A computational procedure for estimation of the mixing time of the random-scan Metropolis algorithm

Abstract

Many situations, especially in Bayesian statistical inference, call for the use of a Markov chain Monte Carlo (MCMC) method as a way to draw approximate samples from an intractable probability distribution. With the use of any MCMC algorithm comes the question of how long the algorithm must run before it can be used to draw an approximate sample from the target distribution. A common method of answering this question involves verifying that the Markov chain satisfies a drift condition and an associated minorization condition (Rosenthal, J Am Stat Assoc 90:558–566, 1995; Jones and Hobert, Stat Sci 16:312–334, 2001). This is often difficult to do analytically, so as an alternative, it is typical to rely on output-based methods of assessing convergence. The work presented here gives a computational method of approximately verifying a drift condition and a minorization condition specifically for the symmetric random-scan Metropolis algorithm. Two examples of the use of the method described in this article are provided, and output-based methods of convergence assessment are presented in each example for comparison with the upper bound on the convergence rate obtained via the simulation-based approach.

Appendices

Appendix 1: Geometric ergodicity of $(\mathsf {X}_t)_{t\ge 0}$

Let $(\mathsf {X}_t)_{t\ge 0}$ be the Markov chain described in Sect. 4.1.

Theorem 3

$(\mathsf {X}_t)_{t\ge 0}$ is geometrically ergodic.

Proof

The approach here is to verify that the three conditions given by Fort et al. (2003) hold for this chain.

In order to verify that Condition 1 holds, it is necessary to ensure that the target density $p(\cdot )$ is positive and continuous over all of $\mathbb {R}^3$. This will ensure that the stationary measure is absolutely continuous with respect to Lebesgue measure. The target density is given by

$$\begin{aligned} p(\mathbf {x}) = \text {exp}\left\{ -\pi \sum \limits _{i=1}^3 \big (x^{(i)} - 5\big )^2\right\} \prod \limits _{i=1}^3 \mathbb {I}_{\mathbb {R}}\big (x^{(i)}\big ). \end{aligned}$$

Note that $\mathbb {I}_{\mathbb {R}}(x^{(i)}) = 1$ for all $x^{(i)}$ $\in$ $\mathbb {R}$ and for all i=1, 2, 3. The quantity $\text {exp}\left\{ -\pi (x^{(i)} - 5)^2\right\}$ is also positive for all $x^{(i)}$ $\in$ $\mathbb {R}$ and for all $i \in \left\{ 1,\,2,\,3 \right\}$. Therefore, $p(\mathbf {x})$ is positive for all $\mathbf {x} \in \mathbb {R}^3$. Furthermore, $(x^{(i)} - 5)^2$ is continuous in $x^{(i)}$ for all real-valued $x^{(i)}$ and for all i=1,2,3, and $\text {exp}\left\{ -\pi (x^{(i)} - 5)^2\right\}$ is a continuous function of $(x^{(i)} - 5)^2$, so $p(\mathbf {x})$ is positive and continuous over $\mathbb {R}^3$. Thus, ${\pi }$ is absolutely continuous with respect to Lebesgue measure.

In order to establish that $(\mathsf {X}_t)_{t\ge 0}$ satisfies Condition 2, it is necessary to examine the increment densities. Recall that for $i = 1,2,3, q_i(y)$ is the normal density with mean 0 and variance $(95/99)^2$. Therefore, if $\delta _i$ is chosen to be 95 / 99 for $i = 1,2,3$, then for all y such that $|y| \le \delta _i, q_i(y) \ge 0.2521$. Thus, choosing $\eta _i = 0.252$ for each $i = 1,2,3$ ensures $(\mathsf {X}_t)_{t\ge 0}$ satisfies Condition 2.

