Noninformative priors for the ratio of variabilities in a bivariate normal population

Abstract

In this paper, we consider some objective priors for the ratio of variabilities in a bivariate normal distribution. We develop the first and second order matching priors and reference priors. We obtain that the second order matching prior matches the alternative coverage probabilities up to the same order. It is also an HPD matching priors. It turns out that the derived reference priors do not satisfy a second order matching criterion. The simulation result shows that the second order matching prior performs better than reference priors based on matching the target coverage probabilities in a frequentist sense. Finally, we show that the second order matching prior and reference priors produce confidence sets with an expected length shorter than the Cox and Reid adjustment.

Appendices

Appendix 1. Proof of impropriety of posterior distribution

The posterior distribution of $(\theta _1,\ldots ,\theta _5)$ under the prior $\pi (\theta _1,\ldots ,\theta _5)=\theta _1^{-1}\theta _2^{-1}\theta _3^{-1}$ is improper.

Proof

Under the prior $\pi (\theta _1,\ldots ,\theta _5)=\theta _1^{-1}\theta _2^{-1}\theta _3^{-1}$, the joint posterior for $\theta _1,\ldots ,\theta _5$ given $\mathbf{Y}$ is

$$\begin{aligned} \pi (\theta _1,\ldots ,\theta _5 \vert {\mathbf{Y}})&\propto \theta _1^{-1}\theta _2^{-1}\theta _3^{-(1+{n\over 2})} \nonumber \\&\times \exp \left\{ -{1\over 2\theta _3} \sum _{i=1}^n \left[ \theta _1(\theta _2^2+\theta _3)^{1/2}(x_i-\theta _4)^2 -2\theta _2(x_i-\theta _4)(y_i-\theta _5)\right. \right. \nonumber \\&\left. \left. \quad +\theta _1^{-1}(\theta _2^2+\theta _3)^{1/2}(y_i-\theta _5)^2\right] \right\} . \end{aligned}$$

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By integrating with respect to $\theta _4$ and $\theta _5$, we obtain

$$\begin{aligned} \pi (\theta _1,\theta _2,\theta _3 \vert {\mathbf{Y}})&\propto c_1 \theta _1^{-1}\theta _2^{-1}\theta _3^{-({n+1\over 2})} \\&\quad \times \exp \left\{ -{1\over 2\theta _3} \left[ \theta _1^{-1}(\theta _2^2+\theta _3)^{1/2}S_2^2 -2\theta _2S_{12}+\theta _1(\theta _2^2+\theta _3)^{1/2}S_1^2\right] \right\} , \end{aligned}$$

where $c_1$ is a constant. Let $\theta _1={\sigma _2 \over \sigma _1}$, $\theta _2=\rho \sigma _1 \sigma _2$, and $\theta _3= \sigma _1^2 \sigma _2^2 (1-\rho ^2)$. Then

$$\begin{aligned}&{\int _0^{\infty }\int _{-\infty }^{\infty }\int _{0}^{\infty } \pi (\theta _1,\theta _2,\theta _3 \vert {\mathbf{Y}}) d\theta _3d\theta _2d\theta _1}\\&\quad =\int _{-1}^1\int _0^{\infty }\int _{0}^{\infty } c_2 \rho ^{-1}(1-\rho ^2)^{-({n+1\over 2})} \sigma _1^{-n} \sigma _2^{-n}\\&\qquad \times \exp \left\{ -{1\over 2(1-\rho ^2)} \left[ {S_1^2\over \sigma _1^2}-2\rho {S_{12}\over \sigma _1\sigma _2} +{S_2^2\over \sigma _2^2}\right] \right\} d\sigma _1 d\sigma _2 d\rho , \end{aligned}$$

where $c_2$ is a constant. Let $u={\sigma _1\sigma _2 \over S_1S_2}$ and $v={\sigma _1/S_1 \over \sigma _2/S_2}$. Then, the above integral reduces to

$$\begin{aligned} \int _{-1}^1\int _0^{\infty }\int _{0}^{\infty } c_3 \rho ^{-1}(1-\rho ^2)^{-({n+1\over 2})} u^{-n} v^{-1} e^{-{1\over 2(1-\rho ^2)u} [{1\over v}+v-2\rho r]} du dv d\rho , \end{aligned}$$

where $c_3$ is a constant and $r={S_{12}\over S_1S_2}$. Thus, if $n-1>0$, then upon integrating out u,

