Bayesian predictive density estimation with parametric constraints for the exponential distribution with unknown location

Abstract

In this paper, we consider prediction for the exponential distribution with unknown location. For the most part, we treat the one-dimensional case and assume that the location parameter is restricted to an interval. The Bayesian predictive densities with respect to prior densities supported on the real line and the restricted space are compared under the Kullback–Leibler divergence. We first consider the case where the scale parameter is known. We obtain general dominance conditions and also minimaxity and admissibility results. Next, we treat the case of unknown scale. In this case, the location parameter is assumed to be less than a known constant and sufficient conditions for domination are obtained. Finally, we treat a multidimensional problem with known scale where the location parameter is restricted to a convex set. The performance of several Bayesian predictive densities is investigated through simulation. Some of the prediction methods are applied to real data.

Appendix

Here, we provide Lemmas 3, 4, and 5 and their proofs as well as those of Lemmas 1 and 2.

Lemma 3

Let $\mu \in \mathbb {R}$ and ${\lambda }\in (0, \infty )$ and let $h :\mathbb {R} \rightarrow \mathbb {R}$. Suppose that $\lim _{z \rightarrow \infty } e^{- {\lambda }z} z h(z) = 0$. Then we have

$$\begin{aligned} E_{( \mu , {\lambda })}^{Z \sim {\mathrm{Ex}} ( \mu , {\lambda })} \left[ \left( Z - \mu - {1 \over {\lambda }} \right) h(Z) \right] = E_{( \mu , {\lambda })}^{Z \sim {\mathrm{Ex}} ( \mu , {\lambda })} \left[ {Z - \mu \over {\lambda }} h' (Z) \right] \text {.} \end{aligned}$$

Proof

By integration by parts,

$$\begin{aligned}&E_{( \mu , {\lambda })}^{Z \sim {\mathrm{Ex}} ( \mu , {\lambda })} [ (Z - \mu ) h(Z) ] \\&\quad = \int _{\mu }^{\infty } {\lambda }e^{- {\lambda }(z - \mu )} (z - \mu ) h(z) dz \\&\quad = \left[ - e^{- {\lambda }(z - \mu )} (z - \mu ) h(z) \right] _{\mu }^{\infty } + \int _{\mu }^{\infty } e^{- {\lambda }(z - \mu )} \{ h(z) + (z - \mu ) h' (z) \} dz \\&\quad = {1 \over {\lambda }} E_{( \mu , {\lambda })}^{Z \sim {\mathrm{Ex}} ( \mu , {\lambda })} [ h(Z) + (Z - \mu ) h' (Z) ] \text {,} \end{aligned}$$

which proves the result. $\square$

Lemma 4

Let $\mu , c \in \mathbb {R}$ and ${\lambda }\in (0, \infty )$ and let $h :\mathbb {R} \rightarrow [0, \infty )$. Suppose that $c > \mu$. Then

$$\begin{aligned} E_{( \mu , {\lambda })}^{Z \sim {\mathrm{Ex}} ( \mu , {\lambda })} [ h( {\lambda }(Z - c)) 1_{(c, \infty )} (Z) ] = e^{{\lambda }( \mu - c)} E^{Z \sim {\mathrm{Ex}} (0, 1)} [ h(Z) ] \text {.} \end{aligned}$$

Proof

We have

$$\begin{aligned} E_{( \mu , {\lambda })}^{Z \sim {\mathrm{Ex}} ( \mu , {\lambda })} [ h( {\lambda }(Z - c)) 1_{(c, \infty )} (Z) ]&= \int _{c}^{\infty } {\lambda }e^{- {\lambda }(z - \mu )} h( {\lambda }(z - c)) dz \\&= e^{{\lambda }( \mu - c)} \int _{0}^{\infty } {\lambda }e^{- {\lambda }z} h( {\lambda }z) dz \\&= e^{{\lambda }( \mu - c)} \int _{0}^{\infty } e^{- z} h(z) dz \\&= e^{{\lambda }( \mu - c)} E^{Z \sim {\mathrm{Ex}} (0, 1)} [ h(Z) ] \text {,} \end{aligned}$$

which proves Lemma 4. $\square$

Part (ii) of the following lemma is used only to establish a strict inequality in the proof of Theorem 7. Although part (ii) is proved below, there will be different proofs.

Lemma 5

Let $p \in \mathbb {N}$ and let $D \subsetneqq \mathbb {R} ^p$ be an open convex set containing the origin. Let ${\omega }_2> {\omega }_1 > 0$. Define ${\omega }D = \{ {\omega }\varvec{\xi }| \varvec{\xi }\in D \}$ for ${\omega }\in \{ {\omega }_1 , {\omega }_2 \}$. Then:

• (i) ${\omega }_1 D \subset {\omega }_2 D$.