Verification that $(\mathsf {X}_t)_{t\ge 0}$ satisfies the third of the Fort et al. (2003) conditions consists of three parts. First, it is necessary to show that there exist constants $\delta$ and $\Delta$ with $0 \le \delta \le \Delta \le +\infty$ such that

$$\begin{aligned} \xi := \inf \limits _{1\le i \le 3}\int _{\delta }^{\Delta } q_i(y)\lambda _1(\text {d}y) > 0. \end{aligned}$$

Since the increment densities are the same for each $i=1,2,3$, it suffices to show that

$$\begin{aligned} \xi = \int _{\delta }^{\Delta } q_1(y)\lambda _1(\text {d}y) > 0. \end{aligned}$$

Let $\delta = 95/99$, and let $\Delta = 2\left( 95/99\right)$. Then $\xi = \varPhi (2) - \varPhi (1)$, where $\varPhi (\cdot )$ denotes the cumulative distribution function (CDF) of the standard normal distribution. Choosing these values of $\delta$ and $\Delta$ yields a value of $\xi$ equal to 0.1358. This establishes that $(\mathsf {X}_t)_{t\ge 0}$ satisfies the first requirement of Condition 3.

The second part of verifying that $(\mathsf {X}_t)_{t\ge 0}$ satisfies Condition 3 requires ensuring that for any sequence $\left\{ \mathbf {x}_n \right\}$ such that $\lim \limits _{n\rightarrow \infty }\Vert \mathbf {x}_n\Vert = \infty$, it is possible to extract a subsequence $\left\{ \tilde{\mathbf {x}}_n\right\}$ such that for some $i \in \left\{ 1,2,3\right\}$,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \frac{p\big (\tilde{\mathbf {x}}_n\big )}{p\big (\tilde{\mathbf {x}}_n - \text {sign}\big (\tilde{x}_n^{(i)}\big )y\mathbf {e}_i\big )} = 0. \end{aligned}$$

In order to establish that this holds, note that if $\lim \limits _{n\rightarrow \infty }\Vert \mathbf {x}_n\Vert$ $=$ $\infty$, then for at least one $i \in \left\{ 1,2,3\right\}$, $\lim \limits _{n\rightarrow \infty }|x_n^{(i)}| = +\infty$. Without loss of generality, assume $\lim \limits _{n\rightarrow \infty }|x_n^{(1)}| = +\infty$. Two cases must be handled here. Let $\left\{ \tilde{\mathbf {x}}_n\right\}$ be a subsequence of $\left\{ \mathbf {x}_n\right\}$ such that $\lim \limits _{n\rightarrow \infty }x_n^{(1)} = +\infty$, and let $\left\{ \hat{\mathbf {x}}_n\right\}$ be a subsequence of $\left\{ \mathbf {x}_n\right\}$ such that $\lim \limits _{n\rightarrow \infty }\hat{x}_n^{(1)} = -\infty$. It is required to show that the above limit is 0 for both subsequences. Beginning with $\left\{ \tilde{\mathbf {x}}_n\right\}$,

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \frac{p\big (\tilde{\mathbf {x}}_n\big )}{p\big (\tilde{\mathbf {x}}_n - \text {sign}\big (\tilde{x}_n^{(1)}\big )y\mathbf {e}_1\big )}= & {} \lim \limits _{n\rightarrow \infty } \frac{p\big (\tilde{\mathbf {x}}_n\big )}{p\big (\tilde{\mathbf {x}}_n - y\mathbf {e}_1\big )}, \end{aligned}$$

since $\tilde{x}_n^{(1)}$ is positive in the tail of $\left\{ \tilde{\mathbf {x}}_n\right\}$. This yields

