A review of Bayesian asymptotics in general insurance applications

Abstract

Over the last two decades, Bayesian methods have been widely used in general insurance applications, ranging from credibility theory to loss-reserves estimation, but this literature rarely addresses questions about the method’s asymptotic properties. In this paper, we review the Bayesian’s notion of posterior consistency in both parametric and nonparametric models and its implication on the sensitivity of the posterior to the actuary’s choice of prior. We review some of the techniques for proving posterior consistency and, for illustration, we apply these results to investigate the asymptotic properties of several recently proposed Bayesian methods in general insurance.

Deferred technical details

1.1 From Example 3.3

Since

$$\begin{aligned} \sup _{\theta >0}T(X_1, \theta )=T(X_1, \theta _{X_1})=\log \Bigl [\Bigl (\frac{\beta }{\theta ^\star }\Bigr )^{\beta }\frac{1}{(1+\beta )^{\beta +1}} \frac{(x+\theta ^\star )^{\beta +1}}{x}\Bigr ], \end{aligned}$$

we have

$$\begin{aligned} E_{\theta ^\star }\Bigl \{\sup _{\theta >0}T(X_1, \theta )\Bigl \} \le \log \Bigl [\Bigl (\frac{\beta }{\theta ^\star }\Bigr )^{\beta }\frac{1}{(\beta +1)^{\beta +1}}\Bigr ]+(1+\beta )|E_{\theta ^\star }[\log (X_1+\theta ^\star )]|+|E_{\theta ^\star }[\log X_1]|. \end{aligned}$$

In view of

$$\begin{aligned} E_{\theta ^\star }[\log X_1]=\beta (\theta ^\star )^{\beta }\int _0^{\infty } \frac{\log x}{(x+\theta ^\star )^{\beta +1}} dx, \end{aligned}$$

and

$$\begin{aligned} E_{\theta ^\star }[\log (x+\theta ^\star )]=\beta (\theta ^\star )^{\beta }\int _0^{\infty }\frac{\log (x+\theta ^\star )}{(x+\theta ^\star )^{\beta +1}}dx, \end{aligned}$$

it suffices to show that both integrals are finite.

To see that the first integral is finite, consider the complex-valued function $f(z)=\frac{\log ^2 z}{(\theta +z)^{\beta +1}}$. It is clear that f has a pole at $z=-\theta$. Take $\epsilon >0$ and $0<\rho<1<R$. Let $L_1$ be the line segment $[\rho +\epsilon i, R+\epsilon i]$, $L_2$ be the line segment $[\rho -\epsilon i, R-\epsilon i]$, $\gamma _{\rho }$ be the part of the circle $|z|=\rho$ from $\rho -\epsilon i$ clockwise to $\rho +\epsilon i$, and $\gamma _R$ be the part of the circle $|z|=R$ from $R+\epsilon i$ counterclockwise to $R-\epsilon i$. Then the Residue Theorem implies

$$\begin{aligned}&\int _{L_1}\frac{\log ^2 x}{(\theta +x)^{\beta +1}}dx+\int _{\gamma _R}\frac{\log z}{(\theta +z)^{\beta +1}}dz+\int _{L_2}\frac{(\log ^2 x+4\pi i\log x-4\pi ^2)}{(\theta +x)^{\beta +1}}dx\\&\quad+ \int _{\gamma _{\rho }}\frac{\log z}{(\theta +z)^{\beta +1}}dz=2\pi i \, \mathrm {Res}(f, -\theta ), \end{aligned}$$

where $\mathrm {Res}(f, -\theta )$ denotes the residue of f at $-\theta$. We have

$$\begin{aligned} \bigg | \int _{\gamma _R}\frac{\log z}{(\theta +z)^{\beta +1}}dz \bigg | \le 2\pi R\frac{|\log R+2\pi |}{(R-\theta )^{\beta +1}}\rightarrow 0 \quad {\text {as}}\,\,R\rightarrow \infty , \end{aligned}$$

and

$$\begin{aligned} \bigg | \int _{\gamma _{\rho }}\frac{\log z}{(\theta +z)^{\beta +1}}dz \bigg | \le 2\pi \rho \frac{|\log \rho +2\pi |}{(\rho -\theta )^{\beta +1}}\rightarrow 0 \quad {\text {as}}\,\,\rho \rightarrow 0. \end{aligned}$$

It follows that

$$\begin{aligned} i\int _0^{\infty } \frac{\log x}{(\theta +x)^{\beta +1}} dx=\frac{i}{2}Res(f, -\theta )+\pi \int _0^{\infty }\frac{1}{(\theta +x)^{\beta +1}}dx. \end{aligned}$$

Therefore, $\int _0^{\infty } \frac{\log x}{(\theta +x)^{\beta +1}} dx$ equals the imaginary part of $\frac{i}{2} \, \mathrm {Res}(f, -\theta )$ which is finite. Similarly, the second integral is also finite.

