2.1. Assumptions and first consequences

Let $\mathcal{H}$ be a real separable Hilbert space with inner product $\langle\cdot,\cdot\rangle$ and denote by $\mathsf S(\mathcal{H}):=\{f\in\mathcal{H}:\|f\|=1\}$ the corresponding unit sphere. For a linear subspace $\mathbb V$ , we write $\mathbb V^\perp$ for its orthogonal complement and define the linear operator $P_{\mathbb{V}}\colon \mathcal{H} \to \mathbb{V}$ as the orthogonal projection from $\mathcal{H}$ to $\mathbb{V}$ .

Suppose we have a densely defined self-adjoint positive semi-definite linear operator $L \colon \mathcal{H} \to \mathcal{H}$ with domain ${\mathcal{D}}(L) \subseteq \mathcal{H}$ ; we will simply write $(L, {\mathcal{D}}(L))$ . Recall that the positive semi-definiteness implies that $\langle Lf,f\rangle \ge 0$ for all $f\in {\mathcal{D}}(L)$ . Then the spectrum of L, denoted by $\sigma(L)$ , is nonempty and nonnegative: $\sigma(L) \subset [0,\infty)$ ; cf. [Reference Hislop and Sigal14, Proposition 5.12]. We define

\begin{equation*} \begin{split} \lambda_0 &:\!= \inf \sigma(L), \\ \lambda_1 &:\!= \inf \left\{\sigma(L)\backslash \{\lambda_0\} \right\}\!, \end{split}\end{equation*}

which are crucial quantities that appear in the following assumption.

Note that under the assumption $\nu>0$ , $\lambda_0$ is indeed an eigenvalue of L (see e.g. [Reference Kulkarni, Nair and Ramesh20, Theorem 4.1]), which we assume to have a one-dimensional eigenspace. Therefore, we have $L\varphi = \lambda_0 \varphi$ , where $\varphi\in \mathsf S(\mathcal{H})$ denotes an eigenfunction of L with eigenvalue $\lambda_0$ . By [Reference Reed and Simon34, Theorem XIII.1], we in fact have the following variational representations of $\lambda_0$ and $\lambda_1$ :

(7) \begin{equation} \begin{split} \lambda_0 &= \inf_{f\in \mathsf S(\mathcal{H})\cap {\mathcal{D}}(L)} \langle f,Lf\rangle,\\ \lambda_1 &= {\sup}_{\mathbb V \in \mathcal V_1}\inf_{f\in \mathbb{V}^\perp \cap \mathsf S(\mathcal{H})\cap {\mathcal{D}}(L)} \langle f, Lf\rangle, \end{split} \end{equation}

where $\mathcal V_1$ denotes the set of one-dimensional linear subspaces of $\mathcal{H}$ . Note that $\lambda_1$ is not necessarily an eigenvalue of L.

Fix a bounded self-adjoint linear operator $H\colon \mathcal{H} \to \mathcal{H}$ , which serves as a ‘small’ perturbation. With this we set

\begin{align*}\widehat L :\!= L+H.\end{align*}

Note that we have ${\mathcal{D}}(\widehat L) = {\mathcal{D}}(L)$ , self-adjointness of $\widehat{L}$ , and $\langle \widehat L f, f\rangle \geq -\Vert H \Vert$ for any $f\in {\mathcal{D}}(L)$ , with $\Vert H \Vert$ being the operator norm of H. Define

(8) \begin{equation} \begin{split} \widehat \lambda_0 &:\!= \inf_{f\in \mathsf S(\mathcal{H})\cap {\mathcal{D}}(L)} \langle f,\widehat Lf\rangle,\\ \widehat\lambda_1 &:\!= {\sup}_{\mathbb V \in \mathcal V_1} \inf_{f\in \mathbb V^\perp \cap \mathsf S(\mathcal{H})\cap {\mathcal{D}}(L)} \langle f, \widehat Lf\rangle. \end{split} \end{equation}

The following lemma is a classical result; see e.g. [Reference Reed and Simon34, Section XIII.15]. For completeness we give a proof in Appendix A.2.

Lemma 1. (Weyl’s lemma.) For $j=0,1$ , we have

(9) \begin{equation} |\lambda_j - \widehat \lambda_j|\le \|H\|. \end{equation}

In the next assumption we formalize the notion that H is a ‘small’ perturbation.

Under the formulated assumptions we have two important consequences which follow from fairly standard functional analytic techniques; see Appendix A.3 for a full proof.

Lemma 2. Under Assumption 1 and 2 we have that $\widehat{\lambda}_0 < \widehat{\lambda}_1$ , and that $\widehat \lambda_0$ is an eigenvalue of $\widehat{L}$ . Furthermore, the perturbed operator $\widehat L$ possesses a unique eigenfunction, up to a sign, $\widehat \varphi \in \mathsf S(\mathcal{H})$ with respect to eigenvalue $\widehat{\lambda}_0$ , that is,

\begin{equation*} \widehat L\widehat \varphi = \widehat \lambda_0\widehat \varphi. \end{equation*}

Without loss of generality, we may assume that $\rho :\!= \langle \varphi, \widehat \varphi\rangle \ge 0$ .

Remark 1. A consequence of the previous results is that

(10) \begin{equation} \widehat \lambda_0 = \langle \widehat \varphi ,\widehat L \widehat \varphi \rangle = \inf_{f\in \mathsf S(\mathcal{H})\cap {\mathcal{D}}(\widehat L)} \langle f,\widehat Lf\rangle, \end{equation}

and the supremum in the definition of $\widehat \lambda_1$ in equation (8) is attained by $\mathbb V = {\mathrm{span}}\{\widehat{\varphi}\}$ .

2.2. Main estimate

The goal now is to estimate $\|\varphi-\widehat \varphi\|$ . We state and prove the main estimate.

Theorem 1. Under Assumption 1 and 2 we have

(11) \begin{equation} \|\varphi-\widehat \varphi\| \leq \frac{\| H\varphi \|}{\nu-2\|H\|}. \end{equation}

Proof. Recall that $\rho :\!= \langle \widehat \varphi, \varphi\rangle \ge 0$ and with $\mathbb{G}:={\mathrm{span}}\{ \varphi\}$ define

\begin{equation*} h :\!= \frac{P_{\mathbb{G}^\perp} \widehat{\varphi}}{\Vert P_{\mathbb{G}^\perp} \widehat{\varphi} \Vert} = (1-\rho^2)^{-1/2} (\widehat \varphi - \rho\, \varphi). \end{equation*}

(If $\rho=1$ , then $\widehat \varphi = \varphi$ and the result trivially holds true.) Note that $\|h\|=1$ and $h\perp \varphi$ as well as $L h \perp \varphi$ by the self-adjointness of L. Consequently, by using the representation of h, $L\varphi = \lambda_0 \varphi$ , and $\widehat L=L+H$ we have that

\begin{align*} \rho \lambda_0 \varphi + \rho H \varphi + \sqrt{1-\rho^2} \widehat L h = \rho \widehat L \varphi + \widehat L (\widehat \varphi -\rho \varphi) = \widehat L\widehat\varphi= \widehat{\lambda}_0 \widehat\varphi. \end{align*}

Taking the inner product with h, employing $\langle \widehat \varphi,h \rangle = \sqrt{1-\rho^2}$ , $\varphi\perp h$ , and rearranging yields

(12) \begin{equation} -\rho \langle H \varphi, h\rangle = \sqrt{1-\rho^2} \left( \langle h,\widehat L h\rangle -\widehat\lambda_0 \right).\end{equation}

Moreover, by $\widehat{L}=L+H$ , $h\perp \varphi$ , and Lemma 1 we have

\begin{align*} B & :\!= \langle h,\widehat{L} h \rangle - \widehat\lambda_0 = \langle h,L h \rangle + \langle h,H h \rangle - \widehat\lambda_0 \geq \lambda_1 + \langle h,H h \rangle - \widehat\lambda_0 \\ &\geq \lambda_1 - \Vert H \Vert - \widehat\lambda_0 \geq \lambda_1 - 2\|H\|-\lambda_0=\nu-2\|H\|,\end{align*}

which is, due to Assumption 2, strictly positive. Squaring (12), rearranging, and taking a square root, with $A:=|\langle H \varphi ,h\rangle|$ we obtain

\begin{equation*} \rho = \frac{B}{\sqrt{A^2 + B^2}}.\end{equation*}

Then

\begin{equation*} 1-\rho = \frac{\sqrt{A^2+B^2}-B}{\sqrt{A^2+B^2}} = \frac{A^2}{(\sqrt{A^2+B^2}+B)\sqrt{A^2+B^2}}\le\frac{A^2}{2B^2},\end{equation*}

from which we have that $\sqrt{2(1-\rho)}\le A/B$ . The fact that we can represent $\widehat \varphi$ by the orthogonal decomposition

\begin{equation*} \widehat \varphi = P_{\mathbb{G}} \widehat \varphi +P_{\mathbb{G}^\perp} \widehat \varphi = \rho\, \varphi + \sqrt{1-\rho^2 } h \end{equation*}

gives, with $\rho = \langle \widehat \varphi, \varphi\rangle \ge 0$ , that $\widehat \varphi - \varphi = (\rho-1)\varphi + \sqrt{1-\rho^2} h.$ Consequently,

\begin{align*}\Vert \widehat\varphi -\varphi \Vert = \sqrt{2(1-\rho)} \leq \frac{A}{B} \leq \frac{\Vert H\varphi \Vert}{\nu - 2\Vert H \Vert},\end{align*}

which finishes the proof.

