Second-order matching prior family parametrized by sample size and matching probability

Abstract

We propose a family of priors that satisfies the second-order probability matching property. The posterior quantile of a probability matching prior is exactly or approximately equal to the frequentist one. Most models lack an exact matching prior. If all quantiles of a prior’s posterior converge to the frequentist ones up to $o(n^{-1/2})$ or $o(n^{-1})$ as the sample size n increases, the prior is called a first-order probability matching prior and a second-order probability matching prior, respectively. Although a second-order matching prior does not necessarily exist, a first-order matching prior always exists. We introduce a class of priors that depend on the sample size and matching probability. We derive the condition under which the family satisfies the second-order probability matching property even when a second-order probability matching prior does not exist. The superiority of the proposed priors is illustrated in several numerical experiments.

Appendix: Proof of Lemma 1

To prove this lemma, we adopt a geometrical invariance viewpoint. We first decompose (7) as

$$\begin{aligned} \partial _{j}\pi _1(\theta ) \frac{I^{j1}}{\sqrt{I^{11}}}+\pi _1(\theta )\partial _j \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) =0. \end{aligned}$$

(20)

Applying the operator $I^{k1}\partial _k/\sqrt{I^{11}}$ to (20), we obtain

$$\begin{aligned}&\partial _j \partial _k\pi _1(\theta )\frac{I^{k1}I^{j1}}{I^{11}} +\partial _j\pi _1(\theta )\partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}} \nonumber \\&\quad +\partial _k\pi _1(\theta )\partial _j \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}} +\pi _1(\theta )\partial _j \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}}=0 . \end{aligned}$$

(21)

Multiplying (20) by $\partial _k ( I^{k1}/\sqrt{I^{11}})$, we also obtain

$$\begin{aligned}&-\partial _j\pi _1(\theta )\partial _k\left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \frac{I^{j1}}{\sqrt{I^{11}}} -\pi _1(\theta ) \partial _j\left( \frac{I ^{j1}}{\sqrt{I^{11}}}\right) \partial _{k}\left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \nonumber \\&-\pi _1(\theta )\partial _j\partial _k\left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}} +\pi _1(\theta )\partial _j\partial _k\left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}}=0, \end{aligned}$$

(22)

where the last two terms (which cancel) have been purposely written out.

It holds that

$$\begin{aligned} \partial _j\left( \frac{I^{j1}I^{k1}}{I^{11}}\right)&= \partial _j\left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}}+\partial _j\left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \frac{I^{j1}}{\sqrt{I^{11}}} \end{aligned}$$

(23)

$$\begin{aligned}&= \left( \partial _j I^{j1}\right) \frac{I^{k1}}{I^{11}} + \left( \partial _j I^{k1}\right) \frac{I^{j1}}{I^{11}} - \left( \partial _j I^{11}\right) \frac{ I^{j1} I^{k1} }{ (I^{11})^2 }. \end{aligned}$$

(24)

Moreover,

$$\begin{aligned} \partial _l I^{mk}&=-I^{mi}I^{kj} \left( \varGamma _{ilj}^{(1)} +\varGamma _{jli}^{(1)}+T_{ijl}\right) , \end{aligned}$$

(25)

$$\begin{aligned} L_{ijk}&=-\varGamma _{ikj}^{(1)}-\varGamma _{jki}^{(1)}- \varGamma _{ijk}^{(1)}-T_{ijk}. \end{aligned}$$

(26)

Condition (8) is then transformed as follows. Expanding (8), we obtain

$$\begin{aligned}&z_{1-\alpha }\biggl [ \frac{1}{3}\partial _{i}\left\{ \pi (\theta ) \frac{I^{j1}I^{k1}}{I^{11}}L_{jkl}\left( 3I^{li}-2\frac{I^{l1}I^{i1}}{I^{11}}\right) \right\} -\partial _j\partial _k \pi (\theta ) \frac{I^{j1}I^{k1}}{I^{11}}\nonumber \\&\quad -\partial _j\partial _k \left( \frac{I^{k1}I^{j1}}{I^{11}}\right) \pi (\theta ) -\partial _j {\pi (\theta )}\partial _k \left( \frac{I^{k1}I^{j1}}{I^{11}}\right) -\partial _k {\pi (\theta )}\partial _j \left( \frac{I^{k1}I^{j1}}{I^{11}}\right) \biggr ]\nonumber \\&\quad + 2\partial _{j}\left( \frac{\pi _2(\theta ;\alpha ) I^{j1}}{\sqrt{I^{11}}}\right) =0. \end{aligned}$$