1.2 From Example 4.3

1.2.1 Verification of Conditions 1 and 2 of Theorem 4.1

To check Conditions 1–2 of Theorem 4.1, we need to identify a candidate sieve $\Theta _n$. Fix an arbitrary $\varepsilon \in (0,1)$ and let $\delta < \varepsilon /4$ be as in Theorem 4.1. Given $b> a > 0$, define the set of gamma scale mixtures $\Theta _{a,b}^\delta =\{\theta _G: G((a,b]) \ge 1-\delta \}$. We will show that $\Theta _n := \Theta _{a_n,b_n}^\delta$, with $a_n=e^{-cn}$ and $b_n=e^{cn}$, satisfies Conditions 1–2 of the theorem, where

$$\begin{aligned} c < \Bigl ( 1 + \log \frac{6 + \delta }{\delta } \Bigr )^{-1} \frac{\varepsilon ^2}{16}. \end{aligned}$$

We begin with checking Condition 2 about the metric entropy. Let d be the $L^1$-distance and let $k_u$ denote the gamma density with scale parameter u (and fixed shape s). Without loss of generality, let $v> u > 0$. Then we get

$$\begin{aligned} d(k_u, k_v)&= \frac{1}{\Gamma (s)} \int _0^{\infty } x^{s-1} |u^{-s} e^{-x/u} - v^{-s} e^{-x/v}| \, dx \\&\le \frac{1}{\Gamma (s)} \Bigl \{u^{-s} \int _0^\infty x^{s-1} |e^{-x/u} - e^{-x/v}| \,dx + (u^{-s} - v^{-s}) \int _0^\infty x^{s-1} e^{-x/v} \,dx \Bigr \} \\&= 2 \Bigl ( \frac{v^s}{u^s} - 1 \Bigr ). \end{aligned}$$

For the $b> a > 0$ and $\delta$ introduced above, let $z=(1 + \delta /2)^{1/s}$ and define

$$\begin{aligned} u_m = a z^m, \quad m=1,\ldots ,M, \end{aligned}$$

(5)

where M is the smallest integer such that $a z^{(M+1)} \ge b$, i.e., $M \le (\log z)^{-1} \log (b/a)$. If we partition (a, b] by the sub-intervals

$$\begin{aligned} E_m = (a z^m, a z^{m+1}], \quad m=1,\ldots ,M, \end{aligned}$$

then it is easy to check, based on the bound on the $L^1$-distance above, that

$$\begin{aligned} u,v \in E_m \implies d(k_u, k_v) \le \delta . \end{aligned}$$

Let $\Delta _M$ be the probability simplex in $\mathbb {R}^{M+1}$ and let $\Delta _M^\delta$ be a $(\delta /6)$-net³ in $\Delta _M$. An argument similar to that in the proofs of Lemmas 1 and 2 of Ghosal et al. [22] shows that

$$\begin{aligned} \Bigl \{\sum _{m=1}^M P_m k_{u_m}: (P_1,\ldots ,P_M) \in \Delta _M^\delta \Bigr \}, \end{aligned}$$

a set of finite scale mixtures of gammas with fixed scales $u_1,\ldots ,u_M$ in (5), is a $\delta$-net in $\Theta _{a,b}^\delta$. Therefore, $\log N(\Theta _{a,b}^\delta , \delta , d) \le \log N(\Delta _M, \delta /6, \Vert \cdot \Vert _1)$, and the argument in the proof of Lemma 1 in Ghosal et al. [22], based on Lemma 8 in Barron et al. [2], shows that

$$\begin{aligned} \log N(\Delta _M, \delta /6, \Vert \cdot \Vert _1) \le \Bigl ( 1 + \log \frac{6 + \delta }{\delta } \Bigr ) M. \end{aligned}$$

