Multisource Bayesian sequential binary hypothesis testing problem

Abstract

We consider the problem of testing two simple hypotheses about unknown local characteristics of several independent Brownian motions and compound Poisson processes. All of the processes may be observed simultaneously as long as desired before a final choice between hypotheses is made. The objective is to find a decision rule that identifies the correct hypothesis and strikes the optimal balance between the expected costs of sampling and choosing the wrong hypothesis. Previous work on Bayesian sequential hypothesis testing in continuous time provides a solution when the characteristics of these processes are tested separately. However, the decision of an observer can improve greatly if multiple information sources are available both in the form of continuously changing signals (Brownian motions) and marked count data (compound Poisson processes). In this paper, we combine and extend those previous efforts by considering the problem in its multisource setting. We identify a Bayes optimal rule by solving an optimal stopping problem for the likelihood-ratio process. Here, the likelihood-ratio process is a jump-diffusion, and the solution of the optimal stopping problem admits a two-sided stopping region. Therefore, instead of using the variational arguments (and smooth-fit principles) directly, we solve the problem by patching the solutions of a sequence of optimal stopping problems for the pure diffusion part of the likelihood-ratio process. We also provide a numerical algorithm and illustrate it on several examples.

Appendix

1.1 A.1 The proof of Lemma 4.2

Let $G_{\ell,r}(\phi,\xi) = \frac{\psi_{\ell}(\phi \land \xi) \eta_{r}(\phi \lor \xi) }{p^{2}(\xi) W_{\ell,r}(\xi)}$, 0<ϕ,ξ<∞ be the Green’s function of the boundary value problem (A ₀ f)(ϕ)−λ ₀ f(ϕ)=−k(ϕ) for every ℓ<ϕ<r and f(ℓ)=f(r)=0. Then $f(\phi)= \int^{r}_{\ell} G_{\ell,r}(\phi,\xi) k(\xi) \,\mathrm {d}\xi = \eta_{r}(\phi) \int^{\phi}_{\ell} \frac{2\psi_{\ell}(\xi)}{p^{2}(\xi) W_{\ell,r}(\xi)} k(\xi) \,\mathrm {d}\xi + \psi_{\ell}(\phi) \int^{r}_{\phi} \frac{2\eta_{r} (\xi)}{p^{2}(\xi) W_{\ell,r}(\xi)} k(\xi) \,\mathrm {d}\xi$, which is also the right-hand side of (4.5), is twice continuously differentiable and solves uniquely the boundary value problem above. Then $e^{-\lambda_{0} (\tau_{\ell,r} \land t)} f(Y^{\varPhi_{0}}_{\tau_{\ell,r} \land t}) = f(Y^{\varPhi_{0}}_{0}) + \int^{\tau_{\ell,r} \land t}_{0} e^{-\lambda_{0} s} (A_{0}f -\lambda_{0}f) ( Y^{\varPhi_{0}}_{s}) \,\mathrm {d}{s} + \int^{\tau_{\ell,r}\land t}_{0} e^{-\lambda_{0} s} f'(Y^{\varPhi_{0}}_{s}) p(Y^{\varPhi_{0}}_{s}) (\mathrm {d}{X}_{s} - \mu_{0} \,\mathrm {d}{s} )$ for every t≥0 by Itô rule. Because f′(⋅) and p(⋅) are continuous on [ℓ,r]⊂(0,∞), they are bounded, and the stochastic integral on the right-hand side is a square-integrable $(\mathbb {P}_{0},\mathbb {F}^{X})$-martingale. Taking firstly the expectations of both sides and then their limits as t→∞, and finally rearranging the terms lead to $f(\phi) = \mathbb {E}^{\phi}_{0} [\int^{\tau_{\ell,r}}_{0} e^{-\lambda_{0} s }k(Y^{\varPhi_{0}}_{s}) \,\mathrm {d}{s} ] + \mathbb {E}^{\varPhi_{0}}_{0} [e^{-\lambda_{0} \tau_{\ell, r}} f(Y^{\varPhi}_{\tau_{\ell,r}})] = \mathbb {E}^{\phi}_{0} [\int^{\tau_{\ell,r}}_{0} e^{-\lambda_{0} s} k(Y^{\varPhi_{0}}_{s}) \,\mathrm {d}{s} ]$, because f(ℓ)=f(r)=0. For the proof of (4.6), note lim _{ℓ↓0,r↑∞} τ _ℓ,r =∞ $\mathbb {P}^{\phi}_{0}$-a.s. for all ϕ∈(0,∞) since 0 and ∞ are natural boundaries of $Y^{\varPhi_{0}}$. Moreover, $\int^{\tau_{\ell,r}}_{0} e^{-\lambda_{0} t} |k(Y^{\varPhi_{0}}_{t})|\,\mathrm {d}{t} \leq c \int^{\tau_{\ell,r}}_{0} e^{-\lambda_{0} t} (1+ Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t} \leq c \int^{\infty}_{0} e^{-\lambda_{0} t} (1+ Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t}$, and since $\mathbb {E}^{\phi}_{0}[Y^{\varPhi_{0}}_{t}] = \phi e^{-(\lambda_{1}-\lambda_{0})t}$ as in (4.4), we have $\mathbb {E}^{\phi}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t} (1+ Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t}] = \frac{1}{\lambda_{0}} + \int^{\infty}_{0} e^{-\lambda_{0} t} \mathbb {E}^{\phi}_{0} [ Y^{\varPhi_{0}}_{t} ]\,\mathrm {d}{t} = \frac{1}{\lambda_{0}} + \frac{\phi}{\lambda_{1}} <\infty$. Therefore, $\lim_{\ell \downarrow 0, r\uparrow \infty} \int^{\tau_{\ell,r}}_{0} e^{-\lambda_{0} t} k(Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t} = \int^{\infty}_{0} e^{-\lambda_{0} t} k(Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t}$-a.s. and the dominated convergence theorem implies $\mathbb {E}^{\phi}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t}\* k(Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t} ]= \lim_{\ell \downarrow 0, r \uparrow \infty} \mathbb {E}^{\phi}_{0} [\int^{\tau_{\ell,r}}_{0} e^{-\lambda_{0} t} k(Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t}]$ equals

