# Differentially Private Bayesian Tests

Abhisek Chakraborty⋆, Saptati Datta⋆

(Department of Statistics, Texas A&M University
3143 TAMU, College Station, TX 77843, USA)

###### Abstract

Differential privacy has emerged as an significant cornerstone in the realm of
scientific hypothesis testing utilizing confidential data. In reporting
scientific discoveries, Bayesian tests are widely adopted since they
effectively circumnavigate the key criticisms of P-values, namely, lack of
interpretability and inability to quantify evidence in support of the
competing hypotheses. We present a novel differentially private Bayesian
hypotheses testing framework that arise naturally under a principled data
generative mechanism, inherently maintaining the interpretability of the
resulting inferences. Furthermore, by focusing on differentially private Bayes
factors based on widely used test statistics, we circumvent the need to model
the complete data generative mechanism and ensure substantial computational
benefits. We also provide a set of sufficient conditions to establish results
on Bayes factor consistency under the proposed framework. The utility of the
devised technology is showcased via several numerical experiments.

Keywords— Bayes Factors, Bayes Factor Consistency, Differential Privacy,
Interpretable ML, Gaussian Mechanism, Laplace Mechanism  111The list of
authors is in alphabetical order. Both the authors contributed equally.

##  1 Introduction

Hypothesis testing is an indispensable tool to answer scientific questions in
the context of clinical trials, bioinformatics, social sciences, etc. The data
within such domains often involves sensitive and private information
pertaining to individuals. Researchers often bear legal obligations to
safeguard the privacy of such data. In this context, differential privacy
(Dwork, 2006) has emerged as a compelling framework for ensuring privacy in
statistical analyses with confidential data. Consequently, differentially
private versions of numerous well-established hypothesis tests have been
developed, although exclusively from a frequentist view point. This
encompasses private adaptations of test of binomial proportions (Awan and
Slavković, 2018), linear regression (Alabi and Vadhan, 2023), goodness of fit
(Kwak et al., 2021), analysis of variance (Swanberg et al., 2019), high-
dimensional normal means (Narayanan, 2022), to name a few. Differentially
private versions of common non-parametric tests (Couch et al., 2019),
permutation tests (Kim and Schrab, 2023), etc., have emerged as well. Another
recent line of work proposes to automatically create private versions of the
existing frequentist tests in a black-box fashion (Kazan et al., 2023, Peña
and Barrientos, 2023).

In this article, we intend to complement the existing literature via providing
a novel generalised framework for differentially private Bayesian testing.
This is genuinely crucial since P-values are routine criticised for it’s lack
of transparency, inability to quantify evidence in support of the null
hypothesis, and inability to measure the importance of a result. Refer to the
American Statistical Association statement on statistical significance and
P-values Wasserstein and Lazar (2016) for further discussions on this. On the
other hand, Bayes factors (Kass and Raftery, 1995, Jeffreys, 1961, Morey and
Rouder, 2011, Rouder et al., 2009, Wagenmakers et al., 2010, Johnson et al.,
2023b) quantify the relative evidence in favor of competing hypotheses,
effectively addressing the persistent criticisms of P-values. This
automatically calls for development of differentially private Bayes Factors to
test the scientific hypotheses of interest with confidentiality guarantees.

Prior to presenting the proposed methodology, we provide a concise overview of
pertinent concepts in Bayesian hypothesis testing and differential privacy,
laying the groundwork for subsequent discussions.

Bayesian Hypothesis Testing. Suppose we intend to test the null hypothesis
$\mbox{H}_{0}:\theta\in\Theta_{0}$ against the alternative hypothesis
$\mbox{H}_{1}:\theta\in\Theta_{1}$. The Bayes Factor (Kass and Raftery, 1995,
Jeffreys, 1961, Morey and Rouder, 2011, Rouder et al., 2009, Wagenmakers et
al., 2010, Johnson et al., 2023b) in favour of the alternative hypothesis
$\mbox{H}_{1}$ against the null hypothesis $\mbox{H}_{0}$ is defined as

$\displaystyle\mbox{BF}_{10}=\frac{\mbox{\rm P}(\mbox{\rm
H}_{1}{\,|\,}x)}{\mbox{\rm P}(\mbox{\rm
H}_{0}{\,|\,}x)}=\frac{m_{1}(x)}{m_{0}(x)}\times\frac{\mbox{\rm P}(\mbox{\rm
H}_{1})}{\mbox{\rm P}(\mbox{\rm H}_{0})},$ (1.1)

where $\mbox{\rm P}(\mbox{\rm H}_{i}{\,|\,}x)$ is the posterior probability of
$\mbox{\rm H}_{i}$ given the data $x$, $\mbox{\rm P}(\mbox{\rm H}_{i})$ is the
prior probability assigned to $\mbox{\rm H}_{i}$, and $m_{i}(x)$ denotes the
marginal probability assigned to the data under $\mbox{\rm H}_{i},\ i=0,1$.
The marginal densities are computed via
$m_{i}(x)=\int_{\Theta_{i}}\pi(x{\,|\,}\theta)\pi_{i}(\theta)d\theta$, where
$\pi(x{\,|\,}\theta)$ is the data generative model given the parameter
$\theta$, and $\pi_{i}(\theta)$ is the prior density for $\theta$ under
$\mbox{\rm H}_{i},\ i=0,1$. Bayes factor effectively addresses the persistent
criticisms regarding lack of interpretability of P-values (Wasserstein and
Lazar, 2016) in reporting outcomes of hypothesis tests, since it directly
quantifies the relative evidence in favor of the competing hypotheses
$\mbox{H}_{0}$ and $\mbox{H}_{1}$. Noteworthy, the value of a Bayes factor
depends on the choices of $\pi_{i}(\theta),\ i=0,1$, and it is generally
difficult to justify any single default choice. Moreover, the expression of
$m_{i}(x),\ i=0,1$ may not be available in closed, and require computing a
potentially high-dimensional numerical integrals.

To circumnavigate the aforementioned issues with Bayes factors, Johnson (2005)
proposed to define Bayes factors based on common test statistics, and put
priors directly on the non-centrality parameter of the test statistics to
specify the various competing hypotheses. For example, suppose we observe data
$X_{1},\ldots,X_{n}\sim\mbox{N}(\mu,\sigma^{2})$, and want to test
$\mbox{H}_{0}:\mu=0$ against $\mbox{H}_{1}:\mu\neq 0$. In this context, the
$t$-statistic $t=\frac{\sqrt{n}\bar{X}}{\hat{\sigma}}$ follows a (non-central)
$t_{n-1}(\delta)$ distribution with non centrality parameter $\delta\geq 0$
and degrees of freedom $n-1$, where $\bar{X}$ and $\hat{\sigma}$ are the mean
and standard deviation of the observed data. The non centrality parameter
$\delta=0$ under $\mbox{H}_{0}$ and $\delta>0$ under $\mbox{H}_{1}$. So, we
can put priors directly on the non-centrality parameter $\delta$ under
$\mbox{H}_{1}$, and conduct a Bayesian testing. Moreover, if we utilise non-
local priors (Johnson and Rossell, 2010b) on $\delta$ under $\mbox{H}_{1}$
instead of local priors, not only the expression of the marginal likelihoods
are often available in closed form, but also it ensures quick accumulation of
evidence in favor of both true $\mbox{H}_{0}$ and true $\mbox{H}_{1}$. Data-
driven approaches for the prior hyper-parameter tuning (Johnson et al., 2023b,
Datta et al., 2023) were subsequently suggested, adding to the objectivity of
such Bayesian testing methodology.

Differential Privacy. Differential privacy (Dwork, 2006, Baraheem and Yao,
2022) presents a principled framework for safeguarding sensitive information
in confidential data utilized for statistical analyses. It achieves this by
ensuring that the probability of any specific output of the statistical
procedure is nearly the same, regardless of the data belonging any one
specific individual in the database. In this context, databases that vary only
with respect to the data corresponding to a single individual are termed
_neighboring_ databases.

######  Definition 1 (Differential Privacy, (Dwork, 2006)).

A randomized algorithm $f:\mathcal{D}\to\mathcal{S}$ is
$\varepsilon$-differentially private if for all possible outcomes $\mbox{\rm
S}\subseteq\mathcal{S}$ and for any two neighbouring databases $\mbox{\rm
D},\mbox{\rm D}^{\prime}\in\mathcal{D}$, we have $\mbox{\rm P}[f(\mbox{\rm
D})\in\mbox{\rm S}]\leq e^{\varepsilon}\mbox{\rm P}[f(\mbox{\rm
D}^{\prime})\in\mbox{\rm S}].$

The _privacy budget_ $\varepsilon$ quantifies the degree of privacy
protection, with smaller values of $\varepsilon$ indicating a stronger privacy
guarantee. In particular, $\varepsilon$-differential privacy ensures that the
privacy loss is constrained by $e^{\varepsilon}$ with a probability of $1$.
Every differentially private algorithm achieves this via incorporating an
element of randomness. The most widely adopted techniques include the Gaussian
mechanism and the Laplace mechanism (Dwork, 2006). These privacy mechanisms
involves introducing a privacy noise sampled from a Gaussian or Laplace
distribution to the result of the query that requires privacy protection. The
measure of how much a the modification of a single row in a database can
influence the output of a query is referred to as _global sensitivity_.

######  Definition 2 (Global sensitivity).

The global sensitivity of a function $f$ is $\mbox{\rm
GS}_{f}=\underset{\mbox{\rm D},\mbox{\rm
D}^{\prime}\in\mathcal{D}}{\textup{max}}\ |f(\mbox{\rm D})-f(\mbox{\rm
D}^{\prime})|,$ where D and $\mbox{\rm D}^{\prime}$ are neighboring databases.

The scale of the Gaussian or Laplace distribution employed to introduce the
privacy noise, is contingent on both the privacy budget $\varepsilon$ and the
global sensitivity $\mbox{\rm GS}_{f}$. Given any function $f$, the _Laplace
mechanism_ is defined as $\tilde{f}(\mbox{\rm D})=f(\mbox{\rm D})+\eta$, where
$\eta$ is a privacy noise drawn from $\mbox{Laplace}(0,\mbox{\rm
GS}_{f}/\varepsilon)$, and $\mbox{\rm GS}_{f}$ is the global sensitivity of
$f$. The Laplace mechanism ensures $\varepsilon$-differentially privacy.
Similarly, given any function $f$, the Gaussian mechanism is defined as
$\tilde{f}(\mbox{\rm D})=f(\mbox{\rm D})+\eta$, where $\eta$ is drawn from
$\mbox{\rm Normal}(0,\ (\mbox{\rm GS}_{f}/\varepsilon)^{2})$, and $\mbox{\rm
GS}_{f}$ is the global sensitivity of $f$. The Gaussian mechanism too ensures
$\varepsilon$-differentially privacy. The Gaussian mechanism finds its utility
in situations where using Laplace noise with heavier tails might not be
appropriate. In this article, we pick the Laplace mechanism for the sake of
expositions, but the proposed methodologies work with Gaussian mechanism too.

Our Contributions. The key contributions in the article are three-fold.
Firstly, to the best of our knowledge, we provide the first-ever
differentially private Bayesian testing framework. This is important in the
light of the American Statistical Association statement on statistical
significance and P-values (Wasserstein and Lazar, 2016) that cautions against
making scientific decisions solely on the basis of P-value. This caution
arises from its lack of transparency, inability to quantify evidence in
support of the null hypothesis, and incapacity to measure the importance of a
result. Although Bayes factor-based tests are instrumental in circumnavigating
the key issues regarding P-values, extending them to differentially private
setup while preserving their inherent interpretability, requires substantial
work. To that end, we carefully embed common strategies in differentially
private testing, such as sub-sample and aggregate, truncated test functions,
etc., within the proposed data generative mechanism. This ensures that the
proposed Bayesian testing procedures arise naturally from the data generative
models, maintaining the interpretability of the inferences.

