A Bayes minimax result for spherically symmetric unimodal distributions

Abstract

We consider Bayesian estimation of the location parameter $\theta$ of a random vector X having a unimodal spherically symmetric density $f(\Vert x - \theta \Vert ^2)$ for a spherically symmetric prior density $\pi (\Vert \theta \Vert ^2)$. In particular, we consider minimaxity of the Bayes estimator $\delta _\pi (X)$ under quadratic loss. When the distribution belongs to the Berger class, we show that minimaxity of $\delta _\pi (X)$ is linked to the superharmonicity of a power of a marginal associated to a primitive of f. This leads to proper Bayes minimax estimators for certain densities $f(\Vert x - \theta \Vert ^2)$.

Appendix

We recall below a Stein-type lemma in the framework of spherically symmetric distributions whose proof can be found in Fourdrinier and Strawderman (2008).

Lemma 2

Let X be a random vector in $\mathbb R^p$ with density as in (1) and let h be a weakly differentiable function from $\mathbb R^p$ into $\mathbb R^p$. Then

$$\begin{aligned} E_\theta \left[ (X - \theta ) \cdot h(X)\right] = E_\theta \left[ \frac{F(\Vert X-\theta \Vert ^2)}{f(\Vert X - \theta \Vert ^2)} \, \mathrm{div} h(X)\right] , \end{aligned}$$

(54)

where F is defined in (7), provided these expectations exist.

Lemma 2 leads directly to the following result.

Lemma 3

Let g be a weakly differentiable function from $\mathbb R^p$ into $\mathbb R$. Then, for f in (1) and F in (7), we have

$$\begin{aligned} \int _{\mathbb R^p} (\theta - x) \, f(\Vert x-\theta \Vert ^{2}) \, g(\theta ) \, \mathrm{d} \theta&= \int _{\mathbb R^p} \nabla _\theta g(\theta ) \, F(\Vert x-\theta \Vert ^{2}) \, \mathrm{d}\theta \nonumber \\&= - \int _{\mathbb R^p} \nabla _\theta F(\Vert x-\theta \Vert ^{2}) \, g(\theta ) \, \mathrm{d}\theta , \end{aligned}$$

(55)

provided that one of these integrals exists. In addition, when the function f is absolutely continuous, we have

$$\begin{aligned} \int _{\mathbb R^p} (\theta - x) \, (- 2 \, f^\prime )(\Vert x-\theta \Vert ^{2}) \, g(\theta ) \, \mathrm{d} \theta&= \int _{\mathbb R^p} \nabla _\theta g(\theta ) \, f(\Vert x-\theta \Vert ^{2}) \, \mathrm{d}\theta \nonumber \\&= - \int _{\mathbb R^p} \nabla _\theta f(\Vert x-\theta \Vert ^{2}) \, g(\theta ) \, \mathrm{d}\theta . \end{aligned}$$

(56)

Proof

We will only prove (55) since this result only relies on the absolute continuity of F. Let $x = (x_1, \ldots ,x_p) \in \mathbb R^p$, $\theta = (\theta _1, \ldots ,\theta _p) \in \mathbb R^p$ and $1 \le i \le p$. Note that

$$\begin{aligned} (\theta _i - x_i) \, g(\theta ) = (\theta - x) \cdot h^{(i)}(\theta ) \end{aligned}$$

where $h^{(i)}(\theta ) = (0, \ldots ,0, g(\theta ), 0, \ldots ,0)$ is the vector in $\mathbb R^p$ whose all components are equal to 0 with the exception of the ith component which is equal to $g(\theta )$. Then we have

$$\begin{aligned} \int _{\mathbb R^p} (\theta _i - x_i) \, f(\Vert x-\theta \Vert ^{2}) \, g(\Vert \theta \Vert ) \, \mathrm{d} \theta= & {} E_x\left[ (\theta - x) \cdot h^{(i)}(\theta )\right] \nonumber \\= & {} E_x\left[ \mathrm{div}_\theta h^{(i)}(\theta ) \, \frac{F(\Vert x-\theta \Vert ^{2})}{f(\Vert x-\theta \Vert ^{2})} \right] \nonumber \\= & {} E_x\left[ \frac{\partial g(\theta )}{\partial \theta _i} \, \frac{F(\Vert x-\theta \Vert ^{2})}{f(\Vert x-\theta \Vert ^{2})} \right] \nonumber \\= & {} \int _{\mathbb R^p} \frac{\partial g(\theta )}{\partial \theta _i} \, F(\Vert x-\theta \Vert ^{2}) \, \mathrm{d} \theta \end{aligned}$$