Specifically, we put

$$\begin{aligned} D=\left\{ (P_1, \ldots , P_m)\in \Delta _M: \sum _{m=1}^MP_m\le 1+\frac{\delta }{6}\right\} . \end{aligned}$$

Let $(P_1, \ldots , P_m)\in \Delta _M$ and $(\widetilde{P}_1, \ldots , \widetilde{P}_m)\in \Delta _M^{\delta }$. Then $||P_m-\widetilde{P}_m||_1<\delta /(6M)$ implies $\sum _{m=1}^M||P_m-\widetilde{P}_m||_1<\delta /6$. Therefore, the $\delta /6$-covering number $N(\Delta _M, \delta /6, ||\cdot ||_1)$ of $\Delta _M$ might be bounded above by the product of the number of cubes of length $\delta /6$ covering $[0, 1]^M$ and the volume of D, which is further bounded above by

$$\begin{aligned} \frac{1}{M!} \left( \frac{6M}{\delta }\right) ^M\left( 1+\frac{\delta }{6}\right) ^M. \end{aligned}$$

It follows that

$$\begin{aligned} \log (\Delta _M, \delta /6, ||\cdot ||_1)\le & {} -\log M!+M\log \left( \frac{6M}{\delta }\right) +M\left( 1+\frac{\delta }{6}\right) \\\le & {} -M\log M + M +M \log M +M\log \frac{6+\delta }{\delta }\\= & {} \left( 1+\log \frac{6+\delta }{\delta }\right) M. \end{aligned}$$

Since M is bounded by a constant (depending only on $\delta$) times $\log (b/a)$, we clearly have that Condition 2 of Theorem 4.1 holds with $a_n = e^{-cn}$ and $b_n = e^{cn}$.

For Condition 1 about the prior mass assigned to the sieve $\Theta _n = \Theta _{a_n,b_n}^\delta$, it suffices to bound the prior probability of $\{G: G((a_n, b_n]) \ge 1-\delta \}$. A fundamental property of the Dirichlet process is that G(A) has a beta distribution with parameters $\alpha G_0(A)$ and $\alpha G_0(A^c)$. In the present case, if we let

$$\begin{aligned} \alpha _n = \alpha G_0((a_n, b_n]) \quad {\text {and}} \quad \beta _n = \alpha \{1 - G_0((a_n, b_n])\}, \end{aligned}$$

then we have

$$\begin{aligned} \Pi (\{G: G((a_n,b_n]) < 1-\delta \}) = \frac{1}{B(\alpha _n,\beta _n)} \int _0^{1-\delta } z^{\alpha _n-1} (1-z)^{\beta _n-1} \,dz, \end{aligned}$$

where $B(a,b) = \Gamma (a)\Gamma (b)/\Gamma (a+b)$ is the beta function. The right-hand side of the previous display is upper-bounded by

$$\begin{aligned} \frac{1}{B(\alpha _n, \beta _n)} (1-\delta )^{\alpha _n-1} \frac{1-\delta ^{\beta _n}}{\beta _n}. \end{aligned}$$

As $n \rightarrow \infty$, it is clear that $\alpha _n \rightarrow \alpha$ and $\beta _n \rightarrow 0$. Some simple analysis shows that the latter two terms in the upper bound have finite and non-zero limits as $n \rightarrow \infty$, so only the beta function term will be relevant. Using some basic properties of the gamma function we have

$$\begin{aligned} B(\alpha _n, \beta _n) = \frac{\Gamma (\alpha _n) \Gamma (\beta _n)}{\Gamma (\alpha _n + \beta _n)} = \{1 + o(1)\} \frac{\Gamma (\beta _n + 1)}{\beta _n} = \frac{O(1)}{\beta _n}. \end{aligned}$$

Therefore, the prior probability of the sieve vanishes at the rate $\beta _n = \alpha \{1-G_0((a_n,b_n])\}$ as $n \rightarrow \infty$. Following Lau et al. [34], if we take $G_0$ to be a gamma distribution with shape parameter t and scale parameter r, then it is easy to see that

$$\begin{aligned} G_0((0,a_n]) = \int _0^{a_n} \frac{1}{r^t \Gamma (t)} x^{t-1} e^{-x/r} \,dx \lesssim a_n^{t-1} \end{aligned}$$

and, by Markov’s inequality,

$$\begin{aligned} G_0((b_n,\infty )) \lesssim b_n^{-1}. \end{aligned}$$

With $a_n = e^{-cn}$ and $b_n = e^{cn}$, it is clear that Condition 1 of Theorem 4.1 holds.