Secondly, we introduced differentially private Bayes factors that rely on
widely used test statistics, extending Johnson (2005) in a private setup. We
place priors directly on the non-centrality parameter of the test statistic
and utilize data-driven methods for tuning the prior hyper-parameters. This
circumvents the need for modeling the complete data generative mechanism and
placing priors on potentially high-dimensional model parameters. This, in
turn, yields key computational benefits when computing marginal likelihoods,
typically involving intractable integrals under a fully parametric formulation
of Bayesian testing. We also provide a set of sufficient conditions to
establish results on Bayes factor consistency (Chib and Kuffner, 2016,
Chatterjee et al., 2020) under the proposed framework.

Thirdly, we provide a general template to develop size $\alpha$ differentially
private Bayesian tests under specific setups and provide numerical schemes for
data-driven hyper-parameter tuning. Examples under particular hypothesis
testing problems are presented in the numerical experiments section. However,
the proposed schemes are applicable beyond the Bayesian tests discussed in the
current article.

Rest of the article is organised as follows. Section 2 introduces the proposed
framework for differentially private Bayesian testing in complete generality,
presents the key features of the methodology, and discusses schemes for the
hyper-parameter tuning under the proposed technology. In Section 3, we
introduce differentially private Bayesian tests based on common test
statistics, and discuss their asymptotic properties under various common
hypothesis testing problems. In Section 4, we study the numerical efficacy of
the proposed technology in different hypothesis testing problems with
practical utility. Finally, we conclude with a discussion.

##  2 General Framework

In this section, the objective is to lay down a hierarchical model equipped
with carefully chosen priors on the parameter of interest to specify competing
hypotheses, which naturally gives rise to the proposed differentially private
Bayesian tests. Notably, this attribute remains elusive in the existing
literature and it hinders the interpretability of the resulting inference. In
particular, many existing differentially private frequentist testing
procedures (Andrés F. Barrientos and Chen, 2019, Peña and Barrientos, 2023,
Kazan et al., 2023) rely on an ad-hoc sampling and aggregation scheme coupled
with a test statistic truncation step, to formulate the desired tests. Despite
the empirical success of these methods, a significant deficiency stems from
the lack of a principled probabilistic interpretation, i.e, these methods does
not enable us to to quantify evidence in favor of the null and alternative
hypotheses. We specifically tackle this issue in the subsequent discussions.

Hierarchical Model and Prior Specification. For a positive integer $t$, denote
$[t]:\,=\\{1,\ldots,t\\}$. Suppose we observe data $x_{1},\ldots,x_{n}$ from a
probability distribution having density $\pi(\cdot\mid\theta),\
\theta\in\Theta$. We then randomly divide the data into $M$ partitions
$\mathbf{x}^{(i)},i\in[M]$. We can assume that the data in each of the $M$
partitions arise from

$\displaystyle\mathbf{x}^{(i)}\sim\prod_{j=1}^{n_{i}}\pi(\mathbf{x}^{(i)}_{j}\mid\theta^{(i)}),\quad\text{independently
for}\ i\in[M],$ (2.1)

such that $\sum_{i=1}^{M}n_{i}=n$. The above data generating scheme comes in
handy to formalize the _sample and aggregate_ scheme under the proposed setup.

Next, we posit the priors on the parameter of interest $\theta$ under the
competing hypotheses. To that end, we assume that

$\displaystyle\theta^{(i)}\mid\tau_{0},\tau_{1}\sim(1-\omega)\
\pi_{0}(\theta^{(i)}\mid\tau_{0}^{2})\ +\ \omega\
\pi_{1}(\theta^{(i)}\mid\tau_{1}^{2})\quad\text{under}\ \mbox{H}_{0},$
$\displaystyle\theta^{(i)}\mid\tau_{0},\tau_{1}\sim\ \omega\
\pi_{0}(\theta^{(i)}\mid\tau_{0}^{2})\ +\ (1-\omega)\
\pi_{1}(\theta^{(i)}\mid\tau_{1}^{2})\quad\text{under}\ \mbox{H}_{1},$ (2.2)

independently for $i\in[M]$, where $\\{\pi_{i}(\cdot),\ i=0,1\\}$ are the
prior probability distributions, $(\tau_{0},\tau_{1})$ are prior hyper-
parameters, and $\omega\in(0,\frac{1}{2})$ is a hyper-parameter that ensures
that the Bayes factor for testing $\mbox{H}_{0}$ against $\mbox{H}_{1}$ is
bounded. If the Bayes factor for testing $\mbox{H}_{0}$ against $\mbox{H}_{1}$
were to be unbounded, the global sensitivity parameter of the Bayes factor
becomes unbounded too. An immediate fix against it would involve truncating
the log Bayes factor beyond an interval of the form $[-a,a],a>0$. But this fix
will ruin the inherent probabilistic interpretation of the Bayes factors. Upon
closer look, truncating the log Bayes factor in principle means that the
maximum evidence that we can obtain in favor of $\mbox{H}_{0}$ (or
$\mbox{H}_{1}$) against $\mbox{H}_{1}$ (or $\mbox{H}_{0}$) is bounded from
above – we precisely impose this constraint a-priori via the hierarchical
specification (2). In summary, the mixture priors on $\\{\theta^{(i)},\
i\in[M]\\}$ under the competing hypotheses are instrumental in formalizing the
_truncation_ scheme necessary to ensure that the global sensitivity parameter
of the privacy mechanism is bounded, while preserving the interpretability of
the resulting Bayes factors. One may criticise the appearance of the parameter
$\omega$ in our formulation. To that end, we carefully introduce an asymptotic
regime to analyse the proposed methodology, so that the effect of $\omega$
washes away as the sample size diverges to $\infty$.

The model and prior specified in equations (2.1)-(2), together with the
assumption that

$\displaystyle\mbox{P}(\mbox{H}_{0})=\mbox{P}(\mbox{H}_{1})=0.5,$ (2.3)

apriori, completely describe our generative model of interest.

Properties. For each partition $\mathbf{x}^{(i)},\ i\in[M]$, we record the
notations for the integrals

$\displaystyle
m_{k}(\mathbf{x}^{(i)}\mid\tau_{k}^{2})=\int_{\Theta}\bigg{[}\prod_{j=1}^{n_{j}}\pi(\mathbf{x}^{(i)}_{j}\mid\theta)\bigg{]}\
\pi_{k}(\theta\mid\tau_{k}^{2})\ d\theta,\quad k=0,1.$

For $k\in\\{0,1\\}$, the quantity $m_{k}(\mathbf{x}^{(i)}\mid\tau_{k}^{2})$
can be interpreted as the marginal likelihood of the data partition
$\mathbf{x}^{(i)}$ under the hypothesis $\mbox{H}_{k}$ when $\omega=0$. Under
this set up, as we eluded to earlier, the Bayes factor for testing
$\mbox{H}_{0}$ against $\mbox{H}_{1}$ can be unbounded, and consequently the
global sensitivity parameter of the privacy mechanism is unbounded too. This
renders the development of differentially private Bayesian tests non-trivial,
and throughout the article we shall assume that the hyper-parameter
$\omega>0$. Under the alternative hypothesis $\mbox{H}_{1}$ as in (2.1)-(2.3),
the marginal distribution of the data partition $\mathbf{x}^{(i)}$ is
expressed as

$m^{t}_{1}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\omega\
m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})\ +\ (1-\omega)\
m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2}),\quad i\in[M];$

and under the null hypothesis $\mbox{H}_{0}$, it is given by:

$m^{t}_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=(1-\omega)\
m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})\ +\ \omega\
m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2}),\quad i\in[M].$

Consequently, in light of the data partition $\mathbf{x}^{(i)}$, the Bayes
factor against the null hypothesis $\mbox{H}_{0}$ is expressed as

$\displaystyle\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\frac{\omega+(1-\omega)\big{[}m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2})\
/\
m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})\big{]}}{(1-\omega)+\omega\big{[}m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2})\
/\ m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})\big{]}},\quad i\in[M].$ (2.4)

We proceed by listing out the key properties of the the Bayes factor against
the null hypothesis
$\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$ for
the data partition $\mathbf{x}^{(i)}$.

######  Proposition 2.1.

For $\omega\in(0,1/2)$, $\mbox{\rm
BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$ is bounded
between $[\frac{\omega}{1-\omega},\frac{1-\omega}{\omega}]$.

###### Proof.

If
$\frac{m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2})}{m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})}=0$,
then
$\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\omega/(1-\omega)$.
Additionally, if
$\frac{m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2})}{m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})}=\infty$,
then
$\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=(1-\omega)/\omega$.
Next, we define a map

$\displaystyle g(t)=\frac{\omega\ +\ (1-\omega)t}{(1-\omega)\ +\ \omega
t},\quad t\in[0,\infty).$

Note that, the derivative of $g(t)$,

$\displaystyle g^{\prime}(t)=\frac{1-2\omega}{(1\ -\ \omega\ +\ \omega
t)^{2}}>0$

for $\omega\in(0,1/2)$. So, $g(t)$ is increasing in $t\in[0,\infty)$, for
$\omega\in(0,1/2)$. This implies that the minimum and maximum attainable
evidence against the null hypothesis is $\frac{\omega}{1-\omega}$ and
$\frac{1-\omega}{\omega}$, respectively. ∎

######  Proposition 2.2.

Under (2.1)-(2.3), the Bayes Factor against the null hypothesis $\mbox{H}_{0}$
combined across all the partitions $\mathbf{x}^{(i)},\ i\in[M]$ is expressed
as

$\displaystyle\mbox{\rm
BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\prod_{i=1}^{M}\mbox{\rm
BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega).$

###### Proof.

Given $\tau_{0}$, $\tau_{1}$ and $\omega\in(0,1/2)$, the prior distributions
on $\theta^{(i)}$ are independent across data partitions $\mathbf{x}^{(i)},\
i\in[M]$. Consequently, the joint marginal
$m_{k}^{t}(\mathbf{x}\mid\tau^{2}_{k},\omega)=\prod_{i=1}^{M}m_{k}^{t}(\mathbf{x}^{(i)}\mid\tau^{2}_{k},\omega)$,
where $k\in\\{0,1\\}$. Further, we note that

$\displaystyle\mbox{BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\frac{m_{1}^{t}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}{m_{0}^{t}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}=\frac{\prod_{i=1}^{M}m_{1}^{t}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}{\prod_{i=1}^{M}m_{0}^{t}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}=\prod_{i=1}^{M}\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega).$

This concludes the proof. ∎

######  Remark 2.3.

Proposition 6.1 together with Proposition 6.2 implies that, the combined Bayes
Factor against the null hypothesis $\mbox{H}_{0}$ across all the partitions
$\mathbf{x}^{(i)},\ i\in[M]$, $\mbox{\rm
BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$ is bounded
between $\left(\frac{\omega}{1-\omega}\right)^{M}$ and
$\left(\frac{1-\omega}{\omega}\right)^{M}$ for $\omega\in(0,1/2)$.

In the sequel, we construct differential private Bayesian tests based on the
randomised mechanism, denoted by $f$, that map a data set $\mathbf{x}$ to the
average log Bayes factor

$\displaystyle
f(\mathbf{x})=\frac{1}{M}\log\mbox{BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\frac{1}{M}\sum_{i=1}^{M}\log\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega).$
(2.5)

We take refuge to common differential privacy mechanism of adding calibrated
privacy noise to the randomised mechanism $f$ to ensure differential privacy
of the inferential procedure. For every data set $\mathbf{x}$, since the log
Bayes factor in each of the partitions
$\log\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$
is bounded between $[-a,a]$ with $a=-\log(\omega/1-\omega)$, the global
sensitivity of
$\log\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$
is $2a$. Consequently, the global sensitivity of $\mbox{GS}_{f}$ of the log
Bayes factor averaged over the partitions, i.e of $f$, is $(2a/M)$. We now
have all the necessary ingredients to introduce our privacy preserving
mechanism.