Remark 2. A similar, and to some extent complementary, bound to (11) can be obtained via the sin $\theta$ theorem of [Reference Davis and Kahan10], which is stated in Section 3.1. Under Assumptions 1 and 2, we obtain

(13) \begin{equation} \|\varphi-\widehat \varphi\| \le \frac{2\sqrt{2}\|H\varphi\|}{\nu}. \end{equation}

For the convenience of the reader we provide in Appendix A.4 arguments to derive (13).

2.3. Additional auxiliary estimates

We now give some additional estimates which are useful for our subsequent examples. Let $\pi,\widetilde \pi$ be probability density functions on ${\mathbb{R}^d}$ . Fix a positive function $\gamma: {\mathbb{R}^d} \to (0,\infty)$ , and let $\Gamma(\mathrm{d} x)=\gamma(x)\mathrm{d} x$ be the corresponding measure on ${\mathbb{R}^d}$ . By $\mathcal L^2({\mathbb{R}^d},\Gamma)$ we denote the space of $\Gamma$ -square-integrable functions $f\colon \mathbb{R}^d \to \mathbb{R}$ equipped with the norm $\Vert f \Vert_2 :\!= \left( \int_{\mathbb{R}^d} \vert f(x) \vert^2 \,\Gamma(\mathrm{d} x) \right)^{1/2}$ . Now define the functions

\begin{equation*} \phi(x) :\!= \frac{\pi(x)}{\gamma(x)},\qquad \widetilde \phi(x) :\!= \frac{\widetilde \pi(x)}{\gamma(x)}, \qquad x\in\mathbb{R}^d.\end{equation*}

We assume that $\phi,\widetilde \phi \in \mathcal L^2(\Gamma)$ : we have that

(14) \begin{equation} \int_{\mathbb{R}^d} \frac{\pi(x)^2}{\gamma(x)}\mathrm{d} x <\infty, \qquad \int_{\mathbb{R}^d} \frac{\widetilde \pi(x)^2}{\gamma(x)}\mathrm{d} x<\infty. \end{equation}

Under this condition we state the first auxiliary estimate.

Lemma 3. Assume that (14) holds, and furthermore that $\Gamma$ is a finite measure, i.e. $\Lambda:\!= \Gamma(\mathbb{R}^d)^{1/2}<\infty$ . Then

\begin{equation*} \int_{\mathbb{R}^d}|\pi(x)-\widetilde \pi(x)|\mathrm{d} x \le \Lambda \|\phi-\widetilde \phi\|_2. \end{equation*}

Proof. We straightforwardly calculate

\begin{equation*} \begin{split} \int_{\mathbb{R}^d}|\pi(x)-\widetilde \pi(x)|\mathrm{d} x&=\int_{\mathbb{R}^d} |\phi-\widetilde \phi|\mathrm{d} \Gamma \\ &\le \left( \int_{\mathbb{R}^d} |\phi-\widetilde \phi|^{2}\mathrm{d} \Gamma \right)^{1/2}\left (\int_{\mathbb{R}^d} \mathrm{d} \Gamma \right )^{1/2} = \Lambda \|\phi-\widetilde \phi\|_2, \end{split} \end{equation*}

where we use the Cauchy–Schwarz inequality.

For the formulation of the next statement we need normalized functions $\phi$ and $\widetilde{\phi}$ with respect to the $\mathcal L^2({\mathbb{R}^d},\Gamma)$ -norm, that is, we define

\begin{equation*} \varphi(x) :\!= \frac{\phi(x)}{Z},\qquad \widetilde \varphi(x) :\!= \frac{\widetilde \phi(x)}{\widetilde Z},\end{equation*}

with $Z:\!= \|\phi\|_2, \widetilde Z:\!= \|\widetilde \phi\|_2$ .

Lemma 4. Suppose that (14) holds and $\Lambda:\!= \Gamma(\mathbb{R}^d)^{1/2}<\infty$ . Additionally, let

(15) \begin{equation} \|\varphi-\widetilde \varphi\|_2 \le \epsilon, \end{equation}

with $0<\epsilon<\Lambda^{-1} Z^{-1}$ being satisfied. Then, we have

(16) \begin{equation} \left \| \phi-\frac{Z}{\widetilde Z}\,\widetilde \phi\, \right \|_2 \le Z\epsilon, \end{equation}

and

(17) \begin{equation} |Z-\widetilde Z|\le C \epsilon, \end{equation}

with $C=Z^2 \Lambda/(1-\Lambda Z \epsilon)$ .

Proof. Note that (15) immediately implies (16). Then, we have

\begin{align*} \left |1-\frac{Z}{\widetilde Z}\right|&=\left | \int_{\mathbb{R}^d} \left(\phi-\frac{Z}{\widetilde Z}\,\widetilde \phi\right)\mathrm{d} \Gamma \right| \le \int \left| \phi - \frac{Z}{\widetilde Z}\,\widetilde \phi \,\right|\mathrm{d} \Gamma\\ &\le \left( \int_{\mathbb{R}^d}\left| \phi - \frac{Z}{\widetilde Z}\,\widetilde \phi \,\right|^2\mathrm{d} \Gamma \right)^{1/2} \left( \int_{\mathbb{R}^d} \mathrm{d} \Gamma\right)^{1/2} \le \Lambda Z \epsilon. \end{align*}

We can then conclude (17), for instance by considering the behaviour of the function $t\mapsto h(t):=Z/t$ . Namely, the former inequality implies $ h(Z)-\Lambda Z\varepsilon \leq h(\widetilde{Z}) \leq h(Z)+\Lambda Z\varepsilon. $ Now, taking the inverse and performing suitable transformations yields the result.

Lemma 5. Assume that (14) and (15) hold, and that $\Lambda:\!= \Gamma(\mathbb{R}^d)^{1/2}<\infty$ . Then, we have

\begin{equation*} \|\phi-\widetilde \phi\|_2 \le \epsilon(Z+C), \end{equation*}

where $C>0$ is defined as in Lemma 4.

Proof. We calculate

\begin{equation*} \begin{split} \|\phi-\widetilde \phi\|_2 &\le \left \|\phi - \frac{Z}{\widetilde Z}\,\widetilde \phi\,\right \|_2+\left \| \frac{Z}{\widetilde Z}\,\widetilde \phi - \widetilde \phi \,\right\|_2 = \left \|\phi - \frac{Z}{\widetilde Z}\,\widetilde \phi\,\right \|_2+|Z-\widetilde Z| \le Z\epsilon + C\epsilon, \end{split} \end{equation*}

where we have used the triangle inequality and Lemma 4.

Putting together the estimates of the former lemmas we obtain the following.

Proposition 2. Assume that (14) and (15) hold, and that $\Gamma$ is a finite measure, with $\Lambda :\!= \Gamma(\mathbb{R}^d)^{1/2}<\infty$ . Then,

\begin{equation*} \int_{\mathbb{R}^d} |\pi(x)-\widetilde \pi(x)|\mathrm{d} x \le \Lambda(Z+C) \epsilon, \end{equation*}

where $C>0$ is defined in Lemma 4.

2.4. Example: logistic regression

We demonstrate how Theorem 1, in combination with Proposition 2, can be applied to the setting of logistic regression, which is currently the most popular application of QSMC methods; see [Reference Kumar21, Reference Pollock, Fearnhead, Johansen and Roberts33]. In that scenario we justify that QSMC methods are robust to small perturbations in the observations, such as could arise from corrupted or missing data.

We consider the following model. There are n observed binary response variables $(y_1,\ldots,y_n)\in \{0,1\}^n$ , and we have a design matrix of covariates $\mathbf X = (\mathbf{X}_{i,j})\in \mathbb{R}^{n\times d}$ , writing $\mathbf X_{[i]}\in {\mathbb{R}^d}$ for the ith row, $i\in \{1,\ldots,n\}$ . We consider the response variables and design matrix to be fixed a priori. For a parameter vector $x\in{\mathbb{R}^d}$ (interpreted as regression coefficients) the likelihood function for the response variables is given as

\begin{align*}\ell(x\mid y_1,\ldots,y_n) =\prod_{i=1}^n p_i(x)^{y_i} \cdot (1-p_i(x))^{1-y_i},\end{align*}

with

\begin{equation*} p_i(x) :\!= \frac{1}{1+\exp (-\mathbf X_{[i]}^\top x)},\quad x\in{\mathbb{R}^d}.\end{equation*}

Following [Reference Kumar21, Reference Pollock, Fearnhead, Johansen and Roberts33], we place an improper uniform prior on the parameters x, and obtain the Lebesgue density $\pi$ of the posterior distribution on ${\mathbb{R}^d}$ , which satisfies

(18) \begin{equation} \pi(x)\propto \prod_{i=1}^n p_i(x)^{y_i} \cdot (1-p_i(x))^{1-y_i}. \end{equation}

We assume that the response and design matrix are such that (18) defines a bona fide probability density function (that is, the right-hand side of (18) is Lebesgue integrable).