(57)

where, for the second equality, Lemma 2 is applied with $h = h^{(i)}$ (the role of x and $\theta$ being interchanged) and, for the third equality, the fact that $\mathrm{div}_\theta h^{(i)}(\theta ) = \partial g(\theta ) / \partial \theta _i$ is used. This gives the first equality in (55). Finally, the second equality in (55) is derived noticing that $(\theta - x) \, f(\Vert x-\theta \Vert ^{2}) = - \nabla _\theta F(\Vert x-\theta \Vert ^{2})$. $\square$

A main feature of Lemma 3 is that the second equality in (55) is valid under weak assumptions on g and F while, in the literature, stronger assumptions are needed. Thus, in Fourdrinier et al. (2012), the function $\theta \longmapsto F\left( \Vert x-\theta \Vert ^{2}\right)$ belongs to a functional space close to the Schwarz space. Here, only the weak differentiability of g is needed.

The next two results are used in Sects. 2 and 3. We use the following notation. For $x \in \mathbb R^p$ and $r \ge 0$, $U_{r,x}$ and $\sigma _{r,x}$ are, respectively, the uniform distribution and the uniform measure on the sphere $S_{r,x} = \{\theta \in \mathbb R^p\ / \Vert \theta - x\Vert = r\}$ of radius r and centered at x. They are related by the following property. If $\gamma$ is a Lebesgue integrable function then

$$\begin{aligned} \int _{\mathbb R^p} \gamma (\theta ) \, \mathrm{d}\theta= & {} \int _0^\infty \int _{S_{r,x}} \gamma (\theta ) \, \mathrm{d}\sigma _{r,x} \, \mathrm{d}r \nonumber \\= & {} \int _0^\infty \sigma _{r,x}(S_{r,x}) \int _{S_{r,x}} \gamma (\theta ) \, \mathrm{d}U_{r,x} \, \mathrm{d}r \end{aligned}$$

(58)

with

$$\begin{aligned} \sigma _{r,x}(S_{r,x}) = \frac{2 \, \pi ^{p/2}}{\Gamma (p/2)} \, \, r^{p-1}. \end{aligned}$$

It follows from (58) that, if $V_{r,x}$ is the uniform distribution on the ball $B_{r,x} = \{\theta \in \mathbb R^p\ / \Vert \theta - x\Vert \le r\}$ of radius r and centered at x and if $\lambda$ is the Lebesgue measure on $\mathbb R^p$, we have

$$\begin{aligned} \int _{B_{r,x}} \gamma (\theta ) \, \mathrm{d}V_{r,x}(\theta )= & {} \frac{1}{\lambda (B_{r,x})} \int _0^r \int _{S_{\tau ,x}} \gamma (\theta ) \, \mathrm{d}\sigma _{\tau ,x}(\theta ) \, \mathrm{d}\tau \nonumber \\= & {} \frac{p}{r^p} \int _0^r \int _{S_{\tau ,x}} \gamma (\theta ) \, \mathrm{d}U_{\tau ,x}(\theta ) \, \tau ^{p-1} \, \mathrm{d}\tau . \end{aligned}$$

(59)

Lemma 4

Let $x \in \mathbb R^p$ be fixed and let $\Theta$ be a random vector in $\mathbb R^p$ with a spherically symmetric distribution around x. Let g be a function from $\mathbb R_+$ into $\mathbb R$. Denote by $E_x$ the expectation with respect to the distribution of $\Theta$.

• (a) If, for $r \ge 0$, $\Theta$ has the uniform distribution $U_{r,x}$ on the sphere $S_{r,x}$ of radius r and centered at x, then there exists a function G from $\mathbb R_+$ into $\mathbb R$ such that

  $$\begin{aligned} E_x \left[ g({\left\| \Theta \right\| }^2) \, \Theta \right] = G(\Vert x\Vert ^2) \, x, \end{aligned}$$

  (60)

  provided this expectation exists. Therefore (60), is valid for any spherically symmetric distribution.

• (b) If $\Theta$ has a unimodal spherically symmetric density $f(\Vert \theta - x\Vert ^2)$ (f is nonincreasing) and if the function g is nonnegative then the function G in (60) is nonnegative.