To that end, let $\mathcal{D}$ denote the collection of all possible of data
sets. We consider a random mechanism
$\mbox{H}:\mathcal{D}\rightarrow\mathbb{R}$ expressed as

$\mbox{H}(\mathbf{x})=f(\mathbf{x})+\eta,\quad\mathbf{x}\in\mathcal{D},$ (2.6)

where $\eta$ is a $\mbox{\rm Normal}(0,\ (\mbox{\rm GS}_{f}/\varepsilon)^{2})$
or a $\mbox{Laplace}(0,\mbox{GS}_{f}/\varepsilon)$ distributed privacy noise.
A test based on $\mbox{H}(\mathbf{x})$ is then obtained as

$\phi(\mathbf{x})=\begin{cases}1&\text{if}\
\mbox{H}(\mathbf{x})\geq\gamma_{\alpha,\varepsilon}(n),\\\
0&\text{otherwise},\end{cases}$ (2.7)

for some cut-off $\gamma_{\alpha,\varepsilon}(n)$, depending on the sample
size $n$ and size $\alpha$ of the test. Here, $\phi(\mathbf{x})=1$ signifies
rejection of $\mbox{H}_{0}$ in favour of $\mbox{H}_{1}$, and
$\phi(\mathbf{x})=0$ accept $\mbox{H}_{0}$.

######  Proposition 2.4.

The function $\phi(\mathbf{x})$ for testing $\mbox{H}_{0}$ against
$\mbox{H}_{1}$, as described in (2.5)-(2.7), is $\varepsilon$-differentially
private.

###### Proof.

The global sensitivity $\mbox{GS}_{f}$ of $f(\mathbf{x})$ is $\frac{2a}{M}$.
Then, defining $\mbox{H}(\mathbf{x})=f(\mathbf{x})+\eta$, where $\eta$ follows
$\mbox{\rm Normal}(0,\ (\mbox{GS}_{f}/\varepsilon)^{2})$ or
$\mbox{Laplace}(0,\mbox{GS}_{f}/\varepsilon)$ completes the proof, by
definition. ∎

Next, given the sample size $n$, privacy mechanism, the user defined
parameters $(\varepsilon,a,M)$, and the size $\alpha$ of the test, we
determine the cut off $\gamma_{\alpha,\varepsilon}(n)$ by satisfying the _size
constraint_

$\displaystyle\Xi_{0}(M,a\mid\varepsilon,\alpha,n)=\mbox{E}\big{[}\phi(\mathbf{x})\mid\mbox{\rm
H}_{0}]=\mbox{P}_{\mbox{\rm H}_{0}}\big{[}\mbox{\rm
H}(\mathbf{x})\geq\gamma_{\alpha,\varepsilon}(n)\big{]}=\alpha.$ (2.8)

Exact calculation of the cut off $\gamma_{\alpha,\varepsilon}(n)$ may not be
amenable under certain cases. However, we can always obtain
$\gamma_{\alpha,\varepsilon}(n)$ via a Monte Carlo simulation. We provide
additional details about this computational aspect in the experiment section.

Hyperparameter Tuning. To apply differentially private Bayesian tests, the
data producer or an analyst permitted on a verification server, initially
establishes the desired privacy level $\varepsilon$. Following this
determination, the task involves selecting appropriate values for the
parameters $M$ and $a$. In the context presented, we consider the case where
the analyst does not have the autonomy to select specific value of
$\varepsilon$, but possess the discretion to choose $(M,a)$. For analysts
endowed with this discretion, they may iteratively employ the described
methodology, varying the values of $\varepsilon$. This iterative process
allows analysts to experiment with diverse combinations of $(M,a)$, thereby
striking a nuanced balance between attaining the desired privacy level and
asymptotic behaviour of the Bayes factors.

Given the sample size $n$, a privacy mechanism, a privacy budget
$\varepsilon$, and the size $\alpha$, we determine the Bayes factor cut off
$\gamma_{\alpha,\varepsilon}(n)$ using (2.8) via a Monte Carlo simulation, for
varying values of $(M,a)$. Now, let $\mathcal{M}\times\mathcal{A}$ denote the
space all possibles pairs $(M,a)$ under consideration. We operate under the
regime that $M=\mathrm{o}(n^{\gamma})$ and
$a=-\log(\omega/1-\omega)=\mathrm{o}(n^{\beta})$ where $0<\beta<\gamma\leq 1$.
The constraint $\beta<\gamma$ ensures that the $\mbox{GS}_{f}$ decreases as
the sample size increases. The order assumption on $a$, or equivalently
$\omega$, enables us to dictate the rate at which the hypothesis specified in
(2.1)-(2) approaches common null hypothesis significance testing framework.
This, in turn, enable us dictate the convergence rate of the privatized Bayes
factor. Specifically, slower the order of $a$, lower would be the rate of
convergence. Refer to the discussion in Section 3 for details. The optimal
choices $(M,a)$ is obtained via _maximising power_ of the test

$\displaystyle\operatorname*{arg\,max}_{M,a\in\mathcal{M}\times\mathcal{A}}\Xi_{1}(M,a\mid\varepsilon,\alpha,n)$
(2.9)

where
$\Xi_{1}(M,a\mid\varepsilon,\alpha,n)=\mbox{E}\big{[}\phi(\mathbf{x})\mid\mbox{\rm
H}_{1}]=\mbox{P}_{\mbox{\rm H}_{1}}\big{[}\mbox{\rm
H}(\mathbf{x})\geq\gamma_{\alpha,\varepsilon}(n)\big{]}$. Exact calculation of
optimal choices of $(M,a)$ may not be amenable under certain cases. However,
we can always obtain it via a Monte Carlo simulation. We provide additional
details about this computational aspect in the experiment section. We now have
all the necessary ingredients to introduce differentially private Bayesian
tests in specific set ups.

##  3 Differentially Private Bayes Factors based on Test Statistics

We can apply the general framework delineated in Section 2 to device fully
parametric Bayesian tests under the differentially private paradigm. However,
considering the limitations associated with fully parametric Bayes factors, as
discussed in the introduction, we suggest making inferences based on Bayes
factors that rely on test statistics instead. Further, in order to study the
asymptotic properties of the proposed Bayes factors, we adopt the widely
recognized notion of “Conventional Bayes factor consistency” (Chib and
Kuffner, 2016, Chatterjee et al., 2020). These asymptotic properties must be
satisfied to ensure that, as the sample size increases, the correct model has
a faster rate of posterior contraction in comparison to an incorrect model.

######  Definition 3 (Conventional Bayes factor consistency (Chib and Kuffner,
2016)).

For two competing models $M_{l}$ and $M_{k}$, the Bayes factor for comparing
models $M_{k}$ and $M_{l}$, denoted as $\mbox{\rm
BF}_{kl}=\frac{m(\mathbf{x}\mid M_{k})}{m(\mathbf{x}\mid M_{l})}$, is
consistent if (i) $\mbox{\rm BF}_{kl}\xrightarrow{p}0$ when $M_{l}$ contains
the true model, and (ii) $\mbox{\rm BF}_{kl}\xrightarrow{p}\infty$ when
$M_{k}$ contains the true model, where $\mathbf{x}$ is the data.

Furthermore, to specify the competing hypotheses as in (2), we use mixtures of
a spike at $0$ and non-local priors (Johnson and Rossell, 2010a). The defining
feature of a non-local priors is that it assigns negligible mass to the
parameters values close to $0$, ensuring faster rate of convergence of Bayes
factors compared to the common local priors (Johnson and Rossell, 2010a,
Johnson et al., 2023a, Datta et al., 2023). Through a hyper-parameter present
in the non-local densities one can specify a substantive deviation from $0$,
in turn specifying the neighbourhood in which the non-local prior assigns
negligible mass. Studies on consistency of Bayes factors based on test
statistics (Johnson, 2008, Datta et al., 2023) is not particularly new.
However, the introduction of differential privacy to this framework introduces
unique challenges. To ensure the convergence of the privatized Bayes factors,
it is essential to incorporate additional meaningful assumptions. This
includes considerations about the distribution of test statistics under the
competing hypotheses (2), order of the truncation parameter $a$ (or
equivalently $\omega$) and number of data partitions $M$ relative to the
sample size, and the characteristics of the global sensitivity parameter.

Next, a few notations are in order. The confluent hyper-geometric function of
order $(m,n)$ (Abramowitz, 1974) is denoted by ${{}_{m}}F_{n}(a,b;z)$. The
Dirac’s delta function is denoted by $\delta_{0}$. Further, $\mbox{N}(a,b)$
denotes normal distribution with a mean of $a$ and variance $b$,
$\mbox{T}_{n}(\theta)$ denotes the t-distribution with degrees of freedom
$\nu$ and non-centrality parameter $\theta$, $\chi^{2}_{n}(\theta)$ denotes
the chi-squared distribution with degrees of freedom $\nu$ and non-centrality
parameter $\theta$, $\mbox{F}_{k,m}(\theta)$ denotes the F-distribution with
degrees of freedom $(k,m)$ and non-centrality parameter $\theta$ is
represented as $\mbox{F}_{k,m}(\theta)$, $\mbox{G}(\alpha,\theta)$ denotes a
gamma distribution with a shape parameter $\alpha$ and a rate parameter
$\theta$, and $\mbox{J}(\tau^{2})$ represents a normal-moment density of order
1 (Johnson and Rossell, 2010a). With these notations, the Bayes Factors based
on test statistics for each partition $\mathbf{x}^{(i)},\ i\in[M]$ of the
observed dataset are presented as follows.

######  Proposition 3.1.

(Two-sided $z$-test) Suppose the generative model for the test statistic $z$
under $\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$ are

$\displaystyle\mbox{\rm H}_{0}:z\mid\lambda$ $\displaystyle\sim$
$\displaystyle\mbox{\rm
N}(\lambda,1),\qquad\lambda\mid\sigma^{2},\tau^{2}\sim(1-\omega)\
\delta_{0}+\omega\ \mbox{\rm J}(\tau^{2})$ $\displaystyle\mbox{\rm
H}_{1}:z\mid\lambda$ $\displaystyle\sim$ $\displaystyle\mbox{\rm
N}(\lambda,1),\qquad\lambda\mid\sigma^{2},\tau^{2}\sim\omega\
\delta_{0}+(1-\omega)\ \mbox{\rm J}(\tau^{2}),\quad\tau>0,$

respectively. Then, the Bayes factor in favor of the $\mbox{\rm H}_{1}$ is of
the form $\mbox{\rm
BF}^{t}_{10}(z\mid\tau^{2},\omega)=\frac{\omega+(1-\omega)\mbox{\rm
R}}{(1-\omega)+\omega\mbox{\rm R}}$, where $\mbox{\rm
R}=m_{1}(z\mid\tau^{2})/m_{0}(z)=(1+\tau^{2})^{-\frac{3}{2}}\
{{}_{1}}\mbox{\rm F}_{1}(3/2,\ 1/2;\ \tau^{2}z^{2}/2(1+\tau^{2}))$.

######  Remark 3.2.

Under the set up in Proposition 3.1, we further assume that (i) the test
statistic $z\sim\mbox{\rm N}(\eta\sqrt{n},1)$, for some $\eta$ ; (ii) the
hyper-parameter $\tau^{2}=\kappa n$, for some $\kappa>0$; (iii)
$(1-{\omega})/{\omega}=\mathrm{o}(\exp(n^{\beta}))$, $0<\beta<1$. Then, the
Bayes Factor $\mbox{\rm
BF}^{t}_{10}(z\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(-n^{\beta})$ under
$\mbox{\rm H}_{0}$, and $\mbox{\rm
BF}^{t}_{10}(z\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(n^{\beta}))$ under
$\mbox{\rm H}_{1}$. This demonstrates that the Bayes factor in Proposition 3.1
is consistent under both $\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$.