(A.1)

We have lim _ℓ↓0↑ψ _ℓ (ξ)=ψ(ξ) and lim _r↑∞↑η _r (ξ)=η(ξ) for every ξ>0. Because α ₀<0, $\int^{\phi}_{\ell} \frac{2 \psi_{\ell}(\xi)}{p^{2}(\xi) W(\xi)} |k(\xi)| \,\mathrm {d}\xi \leq c \int^{\phi}_{0} \frac{2 \psi(\xi)}{p^{2}(\xi) W(\xi)} (1+\xi) \,\mathrm {d}\xi = c\int^{\phi}_{0} \xi^{-1-\alpha_{0}}(1+\xi)\,\mathrm {d}\xi <\nobreak\infty$, and the dominated convergence theorem implies that $\lim_{\ell \downarrow 0} \int^{\phi}_{\ell} \frac{2 \psi_{\ell}(\xi)}{p^{2}(\xi) W(\xi)} k(\xi) \,\mathrm {d}\xi = \int^{\phi}_{0} \frac{2 \psi(\xi)}{p^{2}(\xi) W(\xi)} k(\xi) \,\mathrm {d}\xi$. Similarly, because α ₁>0, we have $\int^{r}_{\phi} \frac{2 \eta_{r}(\xi)}{p^{2}(\xi) W(\xi)} |k(\xi)| \,\mathrm {d}\xi \leq c \int^{\infty}_{\phi}\frac{2 \eta(\xi)}{p^{2}(\xi) W(\xi)} (1+\xi) \,\mathrm {d}\xi = c\int^{\infty}_{\alpha} \xi^{-1-\alpha_{1}}(1+\xi) \,\mathrm {d}\xi < \infty$, and by the dominated convergence, $\lim_{r\uparrow \infty} \int^{r}_{\phi} \frac{2\eta_{r}(\xi)}{p^{2}(\xi) W(\xi)} k(\xi) \,\mathrm {d}\xi = \int^{\infty}_{\phi} \frac{2 \eta(\xi)}{p^{2}(\xi) W(\xi)} k(\xi) \,\mathrm {d}\xi$. Taking the limits on the right-hand side of (A.1) and using (4.12) complete the proof of (4.6), which can be directly shown to satisfy (A ₀ f−λ ₀ f)(ϕ)+k(ϕ)=0 for every ϕ>0.