For our stochastic process, we are using a general reversible diffusion of the form (3), namely

\begin{equation*} \mathrm{d} X_t = \nabla A(X_t)\mathrm{d} t + \mathrm{d} W_t,\end{equation*}

where the drift A is chosen independently of the data. We assume that the relevant technical conditions on A detailed in [Reference Wang, Kolb, Roberts and Steinsaltz43] hold, to ensure that $\pi$ is a valid quasi-stationary distribution; for details see also Section 1.3. As in [Reference Wang, Kolb, Roberts and Steinsaltz43], we work on the Hilbert space $\mathcal H = \mathcal L^2({\mathbb{R}^d},\Gamma)$ , where $\Gamma(\mathrm{d} x)=\gamma(x)\mathrm{d} x$ with $\gamma(x)=\exp\!(2A(x))$ .

Note that the observations $y_1,\ldots,y_n$ and covariates $\mathbf X$ only influence the killed diffusion through the killing rate $\kappa$ and so any perturbation of the data results in a perturbation of the killing rate.

From (6) and following the calculations in [Reference Kumar21, Appendix C], we can express the killing rate as

(19) \begin{equation}\begin{split} \kappa(x) = \frac{1}{2}\bigg( \sum_{j=1}^d\left[ \sum_{i=1}^n \left (y_i-p_i(x)\right )\mathbf X_{i,j} \right]^2 - \sum_{j=1}^d \sum_{i=1}^n p_i(x)\left (1-p_i(x)\right )\mathbf X^2_{i,j} \\ -2\Delta A - 2\left \langle \nabla A, \sum_{i=1}^n \left (y_i - p_i(x)\right ) {\mathbf X_{[i]}} \right \rangle \bigg)-\Phi, \end{split}\end{equation}

where the constant $\Phi \in \mathbb{R}$ is chosen to ensure that $\kappa \ge 0$ everywhere; see Condition (1) of [Reference Pollock, Fearnhead, Johansen and Roberts33] for a further discussion. Let $L^\kappa$ be the generator of the diffusion $X=(X_t)_{t\geq 0}$ killed at rate $\kappa$ , which possesses $\varphi = \pi/\gamma$ as its minimal eigenfunction corresponding to the bottom of the spectrum. The fact that the eigenspace is one-dimensional follows from the Perron–Frobenius theorem; see [Reference Reed and Simon34, Chapter XIII.12]. Let us write ${\lambda_0^\kappa}$ for the bottom of the spectrum of ${L^\kappa}$ . Then in fact it follows that ${\lambda_0^\kappa}=-\Phi$ ; see [Reference Wang, Kolb, Roberts and Steinsaltz43, Section 3.3]. That is, the bottom of the spectrum of the resulting operator is in fact equal to $-\Phi$ .

We first consider the common choice of $A\equiv 0$ , as in [Reference Kumar21, Reference Pollock, Fearnhead, Johansen and Roberts33]. In this case, the expression (19) simplifies, and the resulting killing rate $\kappa$ is uniformly bounded from above. Moreover, since we are assuming that $\pi$ is integrable, it follows that

(20) \begin{equation} \liminf_{\vert x\vert \to \infty} \kappa(x) > {\lambda_0^\kappa},\end{equation}

with $\vert \cdot \vert$ denoting the Euclidean norm.

As already mentioned, we imagine having corrupted or missing data, giving us access only to a perturbed killing rate $\widetilde{\kappa} \colon \mathbb{R}^d \to [0,\infty)$ .

Remark 3. For example, it may be the case that our covariates are corrupted, so we only have access to $\widetilde{\mathbf X}_{i,j}:=\mathbf X_{i,j}+\epsilon_{i,j}$ , where the $(\epsilon_{i,j})\in\mathbb{R}^{n\times d}$ models some corrupting noise (alternatively, the case $\epsilon_{i,j}=-\mathbf X_{i,j}$ could be considered a case of missing data). Assume that the corresponding additive constants are equal, $\Phi=\widetilde \Phi$ (which can be done by setting both equal to the maximum value $\Phi \vee \widetilde \Phi$ ). If we write $\|\mathbf A\|_\infty = {\sup}_{1\leq i\leq n,\; 1\leq j\leq d}|\mathbf A_{i,j}|$ for a matrix $\mathbf A= (\mathbf{A}_{i,j})\in \mathbb{R}^{n\times d}$ , a straightforward calculation shows that for a fixed constant $C>0$ we have

\begin{align*}\|\widetilde\kappa -\kappa\|_\infty \le C n^2 d\|\mathbf X\|_\infty (\|\epsilon\|_\infty^2+\|\epsilon\|_\infty).\end{align*}

Thus, provided the data corruption level $\|\epsilon\|_\infty$ is sufficiently small, $\|\widetilde\kappa -\kappa\|_\infty$ can be made arbitrarily small.

Now we ask whether the difference of the eigenfunctions that determine the quasi-stationary distributions of the killed Markov processes based on $\kappa$ and $\widetilde{\kappa}$ is small when $\Vert \kappa - \widetilde{\kappa} \Vert_\infty$ is sufficiently small. Thus, we consider a perturbation operator $H\colon \mathcal{L}^2({\mathbb{R}^d},\Gamma)\to \mathcal{L}^2({\mathbb{R}^d},\Gamma)$ given by

\begin{equation*} Hf :\!= (\widetilde \kappa - \kappa)f, \quad f \in \mathcal L^2({\mathbb{R}^d},\Gamma). \end{equation*}

Note that since H is just a multiplication operator it is self-adjoint, and the operator norm coincides with $\Vert \kappa - \widetilde{\kappa} \Vert_\infty$ . We obtain the following result.

Proposition 3. We work in the case $A\equiv 0$ and assume that $\lim_{\|x\|\to\infty}\kappa(x)$ exists. Then there exist a $\delta >0$ and constant $C>0$ such that, for any (perturbed) killing rate $\widetilde \kappa$ with $\|\kappa-\tilde \kappa\|_\infty < \delta$ , there is a quasi-stationary distribution of the killed Markov process killed at rate $\widetilde\kappa$ . In particular, this quasi-stationary distribution has a $\Gamma$ -density $\widehat\varphi \in \mathcal L^2({\mathbb{R}^d},\Gamma)$ with $\Vert \widehat{\varphi} \Vert_2 = 1$ , which satisfies

\begin{equation*} \|\varphi-\widehat \varphi \|_2 \le C \|\tilde\kappa-\kappa\|_\infty. \end{equation*}

Proof. We start by arguing that the generator $L^\kappa$ has a spectral gap $\nu>0$ . This fact follows readily by (20) and the fact that the base of the essential spectrum must be strictly larger than $\lambda_0^\kappa$ by [Reference Davies9, Lemma 14.4.1] (see also [Reference Kolb and Steinsaltz19, Lemma 3.3(v)] and [Reference Wang, Kolb, Roberts and Steinsaltz43, Remark 1, Section 1.4]). We choose $\delta$ to be

\begin{equation*} \delta =\frac{1}{2}\min\left \{\nu, \liminf_{\vert x\vert\to \infty}\kappa(x) - \lambda_0^\kappa \right \}. \end{equation*}

Let $\|H\|$ be the $\mathcal L^2({\mathbb{R}^d},\Gamma)$ -operator norm. Then, we have

\begin{align*}\|H\| = \Vert \kappa-\widetilde{\kappa} \Vert_\infty< \frac{1}{2}\left (\liminf_{\vert x\vert \to \infty}\kappa(x) - \lambda_0^\kappa\right).\end{align*}

The latter inequality ensures that the resulting minimal eigenfunction $\widehat \varphi$ is also in $\mathcal L^1({\mathbb{R}^d},\Gamma)$ , by [Reference Champagnat and Villemonais7, Theorem 4.5]. The fact that the quasi-stationary distribution corresponds to the bottom of the spectrum follows from the argument of [Reference Simon39]; see also [Reference Kolb and Steinsaltz19, Section 4.2]. The rest follows by applying our abstract result, Theorem 1.

In light of the above remark, it is of interest to ask whether or not it is possible to construct an alternative diffusion to the case $A\equiv 0$ , which still possesses $\pi$ as a valid quasi-stationary distribution, but such that $\Gamma$ is a finite measure. This would enable us, by an application of the Cauchy–Schwarz inequality, to conclude robustness in $\mathcal L^1({\mathbb{R}^d},\Gamma)$ and not only in $\mathcal L^2({\mathbb{R}^d},\Gamma)$ . This can be indeed done, by choosing A to satisfy the following properties.

This is largely a technical assumption for the following proof, although, in addition, the resulting diffusion process with drift $\nabla A$ will also be relatively straightforward to simulate in the tails due to its simple form. However, such an A exists since $\pi$ is a log-concave density function. With this choice of A, we have the following result.