(27)

Multiplying (21) by $z_{1-\alpha }$ and adding it to (27), we obtain

$$\begin{aligned}&z_{1-\alpha }\biggl [ \frac{1}{3}\partial _{i}\left\{ \pi _1 (\theta ) \frac{I^{j1}I^{k1}}{I^{11}}L_{jkl} \left( 3I^{li}-2\frac{I^{l1}I^{i1}}{I^{11}}\right) \right\} -\partial _j\partial _k \left( \frac{I^{j1}I^{k1}}{I^{11}}\right) \pi _1(\theta ) \nonumber \\&\quad -\partial _j {\pi _1(\theta )}\partial _k \left( \frac{I^{j1}I^{k1}}{I^{11}}\right) -\partial _k {\pi _1(\theta )}\partial _j \left( \frac{I^{j1}I^{k1}}{I^{11}}\right) +\partial _j\pi _1(\theta )\partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}} \nonumber \\&\quad +\partial _k\pi _1(\theta )\partial _j\left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}}+\pi _1(\theta )\partial _j \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}}\biggr ]\nonumber \\&\quad +2\partial _{j}\left( \frac{\pi _2(\theta ;\alpha )I^{j1}}{\sqrt{I^{11}}}\right) =0. \end{aligned}$$

(28)

Note that condition (8) is equivalent to condition (28) when Eq. (7) holds.

Using (23), (28) is expanded as follows:

$$\begin{aligned}&z_{1-\alpha }\biggl [ \frac{1}{3}\partial _{i}\left\{ \pi _1(\theta ) \frac{I^{j1}I^{k1}}{I^{11}}L_{jkl}\left( 3I^{li}-2\frac{I^{l1}I^{i1}}{I^{11}}\right) \right\} -\partial _j \partial _k \left( \frac{I^{k1}}{\sqrt{I^{11}}} \right) \frac{I^{j1}\pi _1(\theta )}{\sqrt{I^{11}}}\nonumber \\&\quad -\partial _j \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \pi _1(\theta ) -\partial _k \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \partial _j \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \pi _1(\theta ) \nonumber \\&\quad -\frac{I^{k1}}{\sqrt{I^{11}}}\partial _j\pi _1(\theta ) \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) -\frac{I^{k1}}{\sqrt{I^{11}}}\partial _k \pi _1(\theta ) \partial _j \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \biggr ]\nonumber \\&\quad + 2\partial _{j}\left( \frac{\pi _2(\theta ;\alpha )I^{j1}}{\sqrt{I^{11}}}\right) =0. \end{aligned}$$

(29)

Using (20), Eq. (29) transforms into

$$\begin{aligned}&z_{1-\alpha }\biggl [ \frac{1}{3}\partial _{i}\left\{ \pi _1(\theta ) \frac{I^{j1}I^{k1}}{I^{11}}L_{jkl}\left( 3I^{li}-2\frac{I^{l1}I^{i1}}{I^{11}}\right) \right\} -\partial _j \partial _k \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \frac{I^{j1}\pi _1(\theta )}{\sqrt{I^{11}}}\nonumber \\&\quad -\partial _j \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \pi _1(\theta ) -\frac{I^{k1}}{\sqrt{I^{11}}}\partial _j\pi _1(\theta )\partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \biggr ]\nonumber \\&\quad + 2\partial _{j}\left( \frac{\pi _2(\theta ;\alpha )I^{j1}}{\sqrt{I^{11}}}\right) =0. \end{aligned}$$

(30)

Note that condition (28) is equivalent to condition (30) when Eq. (7) holds.

