \end{aligned}$$

which implies that

$$\begin{aligned} f(x - 1) \ge f(x) F(x - 1) / F(x) \text {.} \end{aligned}$$

Therefore,

$$\begin{aligned} {F( x_2 ) \over F( x_2 + k)} - {F( x_1 ) \over F( x_1 + k)}&= {F( x_1 + 1) \over F( x_1 + 2)} - {F( x_1 ) \over F( x_1 + 1)} \\&= {f( x_1 + 1) \over F( x_1 + 1)} - {f( x_1 + 2) \over F( x_1 + 2)} \\&\ge {f( x_1 + 2) \over F( x_1 + 2)} - {f( x_1 + 2) \over F( x_1 + 2)} = 0 \text {.} \end{aligned}$$

For part (ii), note that $f(x) / f(x + 1) \ge f(z) / f(z + 1)$ for all $x, z \in \mathbb {N} _0$ with $x \le z$ by the log-convexity of f. Then, for all $x \in \mathbb {N} _0$,

$$\begin{aligned} {f(x + 1) - f(x) \over f(x + 1)} \{ 1 - F(x) \}&= {f(x + 1) - f(x) \over f(x + 1)} \sum _{z = x}^{\infty } f(z + 1) \\&\le \sum _{z = x}^{\infty } {f(z + 1) - f(z) \over f(z + 1)} f(z + 1) = - f(x) \text {,} \end{aligned}$$

where the last equality follows from the assumption that $\lim _{z \rightarrow \infty } f(z) = 0$, and thus we have

$$\begin{aligned} f(x) \ge f(x + 1) \{ 1 - F(x - 1) \} / \{ 1 - F(x) \} \text {.} \end{aligned}$$

Hence,

$$\begin{aligned} {1 - F( x_2 + k) \over 1 - F( x_2)} - {1 - F( x_1 + k) \over 1 - F( x_1)}&= {1 - F( x_1 + 2) \over 1 - F( x_1 + 1)} - {1 - F( x_1 + 1) \over 1 - F( x_1 )} \\&= {f( x_1 + 1) \over 1 - F( x_1 )} - {f( x_1 + 2) \over 1 - F( x_1 + 1)} \\&\ge {f( x_1 + 2) \over 1 - F( x_1 + 1)} - {f( x_1 + 2) \over 1 - F( x_1 + 1)} = 0 \text {.} \end{aligned}$$

For part (iii), let $W = X + Z$ and $T = X - Z$. Let $S = 1(W \in 2 \mathbb {N} _0 + 1)$. Suppose first that $\{ \partial / ( \partial x) \} \{ \log f(x) \}$ is a convex function of $x \in [0, \infty )$ and that $E[ W | S = 0 ] \le E[ W | S = 1 ]$. Then

$$\begin{aligned} E[ W | T \ge 1 ]&= E[ W | |T| \ge 1 ] = \frac{ \displaystyle \sum _{w = 0}^{\infty } w \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \\ |t| \ge 1 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) }{ \displaystyle \sum _{w = 0}^{\infty } \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \\ |t| \ge 1 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) }, \end{aligned}$$

and

$$\begin{aligned} E[ W ]&= \frac{ \displaystyle \sum _{w = 0}^{\infty } w \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) }{ \displaystyle \sum _{w = 0}^{\infty } \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) } \text {.} \end{aligned}$$

Note that for $w \in \mathbb {N} _0$,

$$\begin{aligned}&\sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \\ |t| \ge 1 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) / \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) \\&\quad = {\left\{ \begin{array}{ll} \rho (w) \text {,} &{} \text {if } w \in 2 \mathbb {N} _0 \text {,} \\ 1 \text {,} &{} \text {if } w \in 2 \mathbb {N} _0 + 1 \text {,} \end{array}\right. } \end{aligned}$$

where

$$\begin{aligned} \rho (w)&= \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \\ |t| \ge 2 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) / \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) \\&= 1 - 1 / \sum _{\begin{array}{c} - w \le t \le w \\ w - t \in 2 \mathbb {N} _0 \end{array}} f \Big ( {w + t \over 2} \Big ) f \Big ( {w - t \over 2} \Big ) / \Big \{ f \Big ( {w \over 2} \Big ) \Big \} ^2 \in [0, 1], \end{aligned}$$

is a nondecreasing function of $w \in 2 \mathbb {N} _0$ by assumption. Then it follows from Theorem 2.1 of Bhattacharya (1984) that

$$\begin{aligned} E[ W | T \ge 1 ]&= \frac{ E[ W \{ (1 - S) \rho (W) + S \} ] }{ E[ (1 - S) \rho (W) + S ] } \ge E[ W ] \text {,} \end{aligned}$$