Proposition 4. Assume that the function $A\colon \mathbb{R}^d \to \mathbb{R}$ satisfies Assumption 3 and that $\lim_{\|x\|\to\infty}\kappa(x)$ exists. Then, there exist a $\delta >0$ and $C>0$ , such that, for any perturbed killing rate $\widetilde{\kappa}$ that satisfies $\|\widetilde \kappa - \kappa\|_\infty < \delta$ , there is a quasi-stationary distribution $\widehat \pi \in \mathcal L^1({\mathbb{R}^d},\Gamma)$ of the killed Markov process based on $\widetilde{\kappa}$ . In particular, we have

(21) \begin{equation} \int_{\mathbb{R}^d} |\pi(x)-\widehat \pi(x)|\mathrm{d} x \le C \|\tilde \kappa - \kappa\|_\infty. \end{equation}

Proof. Note that by Assumption 3 we have that the killing rate $\kappa$ from (19), which appears in the unperturbed killed Markov process, is bounded from above. Moreover, (20) is satisfied and therefore there exists similarly by [Reference Davies9, Lemma 14.4.1] a spectral gap $\nu>0$ for the generator $L^\kappa$ . As in the proof of Proposition 3, it follows that

\begin{align*} \|\varphi-\widehat \varphi \|_2 \le \widetilde{C} \|\tilde\kappa-\kappa\|_\infty \end{align*}

for the corresponding $\mathcal L^2({\mathbb{R}^d},\Gamma)$ -normalized eigenfunctions for some $\widetilde{C}>0$ . The desired $\mathcal L^1({\mathbb{R}^d},\Gamma)$ bound (21) follows since $\Gamma$ is assumed to be a finite measure. Therefore, we can apply Proposition 2 and obtain the assertion.

3.1. Projection notation

Following [Reference Davis and Kahan10], we make use of projections onto invariant subspaces. Now we describe the appropriate setting and notation, which will be used throughout Section 3.

For an unbounded self-adjoint operator $(L, {\mathcal{D}}(L))$ on a separable Hilbert space $\mathcal{H}$ , we suppose that we have a one-dimensional L-invariant subspace, given by its orthogonal projection $P=v_0 \langle \cdot, v_0\rangle$ , for some $v_0\in\mathcal{H}$ with $\|v_0\|=1$ . In fact $v_0$ must be an eigenvector of L, with eigenvalue $\lambda_0\in\mathbb{R}$ , say. Let $P^{\perp}$ denote the orthogonal projection onto the complement subspace $({\mathrm{span}}\{v_0\})^\perp$ .

Following [Reference Davis and Kahan10], we can define isometric mappings $E_0: \mathcal K(E_0) \to \mathcal{H}$ and $E_1: \mathcal K (E_1) \to \mathcal{H}$ , with source Hilbert spaces $\mathcal K(E_0), \mathcal K(E_1)$ , which satisfy

\begin{equation*} E_0 E_0^* = P,\quad E_1 E_1^* = P^{\perp},\end{equation*}

where $E_0^*: \mathcal{H} \to \mathcal K(E_0)$ and $E_1^*: \mathcal{H} \to \mathcal K(E_1)$ are the corresponding adjoint operators.

Since P projects onto the one-dimensional subspace ${\mathrm{span}}\{v_0\} \subseteq \mathcal{H}$ , we can identify $\mathcal K(E_0) \equiv \mathbb{R}$ , equipped with the Euclidean inner product. This allows the explicit characterization

\begin{equation*} \begin{split} E_0&: \mathbb{R} \to \mathcal H, \quad \theta \mapsto \theta v_0,\\ E_0^* &: \mathcal H \to \mathbb{R},\quad v \mapsto \langle v, v_0\rangle. \end{split}\end{equation*}

By the L-invariance of ${\mathrm{span}}\{v_0\}$ , we can now decompose $L\colon \mathcal{D}(L) \to \mathcal{H}$ as

\begin{align*} Lv = \left(\begin{array} {c@{\quad}c} E_0 & E_1 \end{array} \right) \left( \begin{array} {c@{\quad}c} L_0 & 0 \\ 0 & L_1 \end{array} \right) \begin{pmatrix} E_0^* v\\ E_1^*v \end{pmatrix},\end{align*}

for $v\in {\mathcal{D}}(L)$ , where $L_0:\mathbb{R} \to \mathbb{R}$ is the bounded linear operator given by $L_0\theta = \lambda_0 \theta$ , and $L_1:{\mathcal{D}}(L_1)\to \mathcal K(E_1)$ is the unbounded operator given by $L_1 = E_1^* L E_1$ with ${\mathcal{D}}(L_1)=E_1^* {\mathcal{D}}(L)$ .

Given another self-adjoint operator $(\widehat L, {\mathcal{D}}(\widehat L))$ on the same Hilbert space $\mathcal{H}$ , which we will later think of as the perturbed operator, we assume that we have a one-dimensional $\widehat L$ -invariant subspace given by the orthogonal projection $Q = \widehat v_0 \langle \cdot, \widehat v_0\rangle$ , where $\|\widehat v_0\|=1$ . So $\widehat v_0$ is an eigenvector of $\widehat L$ , with corresponding eigenvalue $\widehat{\lambda}_0\in\mathbb{R}$ , say. Let us write $Q^{\perp}$ for the projection onto the orthogonal complement of ${\mathrm{span}}\{\widehat{v}_0\}$ .

As above, we can similarly define corresponding isometric mappings $F_0:\mathcal{R}\to \mathcal{H}$ and $F_1:\mathcal K(F_1)\to \mathcal{H}$ with $F_0 \theta = \theta \widehat{v}_0$ and $F_0^* v = \langle v,\widehat{v}_0 \rangle$ , such that $F_0 F_0^* = Q$ and $F_1F_1^*=Q^\perp$ . Moreover, with these definitions, $\widehat{L} \colon {\mathcal{D}}(\widehat{L})\to \mathcal{H} $ can be decomposed for $v\in {\mathcal{D}}(\widehat L)$ as

(22) \begin{align} \widehat L v = \left(\begin{array} {c@{\quad}c} F_0 & F_1 \end{array} \right) \left( \begin{array} {c@{\quad}c} \widehat L_0 & 0 \\ 0 & \widehat L_1 \end{array} \right) \begin{pmatrix} F_0^*v \\ F_1^*v \end{pmatrix}, \end{align}

where $\widehat{L}_0\colon \mathbb{R} \to \mathbb{R}$ is given by $\widehat{L}_0\theta = \widehat{\lambda}_0 \theta $ and $\widehat{L}_1 \colon \mathcal{D}(\widehat{L}_1) \to \mathcal{K}(F_1)$ is the unbounded operator given by $\widehat{L}_1 = F_1^* \widehat{L} F_1$ with $\mathcal{D}(\widehat{L}_1) = F_1^* \mathcal{D}(\widehat{L})$ .

With these definitions we are able to formulate the remarkable sin $\theta$ theorem of [Reference Davis and Kahan10]. For the convenience of the reader, we state it here in our present notation.

In fact, this is a slight modification of the $\sin \,\theta$ theorem as stated in [Reference Davis and Kahan10], since our operators $L, \widehat L$ are in general unbounded; however, see the discussion immediately preceding [Reference Davis and Kahan10, Theorem 5.2]. These subtleties will be addressed in the sequel.

3.2. Further notation and results

We consider the situation described in Section 1.3 and set $\mathcal{H} = \mathcal L^2({\mathbb{R}^d}, \Gamma)$ , which consists of all measurable real-valued functions $f\colon {\mathbb{R}^d} \to \mathbb{R} $ satisfying

\begin{align*} \Vert f \Vert_2:\!= \left( \int_{{\mathbb{R}^d}} f(y)^2 \mathrm{d}\Gamma(y) \right)^{1/2} <\infty,\end{align*}

where the measure $\Gamma$ is given by $\mathrm{d} \Gamma(y)=\gamma(y)\mathrm{d} y$ and $\gamma:\!= \exp\!(2A)$ , with $\mathrm{d} y$ denoting the Lebesgue measure on ${\mathbb{R}^d}$ . In other words, $\mathcal{H}$ is the Hilbert space of functions, equipped with the inner product

\begin{equation*} \langle f,g\rangle = \int_{{\mathbb{R}^d}} f(y)g(y) \mathrm{d}\Gamma(y),\end{equation*}

where $f,g:{\mathbb{R}^d} \to \mathbb{R}$ . As before, by $\mathsf{S}(\mathcal{H})$ we denote the unit sphere in $\mathcal{H}$ with respect to $\Vert\cdot\Vert_2$ .

