Pitman closeness domination in predictive density estimation for two-ordered normal means under $\alpha$-divergence loss

Abstract

We consider Pitman closeness domination in predictive density estimation problems when the underlying loss metric is $\alpha$-divergence, $\{D(\alpha )\}$, a loss introduced by Csiszàr (Stud Sci Math Hung 2:299–318, 1967). The underlying distributions considered are normal location-scale models, including the distribution of the observables, the distribution of the variable, whose density is to be predicted, and the estimated predictive density which will be taken to be of the plug-in type. The scales may be known or unknown. Chang and Strawderman (J Multivar Anal 128:1–9, 2014) have derived a general expression for the $\alpha$-divergence loss in this setup, and have shown that it is a concave monotone function of quadratic loss, and also a function of the variances (predicand, and plug-in). We demonstrate $\{D(\alpha )\}$ Pitman closeness domination of certain plug-in predictive densities over others for the entire class of metrics simultaneously when modified Pitman closeness domination holds in the related problem of estimating the mean. We also establish $\{D(\alpha )\}$ Pitman closeness results for certain generalized Bayesian (best invariant) predictive density estimators. Examples of $\{D(\alpha )\}$ Pitman closeness domination presented relate to the problem of estimating the predictive density of the variable with the larger mean. We also consider the case of two-ordered normal means with a known covariance matrix.

Appendix

Proof of Theorem 3

We make the variable transformation

$$\begin{aligned} Z_i = \frac{ {\bar{X}}_i-\mu _1}{\tau _1}, \quad i=1,2. \end{aligned}$$

Then, $Z_1$ and $Z_2$ are mutually independently distributed as N(0, 1) and $N(\varDelta /\tau _1, \tau _2^2/\tau _1^2)$, respectively.

We note that $\hat{\varvec{\mu }}^{{\rm CS}} \ne \hat{\varvec{\mu }}^{{\rm OS}}$ if and only if ${\bar{X}}_1 > {\bar{X}}_2$ and $s_1^2 > s_2^2$. In this case, putting $c_i=n_i/(n_1+n_2), i=1,2$, we have

$$\begin{aligned}&\sum _{i=1}^2 (\hat{\mu }_i^{{\rm CS}}- \mu _i)^2/\sigma _i^2 \\&\quad =\biggl ( \frac{c_1( {\bar{X}}_1-\mu _1)+c_2({\bar{X}}_2- \mu _1)}{\sqrt{n_1}\tau _1} \biggr )^2 + \biggl ( \frac{c_1( {\bar{X}}_1-\mu _2)+c_2({\bar{X}}_2- \mu _2)}{\sqrt{n_2}\tau _2} \biggr )^2 \nonumber \\&\quad =\frac{1}{n_1} (c_1 Z_1+ c_2Z_2)^2+\frac{1}{n_2} \biggl ( \frac{\tau _1}{\tau _2}(c_1 Z_1+c_2Z_2) - \frac{\varDelta }{\tau _2} \biggr ) ^2 \nonumber \\&\quad = \biggl (\frac{1}{n_1}+ \frac{\tau _1^2}{n_2\tau _2^2} \biggr )(c_1 Z_1+ c_2Z_2)^2- \frac{2\varDelta \tau _1}{n_2\tau _2^2}(c_1 Z_1+ c_2Z_2)+ \frac{\varDelta ^2}{n_2\tau _2^2}. \end{aligned}$$

Similarly

$$\begin{aligned} \sum _{i=1}^2 (\hat{\mu }_i^{{\rm OS}}- \mu _i)^2/\sigma _i^2= \biggl (\frac{1}{n_1}+ \frac{\tau _1^2}{n_2\tau _2^2} \biggr )(\gamma _1 Z_1+ \gamma _2Z_2)^2- \frac{2\varDelta \tau _1}{n_2\tau _2^2} (\gamma _1 Z_1+ \gamma _2Z_2)+ \frac{\varDelta ^2}{n_2\tau _2^2}, \end{aligned}$$

where $\gamma _1=n_1s_2^2/(n_1s_2^2+n_2s_1^2)$ and $\gamma _2=n_2s_1^2/(n_1s_2^2+n_2s_1^2)$. Since $Z_1 - Z_2= ({\bar{X}}_1- {\bar{X}}_2)/ \tau _1 >0$ and $c_1 \ge \gamma _1$, we have $(c_1- \gamma _1)Z_1+(c_2-\gamma _2)Z_2 \ge 0$. Therefore, we see that

$$\begin{aligned}&\sum _{i=1}^2 (\hat{\mu }_i^{{\rm CS}}- \mu _i)^2/\sigma _i^2 \le \sum _{i=1}^2 (\hat{\mu }_i^{{\rm OS}}- \mu _i)^2/\sigma _i^2 \\&\quad \Leftrightarrow [(c_1- \gamma _1)Z_1+ (c_2-\gamma _2)Z_2 ] \biggl [ \biggl (\frac{1}{n_1}+ \frac{\tau _1^2}{n_2\tau _2^2} \biggr ) [ (c_1+ \gamma _1)Z_1+ (c_2+\gamma _2)Z_2] \\&\qquad - \frac{2\varDelta \tau _1}{n_2\tau _2^2} \biggr ] \le 0 \\&\quad \Leftrightarrow \biggl (\tau _1^2+ \frac{n_2\tau _2^2}{n_1} \biggr ) [ (c_1+ \gamma _1)Z_1+ (c_2+\gamma _2)Z_2] - 2\varDelta \tau _1 \le 0. \end{aligned}$$