Finally, suppose that the limit k(0+)=lim _ϕ↓0 k(ϕ) exists. For every 0<ϕ≤1, we have $Y^{\phi}_{t} \leq Y^{1}_{t}$ for every t≥0 $\mathbb {P}^{1}_{0}$-a.s. Note that $\mathbb {E}^{\phi}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t} k(Y^{\varPhi_{0}}_{t})\,\mathrm {d}{t} ] = \mathbb {E}^{1}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t} k(Y^{\phi}_{t})\,\mathrm {d}{t} ]$ for every ϕ>0, where $|k(Y^{\phi}_{t})|\leq c(1+Y^{\phi}_{t})\leq c(1+Y^{1}_{t})$, and $\mathbb {E}^{1}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t} (1+Y^{1}_{t})\,\mathrm {d}{t}] = \int^{\infty}_{0} e^{-\lambda_{0} t} (1+ \mathbb {E}^{1}_{0} Y^{1}_{t})\,\mathrm {d}{t} = \int^{\infty}_{0} e^{-\lambda_{0} t} (1+ e^{-(\lambda_{1}-\lambda_{0}) t}) \,\mathrm {d}{t} = \int^{\infty}_{0} e^{-\lambda_{0} t} \,\mathrm {d}{t} + \int^{\infty}_{0} e^{-\lambda_{1} t}\,\mathrm {d}{t}= \frac{1}{\lambda_{0}} + \frac{1}{\lambda_{1}}<\infty$. Because $\lim_{\phi \downarrow 0} Y^{\phi}_{t} = 0$ and $\lim_{\phi \downarrow 0} k(Y^{\phi}_{t}) = k(0+)$ for every t≥0 ℙ¹-a.s., the dominated convergence implies that $\lim_{\phi \downarrow 0} \mathbb {E}^{\phi}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t} k(Y^{\varPhi_{0}}_{t}) \,\mathrm {d}{t}] = \lim_{\phi \downarrow 0} \mathbb {E}^{1}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t}\* k(Y^{\phi}_{t}) \,\mathrm {d}{t}] = \mathbb {E}^{1}_{0} [\int^{\infty}_{0} e^{-\lambda_{0} t} (\lim_{\phi \downarrow 0} k(Y^{\phi}_{t})) \,\mathrm {d}{t}] =k(0+) \int^{\infty}_{0} e^{-\lambda_{0} t}\,\mathrm {d}{t} =\frac{k(0+)}{\lambda_{0}}$, which completes the proof.

1.2 A.2 The proof of Lemma 4.3

Recall w:ℝ₊↦ℝ is increasing and −b≤w(ϕ)≤h(ϕ), ϕ∈ℝ₊. Then w(0+)<0, since otherwise w(⋅)≡0. Also $(K w)(\phi) = \int_{E} w(\frac{\lambda_{1}}{\lambda_{0}} \frac{\mathrm {d}\nu_{1}}{\mathrm {d}\nu_{0}}(z)\phi) \nu_{0}(\mathrm {d}{z})$ is increasing and −b≤(Kw)(ϕ)≤0, ϕ∈ℝ₊. Then $\lim_{\phi \downarrow 0} \int^{\phi}_{0} \xi^{-1-\alpha_{0}}|(K w)(\xi)| \,\mathrm {d}\xi \leq b\, \lim_{\phi \downarrow 0} \int^{\phi}_{0} \xi^{-1-\alpha_{0}} \,\mathrm {d}\xi = b\, \lim_{\phi \downarrow 0} \frac{\phi^{-\alpha_{0}}}{(-\alpha_{0})}$, since α ₀<0, and (i) follows.

Next notice that lim _ϕ↓0(Kw)(ϕ)=(Kw)(0+)=w(0+) by the bounded convergence theorem. For every fixed ε>0, there exists some δ>0 such that ϕ∈(0,δ) implies that w(0+)≤(Kw)(ϕ)≤w(0+)(1−ε). Then for every ϕ∈(0,δ) $w(0+) \int^{\phi}_{0} \xi^{-1-\alpha_{0}}\,\mathrm {d}\xi \leq \int^{\phi}_{0} \xi^{-1-\alpha_{0}} (K w)(\xi)\,\mathrm {d}\xi \leq w(0+)(1-\varepsilon)\int^{\phi}_{0} \xi^{-1-\alpha_{0}}\,\mathrm {d}\xi$ or $\frac{w(0+)}{(-\alpha_{0})} \leq \phi^{\alpha_{0}} \int^{\phi}_{0} \xi^{-1-\alpha_{0}} (K w)(\xi)\,\mathrm {d}\xi \allowbreak \leq \frac{w(0+)}{(-\alpha_{0})}(1-\varepsilon)$, which proves (ii) after taking limits as ϕ↓0, since ε>0 was arbitrary.