• (ii) $( {\omega }_2 D) \setminus ( {\omega }_1 D)$ has a nonempty interior.

Proof

Part (i) is trivial. We prove part (ii). By assumption, there exists $\varvec{\xi }_0 \in \mathbb {R} ^p$ such that $\varvec{\xi }_0 \notin D$. Clearly, $\varvec{\xi }_0 \ne \varvec{0}$, where $\varvec{0}$ denotes the origin $\varvec{0} ^{(p)}$ in $\mathbb {R} ^p$. Define $I = \{ t \in [0, \infty ) | \tilde{t} \varvec{\xi }_0 \in D \text {for all} \tilde{t} \in [0, t] \} = \{ t \in [0, \infty ) | t \varvec{\xi }_0 \in D \}$, where the second equality follows from the convexity of D. Let $t_0 = \sup I \le 1$. Since $\varvec{0}$ is an interior point of D, we have $t_0 > 0$. By definition, $[0, t_0 ) \subset I \subset [0, t_0 ]$. Since D is an open set, $t_0 \notin D$. Thus, $I = [0, t_0 )$ and $t_0 \varvec{\xi }_0 \notin D$.

Now, ${\omega }_1 D$ and ${\omega }_2 D$ are open convex sets containing $\varvec{0}$. Let $\varvec{\xi }_1 = ( {\omega }_1 t_0 ) \varvec{\xi }_0$ and $\varvec{\xi }_2 = [ \{ ( {\omega }_1 + {\omega }_2 ) / 2 \} t_0 ] \varvec{\xi }_0$. Since $\varvec{\xi }_1 \notin {\omega }_1 D$, there exists $\varvec{0} \ne \varvec{a}\in \mathbb {R} ^p$ such that $\varvec{a}^{\top } \varvec{\xi }\ge \varvec{a}^{\top } \varvec{\xi }_1$ for all $\varvec{\xi }\in {\omega }_1 D$. Since $\varvec{0}$ is an interior point of ${\omega }_1 D$, we have $\varvec{a}^{\top } \varvec{\xi }< 0$ for some $\varvec{\xi }\in {\omega }_1 D$. It follows that $\varvec{a}^{\top } \varvec{\xi }_1 < 0$, that $\varvec{a}^{\top } \varvec{\xi }_2 < \varvec{a}^{\top } \varvec{\xi }_1$, and that $\varvec{\xi }_2$ is not a closure point of ${\omega }_1 D$. On the other hand, $\varvec{\xi }_2 \in {\omega }_2 D$. Therefore, $\varvec{\xi }_2$ is contained in the intersection of ${\omega }_2 D$ with the complement of the closure of ${\omega }_1 D$. Thus, since the intersection is open, we conclude that $\varvec{\xi }_2$ is an interior point of $( {\omega }_2 D) \setminus ( {\omega }_1 D)$. $\square$

Proof of Lemma 1

The expressions for $R( \mu , {\hat{p}}^{( \pi _{{\beta }} )} )$ and for $R( \mu , {\hat{p}}^{( {\pi _{{\beta }}}^{*} )} )$ for $\underline{\mu } = - \infty$ are verified by direct calculation. When $\overline{\mu } = \infty$, we have

$$\begin{aligned}&R\left( \mu , {\hat{p}}^{( {\pi _{{\beta }}}^{*} )} \right) = E_{\mu }^{( \varvec{Y}, \varvec{X})} \Bigg [ n \mu + \log {n + m + {\beta }\over m + {\beta }} - \log {e^{(n + m + {\beta }) ( \underline{Y} \wedge \underline{X} )} \over e^{(m + {\beta }) \underline{X}}} \Bigg .\\&\Bigg .\qquad + (- \log [ 1 - e^{(n + m + {\beta }) \{ \underline{\mu } - ( \underline{Y} \wedge \underline{X} ) \} } ]) - \Bigg [- \log \{ 1 - e^{(m + {\beta }) ( \underline{\mu } - \underline{X} )} \} \Bigg ] \Bigg ] \\&\quad = E_{\mu }^{( \varvec{Y}, \varvec{X})} \left[ n \mu + \log {n + m + {\beta }\over m + {\beta }} - \left\{ (n + m + {\beta }) ( \underline{Y} \wedge \underline{X} ) - (m + {\beta }) \underline{X} \right\} \right. \\&\Bigg .\qquad + \sum _{k = 1}^{\infty } {1 \over k} e^{k \left( n + m + {\beta }\right) \left\{ \underline{\mu } - ( \underline{Y} \wedge \underline{X} ) \right\} } - \sum _{k = 1}^{\infty } {1 \over k} e^{k (m + {\beta }) ( \underline{\mu } - \underline{X} )} \Bigg ] \text {,} \end{aligned}$$