Then, the numerator of (14) becomes

$$\begin{aligned}&P \biggl \{ \sum _{i=1}^2 (\hat{\mu }_i^{{\rm CS}}- \mu _i)^2/\sigma _i^2 \le \sum _{i=1}^2 (\hat{\mu }_i^{{\rm OS}}- \mu _i)^2/\sigma _i^2, {\bar{X}}_1> {\bar{X}}_2, s_1^2> s_2^2 \biggr \} \nonumber \\&\quad = P \biggl \{ \biggl (\tau _1^2+ \frac{n_2\tau _2^2}{n_1} \biggr ) [ (c_1+ \gamma _1)Z_1+ (c_2+\gamma _2)Z_2] - 2\varDelta \tau _1 \le 0 , Z_1> Z_2,\nonumber \\&\qquad s_1^2 > s_2^2 \biggr \}. \end{aligned}$$

(19)

We further make the variable transformation:

$$\begin{aligned} Y_1 = Z_1 - Z_2, \quad Y_2 = Z_1 + (\tau _1^2/ \tau _2^2) Z_2 . \end{aligned}$$

Then, $Y_1$ and $Y_2$ are mutually independently distributed as $N( -\varDelta /\tau _1,$ $(\tau _1^2+\tau _2^2)/\tau _1^2)$ and $N( \tau _1\varDelta /\tau _2^2, (\tau _1^2+\tau _2^2)/\tau _2^2)$, respectively.

Noting that

$$\begin{aligned} Z_1 = \frac{\tau _1^2 Y_1+\tau _2^2 Y_2}{\tau _1^2 + \tau _2^2} \quad \text{ and } \quad Z_2 = \frac{\tau _2^2(Y_2-Y_1)}{\tau _1^2 + \tau _2^2}, \end{aligned}$$

we have

$$\begin{aligned}&\biggl (\tau _1^2+ \frac{n_2\tau _2^2}{n_1} \biggr ) \biggl [(c_1+ \gamma _1) Z_1+(c_2+ \gamma _2)Z_2 \biggr ] -2 \varDelta \tau _1 \\&\quad =\biggl (\frac{\tau _1^2+ n_2\tau _2^2/n_1}{\tau _1^2 + \tau _2^2} \biggr ) \biggl [\{(c_1+ \gamma _1) \tau _1^2-(c_2+ \gamma _2 )\tau _2^2 \}Y_1+ 2\tau _2^2 Y_2 \\&\qquad -2 \varDelta \tau _1 \frac{\tau _1^2 + \tau _2^2}{\tau _1^2+ n_2\tau _2^2/n_1}\biggr ]\\&\quad \le \biggl (\frac{\tau _1^2+ n_2\tau _2^2/n_1}{\tau _1^2 + \tau _2^2} \biggr ) \biggl [\{(c_1+ \gamma _1) \tau _1^2-(c_2+ \gamma _2 )\tau _2^2 \}Y_1+ 2\tau _2^2 Y_2 -2 \varDelta \tau _1 \biggr ] \end{aligned}$$

if $n_1 \ge n_2$. Therefore, we see that

$$\begin{aligned}{}[(c_1+ \gamma _1) \tau _1^2-(c_2+ \gamma _2 )\tau _2^2]Y_1+ 2\tau _2^2 Y_2 -2 \varDelta \tau _1\le 0 \end{aligned}$$

implies

$$\begin{aligned} \biggl (\tau _1^2+ \frac{n_2\tau _2^2}{n_1} \biggr ) \biggl [(c_1+ \gamma _1) Z_1+(c_2+ \gamma _2)Z_2 \biggr ] -2 \varDelta \tau _1 \le 0. \end{aligned}$$

Thus, we have

$$\begin{aligned} (\hbox {19})\ge & {} P\biggl \{ Y_2 \le \frac{2\varDelta \tau _1- [(c_1+ \gamma _1) \tau _1^2 -(c_2+ \gamma _2) \tau _2^2]Y_1}{2 \tau _2^2}, Y_1>0, s_1^2> s_2^2 \biggr \} \\= & {} \int _{s_1^2 > s_2^2}\biggl [\int _0 ^\infty \varPhi \biggl ( \frac{[(c_2+\gamma _2)\tau _2^2-(c_1+\gamma _1)\tau _1^2]y_1 }{2\tau _2\sqrt{\tau _1^2+\tau _2^2} } \biggr ) g(y_1 ) {\text{d}}y_1 \biggr ] h(s_1^2, s_2^2) ds_1^2 ds_2^2, \nonumber \end{aligned}$$

which is larger than $1/2 P \{ Y_1> 0, s_1^2 > s_2^2\},$ because

$$\begin{aligned} (c_2+\gamma _2)\tau _2^2-(c_1+\gamma _1)\tau _1^2> 0 \quad \text{ for } \quad s_1^2 > s_2^2. \end{aligned}$$

This completes the proof.