1.2.2 The best scale-mixture of gammas is a single gamma

Recall that $\theta ^\star$ is a $\mathsf{Gamma}(\alpha ^\star , \lambda ^\star )$ density and $\theta ^\lambda$ is a $\mathsf{Gamma}(\alpha , \lambda )$ density, where $\alpha < \alpha ^\star$. To prove that the best mixture $\theta _G = \int \theta ^\lambda \, G(d\lambda )$ corresponds to just a single gamma density with an appropriately chosen rate $\lambda$, we need to consider the projection of $\theta ^\star$ onto the space of mixtures $\theta _G$. The claim is that the best mixture approximation to $\theta ^\star$, according to the Kullback–Leibler divergence, is one with G equal to a point mass $\delta _\Lambda$ at

$$\begin{aligned} \Lambda = \Lambda (\alpha , \alpha ^\star , \lambda ^\star ) = \frac{\alpha \lambda ^\star }{\alpha ^\star }, \end{aligned}$$

(6)

which is the value of $\lambda$ that makes the mean of $\theta ^\lambda$ the same as that of $\theta ^\star$. In this context, according to Lemma 2.5 in Shyamalkumar [51] and Lemma 2.3 in Kleijn and van der Vaart [30], it suffices to show that

$$\begin{aligned} \sup _G \int \theta ^\star (x) \frac{\theta _G(x)}{\theta ^\Lambda (x)} \,dx \le 1, \end{aligned}$$

where $\Lambda$ is as in (6). Writing out the definition of $\theta _G$ and switching order of integration, we can see that it suffices to show that

$$\begin{aligned} \int \theta ^\star (x) \frac{\theta ^\lambda (x)}{\theta ^\Lambda (x)} \,dx \le 1 \quad \forall \; \lambda > 0. \end{aligned}$$

(7)

A straightforward calculation shows that

$$\begin{aligned} \int \theta ^\star (x) \frac{\theta ^\lambda (x)}{\theta ^\Lambda (x)} \,dx = \Bigl ( \frac{\lambda ^\star }{\lambda ^\star - \Lambda + \lambda } \Bigr )^{\alpha ^\star } \Bigl (\frac{\lambda }{\Lambda } \Bigr )^\alpha = \Bigl ( \frac{\lambda ^\star }{\lambda + \frac{\alpha ^\star -\alpha }{\alpha ^\star } \lambda ^\star } \Bigr )^{\alpha ^\star } \Bigl ( \frac{\alpha ^\star \lambda }{\alpha \lambda ^\star } \Bigr )^\alpha . \end{aligned}$$

The right-hand side clearly equals 1 if $\lambda = \Lambda$; this is also obvious from the definition. Moreover, using the fact that $\alpha < \alpha ^\star$, it is an easy calculus exercise to show that the right-hand side is actually maximized at $\lambda = \Lambda = (\alpha /\alpha ^\star )\lambda ^\star$. This justifies the intuition from Example 4.3 that the best Kullback–Leibler approximation of $\theta ^\star$ (or $\eta ^\star$) by mixtures of $\theta ^\lambda$ (or $\eta ^\lambda$) is just a single $\theta ^\Lambda$ (or $\eta ^\Lambda$), for suitably chosen $\Lambda$. Therefore, $K(\theta ^\star , \theta _G) \ge K(\theta ^\star , \theta ^\Lambda )$, and the lower-bound is strictly positive.

1.3 From Example 4.4

We first want to show that the proposed prior satisfies the Kullback–Leibler property at $\theta ^\star$. Let $p_\theta (x,y)$ be the joint density of (X, Y) under the proposed nonparametric regression model. For any two regression functions $\theta$ and $\eta$, the Kullback–Leibler divergence $K(p_{\theta }, p_{\eta })$ of $p_{\eta }$ from $p_{\theta }$ can be expressed in terms of the $L^2$-distance between $\theta$ and $\eta$. Indeed, since

$$\begin{aligned} \{y-\eta (x)\}^2=\{y-\theta (x)\}^2+2\{\theta (x)-\eta (x)\}\{y-\theta (x)\} +\{\theta (x)-\eta (x)\}^2, \end{aligned}$$

and $\int \{y-\theta (x)\}p_{\theta }(x, y) \, dy=0$ for all x, we have

$$\begin{aligned} K(p_{\theta }, p_{\eta })= & {} \frac{1}{2} \int \int \{y-\eta (x)\}^2-\{y-\theta (x)\}^2 p_{\theta }(x, y) \, dy \, dx\nonumber \\= & {} \frac{1}{2}\int \{\theta (x)-\eta (x)\}^2 \, dx\nonumber \\= & {} \frac{1}{2}\Vert \theta -\eta \Vert _2^2, \end{aligned}$$