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{p\big (\tilde{\mathbf {x}}_n\big )}{p\big (\tilde{\mathbf {x}}_n - y\mathbf {e}_1\big )}= & {} \lim \limits _{n\rightarrow \infty }\frac{e^{-\pi \sum \limits _{i=1}^3 (\tilde{x}_n^{(i)} - 5)^2}}{e^{-\pi \left[ (\tilde{x}_n^{(1)} - y -5)^2 - \sum \limits _{i=2}^3 (\tilde{x}_n^{(i)} - 5)^2\right] }}\\= & {} \lim \limits _{n\rightarrow \infty } e^{-2\pi \tilde{x}_n^{(1)}y + \pi y^2 + 10\pi y}\\\le & {} e^{20\pi \frac{95}{99} + 4\pi \left( \frac{95}{99}\right) ^2}\lim \limits _{\tilde{x}_n^{(1)} \rightarrow \infty }e^{-2\pi \left( \frac{95}{99}\right) \tilde{x}_n^{(1)}}\\= & {} 0. \end{aligned}$$

Now examine the limit for $\left\{ \hat{\mathbf {x}}_n\right\}$. Since $\hat{x}_n^{(1)}$ is negative in the tail of $\left\{ \hat{\mathbf {x}}_n\right\}$, it follows that

$$\begin{aligned} \lim \limits _{n \rightarrow \infty } \frac{p\big (\hat{\mathbf {x}}_n\big )}{p\big (\hat{\mathbf {x}}_n - \text {sign}\big (\hat{x}_n^{(1)}\big )y\mathbf {e}_1\big )}= & {} \lim \limits _{n\rightarrow \infty }\frac{p\big (\hat{\mathbf {x}}_n\big )}{p\big (\hat{\mathbf {x}}_n + y\mathbf {e}_1\big )}. \end{aligned}$$

Examination of the limit of this ratio gives

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{p\big (\hat{\mathbf {x}}_n\big )}{p\big (\hat{\mathbf {x}}_n + y\mathbf {e}_1\big )}= & {} \lim \limits _{n\rightarrow \infty }\frac{e^{-\pi \sum \limits _{i=1}^3 (\hat{x}_n^{(i)} - 5)^2}}{e^{-\pi \left[ (\hat{x}_n^{(1)} + y -5)^2 - \sum \limits _{i=2}^3 (\hat{x}_n^{(i)} - 5)^2\right] }}\\= & {} \lim \limits _{n\rightarrow \infty } e^{\pi \left[ 2\hat{x}_n^{(1)}y + y^2 -10y\right] }\\\le & {} e^{-10\pi \frac{95}{99} + 4\pi \left( \frac{95}{99}\right) ^2}\\&\times \lim \limits _{\hat{x}_n^{(1)}\rightarrow -\infty }e^{4\pi \left( \frac{95}{99}\right) \hat{x}_n^{(1)}}\\= & {} 0. \end{aligned}$$

This establishes that $(\mathsf {X}_t)_{t\ge 0}$ satisfies the second part of condition 3. In order to establish that $(\mathsf {X}_t)_{t\ge 0}$ satisfies the final part of condition 3, it is necessary to show that the limit of the ratio

$$\begin{aligned} \frac{p\bigg (\mathbf {x} + \text {sign}\big (x_n^{(1)}\big )y\mathbf {e}_1\bigg )}{p(\mathbf {x})} \end{aligned}$$

is 0 for both $\left\{ \tilde{\mathbf {x}}_n \right\}$ and $\left\{ \hat{\mathbf {x}}_n \right\}$ as n tends to $\infty$. Beginning with $\left\{ \tilde{\mathbf {x}}_n \right\}$, since $\tilde{x}_n^{(1)}$ is positive in the tails of $\left\{ \tilde{\mathbf {x}}_n\right\}$, it follows that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{p\bigg (\tilde{\mathbf {x}}_n + \text {sign}\big (\tilde{x}_n^{(1)}\big )y\mathbf {e}_1\bigg )}{p\big (\tilde{\mathbf {x}}_n\big )}= & {} \lim \limits _{n\rightarrow \infty } \frac{p\big (\tilde{\mathbf {x}}_n + y\mathbf {e}_1\big )}{p\big (\tilde{\mathbf {x}}_n\big )}. \end{aligned}$$