The infinitesimal generator of the diffusion X killed at rate $\kappa$ , see (3), is given by

(23) \begin{equation} L^\kappa = -\tfrac{1}{2}\Delta - \nabla A \cdot \nabla + \kappa, \end{equation}

where $\Delta$ denotes the Laplacian operator. As in [Reference Wang42, Section 3.3.3], the operator $L^\kappa$ can be realized as a densely defined self-adjoint positive semi-bounded linear operator on $\mathcal{H}$ with domain

(24) \begin{equation} {\mathcal{D}}(L^\kappa) = \{f\in \mathcal{H}: -\tfrac{1}{2}\Delta f - \nabla A \cdot \nabla f + \kappa f \in \mathcal{H}\}, \end{equation}

where the derivatives are understood in the weak sense. Note that ${\mathcal{D}}({L^\kappa})$ contains $C_c^\infty({\mathbb{R}^d})$ , the set of smooth, compactly supported functions on ${\mathbb{R}^d}$ . Therefore, we have a self-adjoint unperturbed operator on $\mathcal{H}$ given by (23) with domain ${\mathcal{D}}(L^\kappa)$ .

For brevity, we write $L^0=-\frac{1}{2}\Delta-\nabla A\cdot \nabla$ for the (formal) unkilled differential operator. We impose the following condition on the killing rate.

In the case where $A\equiv0$ and $\kappa$ is given by (6), i.e. in the setting of QSMC with Brownian motion, Assumption 4 corresponds to a scenario where $\pi$ has tails decaying at a faster-than-exponential rate (such as Gaussian or lighter). A standard consequence of the assumption is the following.

Remark 5. Setting $\lambda_1^\kappa :\!= \inf \{\sigma(L^\kappa)\backslash \{\lambda^\kappa_0\} \}$ , we have by the fact that ${L^\kappa}$ is self-adjoint and bounded from below, see [Reference Reed and Simon34, Theorem XIII.1], the variational representations as in (7), that is,

\begin{equation*} \begin{split} {\lambda_0^\kappa} &= \inf_{f\in \mathsf S(\mathcal{H})\cap {\mathcal{D}}({L^\kappa})} \langle f,{L^\kappa} f\rangle,\\ \lambda_1^\kappa &= {\sup}_{\mathbb V \in \mathcal V_1} \inf_{f\in \mathbb V^\perp \cap \mathsf S(\mathcal{H})\cap {\mathcal{D}}({L^\kappa})} \langle f, {L^\kappa} f\rangle. \end{split} \end{equation*}

We are interested in truncating the unbounded killing rate $\kappa$ and understanding the resulting impact on the quasi-stationary distribution. So, we define, for arbitrary $M\ge 0$ , the truncated killing rate $\kappa_M:{\mathbb{R}^d}\to [0,\infty)$ ; writing $a\wedge b :\!= \min\{a,b\}$ , it is given pointwise by

\begin{equation*} \kappa_M(x) = \kappa(x) \wedge M.\end{equation*}

With this we can define our perturbed operator, $\widehat{L}^M :=L^{\kappa_M}$ on $\mathcal{H}$ , i.e.

(25) \begin{equation} \widehat{L}^M = -\tfrac 1 2 \Delta - \nabla A \cdot \nabla + \kappa_M. \end{equation}

Similarly to ${L^\kappa}$ , $\widehat{L}^M$ can also be realized as a self-adjoint operator, with corresponding domain ${\mathcal{D}} (\widehat{L}^M)$ given analogously to (24). This corresponds to our diffusion X killed at state-dependent rate $\kappa_M$ .

We require some further notation to state the resulting perturbation. For any function $f:{\mathbb{R}^d} \to \mathbb{R}$ , we write $f_+$ for the positive part; $f_+(x)=\max\{f(x),0\}$ for each $x\in {\mathbb{R}^d}$ . Now define our (unbounded) perturbation $H_M$ on $\mathcal{H}$ as the multiplication operator given by

\begin{equation*} H_M f = -(\kappa -M)_+f,\end{equation*}

for $f \in {\mathcal{D}}(H_M)$ with ${\mathcal{D}}(H_M):=\{f\in \mathcal{H}: \int_{{\mathbb{R}^d}} |(\kappa - M)_+ f|^2 \mathrm{d} \Gamma <\infty\} $ . We can characterize ${\mathcal{D}}(H_M)$ .

Lemma 7. For any $M\geq 0$ , we have

(26) \begin{equation} {\mathcal{D}}(H_M) = \{f\in \mathcal{H}: \int_{{\mathbb{R}^d}}|\kappa f|^2\mathrm{d} \Gamma <\infty\}. \end{equation}

Consequently, ${\mathcal{D}}(H_M)$ is independent of M.

To emphasize that the domain of $H_M$ does not depend on M, we define ${\mathcal{D}}(\kappa):\!= {\mathcal{D}}(H_M)$ .

Lemma 8. Suppose $g\in {\mathcal{D}}({L^\kappa})\cap{\mathcal{D}}(\kappa)$ . Then $g\in {\mathcal{D}}(\widehat{L}^M)$ and

\begin{equation*} \widehat{L}^M g = L^\kappa g + H_M g. \end{equation*}

Hence, using the fact that $L^\kappa$ is closed and $\kappa g \in \mathcal H$ , and also the dominated convergence theorem, we have the following convergence in $\mathcal H$ :

\begin{equation*} \widehat{L}^M g_n = \widehat{L}^0 g_n + \kappa g_n - (\kappa-M)_+ g_n \longrightarrow L^\kappa g - (\kappa-M)_+ g \in \mathcal H. \end{equation*}

From now on, we impose the following standard assumption.

Together Lemma 8 and Assumption 5 immediately imply the following.

We now turn to the question of existence of a perturbed quasi-stationary distribution, corresponding to a minimal eigenfunction of $\widehat{L}^M$ . Clearly, if we set $M=0$ there is no killing, and so there is no quasi-stationary distribution to speak of. Thus it is necessary to enforce a minimum truncation level.

Proof. Since $\widehat{L}^M$ is a self-adjoint semi-bounded operator, we have the variational representation [Reference Reed and Simon34, Theorem XIII.1]:

\begin{equation*} \widehat{\lambda}_0^M = \inf_{f \in {\mathcal{D}}(\widehat{L}^M)\cap \mathsf S(\mathcal{H})} \langle f, \widehat{L}^M f\rangle. \end{equation*}

From Lemma 9, we have $\varphi_0 \in {\mathcal{D}}(\widehat{L}^M)\cap \mathsf S(\mathcal{H})$ . Hence, making use of Lemma 8, we have

\begin{equation*} \begin{split} \widehat{\lambda}_0^M &\le \langle \varphi_0, \widehat{L}^M \varphi_0\rangle \\ &= \langle \varphi_0, {L^\kappa} \varphi_0\rangle+\langle \varphi_0, H_M \varphi_0\rangle\\ &= {\lambda_0^\kappa} -\int_{{\mathbb{R}^d}} (\kappa-M)_+ \varphi_0^2\mathrm{d} \Gamma < M, \end{split} \end{equation*}

since we are assuming that ${\lambda_0^\kappa} \le M$ , and the final term is strictly negative, since $\varphi_0$ is strictly positive.

Thus the bottom of the spectrum $\widehat{\lambda}_0^M$ of $\widehat{L}^M$ is strictly smaller than M, which is the limit of the killing at infinity; hence, by [Reference Champagnat and Villemonais7, Theorem 4.5] (and cf. [Reference Kolb and Steinsaltz19, Theorem 4.3]), there is a unique quasi-stationary distribution, and there is a corresponding eigenfunction $\widehat \varphi_0$ with eigenvalue $\widehat{\lambda}_0^M$ .

Remark 8. It follows by [Reference Davies9, Lemma 14.4.1] that, for the bottom of the essential spectrum, we have $\inf \sigma_\mathrm{ess}(\widehat{L}^M)\ge M$ , and hence $\widehat{L}^M$ also possesses a spectral gap. This can also be read off from the proof of Theorem 3. The calculation there shows that

\begin{equation*} \widehat{\lambda}_0^M < \lambda_0^\kappa. \end{equation*}

Now we define, for arbitrary $M\geq 0$ ,

\begin{equation*} \widehat\lambda_1^M :\!= {\sup}_{\mathbb V \in \mathcal V_1} \inf_{f\in \mathbb V^\perp \cap \mathsf S(\mathcal{H})\cap {\mathcal{D}}(\widehat L^M)} \langle f, \widehat L^M f\rangle.\end{equation*}

Then, by [Reference Reed and Simon34, Theorem XIII.1], $\widehat \lambda_1^M$ is either an eigenvalue of $\widehat L^M$ or coincides with the bottom of the essential spectrum $\sigma_\mathrm{ess}(\widehat L^M)$ .