By Eq. (22), we have

$$\begin{aligned}&-\frac{I^{k1}}{\sqrt{I^{11}}}\partial _j\pi _1(\theta )\partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \\&\quad -\partial _j \partial _k \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \frac{I^{j1}\pi _1(\theta )}{\sqrt{I^{11}}} -\partial _j \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \pi _1(\theta )\nonumber \\&=-\frac{I^{k1}}{\sqrt{I^{11}}}\partial _j\pi _1(\theta )\partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \\&\quad - \partial _j\pi _1(\theta )\partial _k\left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \frac{I^{j1}}{\sqrt{I^{11}}} -\partial _j \partial _k \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \frac{I^{j1}\pi _1(\theta )}{\sqrt{I^{11}}}\nonumber \\&\quad -\pi _1(\theta )\partial _j\partial _k\left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \frac{I^{k1}}{\sqrt{I^{11}}} -\partial _j \left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \pi _1(\theta )\nonumber \\&\quad -\pi _1(\theta )\partial _j \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \partial _k\left( \frac{I^{k1}}{\sqrt{I^{11}}}\right) +\pi _1(\theta )\frac{I^{k1}}{\sqrt{I^{11}}}\partial _j\partial _k\left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \nonumber \\&=-\partial _j \pi _1(\theta ) \partial _k\left( \frac{I^{j1}I^{k1}}{I^{11}}\right) -\pi _1(\theta ) \partial _j\partial _k\left( \frac{I^{j1}I^{k1}}{I^{11}}\right) +\pi _1(\theta )\frac{I^{k1}}{\sqrt{I^{11}}}\partial _j \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) \nonumber \\&=-\partial _j \left\{ \pi _1(\theta )\partial _k \left( \frac{I^{j1}I^{k1}}{I^{11}}\right) \right\} +\pi _1(\theta )\frac{I^{k1}}{\sqrt{I^{11}}}\partial _j \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) . \end{aligned}$$

Therefore, (30) reduces to

$$\begin{aligned}&\frac{z_{1-\alpha }}{3}\partial _{i}\left[ \pi _1(\theta )\left\{ \frac{I^{j1}I^{k1}}{I^{11}}L_{jkl}\left( 3I^{li}-2\frac{I^{l1}I^{i1}}{I^{11}}\right) -3 \partial _k \left( \frac{I^{i1}I^{k1}}{I^{11}}\right) \right\} \right] \nonumber \\&\quad +\pi _1(\theta )\frac{z_{1-\alpha }I^{k1}}{\sqrt{I^{11}}}\partial _j \partial _k \left( \frac{I^{j1}}{\sqrt{I^{11}}}\right) + 2\partial _{j}\left( \frac{\pi _2(\theta ;\alpha )I^{j1}}{\sqrt{I^{11}}}\right) =0 . \end{aligned}$$

(31)

Condition (30) is equivalent to condition (31) when Eq. (7) holds.

We now transform the first and second terms of the left-hand side of Eq. (31), respectively. Using (24), (25) and (26), the first term of (31) is equivalent to

$$\begin{aligned}&\frac{z_{1-\alpha }}{3}\partial _k \biggl [ \pi _1(\theta )\biggl \{ \frac{I^{j1}I^{l1}}{I^{11}} \left( -\varGamma _{jml}^{(1)}-\varGamma _{lmj}^{(1)}-\varGamma _{jlm}^{(1)}-T_{jlm} \right) \Bigl (3I^{mk}-2\frac{I^{m1}I^{k1}}{I^{11}} \Bigr ) \nonumber \\&\quad +3\biggl ( \frac{ I^{j1}I^{k1}I^{li} }{I^{11}} \Bigl (\varGamma _{ijl}^{(1)}+\varGamma _{jli}^{(1)}+T_{ijl} \Bigr ) + \frac{ I^{j1}I^{l1}I^{ki} }{I^{11}} \Bigl (\varGamma _{ilj}^{(1)}+\varGamma _{jli}^{(1)}+T_{ijl} \Bigr ) \nonumber \\&\quad -\frac{ I^{i1}I^{j1}I^{k1}I^{l1} }{(I^{11})^2} \Bigl (\varGamma _{ilj}^{(1)}+\varGamma _{jli}^{(1)}+T_{ijl} \Bigr ) \biggr ) \biggr \} \biggr ], \end{aligned}$$

which reduces to

$$\begin{aligned}&\frac{z_{1-\alpha }}{3}\partial _k \biggl [ \pi _1(\theta ) \biggl \{ -\frac{ 3I^{j1}I^{l1}I^{mk} }{I^{11}}\varGamma _{jml}^{(1)}\nonumber \\&\quad -\frac{ I^{j1}I^{k1}I^{l1}I^{m1} }{(I^{11})^2}T_{jlm}+ \frac{ 3I^{j1}I^{k1}I^{lm} }{I^{11}} \Bigl (\varGamma _{mlj}^{(1)}+\varGamma _{jlm}^{(1)}+T_{jlm} \Bigr )\biggr \} \biggr ]. \end{aligned}$$