As n tends to $\infty$, the limit of this ratio is

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } \frac{p\big (\tilde{\mathbf {x}}_n + y\mathbf {e}_1\big )}{p\big (\tilde{\mathbf {x}}_n\big )}= & {} \frac{e^{-\pi (\tilde{x}_n^{(1)} + y-5)^2}}{e^{-\pi (\tilde{x}_n^{(1)} - 5)^2}}\\= & {} \lim \limits _{n\rightarrow \infty }e^{-\pi \left( \tilde{x}_n^{(1)}y-10y+25\right) }\\\le & {} e^{20\pi \left( \frac{95}{99}\right) - 25\pi }\lim \limits _{\tilde{x}_n^{(1)} \rightarrow +\infty }e^{-\pi \left( \frac{95}{99}\right) \tilde{x}_n^{(1)}}\\= & {} 0. \end{aligned}$$

Since $\hat{x}_n^{(1)}$ is negative in the tail of $\left\{ \hat{\mathbf {x}}_n\right\}$, it follows that

$$\begin{aligned} \lim \limits _{n\rightarrow \infty }\frac{p\bigg (\hat{\mathbf {x}}_n + \text {sign}\big (\hat{x}_n^{(1)}\big )y\mathbf {e}_1\bigg )}{p\big (\hat{\mathbf {x}}_n\big )}= & {} \lim \limits _{n\rightarrow \infty } \frac{p\big (\hat{\mathbf {x}}_n - y\mathbf {e}_1\big )}{p\big (\hat{\mathbf {x}}_n\big )}. \end{aligned}$$

Thus, as n tends to $\infty$, the limit of this ratio is

$$\begin{aligned} \lim \limits _{n\rightarrow \infty } \frac{p\big (\hat{\mathbf {x}}_n - y\mathbf {e}_1\big )}{p\big (\hat{\mathbf {x}}_n\big )}= & {} \frac{e^{-\pi (\hat{x}_n^{(1)} - y-5)^2}}{e^{-\pi (\hat{x}_n^{(1)} - 5)^2}}\\= & {} \lim \limits _{n\rightarrow \infty }e^{\pi \left( 2\hat{x}_n^{(1)}y-10y- y^2\right) }\\\le & {} \lim \limits _{\hat{x}_n^{(1)} \rightarrow -\infty }e^{-10\pi \left( \frac{95}{99}\right) - \pi \left( \frac{95}{99}\right) ^2+4\pi \left( \frac{95}{99}\right) \hat{x}_n^{(1)}}\\= & {} 0. \end{aligned}$$

This establishes that $(\mathsf {X}_t)_{t\ge 0}$ satisfies the three conditions given by Fort et al. (2003). Thus, $(\mathsf {X}_t)_{t\ge 0}$ is geometrically ergodic, and $V(\mathbf {x})$ $=$ $\left[ p(\mathbf {x})\right] ^{-s}$ is a useful drift function, where s is chosen so that $s(1-s)^{\frac{1}{s} - 1}$ $<$ 0.01555, in accordance with the hypothesis of Theorem 2. $\square$

Appendix 2: Geometric ergodicity of $({\tau }_t)_{t\ge 0}$

Let $({\tau }_t)_{t\ge 0}$ be the Markov chain described in Sect. 4.2.

Theorem 4

The Markov chain $({\tau }_t)_{t=0}^{\infty }$ is geometrically ergodic.

Proof

The approach here is again to verify that $({\tau }_t)_{t\ge 0}$ satisfies the three conditions of Fort et al. (2003). Recall that the target density for this example is given by

$$\begin{aligned}&p({\tau }|\mathbf {S}) \propto e^{-\frac{1}{10}\sum _{i=1}^{90}(S_i - \mathbf {Z}_i{\tau })^2 - \frac{1}{4}\sum _{k=1}^{9}(\tau ^{(k)} - 80)^2}. \end{aligned}$$

(13)