(8)

where $\Vert \cdot \Vert _2$ denotes the $L^2$-norm on $[0, 1]^q$. In view of (8), to verify the Kullback–Leibler property of $\Pi$ at $\theta ^\star$, it suffices to show that $\Pi (\{\theta : \Vert \theta -\theta _0\Vert _2 \le \varepsilon \}>0$ for all $\varepsilon >0$. Since $\theta \in L^2([0, 1]^q)$, the Bessel inequality and (4) imply that for any $\varepsilon >0$, there exists an integer J such that

$$\begin{aligned} \sum _{j>J}(\sigma _j^2+\theta _j^{\star 2})\le \frac{\varepsilon ^2}{4}. \end{aligned}$$

Following Shen and Wasserman ([49, Lemma 5), we have

$$\begin{aligned} \Pi \{\theta : \Vert \theta -\theta ^\star \Vert _2\le \varepsilon \}&=\Pi \Bigl \{\theta : \sum _j(\theta _j-\theta _j^\star )^2<\varepsilon ^2\Bigr \}\\&\ge \Pi \Bigl \{\theta : \sum _{j=1}^J(\theta _j-\theta _j^\star )^2<\frac{\varepsilon ^2}{2}\Bigr \} \, \Pi \Bigl \{\theta : \sum _{j>J}(\theta _j-\theta _j^\star )^2< \frac{\varepsilon ^2}{2} \Bigr \}\\&= \Pi \Bigl \{\theta : \sum _{j=1}^J(\theta _j-\theta _j^\star )^2< \frac{\varepsilon ^2}{2}\Bigr \} \Bigl [1-\Pi \Bigl \{\theta : \sum _{j>J}(\theta _j-\theta _j^\star )^2> \frac{\varepsilon ^2}{2} \Bigr \}\Bigr ]\\&\ge \Pi \Bigl \{\theta : \sum _{j=1}^J(\theta _j-\theta _j^\star )^2< \frac{\varepsilon ^2}{2} \Bigr \} \Bigl [1-\frac{2}{\varepsilon ^2}\sum _{j>J}E(\theta _j-\theta _j^\star )^2\Bigr ]\\&\ge \frac{1}{2} \Pi \Bigl \{\theta : \sum _{j=1}^J(\theta _j-\theta _j^\star )^2<\frac{\varepsilon ^2}{2} \Bigr \}. \end{aligned}$$

The first inequality is due to monotonicity of $\Pi$ and independent of the $\theta _j$’s under $\Pi$; the second inequality is due to the Markov inequality; and the third inequality follows from the definition of J and the fact $E(\theta _j-\theta _j^\star )^2=\sigma _j^2+\theta _j^{\star 2}$. The lower bound involves a ball probability for a J-dimensional normal distribution and, since such a distribution has a positive density on $\mathbb {R}^J$, the latter probability is positive, proving that $\Pi \{\theta : \Vert \theta -\theta ^\star \Vert _2\le \varepsilon \}>0$ for all $\varepsilon >0$. Therefore, the given $\Pi$, with variance $\sigma _j^2$, satisfies the Kullback–Leibler condition at any $\theta ^\star$ such that (4) holds.

Next, to prove consistency with respect to the $L^2$-distance d, it suffices to work with the $L^\infty$-norm $\Vert \cdot \Vert _\infty$, since $\Vert \cdot \Vert _2 \le \Vert \cdot \Vert _\infty$. For an increasing sequence $M_n$ to be specified later, define the sets

$$\begin{aligned} \Theta _{n0}=\{\theta : ||\theta ||_{\infty }<M_n\}\quad {\text {and}}\,\,\Theta _{n1}=\{\theta : ||\theta '||_{\infty }<M_n\}. \end{aligned}$$