Proof

We will use the orthogonal decomposition $\Theta = x + U = x + \alpha + \beta$ with U spherically symmetric around 0, $\alpha \in \triangle _x$ and $\beta \in \triangle _x^\bot$ where $\triangle _x$ denotes the linear space spanned by x and $\triangle _x^\bot$ is its orthogonal subspace in $\mathbb R^p$. We have

$$\begin{aligned} E_x \left[ \Theta \, g({\left\| \Theta \right\| }^2) \right] = A(x) + B(x) \end{aligned}$$

(61)

where

$$\begin{aligned} A(x) = E_0[(x + \alpha ) \, g(\Vert x + \alpha \Vert ^2 + \Vert \beta \Vert ^2)] \end{aligned}$$

and

$$\begin{aligned} B(x)= E_0\big [E_0[\beta \, g(\Vert x + \alpha \Vert ^2 + \Vert \beta \Vert ^2) | \alpha ] \big ] = 0, \end{aligned}$$

since $\beta | \alpha$ is spherically symmetric around 0. Setting $\alpha = Z \, x / \Vert x\Vert$, we have

$$\begin{aligned} A(x) = x \, G(\Vert x\Vert ^2) \end{aligned}$$

where

$$\begin{aligned} G(\Vert x\Vert ^2) = E_0 \left[ E_0 \left[ \left( 1 + \frac{Z}{\Vert x\Vert }\right) g \left( \Vert x\Vert ^2 \left( 1 + \frac{Z}{\Vert x\Vert }\right) ^2 + \Vert \beta \Vert ^2 \right) \Bigg | \beta \right] \right] . \end{aligned}$$

This proves the first part.

Now assume that the density $f(\Vert \theta - x\Vert ^2)$ is unimodal and consider

$$\begin{aligned}&E_0 \left[ \left( 1 + \frac{Z}{\Vert x\Vert }\right) g \left( \Vert x\Vert ^2 \left( 1 + \frac{Z}{\Vert x\Vert }\right) ^2 + \Vert \beta \Vert ^2 \right) \Bigg | \beta , \left( 1 + \frac{Z}{\Vert x\Vert }\right) ^2 = y^2 \right] \\&= g \left( \Vert x\Vert ^2 \, y^2 + \Vert \beta \Vert ^2 \right) E_0 \left[ \left( 1 + \frac{Z}{\Vert x\Vert }\right) \Bigg | \beta , \left( 1 + \frac{Z}{\Vert x\Vert }\right) ^2 = y^2 \right] . \end{aligned}$$

To finish the proof it suffices to show that the above conditional expectation is nonnegative which can be seen noticing that

$$\begin{aligned}&E_0 \left[ \left( 1 + \frac{Z}{\Vert x\Vert }\right) \Bigg | \beta , \left( 1 + \frac{Z}{\Vert x\Vert }\right) ^2 = y^2 \right] \\&\quad = \displaystyle { |y| \, \frac{f\big (\{[- 1 + |y|]^2 \, \Vert x\Vert ^2\} + \Vert \beta \Vert ^2) - f(\{[- 1 - |y|]^2 \, \Vert x\Vert ^2\} + \Vert \beta \Vert ^2\big )}{f\big (\{[- 1 + |y|]^2 \, \Vert x\Vert ^2\} + \Vert \beta \Vert ^2) + f(\{[- 1 - |y|]^2 \, \Vert x\Vert ^2\} + \Vert \beta \Vert ^2\big )} } \ge 0 , \end{aligned}$$

by monotonicity of f. $\square$

Versions of Lemma 4 have been often used in the literature (for instance, in Cellier et al. 1995 and in Fourdrinier et al. 2012) when dealing with spherical densities. Here, we provide an extension to the entire class of spherically symmetric distributions.

Lemma 5

If the prior $\pi (\Vert \theta \Vert ^2)$ in (3) is unimodal and if its Laplacian $\Delta (\pi (\Vert \theta \Vert ^2))$ is a nondecreasing function of $\Vert \theta \Vert ^2$ then, for any $r \ge 0$ and for any $x \in \mathbb R^p$,

$$\begin{aligned} \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}\sigma _{r,x}(\theta ) = \frac{2 \, \pi ^{p/2}}{\Gamma (p/2)} \, r^{p-1} \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}U_{r,x}(\theta ) \le 0. \end{aligned}$$

(62)

Proof

Under the conditions of the lemma, Fourdrinier and Strawderman (2008) showed that the sphere mean $\int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}U_{r,x}(\theta )$ is a nondecreasing function of r and that the ball mean $\int _{B_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}V_{r,x}(\theta )$ is nonpositive (which may also be derived from Lemma 4). Therefore, for any $r \ge 0$,

$$\begin{aligned} 2 \, \Vert x\Vert ^2 \, \pi ^\prime (\Vert x\Vert ^2) = x \cdot \nabla \pi (\Vert x\Vert ^2) \le \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}U_{r,x}(\theta ) \end{aligned}$$

and, as $\pi ^\prime (\Vert x\Vert ^2) \le 0$ by unimodality of $\pi (\Vert \theta \Vert ^2)$, $\int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}U_{r,x}(\theta )$ is nonpositive in a neighborhood of 0. If there exists $r_0 > 0$ such that