1.1 Verification of Condition 1

In order to verify that $({\tau }_t)_{t\ge 0}$ satisfies Condition 1, it is necessary to show that $p({\tau }|\mathbf {S})$ is positive and continuous over $\mathbb {R}^9$, and that the stationary measure is absolutely continuous with respect to Lebesgue measure. Note first that for all $x \in \mathbb {R}$, the function $g(x) = e^x$ is positive. Since no value of ${\tau }$ exists in $\mathbb {R}^9$ for which the exponent is not real, it follows that $p({\tau }|\mathbf {S})$ is positive over $\mathbb {R}^9$. Note also that the exponent has no points of discontinuity in ${\tau }$, since the exponent is a linear combination of functions that are continuous in ${\tau }$ over $\mathbb {R}^9$. Therefore, $p({\tau }|\mathbf {S})$ is continuous over $\mathbb {R}^9$. This shows that the target density is positive and continuous over $\mathbb {R}^9$. Therefore, if $\pi (\cdot )$ is the stationary measure, then $\pi (B) = 0$ if and only if B is a set of Lebesgue measure 0. This implies that the stationary measure is absolutely continuous with respect to Lebesgue measure, and Condition 1 is satisfied.

1.2 Verification of Condition 2

Condition 2 is easily verified, since the increment density is uniform on the interval $[-2.25, 2.25]$. Hence, for $i=1,2,\ldots , 9$,

$$\begin{aligned} q_i(y) = \frac{1}{4.5}\mathbb {I}_{[-2.25, 2.25]}(y). \end{aligned}$$

Let $\delta _i = 2.25$ for $i=1,2,\ldots , 9$. Then for y such that $|y| \le \delta _i$, $q_i(y) = 1/4.5$. Therefore, $\eta _i = 1/4.5$ for $i=1,2,\ldots , 9$, and $({\tau }_t)_{t\ge 0}$ is satisfied.

1.3 Verification of Condition 3

Using the notation from the statement of Condition 3, let $\delta = 0.1$ and $\Delta = 2.25$. Since for each $i=1,2,\ldots , 9$, the increment density is the $U[-2.25, 2.25]$ density, it follows that

$$\begin{aligned} \xi= & {} \inf _{i=1,2,\ldots , 9}\int _{0.1}^{2.25} q_i(y)\lambda (\text {d}y)\\= & {} \int _{0.1}^{2.25}\frac{1}{4.5}\text { d}y\\= & {} 0.4778. \end{aligned}$$

This establishes that $({\tau }_t)_{t\ge 0}$ satisfies the first part of Condition 3. The next step is to investigate the ratios given in Theorem 2. To begin, let $\left\{ {\tau }_j^+ \right\}$ be a sequence for which $\lim _{j}\Vert {\tau }_j\Vert = +\infty$. Then for at least one $i \in \left\{ 1,2,\ldots , 9\right\}$, $\lim _j |\tau _j^{(i)}| = +\infty .$ This leaves three possibilities:

• $\lim _j \tau _j^{(i)} = +\infty$

• $\lim _j \tau _j^{(i)} = -\infty$

• $\tau _j^{(i)}$ alternates, and it is possible to extract one subsequence $\left\{ {\tau }\right\}$ for which $\tau _j^{(i)} \rightarrow +\infty$ and one subsequence for which $\tau _j^{(i)} \rightarrow -\infty$.

In order to verify that $({\tau }_t)_{t\ge 0}$ satisfies Condition 3, it suffices to show that

$$\begin{aligned}&\lim _j \frac{p({\tau }_j|\mathbf {S})}{p({\tau }_j - \text {sign}(\tau _j^{(i)})y\mathbf {e}_i|S)} = 0 \text { and }\\&\lim _j \frac{p({\tau }_j + \text {sign}(\tau _j^{(i)})y\mathbf {e}_i|\mathbf {S})}{p({\tau }_j|\mathbf {S})} = 0 \end{aligned}$$

for each of the two subsequences of $\left\{ {\tau }_j\right\}$ described in the third situation.