since E[W|S] and $E[ (1 - S) \rho (W) + S | S]$ are nondecreasing functions of S by assumption and since W and $(1 - S) \rho (W) + S$ are nondecreasing functions of W. Next, suppose instead that f is log-convex. Then

$$\begin{aligned} E[ X + Z | T \ge 1 ]&= E[ E[ X + Z | T ] | T \ge 1 ] \\&= E \Big [ \frac{ \sum _{z = 0}^{\infty } (2 z + T) f(z + T) f(z) }{ \sum _{z = 0}^{\infty } f(z + T) f(z) } \Big | T \ge 1 \Big ] \\&\ge E \Big [ \frac{ \sum _{z = 0}^{\infty } (2 z + T) \{ f(z) \} ^2 }{ \sum _{z = 0}^{\infty } \{ f(z) \} ^2 } \Big | T \ge 1 \Big ] \\&\ge E \Big [ \frac{ \sum _{z = 0}^{\infty } 2 z \{ f(z) \} ^2 }{ \sum _{z = 0}^{\infty } \{ f(z) \} ^2 } \Big | T \ge 1 \Big ] = E[ X + Z | T = 0 ] \text {,} \end{aligned}$$

where the first inequality follows from the covariance inequality since $f(z + T) / f(z)$ is a nondecreasing function of $z \in \mathbb {N} _0$ by assumption. Thus,

$$\begin{aligned} E[ W ]&= P(T = 0) E[ W | T = 0 ] + P(T \ge 1) E[ W | T \ge 1 ] + P(T \le - 1) E[ W | T \le - 1 ] \\&= P(T = 0) E[ W | T = 0 ] + \{ 1 - P(T = 0) \} E[ W | T \ge 1 ] \le E[ W | T \ge 1 ] \text {.} \end{aligned}$$

This completes the proof. $\square$

Proof of Proposition 3.1. Let $W = X + Y$ and $R = X / W$. Let ${\Delta }= E [ ( \max \{ Y, (X + Y) / 2 \} - {\alpha }_2 )^2 ] - E[ (Y - {\alpha }_2 )^2 ]$. Note that $W \sim \mathrm{{Ga}} ( {\alpha }_1 + {\alpha }_2 , 1)$ and $R \sim \mathrm{{Beta}} ( {\alpha }_1 , {\alpha }_2 )$ and these are mutually independent.

For part (i), suppose that ${\alpha }_1 = {\alpha }_2$. Then it can be seen that

$$\begin{aligned} {\Delta }&= E \Big [ \Big [ W^2 \Big \{ \Big ( {1 \over 2} \Big ) ^2 - (1 - R)^2 \Big \} - 2 {\alpha }_2 W \Big \{ {1 \over 2} - (1 - R) \Big \} \Big ] 1 \Big ( 1 - R< {1 \over 2} \Big ) \Big ] \\&= E \Big [ \Big [ \{ 2 {\alpha }_2 + (2 {\alpha }_2 )^2 \} \Big \{ \Big ( {1 \over 2} \Big ) ^2 - (1 - R)^2 \Big \} - 2 {\alpha }_2 (2 {\alpha }_2 ) \Big \{ {1 \over 2} - (1 - R) \Big \} \Big ] 1 \Big ( 1 - R < {1 \over 2} \Big ) \Big ] \\&\sim {\alpha }_2 / 4 > 0 \end{aligned}$$

as ${\alpha }_2 \rightarrow 0$.

For part (ii), let $\phi (R) = \max \{ Y, (X + Y) / 2 \} / Y = \max \{ 1, 1 / \{ 2 (1 - R) \} \}$. Then

$$\begin{aligned} {\Delta }&= E[ \{ \phi (R) - 1 \} (1 - R) [ \{ \phi (R) + 1 \} (1 - R) W^2 - 2 {\alpha }_2 W ] ] \\&= E[ \{ \phi (R) - 1 \} (1 - R) [ \{ \phi (R) + 1 \} (1 - R) ( {\alpha }_1 + {\alpha }_2 + 1) - 2 {\alpha }_2 ] ] ( {\alpha }_1 + {\alpha }_2 ) \text {.} \end{aligned}$$

Since $\{ \phi (R) + 1 \} (1 - R) \le 1$ when $\phi (R) - 1 \ne 0$ and since ${\alpha }_1 + {\alpha }_2 + 1 < 2 {\alpha }_2$ by assumption, it follows that ${\Delta }< 0$. $\square$