$$\begin{aligned} \delta = \int _{S_{r_0,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}U_{r_0,x}(\theta ) > 0, \end{aligned}$$

then, using (59), we have

$$\begin{aligned} 0\ge & {} \int _{B_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}V_{r,x}(\theta ) \\= & {} \frac{p}{r^p} \left( \int _0^{r_0} \tau ^{p-1 } \int _{S_{\tau ,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}U_{\tau ,x}(\theta ) \, \mathrm{d}\tau \right. \\&\left. + \int _{r_0}^r \tau ^{p-1 } \int _{S_{\tau ,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}U_{\tau ,x}(\theta ) \, \mathrm{d}\tau \right) \\\ge & {} \frac{p}{r^p} \left( 2 \, \Vert x\Vert ^2 \, \pi ^\prime (\Vert x\Vert ^2) \int _0^{r_0} \tau ^{p-1 } \, \mathrm{d}\tau + \delta \int _{r_0}^r \tau ^{p-1 } \, \mathrm{d}\tau \right) \\= & {} \frac{1}{r^p} \left( 2 \, r_0^p \, \Vert x\Vert ^2 \, \pi ^\prime (\Vert x\Vert ^2) + \delta \, (r^p - r_0^p) \right) . \end{aligned}$$

As this last quantity goes to $\delta$ when r goes to infinity, the nonpositivity of the ball mean $\int _{B_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}V_{r,x}(\theta )$ is contradicted and (62) follows. $\square$

The sphere mean in (62) occurs in Bayesian analysis such as in Fourdrinier and Strawderman (2008). The interest of Lemma 5 is that its sign may be controled although the integrand term changes sign.

Corollary 4

Under the conditions of Lemma 5, for any $y \in \mathbb R_+$, the ratio of the functions defined in (19) and (20) equals, for any $x \in \mathbb R^p$ and for $y = \Vert x\Vert ^2$,

$$\begin{aligned} \frac{\gamma (y)}{\Gamma (y)} = E_x^*\left[ \frac{f(R^2)}{F(R^2)}\right] \end{aligned}$$

(63)

where $E_x^*$ is the expectation with respect to a density proportional to

$$\begin{aligned} F(r^2) \left| \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^2) \, \mathrm{d}\sigma _{r,x}(\theta ) \right| , \end{aligned}$$

provided the function $\pi$ is not constant.

Proof

According to (18), we can write, for any $x \in \mathbb R^p$,

$$\begin{aligned} x \cdot \nabla m(\Vert x\Vert ^2) = \Vert x\Vert ^2 \, \gamma (\Vert x\Vert ^2) \end{aligned}$$

and

$$\begin{aligned} x \cdot \nabla M(\Vert x\Vert ^2) = \Vert x\Vert ^2 \, \Gamma (\Vert x\Vert ^2) \end{aligned}$$

so that

$$\begin{aligned} \frac{\gamma (\Vert x\Vert ^2)}{\Gamma (\Vert x\Vert ^2)} = \frac{x \cdot \nabla m(\Vert x\Vert ^2)}{x \cdot \nabla M(\Vert x\Vert ^2)} . \end{aligned}$$

(64)

Now, by absolute continuity of f and F, interchange of the gradient and the integral sign is valid so that, according to the expressions of m and M in (5) and (6), (64) becomes

$$\begin{aligned} \frac{\gamma (\Vert x\Vert ^2)}{\Gamma (\Vert x\Vert ^2)}= & {} \frac{x \cdot \int _{\mathbb R^{p}} \nabla f(\Vert x-\theta \Vert ^{2}) \, \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta )}{x \cdot \int _{\mathbb R^{p}} \nabla F(\Vert x-\theta \Vert ^{2}) \, \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta )} \nonumber \\= & {} \frac{x \cdot \int _{\mathbb R^{p}} f(\Vert x-\theta \Vert ^{2}) \, \nabla \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta )}{x \cdot \int _{\mathbb R^{p}} F(\Vert x-\theta \Vert ^{2}) \, \nabla \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta )}, \end{aligned}$$

(65)

by Lemma 3. Hence, by linearity of the inner product and integrating over the sphere, it follows from (65) that

$$\begin{aligned} \frac{\gamma (\Vert x\Vert ^2)}{\Gamma (\Vert x\Vert ^2)}= & {} \frac{ \int _0^\infty f(r^{2}) \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\sigma _{r,x}(\theta ) \mathrm{d}r }{ \int _0^\infty F(r^{2}) \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\sigma _{r,x}(\theta ) \mathrm{d}r } \nonumber \\= & {} \frac{ \int _0^\infty f(r^{2}) \left| \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\sigma _{r,x}(\theta ) \right| \mathrm{d}r }{ \int _0^\infty F(r^{2}) \left| \int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\sigma _{r,x}(\theta ) \right| \mathrm{d}r } \, , \end{aligned}$$

(66)

since, according to (62), $\int _{S_{r,x}} x \cdot \nabla \pi (\Vert \theta \Vert ^{2}) \, d\sigma _{r,x}(\theta )$ has constant sign. Hence the result follows. $\square$

The next lemma is used in the proof of Theorem 3.