Because w(0+)≡(Kw)(0+)<0, there exists some δ such that ϕ∈(0,δ) implies that (Kw)(ϕ)<(1/2)(Kw)(0+). Then $\int^{\infty}_{\phi}\xi^{-1-\alpha_{1}}(K w)(\xi)\,\mathrm {d}\xi \leq \int^{\delta}_{\phi} \xi^{-1-\alpha_{1}} (K w)(\xi)\,\mathrm {d}\xi \leq \frac{w(0+)}{2} \int^{\delta}_{\phi} \xi^{-1-\alpha_{1}}\,\mathrm {d}\xi = \frac{w(0+)(\delta^{-\alpha_{1}}-\phi^{-\alpha_{1}})}{2(-\alpha_{1})}$ for every ϕ∈(0,δ), and because w(0+)<0 and α ₁>1, we have $\varlimsup_{\phi \downarrow 0} \int^{\infty}_{\phi} \xi^{-1-\alpha_{1}} (K w)(\xi)\,\mathrm {d}\xi \leq -\infty$, which completes the proof of (iii).

For every fixed ε>0, there exists some δ>0 such that ϕ∈(0,δ) implies that w(0+)≡(Kw)(0+)≤(Kw)(ϕ)≤w(0+)(1−ε). Therefore, for every ϕ∈(0,δ) we have $\frac{w(0+)}{\alpha_{1}}(\phi^{-\alpha_{1}} - \delta^{-\alpha_{1}}) \leq \int^{\delta}_{\phi} \xi^{-1-\alpha_{1}} (K w)(\xi)\,\mathrm {d}\xi \leq \frac{w(0+)}{\alpha_{1}}(1-\varepsilon)(\phi^{-\alpha_{1}}-\delta^{-\alpha_{1}})$. Adding $\int^{\infty}_{\delta} \xi^{-1-\alpha_{1}} (K w)(\xi)\,\mathrm {d}\xi$, which is finite, and multiplying by $\phi^{\alpha_{1}}$ all three sides give

Since α ₁>0, $\frac{w(0+)}{\alpha_{1}} \leq \varliminf_{\phi \downarrow 0} \phi^{\alpha_{1}} \int^{\infty}_{\phi} \xi^{-1-\alpha_{1}} (K w)(\xi) \,\mathrm {d}\xi \leq \varlimsup_{\phi \downarrow 0} \phi^{\alpha_{1}} \int^{\infty}_{\phi} \xi^{-1-\alpha_{1}} (K w)(\xi) \*\,\mathrm {d}\xi \leq \frac{w(0+)}{\alpha_{1}}(1-\varepsilon)$, which completes the proof of (iv) because ε>0 is arbitrary.

For the proof of (v), firstly note that the monotonicity of w(⋅) and the bounded convergence theorem implies that $(K w)(\infty) = \lim_{\phi \uparrow \infty} (K w)(\phi) = \lim_{\phi \uparrow \infty} \int w(\frac{\lambda_{1}}{\lambda_{0}}\frac{\mathrm {d}\nu_{1}}{\mathrm {d}\nu_{0}}(z)\phi) \nu_{0}(\mathrm {d}{z})$ exists and equals w(∞). Then for every ε>0 there exists some M>0 such that ϕ>M implies that w(∞)−ε≤(Kw)(ϕ)≤w(∞). Therefore, for every ϕ>M

Adding $\int^{M}_{0} \xi^{-1-\alpha_{0}} (K w)(\xi) \,\mathrm {d}\xi$, which is finite, and multiplying by $\phi^{\alpha_{0}}$ all three sides give

Letting ϕ↑∞ and recalling that α ₀<0 gives $\frac{w(\infty)-\varepsilon}{(-\alpha_{0})} \leq \varliminf_{\phi \uparrow \infty} \phi^{\alpha_{0}} \int^{\phi}_{M} \xi^{-1-\alpha_{0}} (K w)(\xi) \,\mathrm {d}\xi \leq \varlimsup_{\phi \uparrow\infty} \phi^{\alpha_{0}} \int^{\phi}_{M} \xi^{-1-\alpha_{0}} (K w)(\xi) \,\mathrm {d}\xi \leq \frac{w(\infty)}{(-\alpha_{0})}$. Because ε>0 is arbitrary, this proves (v). And (vi) follows from α ₁>0 and that $\lim_{\phi \uparrow \infty} \int^{\infty}_{\phi} \xi^{-1-\alpha_{1}} |(K w)(\xi)| \,\mathrm {d}\xi \leq b\cdot \lim_{\phi \uparrow \infty} \int^{\infty}_{\phi} \xi^{-1-\alpha_{1}} \,\mathrm {d}\xi\allowbreak = b\cdot \lim_{\phi\uparrow \infty} \frac{\phi^{-\alpha_{1}}}{\alpha_{1}} = 0$. To prove (vii), let ε and M be as in the proof of (v). Then for ϕ>M