The justification for assumption (i) is provided by the order of commonly
utilized $z$ statistics under $\mbox{H}_{1}$. For instance, consider the test
means of an univariate normal population $\mbox{N}(\mu,1)$. The test statistic
of interest $z=\sqrt{n}\bar{X}$ satisfies the assumption, with $\bar{X}$ and
$n$ being respectively the mean and number of observations collected from the
population. Assumption (ii) stems from the choice of $\tau^{2}$ (Johnson et
al., 2023a), wherein the mode of the non-local density is equated with a
function of the standardized effect size. See Table 1 in Johnson et al.
(2023a) further discussion on this assumption. Assumption _(iii)_ arises from
selection of the truncation parameter $a=-\log(\omega/1-\omega)$. Recall that,
we divide $n$ observation into $M$ partitions. We operate under the asymptotic
regime that $M=\mathrm{o}(n^{\gamma})$ and $a=\mathrm{o}(n^{\beta})$ where
$0<\beta<\gamma<1$. This selection ensures that the $\mbox{GS}_{f}$ decreases
as the sample size increases, while maintaining the rapid convergence the
privatized Bayes factor. We shall put it in concrete terms in Remark 3.7. For
the time being, we carry on with the discussion on privatised Bayes factors
based on $t$,$\chi^{2}$ and $F$ test statistics for each partition
$\mathbf{x}^{(i)},\ i\in[M]$ of the observed dataset.

######  Proposition 3.3.

(Two-sided $t$-test) Suppose the generative model for the test statistic $t$
under $\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$ are

$\displaystyle\mbox{\rm H}_{0}:t\mid\lambda$ $\displaystyle\sim$
$\displaystyle\mbox{\rm
T}_{\nu}(\lambda),\qquad\lambda\mid\sigma^{2},\tau^{2}\sim(1-\omega)\
\delta_{0}+\omega\ \mbox{\rm J}(\tau^{2})$ $\displaystyle\mbox{\rm
H}_{1}:t\mid\lambda$ $\displaystyle\sim$ $\displaystyle\mbox{\rm
T}_{\nu}(\lambda),\qquad\lambda\mid\sigma^{2},\tau^{2}\sim\omega\
\delta_{0}+(1-\omega)\ \mbox{\rm J}(\tau^{2}),\quad\tau>0,$

respectively. Then, the Bayes factor in favor of the $\mbox{\rm H}_{1}$ is of
the form $\mbox{\rm
BF}^{t}_{10}(t\mid\tau^{2},\omega)=\frac{\omega+(1-\omega)\mbox{\rm
R}}{(1-\omega)+\omega\mbox{\rm R}}$, where $\mbox{\rm
R}=m_{1}(t\mid\tau^{2})/m_{0}(t)=c\bigg{[}{{}_{2}}\mbox{\rm
F}_{1}\big{(}\frac{\nu+1}{2},\frac{3}{2},\frac{1}{2},y^{2}\big{)}+y\frac{\Gamma(\frac{\nu}{2}+1)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}}\frac{\Gamma(2)}{\Gamma(\frac{3}{2})}\
{{}_{2}}\mbox{\rm
F}_{1}\big{(}\frac{\nu}{2}+1,2,\frac{3}{2};y^{2}\big{)}\bigg{]}$, with $y=\tau
t/\sqrt{(\nu+t^{2})(1+\tau^{2}})$ and $c=(1+\tau^{2})^{-\frac{3}{2}}$.

######  Remark 3.4.

Under the set up in Proposition 3.3, we further assume that (i) the test
statistic $t\sim\mbox{\rm T}_{\nu}(\eta\sqrt{n})$, for some $\eta$; the hyper-
parameters (ii) $\nu=\delta n-u$ and (iii) $\tau^{2}=\kappa n$, for some
$\kappa>0$, (iv) $(1-{\omega})/{\omega}=\mathrm{o}(\exp(n^{\beta}))$,
$0<\beta<1$. Then, the Bayes Factor $\mbox{\rm
BF}^{t}_{10}(t\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(-n^{\beta}))$ under
$\mbox{\rm H}_{0}$, and $\mbox{\rm
BF}^{t}_{10}(t\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(n^{\beta}))$ under
$\mbox{\rm H}_{1}$. This demonstrates that the Bayes factor in Proposition 3.3
is consistent under both $\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$.

Assumption _(ii)_ also derives from the degrees of freedom of frequently used
$t$ statistic. For instance, consider the test means of an univariate normal
population $\mbox{N}(\mu,\sigma^{2})$, where $\sigma^{2}$ is unknown. Then,
the $t$-statistic follows a $t$ distribution with degrees of freedom $n-1$.
Rest of assumptions follow from considerations described after Remark 3.2.

######  Proposition 3.5.

($\chi^{2}$-test) Suppose the generative model for the test statistic $h$
under $\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$ are

$\displaystyle\mbox{\rm H}_{0}:h\mid\lambda$ $\displaystyle\sim$
$\displaystyle\chi^{2}_{k}(\lambda),\quad\lambda\mid\tau^{2}\sim(1-\omega)\
\delta_{0}+\omega\ \ \mbox{\rm G}((k/2)+1,\ 1/2\tau^{2}),$
$\displaystyle\mbox{\rm H}_{1}:h\mid\lambda$ $\displaystyle\sim$
$\displaystyle\chi^{2}_{k}(\lambda),\quad\lambda\mid\tau^{2}\sim\omega\
\delta_{0}+(1-\omega)\ \mbox{\rm G}((k/2)+1,\ 1/2\tau^{2}),\ \tau>0.$

Then, the Bayes factor in favor of the $\mbox{\rm H}_{1}$ is of the form
$\mbox{\rm BF}^{t}_{10}(h\mid\tau^{2},\omega)=\frac{\omega+(1-\omega)\mbox{\rm
R}}{(1-\omega)+\omega\mbox{\rm R}}$, where $\mbox{\rm
R}=m_{1}(h\mid\tau^{2})/m_{0}(h)=(1+\tau^{2})^{-k/2-1}{{}_{1}}\mbox{\rm
F}_{1}((k/2)+1,\ k/2;\ (\tau^{2}h)/(2(1+\tau^{2})).$

Note that gamma densities can be viewed as scaled $\chi^{2}$ densities. A
$\chi^{2}_{\nu}$ density with degrees of freedom $(\nu)>3$ has zero mass at
the origin. This implies $\mbox{\rm G}((k/2)+1,\ 1/2\tau^{2})$ represents a
non-local density for every $k\in\mathrm{N}$.

######  Remark 3.6.

Under the set up in Proposition 3.5, we further assume that (i) the test
statistic $h\sim\chi^{2}_{\nu}(\eta n)$, for some $\eta$; the hyper-parameters
(ii) $\nu=\delta n-u$ and (iii) $\tau^{2}=\kappa n$, for some $\kappa>0$, (iv)
$(1-{\omega})/{\omega}=\mathrm{o}(\exp(n^{\beta}))$ , $0<\beta<1$. Then, the
Bayes Factor $\mbox{\rm
BF}^{t}_{10}(h\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(-n^{\beta}))$ under
$\mbox{\rm H}_{0}$, and $\mbox{\rm
BF}^{t}_{10}(h\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(n^{\beta}))$ under
$\mbox{\rm H}_{1}$. This demonstrates that the Bayes factor in Proposition 3.5
is consistent under both $\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$.

######  Proposition 3.7.

(F\- test) Suppose the generative model for the test statistic $F$ under
$\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$ are

$\displaystyle\mbox{\rm H}_{0}:f\mid\lambda$ $\displaystyle\sim$
$\displaystyle\mbox{\rm
F}_{a,b}(\lambda),\quad\lambda\mid\tau^{2}\sim(1-\omega)\ \delta_{0}+\omega\ \
\mbox{\rm G}((a/2)+1,\ 1/2\tau^{2}),$ $\displaystyle\mbox{\rm
H}_{1}:f\mid\lambda$ $\displaystyle\sim$ $\displaystyle\mbox{\rm
F}_{a,b}(\lambda),\quad\lambda\mid\tau^{2}\sim\omega\ \delta_{0}+(1-\omega)\
\mbox{\rm G}((a/2)+1,\ 1/2\tau^{2}),\ \tau>0.$

Then, the Bayes factor in favor of the $\mbox{\rm H}_{1}$ is of the form
$\mbox{\rm BF}^{t}_{10}(f\mid\tau^{2},\omega)=\frac{\omega+(1-\omega)\mbox{\rm
R}}{(1-\omega)+\omega\mbox{\rm R}}$, where $\mbox{\rm
R}=m_{1}\left(f\mid\tau^{2}\right)/m_{0}\left(f\right)=(1+\tau^{2})^{-\frac{a}{2}-1}{{}_{2}}\mbox{\rm
F}_{1}\left(\frac{a}{2}+1,\ \frac{a+b}{2},\ \frac{a}{2};\
\frac{af\tau^{2}}{(1+\tau^{2})(b+af)}\right)$.

######  Remark 3.8.

Under the set up in Proposition 3.7, we further assume that (i) the test
statistic $f\sim\mbox{\rm F}_{a,b}(\eta n)$, for some $\eta$ ; the hyper-
parameters (ii) $b=\delta n-u$ and (iv)
$(1-{\omega})/{\omega}=\mathrm{o}(\exp(n^{\beta}))$ , $0<\beta<1$. Then, the
Bayes Factor $\mbox{\rm
BF}^{t}_{10}(f\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(-n^{\beta}))$ under
$\mbox{\rm H}_{0}$, and $\mbox{\rm
BF}^{t}_{10}(f\mid\tau^{2},\omega)=\mathrm{o}_{p}(\exp(n^{\beta}))$ under
$\mbox{\rm H}_{1}$. This demonstrates that the Bayes factor in Proposition 3.7
is consistent under both $\mbox{\rm H}_{0}$ and $\mbox{\rm H}_{1}$.

Remarks 3.2, 3.4, 3.6 and 3.8 on the convergence of $\mbox{\rm BF}^{t}_{10}$
based on the each data partition $\mathbf{x}^{(i)},i\in[M]$ imply rapid
convergence of the pooled privatised log-Bayes factor
$f(\mathbf{x})=\frac{1}{M}\log\mbox{BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$
in (2.6), provided appropriate assumptions are met - the specifics of the
result is detailed in Remark 3.9.

######  Remark 3.9.

Additionally, with the assumptions of Remarks 3.2, 3.4, 3.6 and 3.8, assume
that the number of partitions of the data $M=\mathcal{O}(n^{\gamma})$,
$0<\beta<\gamma\leq 1$. This implies the privacy noise
$\eta=\mathcal{O}_{p}(1)$. Let us denote the privatized Bayes factor against
the null hypothesis $\mbox{\rm H}_{0}$ obtained using the random mechanism in
(2.6) by $\mbox{\rm BF}_{10}(S)$, where $S$ is the test statistic under
consideration. When the $\mbox{\rm H}_{0}$ is true, $\mbox{\rm
BF}_{10}(S)=\mathrm{o}_{p}(\exp(-n^{\beta}))$. When the $\mbox{\rm H}_{1}$ is
true $\mbox{\rm BF}_{10}(S)=\mathrm{o}_{p}(\exp(n^{\beta}))$.

Figure 1: Determining size $\alpha$ Bayes factor cut-off ($t$-test). We
present the distribution of log-Bayes factor in Non-private and Private, under
$\mbox{H}_{0}$. Non-private and Private Bayes factor cut-offs obtained via
Monte Carlo simulations, given size of the test $\alpha=0.05$, sample size
$n=100$ and privacy budget $\varepsilon=1$, with fixed hyper-parameters
$(M,a)=(5,3)$.