Lemma 6

Let $\varphi$ be a function from $\mathbb R^p$ into $\mathbb R$ such that its Laplacian exists on $\mathbb R^p$ and let $\beta \in \mathbb R$. Then, for any $x \in \mathbb R^p$, we have

$$\begin{aligned} \Delta \varphi ^\beta (x) = \beta \, \varphi ^\beta (x) \left[ \frac{\Delta \varphi (x)}{\varphi (x)} + (\beta - 1) \left\| \frac{\nabla \varphi (x)}{\varphi (x)} \right\| ^2 \right] . \end{aligned}$$

(67)

Proof

For any $x \in \mathbb R^p$, we have

$$\begin{aligned} \Delta \varphi ^\beta (x)= & {} \mathrm{div} \left( \nabla \varphi ^\beta (x)\right) \\= & {} \mathrm{div} \left( \beta \, \varphi ^{\beta -1}(x) \, \nabla \varphi (x) \right) \\= & {} \beta \left\{ \varphi ^{\beta -1}(x) \, \mathrm{div} \left( \nabla \varphi (x)\right) + \nabla \varphi ^{\beta -1}(x) \cdot \nabla \varphi (x) \right\} \\= & {} \beta \left\{ \varphi ^{\beta -1}(x) \, \Delta \varphi (x) + (\beta - 1) \, \varphi ^{\beta -2}(x) \, \nabla \varphi (x) \cdot \nabla \varphi (x) \right\} \\= & {} \beta \, \varphi ^\beta (x)\left\{ \frac{\Delta \varphi (x)}{\varphi (x)} + (\beta - 1) \, \frac{\Vert \nabla \varphi (x)\Vert ^2}{\varphi ^2(x)} \right\} . \end{aligned}$$

$\square$

The next lemma is used in the proof of Theorem 4.

Lemma 7

Let V and T two random variables such that (V, T) has density $f_{(V,T)}$ of the form

$$\begin{aligned} f_{(V,T)}(v,t) = q(t+v) \, h(t) \, g(v) \end{aligned}$$

(68)

for some functions q, h and g.

Let $U = V + T$. Then the density of T given U does not depend on q.

If $g(u-t)$ has nondecreasing monotone likelihood ratio in t with respect to u and if $\lambda$ is a nondecreasing function then the conditional expectation $E[\lambda (T) \, | \, U = u]$ is nondecreasing in u.

Proof

Clearly (U, T) has density $f_{(U,T)}$ given by

$$\begin{aligned} f_{(U,T)}(u,t) = f_{(V,T)}(u - t,t) \end{aligned}$$

so that the conditional density of T given $U = u$ can be expressed as

$$\begin{aligned} f_T(t \, | \, U = u) = \frac{f_{(U,T)}(u,t)}{\int f_{(U,T)}(u,t^\prime ) \, \mathrm{d}t^\prime } = \frac{f_{(V,T)}(u - t,t)}{\int f_{(V,T)}(u - t^\prime ,t^\prime ) \, \mathrm{d}t^\prime }. \end{aligned}$$

Using (68) this becomes

$$\begin{aligned} f_T(t \, | \, U = u) = \frac{q(u) \, h(t) \, g(u - t)}{\int q(u) \, h(t^\prime ) \, g(u - t^\prime ) \, \mathrm{d}t^\prime } = \frac{h(t) \, g(u - t)}{\int h(t^\prime ) \, g(u - t^\prime ) \, \mathrm{d}t^\prime }. \end{aligned}$$

Hence the first result follows.

Now, according to the monotone likelihood property of g, for fixed $u_1 < u_2$,

$$\begin{aligned} \frac{f_T(t \, | \, U = u_2)}{f_T(t \, | \, U = u_1)} \propto \frac{g(u_2 - t)}{g(u_1 - t)} \end{aligned}$$

is nondecreasing in t. Hence, as $\lambda$ is a nondecreasing function, $E[\lambda (T) \, | \, U = u]$ is nondecreasing in u. $\square$

The two following lemmas are used in the proof of Theorem 4.