Take $\Theta _n=\Theta _{n0} \cap \Theta _{n1}$ as the sieve set. By the relation between $L^2$- and $L^{\infty }$-distance on regression functions, it is clear that the $L^2$ covering numbers for $\Theta _n$ are less than the corresponding $L^{\infty }$ covering numbers. It follows from Theorem 2.7.1 in [54] that the $\delta$-covering number of $\Theta _n$, relative to the $L^{\infty }$-distance, satisfies $\log N(\Theta _n, \delta , \Vert \cdot \Vert _{\infty }) \le CM_n\delta ^{-1}$, where C is a constant that does not depend on n or $\delta$. If we take $\delta <\epsilon /4$, as in Theorem 4.1, then we can get $\log (\Theta _n, \delta , \Vert \cdot \Vert _{\infty })<nr$, for $\beta =\delta ^2<\epsilon ^2/16<\epsilon ^2/8$, by selecting $M_n= r n$, where $r<\delta ^3/C$. This verifies Conditions 2 of Theorem 4.1.

To verify Condition 1 of Theorem 4.1, we need to show that the $\Pi$-probability of $\Theta _n^c$ is exponentially small. To prove this, it suffices to show that both $\Theta ^c_{n0}$ and $\Theta ^c_{n1}$ have exponentially small $\Pi$-probability. Start with $\Theta _{n0}$. Like in Choi and Schervish ([11], Sec. 6.1), we have the following, based on the Chernoff inequality:

$$\begin{aligned} \Pi \{\theta : ||\theta ||_{\infty }>M_n\}\le & {} \Pi \Bigl \{\theta : \sum _j a_j|\theta _j|>M_n\Bigr \}\\\le & {} e^{-tM_n}E\bigl (e^{t\sum _ja_j|\theta _j|}\bigr ), \quad \forall \; t>0\\= & {} e^{-tM_n}\prod _jE\bigl (e^{ta_j\sigma _j|Z_j|}\bigr )\\= & {} e^{-tM_n}e^{(t^2/2)\sum _ja_j^2\sigma _j^2}\prod _j2\Phi (a_j\sigma _jt), \end{aligned}$$

where $Z_j$ is a standard normal random variable with density $\phi$ and distribution function $\Phi$. Here we used the formula for the moment-generating function of the half-normal |Z|:

$$\begin{aligned} \frac{1}{\sqrt{2\pi }}\int _{-\infty }^{\infty }e^{t|z|}e^{-z^2/2}dz= & {} \frac{2}{\sqrt{2\pi }}\int _0^{\infty }e^{tz}e^{-z^2/2}dz \\= & {} \frac{2e^{t^2/2}}{\sqrt{2\pi }}\int _0^{\infty }e^{-(z-t)^2/2}dz\\= & {} \frac{2e^{t^2/2}}{\sqrt{2\pi }}\int _{-t}^{\infty }e^{-u^2/2}du\quad (u=z-t)\\= & {} 2e^{t^2/2}\Phi (t). \end{aligned}$$

Since $\Phi (z)$ is concave on $[0, \infty )$, it can be bounded from above, for small $z>0$, by its first-order Taylor approximation at $z=0$. This implies that $\log \{2\Phi (z)\}\le \log \{1+2\phi (0)z\}$ which, in our cases, gives

$$\begin{aligned} \log \prod _j2\Phi (a_j\sigma _jt)\le \sum _j\log \left( 1+\frac{2a_j\sigma _jt}{\sqrt{2\pi }}\right) \le \frac{2t}{\sqrt{2\pi }}\sum _j a_j\sigma _j. \end{aligned}$$

By assumption $\sum _j a_j\sigma _j<\infty$, we also have $\sum _ja_j^2\sigma ^2_j<\infty$. In addition, we have $M_n=O(n)$. Thus, the upper bound for $\Pi (\Theta ^c_{n0})$ is of the form $c_1e^{-c_2n}$ for some constants $c_1$ and $c_2$. The exact same calculation, with $b_j$ in place of $a_j$, gives $\Pi (\Theta ^c_{n1})\le c_1e^{-c_2n}$ for some different $c_1$ and $c_2$. Since $\Pi (\Theta ^c_{n})\le \Pi (\Theta ^c_{n0})+\Pi (\Theta ^c_{n1})$, Condition 2 of Theorem 4.1 has been verified. Therefore, the posterior distribution is consistent with respect to $L^2$-distance at any $\theta ^\star$ satisfying (4).