Proof. This follows from a routine calculation, which we include for completeness. Since the set of smooth, compactly supported functions $C_\mathrm c^\infty({\mathbb{R}^d})\subset \mathcal{H}$ is a core for ${L^\kappa}$ and for $\widehat L^M$ for any $M>0$ , in the variational representation of eigenvalues we can restrict to functions $f\in C_\mathrm c^\infty({\mathbb{R}^d})$ . Then,

\begin{equation*} \begin{split} \widehat \lambda_1^{M_2} &= {\sup}_{\mathbb V \in \mathcal V_1} \inf_{f\in \mathbb V^\perp \cap \mathsf S(\mathcal{H})\cap C_\mathrm c^\infty} \left (\langle f,\widehat L^0 f\rangle + \langle f,(\kappa \wedge M_2)f\rangle \right )\\ &= {\sup}_{\mathbb V \in \mathcal V_1} \inf_{f\in \mathbb V^\perp \cap \mathsf S(\mathcal{H})\cap C_\mathrm c^\infty} \left ( \langle f,\widehat L^0 f\rangle + \langle f,(\kappa \wedge M_1)f\rangle + \langle f, (\kappa\wedge M_2 - \kappa \wedge M_1)f\rangle \right)\\ &\ge {\sup}_{\mathbb V \in \mathcal V_1} \inf_{f\in \mathbb V^\perp \cap \mathsf S(\mathcal{H})\cap C_\mathrm c^\infty} \left ( \langle f,\widehat L^0 f\rangle + \langle f,(\kappa \wedge M_1)f\rangle \right)\\ &= \widehat \lambda_1^{M_1}. \end{split} \end{equation*}

An analogous argument also gives $\lambda_1^\kappa\ge\widehat \lambda_1^{M_2}$ .

We have established in Theorem 3 the existence of a minimal eigenfunction $\widehat \varphi_0$ of $\widehat L^M$ , under Assumptions 4, 5, and 6 which we assume to hold in the following. We can thus define one-dimensional invariant subspaces of ${L^\kappa}$ and $\widehat L^M$ , respectively ${\mathrm{span}}\{\varphi_0\}$ and ${\mathrm{span}}\{\widehat \varphi_0\}$ , and make use of the projection operators as previously defined in Section 3.1. Thus, with $E_0: \mathbb{R} \to \mathcal{H}$ and $E_1: \mathcal K(E_1)\to \mathcal{H}$ we have for $f\in {\mathcal{D}}({L^\kappa})$ that

\begin{align*} {L^\kappa} f= \left(\begin{array} {c@{\quad}c} E_0 & E_1 \end{array} \right) \left( \begin{array} {c@{\quad}c} \Lambda_0 & 0\\ 0 & \Lambda_1 \end{array} \right) \begin{pmatrix} E_0^*f\\ E_1^*f \end{pmatrix}.\end{align*}

In particular, $\Lambda_0:\mathbb{R}\to\mathbb{R}$ is the bounded linear operator given by

\begin{equation*} \Lambda_0: a \mapsto {\lambda_0^\kappa} a, \quad a\in\mathbb{R},\end{equation*}

and $\Lambda_1$ is the self-adjoint operator given by $\Lambda_1=E_1^* {L^\kappa} E_1$ with domain ${\mathcal{D}}(\Lambda_1) = E_1^* {\mathcal{D}}({L^\kappa})$ . For our perturbed operator $\widehat L^M$ , we have with $F_0: \mathbb{R}\to \mathcal{H}$ and $F_1: \mathcal K(F_1)\to \mathcal{H}$ the decomposition

\begin{align*} \widehat L^M f= \left( \begin{array} {c@{\quad}c} F_0 & F_1 \end{array} \right) \left( \begin{array} {c@{\quad}c} \widehat \Lambda_0 & 0 \\ 0 & \widehat \Lambda_1 \end{array} \right) \begin{pmatrix} F_0^*f \\ F_1^*f \end{pmatrix},\end{align*}

for $f \in {\mathcal{D}}(\widehat L^M)$ , where $\widehat{\Lambda}_0 \colon \mathbb{R} \to\mathbb{R}$ with $\widehat{\Lambda}_0 a = \widehat{\lambda}_0^M a$ and $\widehat \Lambda_1=F_1^* \widehat L^M F_1$ are self-adjoint operators. Note that throughout the norm of an operator always refers to the corresponding operator norm.

Our goal is to bound $\|\varphi_0-\widehat\varphi_0\|_2$ . For this we aim to apply the results of [Reference Davis and Kahan10]. However, we cannot immediately use the sin $\theta$ theorem as described in [Reference Davis and Kahan10], because of the unbounded nature of our operators. In order to overcome this issue, we begin by defining an operator $R: \mathbb{R} \to \mathcal{H}$ , given by

(27) \begin{equation} R = \widehat L^M E_0 - E_0 \Lambda_0.\end{equation}

Lemma 10. The operator $R \colon \mathbb{R}\to\mathcal{H}$ as defined in (27) is a bounded linear operator, which can be represented as

\begin{equation*} Ra = H_M E_0 a = a (H_M\varphi_0), \quad a\in\mathbb{R}. \end{equation*}

Moreover, the adjoint operator $R^*: \mathcal{H} \to \mathbb{R}$ is given by

(28) \begin{equation} R^* f = \langle f, H_M\varphi_0\rangle, \quad f\in \mathcal{H}. \end{equation}

Furthermore, we can directly compute that

\begin{equation*} \begin{split} Ra &= \widehat L^M a \varphi_0 - \lambda_0^\kappa a \varphi_0\\ &= a [L^\kappa \varphi_0 + H_M\varphi_0- \lambda_0^\kappa \varphi_0]\\ &= a[ \lambda_0^\kappa \varphi_0+H_M\varphi_0- \lambda_0^\kappa \varphi_0] = a H_M \varphi_0, \end{split} \end{equation*}

where we have made use of Lemma 8.

For the adjoint, recall that $R^*: \mathcal{H} \to \mathbb{R}$ is the unique operator with the property that, for arbitrary $f\in\mathcal H$ and $a \in \mathbb{R}$ , we have that

\begin{equation*} \langle f, R a \rangle = a (R^* f) . \end{equation*}

A simple calculation then gives (28).

Now consider the bounded operator $R^* F_1: \mathcal K(F_1) \to \mathbb{R}$ . By (28), we have, for $h \in \mathcal K(F_1)$ ,

\begin{equation*} R^* F_1 h = \langle F_1 h, H_M \varphi_0 \rangle.\end{equation*}

Lemma 11. The bounded operator $R^* F_1$ is the closure of the densely defined operator

\begin{equation*} E_0^* F_1 \widehat\Lambda_1 - \Lambda_0 E_0^* F_1, \end{equation*}

on smooth compactly supported functions.

Proof. Let $f\in \mathcal{H}$ be a smooth compactly supported function and consider $h :\!= F_1^* f\in \mathcal K(F_1)$ . Let $Q^\perp=F_1^*F_1$ be the projection onto ${\mathrm{span}}\{\widehat\varphi_0\}^\perp$ , P be the projection onto ${\mathrm{span}}\{\varphi_0\}$ , and $P^{\perp}$ be the projection onto ${\mathrm{span}}\{\varphi_0\}^\perp$ . We can straightforwardly compute the following, using the fact that $\widehat L^M$ is invariant on $\mathrm{Ran}(F_1)$ and the fact that ${L^\kappa}$ is invariant on ${\mathrm{span}}\{\varphi_0\}$ and its orthogonal complement:

\begin{equation*} \begin{split} E^*_0 F_1 \widehat\Lambda_1 h - \Lambda_0 E_0^* F_1 h &= E_0^* (Q^\perp\widehat L^M F_1 h) - \Lambda_0 E_0^* F_1 h\\ &= E_0^* (\widehat L^M F_1 h) - \Lambda_0 E_0^* F_1 h\\ &= \langle \widehat L^M F_1 h - \lambda_0^\kappa F_1 h , \varphi_0\rangle \\ &= \langle L^\kappa F_1 h + H_M F_1 h - \lambda_0^\kappa F_1 h , \varphi_0\rangle\\ &= \langle L^\kappa (P F_1 h + P^\perp F_1 h) + H_M F_1 h- \lambda_0^\kappa F_1 h , \varphi_0\rangle\\ &= \langle \lambda_0^\kappa PF_1 h +0 + H_M F_1 h - \lambda_0^\kappa F_1 h , \varphi_0\rangle\\ &= \lambda_0^\kappa\langle F_1 h,\varphi_0\rangle \langle \varphi_0,\varphi_0\rangle +\langle H_M F_1 h, \varphi_0\rangle - \langle\lambda_0^\kappa F_1 h , \varphi_0\rangle\\ &= \langle F_1 h, H_M\varphi_0\rangle = R^* F_1 h. \end{split} \end{equation*}

At the end we also used self-adjointness of $H_M$ and (28).

We require an additional technical condition.

The following auxiliary estimate is a consequence of the latter assumption, Lemma 11, i.e. $R^* F_1 = E_0^* F_1 \widehat \Lambda_1 - \Lambda_0 E_0^* F_1$ on a dense subset of $\mathcal{H}$ , and [Reference Davis and Kahan10, Theorem 5.1]; see their comments preceding [Reference Davis and Kahan10, Theorem 5.2] concerning the unbounded case.

Lemma 12. Suppose that Assumption 7 holds. Then

(29) \begin{equation} \delta \|E_0^* F_1\|\le \|R^* F_1\|. \end{equation}

Now we state our perturbation result, whose proof follows that of the sin $\theta$ theorem of [Reference Davis and Kahan10].