Lemma 8

Assume that the sampling density in (1) is a variance mixture of normals as in (33), that is,

$$\begin{aligned} f(t) = \int ^{\infty }_{0} \, \frac{1}{(2 \, \pi \, v)^{p/2}} \, \exp \left( -\frac{t}{2 \, v}\right) \mathrm{d}G(v) , \end{aligned}$$

(69)

where G is the distribution of a random variable V on $\mathbb R_{+}$. Denote by E the expectation with respect to G. Assume also that the prior density is as in (3) and is unimodal, that is, $\pi ^\prime (\left\| \theta \right\| ^{2}) \le 0$. For $\lambda > 0$ and for a random variable W having a noncentral chi-squared distribution $\chi _p^2(\lambda )$ with p degrees of freedom and noncentral parameter $\lambda$, let

$$\begin{aligned} H^\prime (\lambda ,V) = \frac{\partial }{\partial \lambda } E_\lambda [\pi (V \, W)] . \end{aligned}$$

(70)

Then, for any $y \in \mathbb R_+$, the ratio of the functions defined in (19) and (20) equals

$$\begin{aligned} \frac{\gamma (y)}{\Gamma (y)} = \frac{E\left[ \frac{1}{V} \, |H^\prime (\frac{y}{2 \, V},V)|\right] }{E\left[ |H^\prime (\frac{y}{2 \, V},V)|\right] } = E_y^*[V^{-1}] \end{aligned}$$

(71)

where $E_y^*$ is the expectation defined through the second equality in (71).

Proof

According to (18), we can write, for any $x \in \mathbb R^p$,

$$\begin{aligned} 2 \, \Vert x\Vert ^2 \, m^\prime (\Vert x\Vert ^2) = x \cdot \nabla m(\Vert x\Vert ^2) = \Vert x\Vert ^2 \, \gamma (\Vert x\Vert ^2) \end{aligned}$$

and

$$\begin{aligned} 2 \, \Vert x\Vert ^2 \, M^\prime (\Vert x\Vert ^2) = x \cdot \nabla M(\Vert x\Vert ^2) = \Vert x\Vert ^2 \, \Gamma (\Vert x\Vert ^2) \end{aligned}$$

so that

$$\begin{aligned} \frac{\gamma (\Vert x\Vert ^2)}{\Gamma (\Vert x\Vert ^2)} = \frac{m^\prime (\Vert x\Vert ^2)}{M^\prime (\Vert x\Vert ^2)} . \end{aligned}$$

(72)

Now, thanks to (5) and to (69), we have

$$\begin{aligned} m(\Vert x\Vert ^2)= & {} \int _{\mathbb R^{p}} \int ^{\infty }_{0} \frac{1}{(2 \, \pi \, v)^{p/2}} \, \exp \left( -\frac{\Vert x-\theta \Vert ^{2}}{2 \, v}\right) \mathrm{d}G(v) \, \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta \nonumber \\= & {} \int ^{\infty }_{0} \int _{\mathbb R^{p}} \frac{1}{(2 \, \pi \, v)^{p/2}} \, \exp \left( -\frac{\Vert x-\theta \Vert ^{2}}{2 \, v}\right) \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta \, \mathrm{d}G(v) \nonumber \\= & {} E \left[ E_{\Vert x\Vert ^2 / 2 V} [\pi (V \, W)] \right] , \end{aligned}$$

(73)

by Fubini’s Theorem for the second equality and noticing, for the third equality, that $W = \Vert \theta \Vert ^{2} / V | V \sim \chi _p^2(\lambda )$ with $\lambda = \Vert x\Vert ^2 / 2 V$. Similarly, thanks to (6) and to (69) and by definition of F in (7), we have

$$\begin{aligned} M(\Vert x\Vert ^2)= & {} \int _{\mathbb R^{p}} \frac{1}{2} \int ^{\infty }_{\Vert x-\theta \Vert ^{2}} \int ^{\infty }_{0} \frac{1}{(2 \, \pi \, v)^{p/2}} \, \exp \left( -\frac{u}{2 \, v}\right) \mathrm{d}G(v) \, \mathrm{d}u \, \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta \nonumber \\= & {} \int ^{\infty }_{0} \frac{1}{(2 \, \pi \, v)^{p/2}} \int _{\mathbb R^{p}} \int ^{\infty }_{\Vert x-\theta \Vert ^{2}} \frac{1}{2} \, \exp \left( -\frac{u}{2 \, v}\right) \mathrm{d}u \, \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta \, \mathrm{d}G(v) \nonumber \\= & {} \int ^{\infty }_{0} \int _{\mathbb R^{p}} \frac{v}{(2 \, \pi \, v)^{p/2}} \, \exp \left( -\frac{\Vert x-\theta \Vert ^{2}}{2 \, v}\right) \, \pi (\Vert \theta \Vert ^{2}) \, \mathrm{d}\theta \, \mathrm{d}G(v) \nonumber \\= & {} E \left[ V \, E_{\Vert x\Vert ^2 / 2 V} [\pi (V \, W)] \right] . \end{aligned}$$