Proof of Theorem 4

We first show that $\hat{\mu }_2^{{\rm MLE}}$ is Pitman closer to $\mu _2$ than ${\bar{X}}_2$.

We see that $\hat{\mu }_2^{{\rm MLE}} \ne {\bar{X}}_2$ if and only if ${\bar{X}}_1 > {\bar{X}}_2$. In this case $\hat{\mu }_2^{{\rm MLE}}=c_1 {\bar{X}}_1 + c_2 {\bar{X}}_2$, where $c_1=n_1\sigma _2^2/(n_1\sigma _2^2+n_2\sigma _1^2)$ and $c_2=n_2\sigma _1^2/(n_1\sigma _2^2+n_2\sigma _1^2)$.

We have

$$\begin{aligned}&P\{ | \hat{\mu }_2^{{\rm MLE}} - \mu _2 |< | {\bar{X}}_2 - \mu _2 | \} \nonumber \\&\quad = P\{ \hat{\mu }_2^{{\rm MLE}} + {\bar{X}}_2< 2 \mu _2 ,{\bar{X}}_2< \hat{\mu }_2^{{\rm MLE}} \} \nonumber \\&\quad = P \{ \hat{\mu }_2^{{\rm MLE}} + {\bar{X}}_2< 2 \mu _2, {\bar{X}}_1> {\bar{X}}_2 \} \nonumber \\&\quad \ge P_r \{ \hat{\mu }_2^{{\rm MLE}} - \mu _2< 0 , {\bar{X}}_1> {\bar{X}}_2 \} \; ( \text{ since } \; {\bar{X}}_2< \mu _2) \nonumber \\&\quad = P \{ c_1 ({\bar{X}}_1 - \mu _2) + c_2 ({\bar{X}}_2- \mu _2)< 0 , {\bar{X}}_1 > {\bar{X}}_2 \}. \end{aligned}$$

(20)

Next, we make the variable transformation

$$\begin{aligned} W_i= {\bar{X}}_i - \mu _2, \quad i=1,2. \end{aligned}$$

Then, $W_1$ and $W_2$ are distributed as $N(-\varDelta , \tau _1^2)$ and $N(0 , \tau _2^2)$, respectively, and $W_1$ and $W_2$ are mutually independent, where $\varDelta = \mu _2- \mu _1$ and $\tau _i^2= \sigma _i^2 /n_i, i=1,2$. Then, (20) becomes

$$\begin{aligned} P\{ c_1 W_1 + c_2 W_2 < 0 , W_1 > W_2 \}. \end{aligned}$$

(21)

We further make the variable transformation:

$$\begin{aligned} V_1 = W_1 - W_2, \quad V_2 = (\tau _2^2/ \tau _1^2) W_1+ W_2. \end{aligned}$$

Then, $V_1$ and $V_2$ are mutually independent and

$$\begin{aligned} V_1 \sim N(-\varDelta , \tau _1^2 + \tau _2^2) \quad \text{ and } \quad V_2 \sim N( -(\tau _2^2/ \tau _1^2)\varDelta , \tau _2^2(\tau _1^2+\tau _2^2)/\tau _1^2). \end{aligned}$$

Noting that

$$\begin{aligned} W_1=\frac{\tau _1^2(V_1+V_2)}{\tau _1^2+\tau _2^2} \quad \text{ and }\quad W_2= \frac{\tau _1^2V_2-\tau _2^2V_1}{\tau _1^2+\tau _2^2}, \end{aligned}$$

and denoting the p.d.f. of $V_1$ by $g(\cdot )$, (21) becomes

$$\begin{aligned}&P \{(c_1\tau _1^2-c_2\tau _2^2)V_1+ \tau _1^2V_2 <0, V_1 >0 \} \\&\quad = \int _0 ^\infty \varPhi \biggl ( \frac{((c_2\tau _2^2-c_1\tau _1^2)v_1+ \tau _2^2\varDelta }{\tau _1\tau _2\sqrt{\tau _1^2+\tau _2^2} } \biggr ) g(v_1 ) dv_1\\&\quad \ge 1/2, \end{aligned}$$

since $c_2\tau _2^2-c_1\tau _1^2= 0$ and $\varDelta \ge 0$. Thus, we have

$$\begin{aligned} {{\rm MPN}}_{\mu _2}(\hat{\mu }_2^{{\rm MLE}}, {\bar{X}}_2) \ge 1/2, \end{aligned}$$

with strict inequality for $\mu _2 > \mu _1$. This completes the proof.

The analogous result for $\hat{\mu }_1^{{\rm MLE}}$ is Pitman closer to $\mu _1$ than ${\bar{X}}_1$.