##  4 Experiments

###  4.1 Test for Normal Mean ($z$ or $t$-test)

Suppose a confidential database has access to samples $x_{1},\ldots,x_{n}$
from a univariate normal $\mbox{N}(\mu,\sigma^{2})$ with unknown variance. The
database releases response to an analyst’s queries, upon privatization via
Laplace mechanism with a fixed privacy budget $\varepsilon$. Without the loss
of generality, we assume $\varepsilon=1$. In this context, we intend to
utilise differentially private Bayes factors based on $t$-statistics, to test
the hypothesis $\mu=0$ against $\mu\neq 0$. The goal is to first showcase the
numerical schemes to determine the size $\alpha$ cut-off for the Bayes factor.
Given the sample size $n$, privacy mechanism, the user defined parameters
$(\varepsilon,a,M)$, and the size $\alpha$ of the test, we determine the cut
off $\gamma_{\alpha,\varepsilon}(n)$ by satisfying the size constraint in
(2.8). We accomplish this numerically via first obtaining $10^{3}$ samples of
size $n$ under the hypothesis that $\mu=0$, and then setting
$\gamma_{\alpha,\varepsilon}(n)$ equal to the $(1-\alpha)$-th quantile of
observed Bayes factors - refer to Figure 1 for an illustration.

Figure 2: Hyper-parameter tuning in size $\alpha$ Bayesian test ($t$-test).
The power of the size $\alpha$ privatised Bayesian test for sample size
$n=100$ and privacy budget $\varepsilon=1$, with varying values of hyper-
parameters $(M,a)$.

Next, we discuss the numerical scheme to tune the hyperparameters $(M,a)$.
Given $n$, privacy mechanism, $\varepsilon$, and $\alpha$, we determine the
Bayes factor cut off $\gamma_{\alpha,\varepsilon}(n)$, for varying values of
the user defined parameters $(M,a)$, as described earlier. Then, the optimal
choices of the hyperparameters $(M,a)$ is obtained via maximising power of the
test over
$\mathcal{M}\times\mathcal{A}=\\{2,\ldots,10\\}\times\\{1,\ldots,5\\}$. We
accomplish this numerically via first obtaining $10^{3}$ samples of size $n$
under the hypothesis that $\mu\neq 0$ for each
$(M,a)\in\mathcal{M}\times\mathcal{A}$, and then choose the
$(\hat{M},\hat{a})$ that maximises power. We assume $\mu\in\\{\pm 0.01,\pm
0.02,\ldots,\pm 1\\}$ describes the values of $\mu$ under the hypothesis that
$\mu\neq 0$. In specific applications, the analyst usually have a pretty good
idea about the effect size of interest, and the set of values of $\mu$ that
describes the hypothesis that $\mu\neq 0$ should be updated accordingly. Refer
to Figure 2 for an example of the hyper-parameter tuning exercise.

Finally, we wish to compare the size $\alpha$ private Bayesian test with the
the size $\alpha$ non-private Bayes factor based on $t$-statistics (Johnson et
al., 2023b), with respect to power. We consider an increasing grid of sample
sizes $n\in\\{25,50,100,200,500\\}$. For each sample size, we proceed as
described previously for determining the size $\alpha$ cut-offs and tuning the
hyper-parameters. The power comarison results are presented in Figure 3. As
expected, we do loose on power to ensure privacy, but the difference in power
of the private and non-private Bayesian tests diminishes as we increase sample
size. More comparison?

Similarly, suppose the confidential database had access to samples
$x_{1},\ldots,x_{n}$ from a univariate normal distribution $\mbox{N}(\mu,1)$
with a known variance. In this context, we would utilize differentially
private Bayes factors based on $z$-statistics to test the hypothesis $\mu=0$
against $\mu\neq 0$. Due to the similarity of this setup with the previous
one, we do not provide numerical results in this context.

Figure 3: Power analysis of size $\alpha$ non-private and private Bayesian
test ($t$-test). Comparison of the size $\alpha$ non-private Bayes factor
based on $t$-statistic, and size $\alpha$ private Bayes factor based on
$t$-statistic with hyper-parameters set at $(\hat{M},\hat{a})$.

###  4.2 Test for Independence in Contingency Tables ($\chi^{2}$-test)

Suppose a confidential database has access to paired binary features
$(x_{1},y_{1}),\ldots,(x_{n},y_{n})\in\\{0,1\\}\times\\{0,1\\}$ on $n$
individuals. The database releases response to an analyst’s queries, upon
privatization via Laplace mechanism with a fixed privacy budget $\varepsilon$.
Without the loss of generality, we assume $\varepsilon=1$. In this context,
suppose we intend to test the hypothesis that the binary features $(X,Y)$ are
independent against the hypothesis that $(X,Y)$ are not independent. We
utilise differentially private Bayes factors based on $\chi^{2}$-statistic to
accomplish this.

We wish to compare the size $\alpha$ private Bayesian test with the size
$\alpha$ non-private Bayes factor based on $\chi^{2}$-statistic (Johnson et
al., 2023b), with respect to power. We consider an increasing grid of sample
sizes $n\in\\{500,1000,1500,2000\\}$. For each sample size, under the
hypothesis that the binary features $(X,Y)$ are independent, we generate $n$
observations independently from a Multinomial distribution with $4$ classes
$\\{1,2,3,4\\}$ with cell probabilities
$\pi_{0}=(1/4,1/4,1/4,1/4)^{{\mathrm{\scriptscriptstyle T}}}$. On the other
hand, under the hypothesis that the binary features $(X,Y)$ are not
independent, we generate $n$ observations independently from a Multinomial
distribution with $4$ classes $\\{1,2,3,4\\}$ with cell probabilities
$\pi_{1}=\pi_{0}+\Delta(1,-1,-1,1)$ where
$\Delta\in\\{0.05,0.10,0.15,0.20\\}$. Under both the hypotheses, the classes
$\\{1,2,3,4\\}$ denote $(X,Y)=(0,0),(0,1),(1,0)$ and $(1,1)$ respectively. For
each sample size, first, we determine the size $\alpha$ cut-off for the Bayes
factors and carry out the hyperparameter tuning step numerically, as the
previous experiment. We skip the details to avoid repetition in our
exposition. The power comparisons of the non-private and private tests are
presented in Figure 4. As expected, we do loose on power to ensure privacy,
but the difference in power of the private and non-private Bayesian tests
diminishes as we increase sample size.

Figure 4: Power analysis of size $\alpha$ non-private and private Bayesian
test ($\chi^{2}$-test). Comparison of the size $\alpha$ non-private Bayes
factor based on $\chi^{2}$-statistic, and size $\alpha$ private Bayes factor
based on $\chi 2$-statistic with hyper-parameters set at $(\hat{M},\hat{a})$.

###  4.3 Test for Significance of Linear Regression ($F$-test)

Suppose a confidential database has access to information on a $p$-dimensional
feature vector and a response of interest
$(x_{1},y_{1}),\ldots,(x_{n},y_{n})\in\mathbf{R}^{p}\times\mathbf{R}$ for $n$
individuals. The database releases response to an analyst’s queries, upon
privatization via Laplace mechanism with a fixed privacy budget $\varepsilon$.
Without the loss of generality, we assume $\varepsilon=1$. Suppose we fit a
linear regression model $y_{i}=\beta_{0}+\beta^{{\mathrm{\scriptscriptstyle
T}}}x_{i}+e_{i},\ i\in[n]$ on the data, where the errors $e_{i}$ are
independently distributed as $\mbox{N}(0,\sigma^{2})$. In this context,
suppose we intend to test whether the linear regression is significant, i.e we
want to test the hypothesis that
$\beta=(\beta_{1},\ldots,\beta_{p})=(0,0,\ldots,0)$ versus the hypothesis that
at least one of $\beta_{j},\ j\in[p]$ is non-zero. We utilise differentially
private Bayes factors based on $F$-statistic to accomplish this.

We wish to compare the size $\alpha$ private Bayesian test with the non-
private Bayes factor based on $F$-statistic (Johnson et al., 2023b) using the
size $\alpha$ cut-off, with respect to power. We consider an increasing grid
of sample sizes $n\in\\{50,100,200,500\\}$. We set the number of covariates
$p=2$, set the intercept parameter $\beta_{0}=1$, and set the regression noise
variance $\sigma^{2}=0.1^{2}$. For each sample size, we generate the each of
covariates from univariate $\mbox{N}(0,1)$ independently. Further, under the
hypothesis that the regression is not significant, we have
$\beta_{1}=\beta_{2}=0$. On the other hand, under the hypothesis that the
regression is significant, we set $(\beta_{1},\beta_{2})\in\\{\pm 0.25,\pm
0.5,\pm 0.75,\pm 1\\}^{2}$. For each sample size, first, we determine the size
$\alpha$ cut-off for the Bayes factors and carry out the hyperparameter tuning
step numerically. We skip the details for to avoid repetition in our
exposition. The power comparisons of the non-private and private tests are
presented in Figure 5. As expected, we do loose on power to ensure privacy,
but the difference in power of the private and non-private Bayesian tests
diminishes as we increase sample size.

Figure 5: Power analysis of size $\alpha$ non-private and private Bayesian
test ($F$-test). Comparison of the size $\alpha$ non-private Bayes factor
based on $\chi^{2}$-statistic, and size $\alpha$ private Bayes factor based on
$F$-statistic with hyper-parameters set at $(\hat{M},\hat{a})$.

##  5 Discussion

We presented a novel differentially private Bayesian hypothesis testing
methodology, coherently embedding popular tools from existing literature
(e.g., subsample and aggregate, truncation of the test function, etc.) within
a generative model framework. In contrast to existing approaches, the proposed
method enables us to systematically accumulate evidence in support of the true
hypothesis—a feature desirable in reporting scientific discoveries utilizing
confidential data. To mitigate the sensitivity of the Bayesian testing
framework to the choice of priors and nuisance parameters, we devised our
differentially private Bayes factors based on commonly used test statistics.
Under an appropriate asymptotic regime, we derive the rates of consistency of
the proposed differentially private Bayes factors. The theoretical findings on
the frequentist operating characteristics of the proposed tests were further
substantiated via numerical experiments.

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## Supplementary material to
“Differentially Private Bayesian Tests”

Supplementary section 6 contains proof of propositions and remarks from
Section 2 in the main document. Supplementary section 7 contains proof of
propositions and remarks from Section 3 in the main document.

##  6 Proofs from Section 2 in the main document

######  Proposition 6.1.

For $\omega\in(0,1/2)$, $\mbox{\rm
BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$ is bounded
between $[\frac{\omega}{1-\omega},\frac{1-\omega}{\omega}]$.