(74)

Therefore, setting $y = \Vert x\Vert ^2$ and differentiating with respect to y, we have

$$\begin{aligned} m^\prime (y) = E\left[ \frac{1}{2 \, V} \, H^\prime \left( \frac{y}{2 \, V},V\right) \right] \end{aligned}$$

(75)

and

$$\begin{aligned} M^\prime (y) = E\left[ \frac{1}{2} \, H^\prime \left( \frac{y}{2 \, V},V\right) \right] \end{aligned}$$

(76)

with $H^\prime (\lambda ,V)$ defined in (70). Note that, as $\pi ^\prime (\left\| \theta \right\| ^{2}) \le 0$, we have $H^\prime (\lambda ,V) \le 0$, for any $\lambda > 0$, since $\chi _p^2(\lambda )$ has increasing monotone likelihood ratio in $\lambda$. Therefore, it follows from (72), (75) and (76) that

$$\begin{aligned} \frac{\gamma (y)}{\Gamma (y)} = \frac{E[\frac{1}{V} \, H^\prime (\frac{y}{2 \, V},V)]}{E\left[ H^\prime (\frac{y}{2 \, V},V)\right] } = \frac{E[\frac{1}{V} \, |H^\prime (\frac{y}{2 \, V},V)|]}{E\left[ |H^\prime (\frac{y}{2 \, V},V)|\right] } = E_y^*[V^{-1}] \, , \end{aligned}$$

which is the desired result. $\square$

Lemma 9

Assume that the sampling density in (1) and the prior density in (3) are variance mixtures of normals as in (33) and (34), respectively, that is,

$$\begin{aligned} f(t) = \int ^{\infty }_{0} \, \frac{1}{(2 \, \pi \, v)^{p/2}} \, \exp \left( -\frac{t}{2 \, v}\right) \mathrm{d}G(v), \end{aligned}$$

(77)

where G is the distribution of a random variable V on $\mathbb R_{+}$, and

$$\begin{aligned} \pi (\left\| \theta \right\| ^{2}) = \int ^{\infty }_{0} \frac{1}{(2 \, \pi \, t)^{p/2}} \, \exp \left( -\frac{\left\| \theta \right\| ^{2}}{2 \, t}\right) h(t) \, \mathrm{d}t, \end{aligned}$$

(78)

for a certain (possibly improper) mixing density h. Then the superharmonicity of $M^\beta$ can be expressed as

$$\begin{aligned} Q + R \, \le \, (1 - \beta ) \, \frac{p-2}{2}, \end{aligned}$$

(79)

where

$$\begin{aligned} Q = - h(0) \left[ \frac{\int ^{\infty }_{0} v^{1-p/2} \, \exp \left( \frac{-s}{v}\right) \mathrm{d}G(v)}{\int ^{\infty }_{0} J_{p/2+1}(v) \, v \, \mathrm{d}G(v)} - (1 - \beta ) \, \frac{\int ^{\infty }_{0} v^{2-p/2} \, \exp \left( \frac{-s}{v}\right) \mathrm{d}G(v)}{\int ^{\infty }_{0} J_{p/2}(v) \, v \, \mathrm{d}G(v)} \right] \end{aligned}$$

(80)

and

$$\begin{aligned} R= & {} \frac{\int ^{\infty }_{0} \int ^{\infty }_{0} - \frac{h^{\prime }(t)}{(v+t)^{p/2}} \, \exp \left( \frac{-s}{v+t}\right) \, \mathrm{d}t \, v \, \mathrm{d}G(v)}{\int ^{\infty }_{0} J_{1+p/2}(v) \, v \, \mathrm{d}G(v)}\nonumber \\&- (1 - \beta ) \, \frac{\int ^{\infty }_{0} \int ^{\infty }_{0} - \frac{h^{'}(t)}{(v+t)^{p/2 - 1}} \, \exp \left( \frac{-s}{v+t}\right) \, \mathrm{d}t \, v \, \mathrm{d}G(v)}{\int ^{\infty }_{0} J_{p/2}(v) \, v \, \mathrm{d}G(v)} \end{aligned}$$