$$\begin{aligned} \int _{-1}^1\int _{0}^{\infty } c_4 \rho ^{-1}(1-\rho ^2)^{n-3\over 2} v^{-1} \left( {1\over v}+v-2\rho r \right) ^{-(n-1)}dv d\rho , \end{aligned}$$

where $c_4$ is a constant. Let $z={{1\over 2}[v+(1/v)]-1\over {1\over 2}[v+(1/v)]-\rho r}$. Then, the above integral implies

$$\begin{aligned}&\int _{-1}^1\int _{0}^{1} c_5 \rho ^{-1}(1-\rho ^2)^{n-3\over 2} (1-r\rho )^{-(n-{3\over 2})} (1-z)^{n-2}\left[ z-{1\over 2}(1+r\rho )z^2\right] ^{-1/2} dz d\rho \\&=\int _{-1}^1 c_6 \rho ^{-1}(1-\rho ^2)^{n-3\over 2} (1-r\rho )^{-(n-{3\over 2})} H\left( {1\over 2};{1\over 2};n-{1\over 2};{1\over 2}(1+r\rho )\right) d\rho , \end{aligned}$$

where $c_5$ and $c_6$ are constants, and H is a hypergeometric function. The last integral does not converge on $(-1,1)$ because of $\rho ^{-1}$. This completes the proof. $\square$

Apprndix 2. Derivation of reference prior

In this appendix, we derive the reference prior for the parameter grouping $\{ \theta _1, \theta _3, \theta _2, \theta _4, \theta _5 \}$. We follow the algorithm given by Berger and Bernardo (1992).

The reference prior is developed by considering a sequence of compact subsets of the parameter space, and taking the limit of a sequence of priors as these compact subsets fill out the parameter space. The compact subsets were taken to be Cartesian products of sets of the form

$$\begin{aligned} \theta _1 \in [a_1,b_1], \theta _2 \in [a_2,b_2],\theta _3 \in [a_3,b_3],\theta _4 \in [a_4,b_4],\theta _5 \in [a_5,b_5]. \end{aligned}$$

Suppose the limit $a_1, a_3$ tend to 0, then $a_2, a_4, a_5$ will tend to $-\infty$, and $b_1, b_2, b_3, b_4, b_5$ will tend to $\infty$. For the derivation of the reference, from the inverse of the Fisher information, we obtain

$$\begin{aligned} h_1= & {} \theta _1^{-2}\theta _3^{-1}(\theta _2^2+\theta _3), h_2=\theta _3^{-2}/4, h_3=(\theta _2^2+\theta _3)^{-1},\\ h_4= & {} \theta _1(\theta _2^2+\theta _3)^{-1/2}, h_5=\theta _1^{-1}\theta _3^{-1}(\theta _2^2+\theta _3)^{1/2}. \end{aligned}$$

In what follows, a subscripted K denotes a constant function independent of any parameters. However, any K may depend on the ranges of the parameters.

Step 1. Note that

$$\begin{aligned} \int _{a_5}^{b_5} h_5^{1/2} d\theta _5= & {} \int _{a_5}^{b_5}[\theta _1^{-1}\theta _3^{-1}(\theta _2^2+\theta _3)^{1/2}]^{1/2}d\theta _5\\= & {} [\theta _1^{-1}\theta _3^{-1}(\theta _2^2+\theta _3)^{1/2}]^{1/2}(b_5-a_5), \end{aligned}$$

so $\pi _5^l \{\theta _5 \vert \theta _1,\theta _3,\theta _2,\theta _4\}=K_1$, where $K_1=(b_5-a_5)^{-1}$.

Step 2. Now,

$$\begin{aligned} E_{\theta }^l\{\log h_4 \vert \theta _1,\theta _3,\theta _2,\theta _4 \}= & {} \int _{a_5}^{b_5} K_1 \log \{\theta _1(\theta _2^2+\theta _3)^{-1/2}\}d\theta _5\\= & {} \log \{\theta _1(\theta _2^2+\theta _3)^{-1/2}\}=K_2(\theta _1,\theta _2,\theta _3). \end{aligned}$$

It follows that

$$\begin{aligned} \int _{a_4}^{b_4} \exp [E_{\theta }^l\{\log h_4 \vert \theta _1,\theta _3,\theta _2,\theta _4 \}/2] d\theta _4 =\exp \{K_2(\theta _1,\theta _2,\theta _3)/2\}(b_4-a_4). \end{aligned}$$