###### Proof.

If
$\frac{m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2})}{m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})}=0$,
then
$\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\omega/(1-\omega)$.
Additionally, if
$\frac{m_{1}(\mathbf{x}^{(i)}\mid\tau_{1}^{2})}{m_{0}(\mathbf{x}^{(i)}\mid\tau_{0}^{2})}=\infty$,
then
$\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=(1-\omega)/\omega$.
Next, we define a map

$\displaystyle g(t)=\frac{\omega\ +\ (1-\omega)t}{(1-\omega)\ +\ \omega
t},\quad t\in[0,\infty).$

Note that, the derivative of $g(t)$,

$\displaystyle g^{\prime}(t)=\frac{1-2\omega}{(1\ -\ \omega\ +\ \omega
t)^{2}}>0$

for $\omega\in(0,1/2)$. So, $g(t)$ is increasing in $t\in[0,\infty)$, for
$\omega\in(0,1/2)$. This implies that the minimum and maximum attainable
evidence against the null hypothesis is $\frac{\omega}{1-\omega}$ and
$\frac{1-\omega}{\omega}$, respectively. ∎

######  Proposition 6.2.

Under (2.1)-(2.3), the Bayes Factor against the null hypothesis $\mbox{H}_{0}$
combined across all the partitions $\mathbf{x}^{(i)},\ i\in[M]$ is expressed
as

$\displaystyle\mbox{\rm
BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\prod_{i=1}^{M}\mbox{\rm
BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega).$

###### Proof.

Given $\tau_{0}$, $\tau_{1}$ and $\omega\in(0,1/2)$, the prior distributions
on $\theta^{(i)}$ are independent across data partitions $\mathbf{x}^{(i)},\
i\in[M]$. Consequently, the joint marginal
$m_{k}^{t}(\mathbf{x}\mid\tau^{2}_{k},\omega)=\prod_{i=1}^{M}m_{k}^{t}(\mathbf{x}^{(i)}\mid\tau^{2}_{k},\omega)$,
where $k\in\\{0,1\\}$. Further, we note that

$\displaystyle\mbox{BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)=\frac{m_{1}^{t}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}{m_{0}^{t}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}=\frac{\prod_{i=1}^{M}m_{1}^{t}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}{\prod_{i=1}^{M}m_{0}^{t}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)}=\prod_{i=1}^{M}\mbox{BF}^{t}_{10}(\mathbf{x}^{(i)}\mid\tau_{0}^{2},\tau_{1}^{2},\omega).$

This concludes the proof. ∎

######  Remark 6.3.

Proposition 6.1 together with Proposition 6.2 implies that, the combined Bayes
Factor against the null hypothesis $\mbox{H}_{0}$ across all the partitions
$\mathbf{x}^{(i)},\ i\in[M]$, $\mbox{\rm
BF}^{t}_{10}(\mathbf{x}\mid\tau_{0}^{2},\tau_{1}^{2},\omega)$ is bounded
between $\left(\frac{\omega}{1-\omega}\right)^{M}$ and
$\left(\frac{1-\omega}{\omega}\right)^{M}$ for $\omega\in(0,1/2)$.

######  Proposition 6.4.

The function $\phi(\mathbf{x})$ for testing $\mbox{H}_{0}$ against
$\mbox{H}_{1}$, as described in (2.5)-(2.7), is $\varepsilon$-differentially
private.

###### Proof.

The global sensitivity $\mbox{GS}_{f}$ of $f(\mathbf{x})$ is $\frac{2a}{M}$.
Then, defining $\mbox{H}(\mathbf{x})=f(\mathbf{x})+\eta$, where $\eta$ follows
$\mbox{\rm Normal}(0,\ (\mbox{GS}_{f}/\varepsilon)^{2})$ or
$\mbox{Laplace}(0,\mbox{GS}_{f}/\varepsilon)$ completes the proof, by
definition. ∎

##  7 Proofs from Section 3 in the main document

###### Proposition 3.1.

According to the prior specification in Proposition 3.1,

$\displaystyle
m_{0}(z)=\frac{1}{\sqrt{2\pi}}\exp\big{(}-\frac{z^{2}}{2}\big{)},$ (7.1)

and,

$\displaystyle m_{1}(z\mid\tau^{2})$ $\displaystyle=$
$\displaystyle\frac{\exp\big{(}-\frac{z^{2}}{2(1+\tau^{2})}\big{)}}{\sqrt{2\pi}(2\tau^{2})^{\frac{3}{2}}\Gamma(\frac{3}{2})}\int_{-\infty}^{\infty}\lambda^{2}\exp\big{(}-\frac{1}{2a}(\lambda-
az)^{2}\big{)}\ d\lambda$ (7.2) $\displaystyle=$
$\displaystyle\frac{\exp\big{(}-\frac{z^{2}}{2(1+\tau^{2})}\big{)}}{(2\tau^{2})^{\frac{3}{2}}\Gamma(\frac{3}{2})\sqrt{\pi}}a^{\frac{3}{2}}\Gamma(r+\frac{1}{2}){{}_{1}}\mbox{F}_{1}\Big{(}-1,\frac{1}{2},-\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Big{)}$
$\displaystyle=$
$\displaystyle\frac{\exp\big{(}-\frac{z^{2}}{2(1+\tau^{2})}\big{)}}{\sqrt{2\pi}(1+\tau^{2})^{\frac{3}{2}}}{{}_{1}}\mbox{F}_{1}\Big{(}-1,\frac{1}{2},-\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Big{)}$
$\displaystyle=$
$\displaystyle\frac{\exp\big{(}-\frac{z^{2}}{2}\big{)}}{\sqrt{2\pi}(1+\tau^{2})^{\frac{3}{2}}}{{}_{1}}\mbox{F}_{1}\Big{(}\frac{3}{2},\frac{1}{2},\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Big{)}.$

Using (2.4), (7.1) and (7.2), we get the Bayes factor in Proposition 3.1. ∎

###### Remark 3.2.

Consider testing $\mbox{H}_{0}^{{}^{\prime}}:z\sim\mbox{N}(0,1)$ versus
$\mbox{H}_{1}^{{}^{\prime}}:z\mid\lambda\sim\mbox{N}(\lambda,1),\
\lambda\mid\sigma^{2},\tau^{2}\sim\mbox{\rm J}(\tau^{2})$. See supplementary
material of Datta et al. (2023).

When $\mbox{H}_{1}^{{}^{\prime}}$ is true,

$\displaystyle\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{3}{2}}}\sum_{i=0}^{\infty}\frac{(r+\frac{1}{2})^{(i)}}{(\frac{1}{2})^{(i)}i!}\Bigg{(}\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Bigg{)}^{i}$
$\displaystyle\geq$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{3}{2}}}\sum_{i=0}^{\infty}\frac{1}{i!}\Bigg{(}\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Bigg{)}^{i}$
$\displaystyle=$ $\displaystyle\frac{1}{(1+\kappa
n)^{\frac{3}{2}}}\exp\Bigg{(}\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Bigg{)}.$

This implies,

$\frac{m_{0}(z)}{m_{1}(z\mid\tau^{2})}=\mathcal{O}_{p}(\exp(-cn)),\quad\text{for
some }c>0.$ (7.3)

Hence, when the alternative under our privatised framework is true,

$\displaystyle\mbox{BF}^{t}_{10}(z\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}}{(1-\omega)+\omega\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}}$
(7.4) $\displaystyle=$
$\displaystyle\frac{1+\frac{(1-\omega)}{\omega}(\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)})}{\frac{(1-\omega)}{\omega}+\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(n^{\beta}))),\ 0<\beta<1.$

When the $\mbox{H}_{0}^{{}^{\prime}}$ is true,

$\displaystyle\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1+\kappa
n)^{\frac{3}{2}}}{{}_{1}}\mbox{F}_{1}\Bigg{(}\frac{3}{2},\frac{1}{2};\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Bigg{)}$
(7.5) $\displaystyle=$ $\displaystyle O_{p}(n^{-\frac{3}{2}}).$

When the null under our privatised framework is true,

$\displaystyle\mbox{BF}^{t}_{10}(z\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}}{(1-\omega)+\omega\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}}$
(7.6) $\displaystyle=$
$\displaystyle\frac{\frac{\omega}{(1-\omega)}+(\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)})}{1+\frac{\omega}{(1-\omega)}\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(-n^{\beta}))),\ 0<\beta<1.$

∎

###### Proposition 3.3.

Under the prior specifications of Proposition 3.3,

$m_{0}(t)=\frac{\Gamma\left(\frac{\nu+1}{2}\right)}{\sqrt{\nu\pi}\Gamma\left(\frac{\nu}{2}\right)}\left(1+\frac{t^{2}}{\nu}\right)^{-\frac{\nu+1}{2}},$
(7.7)

and,

$\displaystyle m_{1}(t\mid\tau^{2})$
$\displaystyle=m_{0}(t)\Bigg{[}c\sum_{k=0}^{\infty}\frac{\Big{(}\frac{\nu+1}{2}\Big{)}^{(k)}}{(\frac{1}{2})^{(k)}k!}\Bigg{(}\frac{t^{2}}{2(t^{2}+\nu)}\Bigg{)}^{k}\int_{-\infty}^{\infty}\lambda^{2(1+k)}\exp\Big{(}-\frac{\lambda^{2}(\tau^{2}+1)}{2\tau^{2}}\Big{)}\,d\lambda\
+$ $\displaystyle\quad
c\frac{\sqrt{2}t\Gamma\Big{(}\frac{\nu}{2}+1\Big{)}}{\sqrt{t^{2}+\nu}\Gamma\Big{(}\frac{\nu+1}{2}\Big{)}}\sum_{k=0}^{\infty}\frac{\Big{(}\frac{\nu+1}{2}\Big{)}^{(k)}}{\big{(}\frac{3}{2}\big{)}^{(k)}k!}\Bigg{(}\frac{t^{2}}{2(t^{2}+\nu)}\Bigg{)}^{k}\int_{-\infty}^{\infty}\lambda^{2(k+1)+1}\exp\Big{(}-\frac{\lambda^{2}(1+\tau^{2})}{2\tau^{2}}\Big{)}\,d\lambda\Bigg{]}.$
(7.8)

See the supplementary material of Datta et al. (2023).

Let,

$\displaystyle I_{1}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\lambda^{2(1+k)}\exp\Big{(}-\frac{\lambda^{2}(\tau^{2}+1)}{2\tau^{2}}\Big{)}\
d\lambda$ (7.9) $\displaystyle=$ $\displaystyle
2\int_{0}^{\infty}\lambda^{2(1+k)}\exp\Big{(}-\frac{\lambda^{2}(\tau^{2}+1)}{2\tau^{2}}\Big{)}\
d\lambda$ $\displaystyle=$
$\displaystyle\Gamma\Big{(}k+\frac{3}{2}\Big{)}\Big{(}\frac{2\tau^{2}}{1+\tau^{2}}\Big{)}^{k+\frac{3}{2}}.$

Assuming $\lambda^{2}=u$,

$\displaystyle I_{2}$ $\displaystyle=$
$\displaystyle\int_{-\infty}^{\infty}\lambda^{2+2k+1}\exp\Big{(}-\frac{\lambda^{2}(1+\tau^{2})}{2\tau^{2}}\Big{)}\
d\lambda$ (7.10) $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{\infty}u^{k+1}\exp\Big{(}-\frac{u(1+\tau^{2})}{2\tau^{2}}\Big{)}\
du$ $\displaystyle=$
$\displaystyle\frac{1}{2}\Gamma(k+2)\Big{(}\frac{2\tau^{2}}{1+\tau^{2}}\Big{)}^{k+2}$
$\displaystyle=$
$\displaystyle\Gamma(k+2)2^{k+1}\Big{(}\frac{\tau^{2}}{1+\tau^{2}}\Big{)}^{k+2}.$

Substituting in (7), the marginal density is

$\displaystyle m_{1}(t\mid\tau^{2})$ $\displaystyle=$ $\displaystyle
m_{0}(t)c\left[{{}_{2}}\mbox{F}_{1}\Big{(}\frac{\nu+1}{2},\frac{3}{2},\frac{1}{2},y^{2}\Big{)}\right.$
(7.11) $\displaystyle\left.+y\
\frac{\Gamma(\frac{\nu}{2}+1)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}}\frac{\Gamma(2)}{\Gamma(\frac{3}{2})}\
\
{{}_{2}}\mbox{F}_{1}\Big{(}\frac{\nu}{2}+1,2,\frac{3}{2},y^{2}\Big{)}\right],$

where

$y=\frac{\tau t}{\sqrt{(\nu+t^{2})(1+\tau^{2}})}\qquad\mbox{and}\qquad
c=\frac{1}{(1+\tau^{2})^{\frac{3}{2}}}.$ (7.12)

Using (2.4), (7.7) and (7.11), we get the Bayes factor in Proposition 3.3. ∎

###### Remark 3.4.

Consider testing $\mbox{H}_{0}^{{}^{\prime}}:t\mid\lambda\sim\mbox{\rm
T}_{\nu}(0)$ versus $\mbox{H}_{1}^{{}^{\prime}}:t\mid\lambda\sim\mbox{\rm
T}_{\nu}(\lambda),\ \lambda\mid\sigma^{2},\tau^{2}\sim\mbox{\rm J}(\tau^{2})$
(Datta et al., 2023).