Multiplying all sides by $\phi^{\alpha_{1}}$ gives $\frac{w(\infty)-\varepsilon}{\alpha_{1}} \leq \phi^{\alpha_{1}} \int^{\infty}_{\phi} \xi^{-1-\alpha_{1}} (K w)(\xi) \,\mathrm {d}\xi \leq \frac{w(\infty)}{\alpha_{1}}$ for all ϕ>M, which proves (vii) and Lemma 4.3 after taking limit as ϕ↑∞ because ε>0 is arbitrary.

1.3 A.3 The proof of Lemma 4.5

Because $Y^{\varPhi_{0}}$ is a regular diffusion on ℝ₊, $\mathbb {P}^{r}_{0}\{\tau_{\ell}<\infty \}>0$ and there exists some 0<t<∞ such that $\mathbb {P}^{r}_{0}\{\tau_{\ell}<t\}>0$. On the other hand, the sample-path decomposition in (2.16) of jump-diffusion process Φ into diffusion part $Y^{\varPhi_{0}}$ and jump part, which are ℙ₀-independent, implies that $\delta := \mathbb {P}^{r}_{0}\{\widetilde {\tau }_{\ell,\infty} \leq t\} \geq \mathbb {P}^{r}_{0}\{\widetilde {\tau }_{\ell,\infty} \leq t, T_{1}>t\} = \mathbb {P}^{r}_{0}\{\tau_{\ell} \leq t, T_{1}>t\} =\allowbreak \mathbb {P}^{r}_{0}\{\tau_{\ell} \leq \nobreak t\} \mathbb {P}^{r}_{0}\{T_{1}>t\} = \mathbb {P}^{r}_{0}\{\tau_{\ell} \leq t\}e^{-\lambda_{0} t}>0$. Next for every ϕ∈(ℓ,r),

Hence, $\sup_{\phi \in [\ell,r]} \mathbb {P}^{\phi}_{0}\{\widetilde {\tau }_{\ell,r} > t\} \leq 1-\delta<1$, and $\mathbb {E}^{\phi}_{0} \widetilde {\tau }^{k}_{\ell,r} = \sum^{\infty}_{m=0} \mathbb {E}^{\phi}_{0} [\widetilde {\tau }^{k}_{\ell,r}1_{\{ m t< \widetilde {\tau }_{\ell,r}\leq (m+1)t\}} ] \leq t^{k} \sum^{\infty}_{m=0} (m+1)^{k} \mathbb {P}^{\phi}_{0} \{\widetilde {\tau }_{\ell,r} > m t\}$ for all k>0. Since Φ is a strong $(\mathbb {P}_{0},\mathbb {F})$-Markov process,

$$ \begin{aligned} \mathbb {P}^{\phi}_0\{\widetilde {\tau }_{\ell,r} > m t\} &= \mathbb {P}^{\phi}_0\bigl\{\widetilde {\tau }_{\ell,r} > (m-1) t, \widetilde {\tau }_{\ell,r} > m t\bigr\} = \mathbb {E}^{\phi}_0 \bigl[ 1_{\{\widetilde {\tau }_{\ell,r} > (m-1) t\}} \mathbb {P}^{\varPhi_{(m-1)t}}_0 \{\widetilde {\tau }_{\ell,r} > t\} \bigr]\\ &\leq (1-\delta)\, \mathbb {P}^{\phi}_0 \bigl\{\widetilde {\tau }_{\ell,r} > (m-1) t\bigr\} \leq \cdots \leq (1-\delta)^m \quad \mbox{for every }m \geq 1, \end{aligned} $$

and $\mathbb {E}^{\phi}_{0} \widetilde {\tau }^{k}_{\ell,r} \leq t^{k} \sum^{\infty}_{m=0} (m+1)^{k} (1-\delta)^{m} <\infty$ for every ϕ∈[ℓ,r], which proves Lemma 4.5.