(32)

The second term of (31) transforms into

$$\begin{aligned}&\pi _1(\theta )\frac{z_{1-\alpha }I^{k1}}{\sqrt{I^{11}}}\partial _k \biggl \{ -\frac{ I^{j1}I^{lm} }{\sqrt{I^{11}}} \Bigl (\varGamma _{mlj}^{(1)}+\varGamma _{jlm}^{(1)}+T_{jlm} \Bigr )\nonumber \\&\quad + \frac{ I^{j1}I^{l1}I^{m1} }{ 2(I^{11})^{3/2}} \Bigl (\varGamma _{ljm}^{(1)}+\varGamma _{mjl}^{(1)}+T_{jlm} \Bigr ) \biggr \}. \end{aligned}$$

By Leibniz’s rule and Eq. (7), it holds that

$$\begin{aligned}&\pi _1(\theta )\frac{z_{1-\alpha }I^{k1}}{\sqrt{I^{11}}}\partial _k \biggl \{ -\frac{ I^{j1}I^{lm} }{\sqrt{I^{11}}} \Bigl (\varGamma _{mlj}^{(1)}+\varGamma _{jlm}^{(1)}+T_{jlm} \Bigr ) + \frac{ I^{j1}I^{l1}I^{m1} }{ 2(I^{11})^{3/2}} \Bigl (\varGamma _{ljm}^{(1)}+\varGamma _{mjl}^{(1)} \Bigr ) \biggr \} \nonumber \\&= z_{1-\alpha } \partial _k \biggl \{ -\frac{\pi _1(\theta )I^{j1}I^{k1}I^{lm}}{I^{11}} \Bigl (\varGamma _{mlj}^{(1)}+\varGamma _{jlm}^{(1)}+T_{jlm} \Bigr ) + \frac{\pi _1(\theta )I^{j1}I^{k1}I^{l1}I^{m1} }{(I^{11})^{2}} \varGamma _{ljm}^{(1)} \biggr \}. \end{aligned}$$

The second term of (31) then reduces to

$$\begin{aligned}&z_{1-\alpha } \partial _k \biggl \{ -\frac{\pi _1(\theta )I^{j1}I^{k1}I^{lm}}{I^{11}} \Bigl (\varGamma _{mlj}^{(1)}+\varGamma _{jlm}^{(1)}+T_{jlm} \Bigr ) + \frac{\pi _1(\theta )I^{j1}I^{k1}I^{l1}I^{m1} }{(I^{11})^{2}} \varGamma _{ljm}^{(1)} \biggr \} \nonumber \\&\quad + \pi _1(\theta )\frac{z_{1-\alpha }I^{k1}}{\sqrt{I^{11}}} \partial _k \left\{ \frac{I^{m1}I^{l1} I^{j1}}{2(I^{11})^{3/2}}T_{jlm}\right\} . \end{aligned}$$

(33)

Dividing both sides of Eq. (31) by $\sqrt{\mathrm {det}I}$ and rearranging through Eqs. (32) and (33), Eq. (31) reduces to

$$\begin{aligned}&\frac{z_{1-\alpha }}{2\sqrt{\mathrm {det}I}} \partial _{i}\left\{ \pi _1(\theta )\varGamma ^{(1)}_{jlk} \frac{I^{j1}I^{k1}}{I^{11}}\left( -I^{li}+\frac{I^{l1} I^{i1}}{I^{11}}\right) \right\} \nonumber \\&\quad +\frac{z_{1-\alpha } \pi _1(\theta )I^{i1}}{12\sqrt{I^{11}\mathrm {det}I}}\partial _{i}\left\{ T_{jkl}\frac{I^{j1}I^{k1}I^{l1}}{(I^{11})^{3/2}}\right\} + \frac{1}{\sqrt{\mathrm {det}I}}\partial _{j}\left( \frac{ \pi _2 ( \theta ;\alpha )I^{j1}}{\sqrt{I^{11}}} \right) =0. \end{aligned}$$

(34)

Thus, condition (8) is equivalent to condition (34) when Eq. (7) holds. The latter is geometrically invariant and contains only first-order differential terms.