$\displaystyle\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}=\frac{1}{(1+\tau^{2})^{r+\frac{1}{2}}}{{}_{2}}\mbox{F}_{1}\Bigg{(}r+\frac{1}{2},\frac{v+1}{2},\frac{1}{2},\frac{\tau^{2}t^{2}}{(t^{2}+\nu)(1+\tau^{2})}\Bigg{)}+$
$\displaystyle\quad\quad\quad\quad\quad\frac{\tau
t}{\sqrt{t^{2}+\nu}\sqrt{1+\tau^{2}}(1+\tau^{2})^{r+\frac{1}{2}}}\frac{\Gamma\big{(}\frac{\nu}{2}+1\big{)}\Gamma(r+1)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}\Gamma\big{(}r+\frac{1}{2}\big{)}}{{}_{2}}\mbox{F}_{1}\Bigg{(}r+1,\frac{\nu}{2}+1,\frac{3}{2},\frac{\tau^{2}t^{2}}{(1+\tau^{2})(t^{2}+\nu)}\Bigg{)}.$

It is known that,
${{}_{1}}\mbox{F}_{1}(a,c,z)=\lim_{b\to\infty}{{}_{2}}F_{1}(a,b,c,z/b)$. Let

$b_{1}=\frac{\nu+1}{2},\quad
z_{1}=\frac{(\nu+1)\tau^{2}t^{2}}{2(1+\tau^{2})(t^{2}+\nu)},\quad
a_{1}=r+\frac{1}{2},c_{1}=\frac{1}{2},$

,

$b_{2}=\frac{\nu}{2}+1,\quad
z_{2}=\frac{(\frac{\nu}{2}+1)\tau^{2}t^{2}}{(1+\tau^{2})(t^{2}+\nu)},\quad
a_{2}=r+1,\quad\mbox{and}\quad c_{2}=3/2$

.

When the $\mbox{H}_{1}^{{}^{\prime}}$ is true, $z_{1},z_{2}$ are $O_{p}({n})$.
As $b_{1},b_{2}\rightarrow\infty$,

$\displaystyle\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}$ $\displaystyle\approx$
$\displaystyle\frac{1}{(1+\tau^{2})^{r+\frac{1}{2}}}{{}_{1}}\mbox{F}_{1}(a_{1},c_{1},z_{1})+$
$\displaystyle\frac{\tau
t}{\sqrt{t^{2}+\nu}\sqrt{1+\tau^{2}}(1+\tau^{2})^{\frac{3}{2}}}\frac{\Gamma\big{(}\frac{\nu}{2}+1\big{)}\Gamma(2)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}\Gamma\big{(}\frac{3}{2}\big{)}}{{}_{1}}\mbox{F}_{1}(a_{2},b_{2},z_{2})$
$\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{3}{2}}}\sum_{i=0}^{\infty}\frac{(\frac{3}{2})^{(i)}}{(\frac{1}{2})^{(i)}i!}\Bigg{(}\frac{(\nu+1)\tau^{2}t^{2}}{2(1+\tau^{2})(t^{2}+\nu)}\Bigg{)}^{i}+$
$\displaystyle\frac{\tau
t}{\sqrt{t^{2}+\nu}\sqrt{1+\tau^{2}}(1+\tau^{2})^{\frac{3}{2}}}\frac{\Gamma\big{(}\frac{\nu}{2}+1\big{)}\Gamma(2)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}\Gamma\big{(}r+\frac{1}{2}\big{)}}\sum_{i=0}^{\infty}\frac{(2)^{(i)}}{(\frac{3}{2})^{(i)}i!}\Bigg{(}\frac{(\frac{\nu}{2}+1)\tau^{2}t^{2}}{(1+\tau^{2})(t^{2}+\nu)}\Bigg{)}^{i}$
$\displaystyle\geq$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{3}{2}}}\sum_{i=0}^{\infty}\Bigg{(}\frac{(\nu+1)\tau^{2}t^{2}}{2(1+\tau^{2})(t^{2}+\nu)}\Bigg{)}^{i}\frac{1}{i!}+$
$\displaystyle\frac{\tau
t}{\sqrt{t^{2}+\nu}\sqrt{1+\tau^{2}}(1+\tau^{2})^{\frac{3}{2}}}\frac{\Gamma\Bigg{(}\frac{\nu}{2}+1\Bigg{)}\Gamma(2)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}\Gamma\big{(}\frac{3}{2}\big{)}}\sum_{i=0}^{\infty}\Bigg{(}\frac{(\frac{\nu}{2}+1)\tau^{2}t^{2}}{(1+\tau^{2})(t^{2}+\nu)}\Bigg{)}^{i}\frac{1}{i!}.$

This implies

$\frac{m_{0}(t)}{m_{1}(t\mid\tau^{2})}=\mathcal{O}_{p}(\exp(-cn)),\quad\text{for
some }c>0.$ (7.14)

Hence,

$\displaystyle\mbox{BF}^{t}_{10}(t\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}}{(1-\omega)+\omega\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}}$
(7.15) $\displaystyle=$
$\displaystyle\frac{1+\frac{(1-\omega)}{\omega}(\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)})}{\frac{(1-\omega)}{\omega}+\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(n^{\beta}))),\ 0<\beta<1.$

When the $\mbox{H}_{0}^{{}^{\prime}}$ is true, $z_{1}$ and $z_{2}$ are
$O_{p}(1)$. Then,

$\displaystyle\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}$ $\displaystyle\approx$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{3}{2}}}{{}_{1}}\mbox{F}_{1}(a_{1},c_{1},z_{1})+$
$\displaystyle\frac{\tau
t}{\sqrt{t^{2}+\nu}\sqrt{1+\tau^{2}}(1+\tau^{2})^{\frac{3}{2}}}\frac{\Gamma\big{(}\frac{\nu}{2}+1\big{)}\Gamma(2)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}\Gamma\big{(}\frac{3}{2}\big{)}}{{}_{1}}\mbox{F}_{1}(a_{2},b_{2},z_{2})$
$\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{3}{2}}}\sum_{i=0}^{\infty}\frac{(\frac{3}{2})^{(i)}}{(\frac{1}{2})^{(i)}i!}\Bigg{(}\frac{(\nu+1)\tau^{2}t^{2}}{2(1+\tau^{2})(t^{2}+\nu)}\Bigg{)}^{i}+$
$\displaystyle\frac{\tau
t}{\sqrt{t^{2}+\nu}\sqrt{1+\tau^{2}}(1+\tau^{2})^{\frac{3}{2}}}\frac{\Gamma\big{(}\frac{\nu}{2}+1\big{)}\Gamma(2)}{\Gamma\big{(}\frac{\nu+1}{2}\big{)}\Gamma\big{(}\frac{3}{2}\big{)}}\sum_{i=0}^{\infty}\frac{(r+1)^{(i)}}{(\frac{3}{2})^{(i)}i!}\Bigg{(}\frac{(\frac{\nu}{2}+1)\tau^{2}t^{2}}{(1+\tau^{2})(t^{2}+\nu)}\Bigg{)}^{i}$
$\displaystyle=$ $\displaystyle
O_{p}(n^{-\frac{3}{2}})+O_{p}(n^{-\frac{3}{2}})$ $\displaystyle=$
$\displaystyle O_{p}(n^{-\frac{3}{2}}).$

Hence,

$\displaystyle\mbox{BF}^{t}_{10}(t\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}}{(1-\omega)+\omega\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}}$
(7.16) $\displaystyle=$
$\displaystyle\frac{\frac{\omega}{(1-\omega)}+(\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)})}{1+\frac{\omega}{(1-\omega)}\frac{m_{1}(t\mid\tau^{2})}{m_{0}(t)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(-n^{\beta}))),\ 0<\beta<1.$

∎

###### Proposition 3.5.

Note that, under the prior specifications of Proposition 3.5,

$\displaystyle
m_{0}(h)=\frac{1}{2^{\frac{k}{2}}\Gamma(\frac{k}{2})}h^{\frac{k}{2}-1}\exp(-\frac{h}{2}),$
(7.17)

and,

$\displaystyle m_{1}(h\mid\tau^{2})$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{\infty}\int\frac{h^{\frac{k}{2}+i-1}\exp(-h/2)2^{-i}}{i!2^{\frac{k}{2}+i}\Gamma\left(\frac{k}{2}+i\right)\left(2\tau^{2}\right)^{\frac{k}{2}+1}\Gamma\left(\frac{k}{2}+1\right)}\lambda^{i+\frac{k}{2}+1-1}\exp\left(-\frac{\lambda}{2\left(1+\tau^{2}\right)}\right)d\lambda$
(7.18) $\displaystyle=$
$\displaystyle\sum_{i=0}^{\infty}\frac{h^{\frac{k}{2}+i-1}\exp(-h/2)2^{-i}}{i!2^{\frac{k}{2}+i}\Gamma\left(\frac{k}{2}+i\right)\left(2\tau^{2}\right)^{\frac{k}{2}+1}\Gamma\left(\frac{k}{2}+1\right)}\frac{\Gamma\left(\frac{k}{2}+1+i\right)2^{\frac{k}{2}+1+i}}{\left(1+1/\tau^{2}\right)^{\frac{k}{2}+1+i}}$
$\displaystyle=$
$\displaystyle\frac{h^{\frac{k}{2}-1}\exp(-h/2)}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}+1\right)\left(1+\tau^{2}\right)^{\frac{k}{2}+1}}\sum_{i=0}^{\infty}\left[\frac{\tau^{2}h}{2\left(1+\tau^{2}\right)}\right]^{i}\frac{1}{i!}\frac{\Gamma\Big{(}\frac{k}{2}+1+i\Big{)}}{\Gamma\Big{(}\frac{k}{2}+i\Big{)}}$
$\displaystyle=$
$\displaystyle\frac{h^{\frac{k}{2}-1}\exp(-h/2)}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)\left(1+\tau^{2}\right)^{\frac{k}{2}+1}}\sum_{i=0}^{\infty}\frac{(\frac{k}{2}+r)^{(i)}}{(\frac{k}{2})^{(i)}}\left[\frac{\tau^{2}h}{2\left(1+\tau^{2}\right)}\right]^{i}\frac{1}{i!}$
$\displaystyle=$
$\displaystyle\frac{m_{0}(h)}{\left(1+\tau^{2}\right)^{\frac{k}{2}+1}}{{}_{1}}\mbox{F}_{1}\Bigg{(}\frac{k}{2}+1,\frac{k}{2},\frac{\tau^{2}h}{2(1+\tau^{2})}\Bigg{)}.$

See the supplementary material of Datta et al. (2023). Using (2.4), (7.17) and
(7.18), we get the desired Bayes factor.

∎

###### Remark 3.6.

Consider testing $\mbox{H}_{0}^{{}^{\prime}}:h\sim\chi^{2}_{\nu}(0)$ vs
$\mbox{H}_{1}^{{}^{\prime}}:h\mid\lambda\sim\chi^{2}_{\nu}(\lambda),\lambda\mid\tau^{2}\sim\mbox{\rm
G}((k/2)+1,\ 1/2\tau^{2})$(Datta et al., 2023).