Proof of Proposition 3.2. Let $W = X + Y$ and $R = X / W$. For part (i), let $\phi _1 (R) = \min \{ X, (X + Y) / 2 \} / X = \min \{ 1, 1 / (2 R) \}$. Then we have

$$\begin{aligned}&E[ X \phi _1 (R) / {\alpha }_1 - 1 - \log \{ X \phi _1 (R) / {\alpha }_1 \} ] - E[ X / {\alpha }_1 - 1 - \log (X / {\alpha }_1 ) ] \\&= E[ - W R \{ 1 - \phi _1 (R) \} / {\alpha }_1 - \log \phi _1 (R) ]\\&= E[ - R \{ 1 - \phi _1 (R) \} ( {\alpha }_1 + {\alpha }_2 ) / {\alpha }_1 - \log \phi _1 (R) ] \text {,} \end{aligned}$$

which is negative since $( {\alpha }_1 + {\alpha }_2 ) / {\alpha }_1 \ge 2$ by assumption and since $2 R \{ 1 - \phi _1 (R) \} + \log \phi _1 (R) > 0$ when $\phi _1 (R) \ne 1$. For part (ii), let $\phi _2 (R) = \max \{ 1, 1 / \{ 2 (1 - R) \} \}$. Then

$$\begin{aligned}&E[ Y \phi _2 (R) / {\alpha }_2 - 1 - \log \{ Y \phi _2 (R) / {\alpha }_2 \} ] - E[ Y / {\alpha }_2 - 1 - \log (Y / {\alpha }_2 ) ] \\&= E[ W (1 - R) \{ \phi _2 (R) - 1 \} / {\alpha }_2 - \log \phi _2 (R) ]\\&= E[ (1 - R) \{ \phi _2 (R) - 1 \} ( {\alpha }_1 + {\alpha }_2 ) / {\alpha }_2 - \log \phi _2 (R) ] \text {,} \end{aligned}$$

which is negative since $( {\alpha }_1 + {\alpha }_2 ) / {\alpha }_1 \le 2$ and since $2 (1 - R) \{ \phi _2 (R) - 1 \} - \log \phi _2 (R) < 0$ when $\phi _2 (R) \ne 1$. $\square$

1.2 Details of the approximation algorithm used in Sect. 4.5

First, the density proportional to

$$\begin{aligned} \Big ( {1 \over 2} \Big ) ^{\mu } \sqrt{\sum _{k = 0}^{\infty } \mathrm{{NB}} (k | \mu , 1 / 2) \{ \psi _1 ( \mu ) - \psi _1 ( \mu + k) \} } \text {,} \quad \mu \in (0, \infty ) \text {,} \end{aligned}$$

is approximated by $\mathrm{{Ga}} (1 / 2, \log 2)$ because for all $k \in \mathbb {N} _0$ and all $\mu \in (0, \infty )$,

$$\begin{aligned} \Big \{ \psi _1 ( \mu ) - \psi _1 ( \mu + k), {1 \over \mu } {k \over \mu + k - 1} \Big \} \subset \Big [ {1 \over \mu } {k \over \mu + k}, {1 \over \mu - 1} {k \over \mu + k - 1} \Big ] \text {,} \end{aligned}$$

provided that $\mu > 1$, and because

$$\begin{aligned} \sqrt{\sum _{k = 0}^{\infty } \mathrm{{NB}} (k | \mu , 1 / 2) {1 \over \mu } {k \over \mu + k - 1}} = \sqrt{{1 \over \mu } {1 \over 2}} \propto \mu ^{1 / 2 - 1}, \end{aligned}$$

for all $\mu \in (0, \infty )$ (see Hudson (1978) for the equality). Next, note that for all $K \in \mathbb {N}$ and all $\mu \in (0, \infty )$,

$$\begin{aligned}&\sqrt{\sum _{k = 0}^{K} \mathrm{{NB}} (k | \mu , 1 / 2) \{ \psi _1 ( \mu ) - \psi _1 ( \mu + k) \} } \le \pi ^\mathrm{{J}} ( \mu ) \\&\le \sqrt{\psi _1 ( \mu ) + \sum _{k = 0}^{K} \mathrm{{NB}} (k | \mu , 1 / 2) \{ - \psi _1 ( \mu + k) \} + P( {\widetilde{X}}> K) \{ - \psi _1 ( \mu + E[ {\widetilde{X}}| {\widetilde{X}}> K ] ) \} }, \end{aligned}$$

by Jensen’s inequality, where ${\widetilde{X}}\sim \mathrm{{NB}} ( \mu , 1 / 2)$ and where $E[ {\widetilde{X}}| {\widetilde{X}}> K ]$ is expressible in closed form (see Shonkwiler (2016)). Then, based on these inequalities, the acceptance ratio can be bounded below and above with arbitrary accuracy in MH steps (such an approach is used, for example, by Polson et al. (2013)).