from which the desired result follows. $\square$

Proof of Lemma 2

By direct calculation, we have that

$$\begin{aligned} \tilde{p} ^{( {\tilde{\pi }}_{{\alpha }} )} ( \varvec{y}; \varvec{X})&= E_{{\tilde{\pi }}_{{\alpha }}}^{( \mu , {\lambda }) | \varvec{X}} \left[ p( \varvec{y}| \mu , {\lambda }) | \varvec{X}\right] \\&= \frac{ \int _{0}^{\infty } \left\{ \int _{- \infty }^{\underline{y} \wedge \underline{X}} {\tilde{\pi }}_{{\alpha }} ( \mu , {\lambda }) {\lambda }^{n + m} e^{{\lambda }(n \mu - y_{\cdot } + m \mu - X_{\cdot } )} d\mu \right\} d{\lambda }}{ \int _{0}^{\infty } \left\{ \int _{- \infty }^{\underline{X}} {\tilde{\pi }}_{{\alpha }} ( \mu , {\lambda }) {\lambda }^m e^{{\lambda }(m \mu - X_{\cdot } )} d\mu \right\} d{\lambda }} \\&= \frac{ \int _{0}^{\infty } \left\{ \int _{- \infty }^{\underline{y} \wedge \underline{X}} {\lambda }^{n + m + {\alpha }- 1} e^{- {\lambda }( y_{\cdot } + X_{\cdot } )} e^{{\lambda }(n + m) \mu } d\mu \right\} d{\lambda }}{ \int _{0}^{\infty } \left( \int _{- \infty }^{\underline{X}} {\lambda }^{m + {\alpha }- 1} e^{- {\lambda }X_{\cdot }} e^{{\lambda }m \mu } d\mu \right) d{\lambda }} \\&= {{\varGamma }(n + m + {\alpha }- 1) / (n + m) \over {\varGamma }(m + {\alpha }- 1) / m} {( X_{\cdot } - m \underline{X} )^{m + {\alpha }- 1} \over \{ y_{\cdot } + X_{\cdot } - (n + m) ( \underline{y} \wedge \underline{X} ) \} ^{n + m + {\alpha }- 1}} \end{aligned}$$

and that

$$\begin{aligned} \tilde{p} ^{( {{\tilde{\pi }}_{{\alpha }}}{}^{*} )} ( \varvec{y}; \varvec{X})&= E_{{{\tilde{\pi }}_{{\alpha }}}{}^{*}}^{( \mu , {\lambda }) | \varvec{X}} [ p( \varvec{y}| \mu , {\lambda }) | \varvec{X}]\\&= \frac{ \int _{0}^{\infty } \left\{ \int _{- \infty }^{\underline{y} \wedge \underline{X}} {{\tilde{\pi }}_{{\alpha }}}{}^{*} ( \mu , {\lambda }) {\lambda }^{n + m} e^{{\lambda }(n \mu - y_{\cdot } + m \mu - X_{\cdot } )} d\mu \right\} d{\lambda }}{ \int _{0}^{\infty } \left\{ \int _{- \infty }^{\underline{X}} {{\tilde{\pi }}_{{\alpha }}}{}^{*} ( \mu , {\lambda }) {\lambda }^m e^{{\lambda }(m \mu - X_{\cdot } )} d\mu \right\} d{\lambda }} \\&= \frac{ \int _{0}^{\infty } \big \{ \int _{- \infty }^{\underline{y} \wedge \underline{X} \wedge \overline{\mu }} {\lambda }^{n + m + {\alpha }- 1} e^{- {\lambda }( y_{\cdot } + X_{\cdot } )} e^{{\lambda }(n + m) \mu } d\mu \big \} d{\lambda }}{ \int _{0}^{\infty } \big ( \int _{- \infty }^{\underline{X} \wedge \overline{\mu }} {\lambda }^{m + {\alpha }- 1} e^{- {\lambda }X_{\cdot }} e^{{\lambda }m \mu } d\mu \big ) d{\lambda }} \nonumber \\&= {{\varGamma }(n + m + {\alpha }- 1) / (n + m) \over {\varGamma }(m + {\alpha }- 1) / m} {\{ X_{\cdot } - m ( \underline{X} \wedge \overline{\mu } ) \} ^{m + {\alpha }- 1} \over \{ y_{\cdot } + X_{\cdot } - (n + m) ( \underline{y} \wedge \underline{X} \wedge \overline{\mu } ) \} ^{n + m + {\alpha }- 1}} \text {.} \end{aligned}$$

This completes the proof. $\square$