When $\mbox{H}_{1}^{{}^{\prime}}$ is true,

$\displaystyle\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{1+\frac{k}{2}}}\sum_{i=0}^{\infty}\frac{(1+\frac{k}{2})^{(i)}}{(\frac{k}{2})^{(i)}i!}\Bigg{(}\frac{\tau^{2}h}{2(1+\tau^{2})}\Bigg{)}^{i}$
$\displaystyle\geq$
$\displaystyle\frac{1}{(1+\tau^{2})^{1+\frac{k}{2}}}\sum_{i=0}^{\infty}\frac{1}{i!}\Bigg{(}\frac{\tau^{2}h}{2(1+\tau^{2})}\Bigg{)}^{i}$
$\displaystyle=$ $\displaystyle\frac{1}{(1+\kappa
n)^{1+\frac{k}{2}}}\exp\Bigg{(}\frac{\tau^{2}h}{2(1+\tau^{2})}\Bigg{)}.$

This implies

$\frac{m_{0}(h)}{m_{1}(h\mid\tau^{2})}=\mathcal{O}_{p}(\exp(-cn)),\quad\text{for
some }c>0.$ (7.20)

Hence, when the alternative is true,

$\displaystyle\mbox{BF}^{t}_{10}(h\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}}{(1-\omega)+\omega\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}}$
(7.21) $\displaystyle=$
$\displaystyle\frac{1+\frac{(1-\omega)}{\omega}(\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)})}{\frac{(1-\omega)}{\omega}+\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(n^{\beta}))),\ 0<\beta<1.$

When the $\mbox{H}_{0}^{{}^{\prime}}$ is true,

$\displaystyle\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1+\kappa
n)^{1+\frac{k}{2}}}{{}_{1}}\mbox{F}_{1}\Bigg{(}1+\frac{k}{2},\frac{k}{2};\frac{\tau^{2}z^{2}}{2(1+\tau^{2})}\Bigg{)}$
(7.22) $\displaystyle=$ $\displaystyle O_{p}(n^{-(1+\frac{k}{2})}).$

When the null is true,

$\displaystyle\mbox{BF}^{t}_{10}(h\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}}{(1-\omega)+\omega\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}}$
(7.23) $\displaystyle=$
$\displaystyle\frac{\frac{\omega}{(1-\omega)}+(\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)})}{1+\frac{\omega}{(1-\omega)}\frac{m_{1}(h\mid\tau^{2})}{m_{0}(h)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(-n^{\beta}))),\ 0<\beta<1.$

∎

###### Proposition 3.7.

Note that, under the prior specifications of Proposition 3.7,

$\displaystyle
m_{0}(f)=\frac{1}{\beta(\frac{a}{2},\frac{b}{2})}\Big{(}\frac{a}{b}\Big{)}^{\frac{a}{2}}f^{\frac{a}{2}-1}\Big{(}1+\frac{a}{b}f\Big{)}^{-\frac{a+b}{2}}$
(7.24)

and,

$\displaystyle m_{1}(f\mid\tau^{2})$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\sum_{i=0}^{\infty}\frac{\exp\Big{[}-\frac{\lambda}{2}\Big{]}\Big{(}\frac{\lambda}{2}\Big{)}^{i}\Gamma(\frac{a}{2}+\frac{b}{2}+i)}{i!\Gamma(\frac{a}{2}+i)\Gamma(\frac{b}{2})}\Big{(}\frac{a}{b}\Big{)}^{\frac{a}{2}+i}\Big{(}\frac{b}{b+af}\Big{)}^{\frac{a}{2}+\frac{b}{2}+i}\times$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
f^{\frac{a}{2}+i-1}\frac{\lambda^{\frac{b}{2}+1-1}\exp\Big{[}-\frac{\lambda}{2\tau^{2}}\Big{]}}{\Gamma(\frac{a}{2}+1)(2\tau^{2})^{\frac{a}{2}+1}}\
d\lambda$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{\infty}\frac{\Gamma(\frac{a}{2}+\frac{b}{2}+i)}{2^{i}i!\Gamma(\frac{a}{2}+i)\Gamma(\frac{b}{2})}\Big{(}\frac{a}{b}\Big{)}^{\frac{a}{2}+i}\Big{(}\frac{b}{b+af}\Big{)}^{\frac{a}{2}+\frac{b}{2}+i}\times$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad
f^{\frac{a}{2}+i-1}\frac{\Gamma(i+\frac{a}{2}+1)}{\Gamma(\frac{a}{2}+1)(2\tau^{2})^{\frac{a}{2}+1}}\Big{(}\frac{2\tau^{2}}{\tau^{2}+1}\Big{)}^{i+\frac{a}{2}+1}$
$\displaystyle=$
$\displaystyle\frac{\Gamma\Big{(}\frac{a+b}{2}\Big{)}}{\Gamma\big{(}\frac{a}{2}\big{)}\Gamma\big{(}\frac{b}{2}\big{)}}\big{(}\frac{a}{b}\big{)}^{\frac{a}{2}}\big{(}1+\frac{af}{b}\big{)}^{\frac{a+b}{2}}f^{\frac{a}{2}-1}\big{(}\frac{1}{1+\tau^{2}}\big{)}^{\frac{a}{2}+1}\times$
$\displaystyle\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\sum_{i=0}^{\infty}\frac{\big{(}\frac{a+b}{2}\big{)}^{(i)}\big{(}\frac{a}{2}+1\big{)}^{(i)}}{\big{(}\frac{a}{2}\big{)}^{(i)}i!}\Big{(}\frac{af\tau^{2}}{(b+af)(1+\tau^{2})}\Big{)}^{i}$
$\displaystyle=$
$\displaystyle\frac{m_{0}(f)}{(1+\tau^{2})^{\frac{a}{2}+1}}{{}_{2}}F_{1}\Big{(}\frac{a+b}{2},\frac{a}{2}+1,\frac{a}{2},\frac{af\tau^{2}}{(b+af)(1+\tau^{2})}\Big{)}.$

See the supplementary material of Datta et al. (2023). Using equations (2.4),
(7.24) and (7), we get the desired Bayes factor. ∎

###### Remark 3.8.

Consider testing $\mbox{H}_{0}^{{}^{\prime}}:f\sim F_{a,b}(0)$ vs
$\mbox{H}_{1}^{{}^{\prime}}:f\mid\lambda\sim F_{a,b}(\lambda),\
\lambda\mid\tau^{2}\sim\mbox{\rm G}((k/2)+1,\ 1/2\tau^{2})$(Datta et al.,
2023). Define $\psi=\frac{(a+b)af\tau^{2}}{2(b+af)(1+\tau^{2})}$. When the
$\mbox{H}_{1}^{{}^{\prime}}$ is true, $\psi=\mathcal{O}_{p}(cn),c>0$. This
implies

$\displaystyle\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}{{}_{2}}F_{1}\Bigg{(}\frac{a}{2}+1,\frac{a+b}{2},\frac{a}{2},\frac{af\tau^{2}}{(b+af)(1+\tau^{2})}\Bigg{)}$
$\displaystyle\approx$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}{{}_{1}}F_{1}\Bigg{(}\frac{a}{2}+1,\frac{a}{2},z\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}\sum_{i=0}^{\infty}\frac{(\frac{a}{2}+1)^{(i)}}{(\frac{a}{2})^{(i)}i!}\Bigg{(}\frac{(a+b)af\tau^{2}}{2(b+af)(1+\tau^{2})}\Bigg{)}^{i}$
$\displaystyle\geq$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}\sum_{i=0}^{\infty}\Bigg{(}\frac{(a+b)af\tau^{2}}{2(b+af)(1+\tau^{2})}\Bigg{)}^{i}\frac{1}{i!}.$

This implies

$\frac{m_{0}(f)}{m_{1}(f\mid\tau^{2})}=\mathcal{O}_{p}(\exp(-cn)),\quad\text{for
some }c>0.$ (7.26)

Hence,

$\displaystyle\mbox{BF}^{t}_{10}(f\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)}}{(1-\omega)+\omega\frac{m_{1}(z\mid\tau^{2})}{m_{0}(z)}}$
(7.27) $\displaystyle=$
$\displaystyle\frac{1+\frac{(1-\omega)}{\omega}(\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)})}{\frac{(1-\omega)}{\omega}+\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(n^{\beta}))),\ 0<\beta<1.$

When the $\mbox{H}_{0}^{{}^{\prime}}$ is true, $\psi=O_{p}(1)$. Then,

$\displaystyle\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)}$ $\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}{{}_{2}}F_{1}\Bigg{(}\frac{a+b}{2},\frac{a}{2}+1,\frac{a}{2},\frac{af\tau^{2}}{(b+af)(1+\tau^{2})}\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}{{}_{2}}F_{1}\Bigg{(}\frac{a}{2}+1,\frac{a+b}{2},\frac{a}{2},\frac{af\tau^{2}}{(b+af)(1+\tau^{2})}\Bigg{)}$
$\displaystyle\approx$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}{{}_{1}}F_{1}\Bigg{(}\frac{a}{2}+1,\frac{a}{2},\psi\Bigg{)}$
$\displaystyle=$
$\displaystyle\frac{1}{(1+\tau^{2})^{\frac{a}{2}+1}}\sum_{i=0}^{\infty}\frac{(\frac{a}{2}+1)^{(i)}}{(\frac{a}{2})^{(i)}i!}\Bigg{(}\frac{(a+b)af\tau^{2}}{2(b+af)(1+\tau^{2})}\Bigg{)}^{i}$
$\displaystyle=$ $\displaystyle O(n^{-\frac{a}{2}-1})O_{p}(1)$
$\displaystyle=$ $\displaystyle O_{p}(n^{-\frac{a}{2}-1}).$

When the null is true,

$\displaystyle\mbox{BF}^{t}_{10}(f\mid\tau^{2},\omega)$ $\displaystyle=$
$\displaystyle\frac{\omega+(1-\omega)\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)}}{(1-\omega)+\omega\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)}}$
(7.28) $\displaystyle=$
$\displaystyle\frac{\frac{\omega}{(1-\omega)}+(\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)})}{1+\frac{\omega}{(1-\omega)}\frac{m_{1}(f\mid\tau^{2})}{m_{0}(f)}}$
$\displaystyle=$ $\displaystyle\mathrm{o}_{p}(\exp(-n^{\beta}))),\ 0<\beta<1.$

∎

###### Remark 3.9.

Let $S^{(i)}$ be the observed test statistic obtained from the $i$-th
partition of the data. Using the randomized mechanism in (2.6), the privatized
weight of evidence against the null is,

$\log\left(\mbox{\rm
BF}_{10}(S^{(i)})\right)=\frac{1}{M}\sum_{i=1}^{M}\log\left(\mbox{\rm
BF}_{10}^{t}(S^{(i)})\right)+\eta.$ (7.29)

This implies, when the alternative is true

$\displaystyle\mbox{\rm BF}_{10}(S^{(i)})$ $\displaystyle=$
$\displaystyle\exp\left(\frac{1}{M}\sum_{i=1}^{M}\log\left(\mbox{\rm
BF}_{10}^{t}(S^{(i)})\right)\right)\exp(\eta)$ (7.30) $\displaystyle=$
$\displaystyle\exp\left(\log\left(\prod_{i=1}^{M}\mbox{\rm
BF}_{10}^{t}(S^{(i)})\right)^{\frac{1}{M}}\right)\exp(\eta)$ $\displaystyle=$
$\displaystyle\left(\prod_{i=1}^{M}\mbox{\rm
BF}_{10}^{t}(S^{(i)})\right)^{\frac{1}{M}}\exp(\eta).$ $\displaystyle=$
$\displaystyle\mathrm{o}_{p}(\exp(n^{\beta}))),\ 0<\beta<1.$

Similarly, when the null is true,

$\displaystyle\mbox{\rm BF}_{10}(S^{(i)})$ $\displaystyle=$
$\displaystyle\exp\left(\frac{1}{M}\sum_{i=1}^{M}\log\left(\mbox{\rm
BF}_{10}^{t}(S^{(i)})\right)\right)\exp(\eta)$ (7.31) $\displaystyle=$
$\displaystyle\exp\left(\log\left(\prod_{i=1}^{M}\mbox{\rm
BF}_{10}^{t}(S^{(i)})\right)^{\frac{1}{M}}\right)\exp(\eta)$ $\displaystyle=$
$\displaystyle\left(\prod_{i=1}^{M}\mbox{\rm
BF}_{10}^{t}(S^{(i)})\right)^{\frac{1}{M}}\exp(\eta)$ $\displaystyle=$
$\displaystyle\mathrm{o}_{p}(\exp(-n^{\beta}))),\ 0<\beta<1.$

∎