(81)

with

$$\begin{aligned} J_{k}(v) = \int ^{\infty }_{0} (t + v)^{-k} \, \exp \left( \frac{-s}{t+v}\right) \, h(t) \, \mathrm{d}t. \end{aligned}$$

(82)

Proof

By the last equality in (31), $\Delta M^\beta (\Vert x\Vert ^2) \le 0$ is equivalent to

$$\begin{aligned} \frac{\Delta M(\left\| x\right\| ^{2})}{\left\| \nabla M(\left\| x\right\| ^{2})\right\| } \, - \, (1 - \beta ) \frac{\left\| \nabla M(\left\| x\right\| ^{2})\right\| }{M(\left\| x\right\| ^{2})} \le 0, \end{aligned}$$

(83)

Now, for a sampling density (77) and a prior (78), it easy to derive

$$\begin{aligned} M(\left\| x\right\| ^{2}) = \frac{1}{(2 \pi )^{p/2}} \int ^{\infty }_{0} \int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2}} \, \exp \left( -\frac{1}{2} \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v), \end{aligned}$$

so that

$$\begin{aligned} \nabla M(\left\| x\right\| ^{2}) = - \frac{1}{(2 \pi )^{p/2}} \int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 1}} \, \exp \left( -\frac{1}{2} \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v) \; x, \end{aligned}$$

and

$$\begin{aligned} \Delta M(\left\| x\right\| ^{2})= & {} \frac{1}{(2 \pi )^{p/2}} \int ^{\infty }_{0} \int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 1}} \left( \frac{\left\| x\right\| ^{2}}{t+v} \, - \, p\right) \\&\times \exp \left( -\frac{1}{2} \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v). \end{aligned}$$

Hence, Inequality (83) becomes

$$\begin{aligned}&\frac{\int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 1}} \left( \frac{\left\| x\right\| ^{2}}{t+v} \, - \, p\right) \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)}{\left\| x\right\| \, \int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 1}} \, \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)} \\&\quad - (1 - \beta ) \, \frac{\left\| x\right\| \, \int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 1}} \, \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)}{\int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2}} \, \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)} \le 0, \end{aligned}$$

which is equivalent to

$$\begin{aligned}&\frac{\int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 2}} \, \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)}{\int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 1}} \, \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)} \\&\quad - (1 - \beta ) \, \frac{\int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2 + 1}} \, \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)}{\int ^{\infty }_{0}\int ^{\infty }_{0} \, \frac{v}{(t+v)^{p/2}} \, \exp \left( -\frac{1}{2} \, \frac{\left\| x\right\| ^{2}}{t+v}\right) \, h(t) \, \mathrm{d}t \, \mathrm{d}G(v)} \le \frac{p}{\Vert x\Vert ^{2}} \end{aligned}$$

and can be conveniently expressed as

$$\begin{aligned} \frac{\int ^{\infty }_{0} J_{p/2+2}(v) \, v \, \mathrm{d}G(v)}{\int ^{\infty }_{0} J_{p/2+1}(v) \, v \, \mathrm{d}G(v)} - (1 - \beta ) \, \frac{\int ^{\infty }_{0} J_{p/2+1}(v) \, v \, \mathrm{d}G(v)}{\int ^{\infty }_{0} J_{p/2}(v) \, v \, \mathrm{d}G(v)} \le \frac{p}{2 \, s} \end{aligned}$$

(84)

where $J_{k}(v)$ is in (82) and $s = \Vert x\Vert ^{2} / 2$. Setting $\mathrm{d}w = (v + t)^{-2}\, \exp \left( -s / (t+v)\right) \, \mathrm{d}t$ and $u = (v + t)^{-k+2} \, h(t)$, so that $w = 1 / s \, \exp \left( -s / (t+v)\right)$ and $\mathrm{d}u = [(v + t)^{1-k} \, (2 - k) \, h(t) + (v + t)^{2-k} \, h^{'}(t)] \, \mathrm{d}t$, we have

$$\begin{aligned} J_{k}(v)= & {} \frac{-1}{s} \, v^{2-k} \, h(0) \, \exp \left( \frac{-s}{v}\right) \, + \, \frac{k - 2}{s} \, J_{k-1}(v)\\&- \frac{1}{s} \int ^{\infty }_{0} \, (v + t)^{2-k} h^{'}(t) \, \exp \left( \frac{-s}{t+v}\right) \, \mathrm{d}t, \end{aligned}$$

since

$$\begin{aligned} \lim _{t \rightarrow \infty } \, (v+t)^{2-k} \, h(t) = \,0. \end{aligned}$$

Then using this representation of $J_{k}(v)$ in (84) gives rise to (79) with Q in (80) and R in (81). $\square$