Hence,

$$\begin{aligned} \pi _4^l \{\theta _4,\theta _5 \vert \theta _1,\theta _3,\theta _2 \}=K_1 (b_4-a_4)^{-1}=K_3. \end{aligned}$$

Step 3. Now,

$$\begin{aligned} E_{\theta }^l\{\log h_3 \vert \theta _1,\theta _3,\theta _2 \}= & {} \int _{a_4}^{b_4}\int _{a_5}^{b_5} K_3 \log \{ (\theta _2^2+\theta _3)^{-1} \}d\theta _5 d\theta _4\\= & {} \log \{ (\theta _2^2+\theta _3)^{-1}\}. \end{aligned}$$

Therefore,

$$\begin{aligned} \int _{a_2}^{b_2} \exp [E_{\theta }^l\{\log h_3 \vert \theta _1,\theta _3,\theta _2 \}/2] d\theta _2 =\log \left\{ {b_2+(b_2^2+\theta _3)^{1/2}\over a_2+(a_2^2+\theta _3)^{1/2}}\right\} =K_4(\theta _3). \end{aligned}$$

Thus

$$\begin{aligned} \pi _3^l \{\theta _2,\theta _4,\theta _5 \vert \theta _1,\theta _3 \}=[K_3/K_4(\theta _3)](\theta _2^2+\theta _3)^{-1/2}. \end{aligned}$$

Step 4. Now,

$$\begin{aligned} E_{\theta }^l\{\log h_2 \vert \theta _1,\theta _3 \}= & {} \int _{a_2}^{b_2}\int _{a_4}^{b_4}\int _{a_5}^{b_5} [K_3/K_4(\theta _3)](\theta _2^2+\theta _3)^{-1/2} \log \{ \theta _3^{-2}/4 \}d\theta _5 d\theta _4 d\theta _2\\= & {} \log \{ \theta _3^{-2}/4 \}. \end{aligned}$$

It follows that

$$\begin{aligned} \int _{a_3}^{b_3} \exp [E_{\theta }^l\{\log h_2 \vert \theta _1,\theta _3 \}/2] d\theta _3 = {1\over 2}\log \left\{ {b_3\over a_3} \right\} . \end{aligned}$$

Hence,

$$\begin{aligned} \pi _2^l \{\theta _3,\theta _2,\theta _4,\theta _5 \vert \theta _1 \}=[K_5/K_4(\theta _3)]\theta _3^{-1}(\theta _2^2+\theta _3)^{-1/2}, \end{aligned}$$

where $K_5=K_3/\log \left\{ {b_3\over a_3}\right\}$.

Step 5. Now,

$$\begin{aligned} E_{\theta }^l\{\log h_1 \vert \theta _1 \}= & {} \int _{a_3}^{b_3}\int _{a_2}^{b_2}\int _{a_4}^{b_4}\int _{a_5}^{b_5} [K_5/K_4(\theta _3)]\theta _3^{-1}(\theta _2^2+\theta _3)^{-1/2}\\&\times \log \{ \theta _1^{-2}\theta _3^{-1}(\theta _2^2+\theta _3) \} d\theta _5 d\theta _4 d\theta _2 d\theta _3\\= & {} K_6 +\log \{\theta _1^{-2}\}. \end{aligned}$$

So,

$$\begin{aligned} \int _{a_1}^{b_1} \exp \left[ E_{\theta }^l\{\log h_1 \vert \theta _1 \}/2\right] d\theta _1 = K_7\log \left\{ {b_1\over a_1} \right\} . \end{aligned}$$

Then,

$$\begin{aligned} \pi _1^l \{\theta _1,\theta _3,\theta _2,\theta _4,\theta _5 \} =[K_8/K_4(\theta _3)]\theta _1^{-1}\theta _3^{-1}(\theta _2^2+\theta _3)^{-1/2}, \end{aligned}$$

where $K_8=K_5{\exp \{K_6/2 \}\over K_7 \log \{{b_1/ a_1}\}}$. Therefore, the prior reference is given by

$$\begin{aligned} \pi _1 \{\theta _1,\theta _3,\theta _2,\theta _4,\theta _5 \} \propto \theta _1^{-1}\theta _3^{-1}(\theta _2^2+\theta _3)^{-1/2}. \end{aligned}$$