Theorem 4. Suppose that Assumption 7 holds and without loss of generality let $\widehat \varphi_0$ be chosen in such a way that $\langle \varphi_0, \widehat \varphi_0\rangle \geq 0$ . Then

\begin{equation*} \|R\|\ge \delta\, \sqrt{ 1- \langle \varphi_0, \widehat \varphi_0\rangle^2 }. \end{equation*}

Proof. We set $\sin \Theta_0 :\!= \sqrt{1-\langle \varphi_0, \widehat \varphi_0\rangle^2}$ . As in the proof of the sin $\theta$ theorem of [Reference Davis and Kahan10], our operator $E^*_0 F_1$ corresponds to a rotation from $\widehat \varphi$ to $\varphi$ , and thus its angle corresponds to $\sin \Theta_0$ . Hence, making use of (29), we see that

\begin{equation*} \delta |\sin\Theta_0| = \delta \|E^*_0 F_1\| \le\|R^* F_1\| \le \|R^*\| = \|R\|. \end{equation*}

We can then complete the argument as in the proof of Theorem 1 to get the following bound.

Theorem 5. Again with $\widehat \varphi_0$ be chosen in such a way that $\langle \varphi_0, \widehat \varphi_0\rangle \geq 0$ , under Assumptions 4, 5, 6, and 7 we have

\begin{equation*} \begin{split} \Vert \widehat \varphi_0 - \varphi_0\Vert_2 \le \frac{\sqrt 2 }{\sqrt{1+\langle \varphi_0, \widehat \varphi_0\rangle}} \cdot \frac{\|H_M E_0\|}{\delta} \le \frac{\sqrt 2 \,\|H_M E_0\|}{\delta}. \end{split} \end{equation*}

Note that

\begin{equation*} \|H_M E_0\| = \|H_M \varphi_0\|_2,\end{equation*}

where on the left-hand side we have the corresponding operator norm and on the right-hand side we have the Hilbert space norm. Moreover, observe that since we have Assumption 5, as the truncation $M\to \infty$ , we obtain $\|H_M \varphi_0\|_2 \to 0$ and at the same time, by Proposition 5, we have that $\delta$ remains constant with M.

3.3. Example: unbounded case

We now illustrate our former truncation result on a Gaussian example. In this setting, many relevant quantities are analytically available, so we can carefully isolate the effect of the truncation. We consider a similar example to [Reference Wang, Kolb, Roberts and Steinsaltz43], namely an Ornstein–Uhlenbeck diffusion targeting a Gaussian quasi-stationary distribution. We will see that in any dimension $d\ge 1$ , for a truncation level M, the error decays like $\exp\!(\!-M)$ up to some (dimension-dependent) multiplicative constant. In the univariate setting $d=1$ , we will be able to explicitly compute this constant. Thus truncating the killing rate in this example appears to be robust with respect to dimension, in that the error decays exponentially for any dimension.

Throughout, we write $\mathcal N(\mu, \sigma^2)$ for the univariate Gaussian distribution with mean $\mu\in\mathbb{R}$ and variance $\sigma^2>0$ , and $\mathcal N(\mu, \Sigma)$ for the corresponding multivariate Gaussian distribution with mean vector $\mu \in {\mathbb{R}^d}$ and covariance matrix $\Sigma \in \mathbb{R}^{d\times d}$ . Moreover, $I_d$ denotes the identity matrix on ${\mathbb{R}^d}$ . We take the diffusion to be a multivariate Ornstein–Uhlenbeck process on ${\mathbb{R}^d}$ determined by

(30) \begin{equation} \mathrm{d} X_t = \tfrac{1}{2}(\!-X_t) \mathrm{d} t + \mathrm{d} W_t, \end{equation}

which possesses an invariant measure $\Gamma(\mathrm{d} x) = \exp{\!(\!-\vert x\vert ^2/2)}\mathrm{d} x$ proportional to the $\mathcal N(0,I_d)$ distribution.

We take the target to be $\pi=\mathcal N(0,I_d/2)$ , in which case the killing rate has a particularly simple form,

(31) \begin{equation} \kappa(y) = \vert y\vert ^2, \quad y \in {\mathbb{R}^d}. \end{equation}

It follows by [Reference Wang, Kolb, Roberts and Steinsaltz43] that $\pi$ is the quasi-stationary distribution of the killed Markov process based on $X=(X_t)_{t\geq0}$ , see also Section 1.3. We are now interested in deriving quantitative bounds on the effect of truncating this divergent killing rate at a level M. In order to obtain explicit constants, we now focus on the univariate case $d=1$ , where all necessary calculations are analytically tractable. We will return to the general multivariate case in Proposition 7.

Having this, the normalized (in $\mathcal L^2(\mathbb{R},\Gamma)$ ) eigenfunction $\varphi_0\colon\mathbb{R}\to\mathbb{R}$ of the corresponding unperturbed generator $L^\kappa$ is given by

\begin{equation*} \varphi_0(x) = C \mathrm e^{-{x^2}/{2}},\end{equation*}

with $C=(\frac{3}{2\pi})^{1/4}$ .

Moreover, the bottom of the spectrum of the unperturbed operator is given by

(32) \begin{equation} \lambda_0^\kappa = \tfrac{1}{2},\end{equation}

as in [Reference Wang, Kolb, Roberts and Steinsaltz43, (2.4)], and the entire spectrum can be represented as

(33) \begin{equation} \sigma(L^\kappa) = \{\lambda_0^\kappa+ n \sigma_\mathrm{gap}:n\in \mathbb N_0\}, \end{equation}

with spectral gap

\begin{equation*} \sigma_\mathrm{gap} = \tfrac{3}{2};\end{equation*}

see [Reference Metafune, Pallara and Priola28].

Clearly our killing rate $\kappa$ as defined in (31) is unbounded, and we are interested in studying the effect of truncating this killing rate upon the quasi-stationary distribution. For $M\geq 0$ we consider a truncation at level M and set

\begin{equation*} \kappa_M(x) :\!= \kappa(x) \wedge M, \quad x\in\mathbb{R}.\end{equation*}

This gives rise to our perturbed Markov process with killing rate $\kappa_M$ with generator $\widehat{L}^M$ . Note that we use the notation of the previous section.

In order to utilize the estimate of Theorem 5 to obtain a quantitative result, we need to calculate the $\delta$ -separation between ${\lambda_0^\kappa}$ and $\widehat \lambda_1^M$ as in Assumption 7, since $\delta$ appears in the bounds. Calculating the spectrum in general is a difficult problem, so we resort to a scheme which was utilized in [Reference Mielnik and Reyes29] to numerically calculate the energy levels of the quantum truncated harmonic oscillator (which correspond to a killed Brownian motion). Since we are interested in a killed Ornstein–Uhlenbeck process (30) rather than a killed Brownian motion, we extend their technique into our present setting.

Considering (25), we obtain the eigenvalue problem

(34) \begin{equation} -\frac{1}{2}\varphi''(x) + \frac{x}{2} \varphi'(x) + \kappa_M(x) \varphi(x) = \lambda \varphi(x), \end{equation}

that is, we seek $\lambda\in\mathbb{R}$ and $\varphi\in\mathcal{L}^2(\mathbb{R},\Gamma)$ such that (34) is satisfied. By the fact that the killing rate $\kappa_M$ is constant for $|x|>\sqrt M$ , using [Reference Taylor40], we can explicitly write down the form of the eigenfunctions in the tails. Define

\begin{equation*} K_\beta^\pm(x) :\!= \int_0^\infty r^{\beta -1} \exp\!(\!-r^2 \pm \sqrt 2 rx) \mathrm{d} r, \quad x \in \mathbb{R}.\end{equation*}

Then, given $\lambda\in\mathbb{R}$ , the solution of (34) when $|x|>\sqrt M$ is

(35) \begin{equation} \varphi(x) = \varphi^{(\lambda)} (x) = \begin{cases} c_1 K_{2(M-\lambda)}^+(x), \quad x< -\sqrt M, \\ c_1 K_{2(M-\lambda)}^- (x),\quad x> \sqrt M, \end{cases} \end{equation}

since the eigenfunction must be square integrable. Note that here we have a common constant $c_1>0$ by symmetry. Now, following [Reference Mielnik and Reyes29], we substitute

\begin{equation*} q:\!= \varphi ,\quad p:\!= \varphi',\end{equation*}

and change to polar coordinates: for suitable functions $\rho\colon\mathbb{R}\to(0,\infty)$ and $\alpha\colon\mathbb{R} \to \mathbb{R}$ we have

\begin{equation*} q=\rho \cos \alpha ,\quad p = \rho \sin \alpha.\end{equation*}

With this substitution, we can modify (34) to the following equations:

(36) \begin{align} \rho' \cos \alpha - \alpha' \rho \sin \alpha &= \rho \sin \alpha ,\\[-38pt]\nonumber\end{align}

(37) \begin{align} \rho' \sin \alpha + \alpha' \rho \cos \alpha &=2(\kappa_M-\lambda) \rho \cos \alpha + x \rho \sin\alpha .\end{align}

Note that in contrast to [Reference Mielnik and Reyes29, (17)] we require the additional final term in (37).

Then by multiplying (36) by $-\sin \alpha$ and adding it to $\cos \alpha$ times (37), we obtain a first-order differential equation for $\alpha$ , given by

(38) \begin{equation} \alpha' = 2(\kappa_M-\lambda) \cos^2 \alpha -\sin^2 \alpha +x \sin \alpha \cos \alpha. \end{equation}

Noting that

\begin{equation*} \frac{\mathrm{d} }{\mathrm{d} x}K_\beta^\pm(x) = \pm \sqrt 2\int_0^\infty r^\beta \exp\!(\!-r^2 \pm \sqrt 2 rx)\mathrm{d} r,\end{equation*}

we can compute that in the tails,

(39) \begin{equation} \tan \alpha(x) = \frac{p(x)}{q(x)}= \begin{cases} \frac{\sqrt 2 \int_0^\infty r^{\beta^*} \exp\!(-r^2 + \sqrt 2 rx)\mathrm{d} r }{\int_0^\infty r^{{\beta^*}-1} \exp\!(-r^2 + \sqrt 2 rx)\mathrm{d} r} , \quad x\le -\sqrt M, \\ \frac{-\sqrt 2 \int_0^\infty r^{\beta^*} \exp\!(-r^2 - \sqrt 2 rx)\mathrm{d} r }{\int_0^\infty r^{{\beta^*}-1} \exp\!(-r^2 - \sqrt 2 rx)\mathrm{d} r},\quad x\ge \sqrt M, \end{cases} \end{equation}

with ${\beta^*} = 2(M-\lambda)$ .

Since we know that the evolution of $\alpha(x)$ is given by (38), and we have boundary values given by (39), in order for $\lambda$ to be a valid eigenvalue, we require the following compatibility condition, analogously to [Reference Mielnik and Reyes29]: for some integer n,

\begin{equation*} \alpha(\sqrt M) = \alpha(\!-\sqrt M)+ \int_{-\sqrt{M}}^{\sqrt{M}} \alpha'(x) \mathrm{d} x +n \pi,\end{equation*}

where $\alpha'$ is defined by (38) and $\alpha (\pm \sqrt M)$ are defined by (39).

Thus, again following the approach of [Reference Mielnik and Reyes29], we define

(40) \begin{equation} s_M(\lambda):\!= \alpha(\sqrt M) - \alpha(\!-\sqrt M)- \int_{-\sqrt{M}}^{\sqrt{M}} \alpha'(x) \mathrm{d} x, \end{equation}

where the right-hand side can be computed numerically to arbitrary accuracy using a numerical ordinary differential equation solver. Thus we need to find values of $\lambda$ such that

(41) \begin{equation} s_M(\lambda) = n\pi, \end{equation}

for integers n. The corresponding values of $\lambda$ for which (41) hold are precisely the eigenvalues for the truncated operator with truncation level M. We have plotted the function $\lambda\mapsto s_M(\lambda)$ in the cases $M=2,10,40$ in Figure 4.

For analogous plots in the case of Brownian motion (rather than our Ornstein–Uhlenbeck process (30)), see Figure 3 of [Reference Mielnik and Reyes29]. We can see that our truncated eigenvalues are all strictly smaller than their untruncated counterparts given by (33), and this difference vanishes as $M\to \infty$ , as is intuitively reasonable.

Thus we can utilize Theorem 5 to derive the following bounds in $\mathcal L^2(\mathbb{R},\Gamma)$ .

Lemma 13. For our Ornstein–Uhlenbeck diffusion (30) with quasi-stationary distribution $\pi=\mathcal N(0,1/2)$ , for any $M\ge 2$ , there exists a perturbed quasi-stationary distribution $\pi_M$ , and writing $\varphi_0, \widehat{\varphi}_{0,M}$ for the corresponding normalized (in $\mathcal L^2(\mathbb{R},\Gamma)$ ) eigenfunctions, we have

\begin{equation*} \|\varphi_0 - \widehat{\varphi}_{0,M}\|_2 \le c_2 \exp\!(\!-M/2), \end{equation*}

with $c_2=\frac{\sqrt{3}\cdot 35^{1/4}}{2\delta}$ .

Note that we do not prove Assumption 7, but rather, as detailed above, we use the numerical method of [Reference Mielnik and Reyes29] to verify that indeed for this example $\delta >0$ as required. For computation of its value, see Remark 10.

Proof. From Theorem 5, we only need to compute $\|H_M E_0\|=\|H_M \varphi_0\|_2$ . With $C=\left( \frac{3}{2\pi}\right)^{1/4}$ we obtain

\begin{align*} \|H_M \varphi_0\|^2_2 &= C^2 \int_{-\infty}^\infty (\kappa(x)-M)_+^2 \exp\!(\!-x^2) \mathrm{d} \Gamma(x)\\ &= 2C^2 \int_{\sqrt M}^\infty (x^2-M)^2 \exp\!(\!-3x^2/2)\mathrm{d} x\\ &\le C^2 \left (\int_{\mathbb{R}} x^8 \exp\!(\!-x^2) \mathrm{d} x\right)^{1/2} \left( \int_{\sqrt M}^\infty \exp\!(\!-2x^2)\mathrm{d} x \right)^{1/2}\\ &\le C^2 (105 \sqrt{\pi}/16)^{1/2} \left( \sqrt{\pi/2} \exp\!(\!-2M) \right)^{1/2}, \end{align*}

where we have used the Cauchy–Schwarz inequality, the fact that for $\mathcal N(0,\sigma^2)$ the eighth moment is $105\sigma^8$ , and standard Gaussian tail bounds; see e.g. [Reference Wainwright41, Proposition 2.5]. Tidying up the constants, the lemma is proved.

We now translate our previous $\mathcal L^2(\mathbb{R},\Gamma)$ estimate into a bound on the corresponding (normalized) quasi-stationary distributions in $\mathcal L^1(\mathbb{R},\Gamma)$ .

Proposition 6. In the same setting as Lemma 13 we obtain, for any $M> 2\log(2(2\pi)^{1/4} Zc_1)\approx 5.58$ , that there exists a perturbed quasi-stationary distribution $\pi_M$ , and

(42) \begin{equation} \int_{\mathbb{R}} |\pi(x)-\pi_M(x)|\mathrm{d} x\le c_3\exp\!(\!-M/2), \end{equation}

for $c_3\approx 5.49$ defined explicitly in (44).

Figure 1.

Plots of the function $\lambda\mapsto s_M(\lambda)$ , as defined in (40), for $M=2,10,40$ .

Thus, with $(\int_{\mathbb{R}} \mathrm{d} \Gamma)^{1/2}=(2\pi)^{1/4}$ , we have from Lemma 3 that

\begin{align*} \int_{\mathbb{R}} |\pi(x)-\pi_M(x)|\mathrm{d} x \le (2\pi)^{1/4} \|\phi-\widehat\phi_M\|_2 . \end{align*}

From Lemma 13, we obtain

(43) \begin{equation} \left \|\phi - \frac{Z\widehat\phi_M}{Z_M}\right \|_2 \le Zc_2 \exp\!(\!-M/2). \end{equation}

By Lemma 4 this gives

\begin{equation*} |Z_M-Z|< (2\pi)^{1/4}Z^2 c_2 \frac{\exp\!(\!-M/2)}{1-(2\pi)^{1/4} Zc_2\exp\!(\!-M/2)}, \end{equation*}

provided that $M>2\log((2\pi)^{1/4} Zc_2)$ . By using Lemma 5 we also have

\begin{align*} \|\phi-\widehat\phi_M\|_2 &\le \exp\!(\!-M/2) \left( Zc_2+ \frac {(2\pi)^{1/4}Z^2 c_2}{1-(2\pi)^{1/4} Zc_2\exp\!(\!-M/2)} \right). \end{align*}

Furthermore, if $M>2\log(2(2\pi)^{1/4} Zc_2)$ , the final term of the former bracket expression is bounded from above by $2(2\pi)^{1/4} Z^2 c_2$ . Putting the pieces together, for all $M>2\log(2(2\pi)^{1/4} Zc_2)$ , by applying Proposition 2 we have

\begin{equation*} \int_{\mathbb{R}}|\pi-\pi_M|\mathrm{d} x \le c_3 \exp\!(\!-M/2), \end{equation*}

with

(44) \begin{equation} c_3 :\!= Zc_2(2\pi)^{1/4}( 1+ {2Z(2\pi)^{1/4} } ). \end{equation}

The above approach can be applied to the more general multivariate Gaussian case at the outset of this subsection. In this more general case, we are unable to apply our extension of [Reference Mielnik and Reyes29], so can no longer obtain explicit constants as in Remark 10, but can still bound the effect of truncating at level M up to a constant.

Proposition 7. In the multivariate setting $d>1$ , given some $\delta$ -separation between ${\lambda_0^\kappa}$ and $\widehat \lambda_1^M$ , we have the bound for some constant $C_d>0$ ,

\begin{equation*} \int_{{\mathbb{R}^d}}|\pi(x) - \pi_M(x)| \mathrm{d} x \le C_d \delta^{-1} \exp\!(\!-M/2). \end{equation*}