To begin, let $\left\{ {\tau }_j^+\right\}$ be a subsequence of $\left\{ {\tau }_j\right\}$ for which $\lim _j |\tau _j^{(i)}| = +\infty$, and assume without loss of generality that $i=1$. Since $\tau _j^{+(1)} > 0$ in the tail of $\left\{ {\tau }_j^+\right\}$,

$$\begin{aligned} \lim _{j} \frac{p({\tau }_j^+|\mathbf {S})}{p({\tau }^+ - \text {sign}(\tau _j^{+(1)})y\mathbf {e}_1|\mathbf {S})}= & {} \lim _j \frac{p({\tau }_j^+|\mathbf {S})}{p(\mathbf {\tau }_j^+ - y\mathbf {e}_1|\mathbf {S})} \text { and }\\ \lim _j \frac{p({\tau }_j^+ + \text {sign}(\tau _j^{+(1)})y\mathbf {e}_1|\mathbf {S})}{p({\tau }_j^+|\mathbf {S})}= & {} \lim _j \frac{p({\tau }_j^+ + y\mathbf {e}_1|\mathbf {S})}{p({\tau }_j^+|\mathbf {S})}. \end{aligned}$$

For the first ratio,

$$\begin{aligned}&\lim _j \frac{p({\tau }_j^+|\mathbf {S})}{p({\tau }_j^+ - y\mathbf {e}_1|\mathbf {S})}\\&\quad = \lim _j \frac{e^{-\frac{1}{10}\sum _{i=1}^{90}(S_i - \mathbf {Z}_i{\tau }_j^+)^2 - \frac{1}{4}\sum _{k=1}^9(\tau _j^{+(k)} - 80)^2}}{e^{-\frac{1}{10}\sum _{i=1}^{90}(S_i - \mathbf {Z}_i({\tau }_j^+ - y\mathbf {e}_1)^2 - \frac{1}{4}\sum _{k=2}^9(\tau _j^{+(k)} - 80)^2 - \frac{1}{4}(\tau _j^{+(k)}-80)-y)^2}}. \end{aligned}$$

Recall that $Z_{ij}=1$ if student i had instructor j and $Z_{ij} = 0$ otherwise. Let $G=\left\{ i: \text { Student }i \text {had instructor 1}\right\}$. Then, after cancelation, the ratio can be written as

$$\begin{aligned}&\lim _j e^{\frac{y}{5}\sum _{i\in G}S_i - 2y\tau _{j}^{+(1)} + y^2- \frac{y}{2}\tau _j^{+(1)} + 40y + \frac{1}{4}y^2} \text { since }G\\&\text {has 10 elements}\\&\le e^{\frac{2.25}{5}(1000) + 2.25^2 + 90 + \frac{2.25^2}{4}}\lim _j e^{-0.2\tau _j^{+(1)} - 0.05\tau _j^{+(1)}}\\&\text {since }S_i \le 100 \text {for each student and }y \in [0.1,2.25].\\&\text {Then the ratio is bounded by}\\&e^{546.328125}\lim _j e^{-0.25\tau _j^{+(1)}} \rightarrow 0\text { as }\text {j }\rightarrow \infty . \end{aligned}$$

Examining the second ratio for this subsequence and making use of a similar mathematical argument to the one that was used to establish the limit of the first ratio gives

$$\begin{aligned}&\lim _j \frac{p\bigg ({\tau }_j + \text {sign}\big (\tau _j^{(i)}\big )y\mathbf {e}_i|\mathbf {S}\bigg )}{p\big ({\tau }_j|\mathbf {S}\big )}\\&\quad = \lim _j e^{-\frac{y}{5}\sum _{i\in G}S_i - \frac{5}{2}y\tau _j^{+(1)} - \frac{5}{4}y^2 + 40y}\\&\quad \le e^{540}\lim _j e^{-0.25\tau _j^{+(1)}}, \end{aligned}$$

which tends to 0 as j tends to $+\infty$. This proves that the two limits in Condition 3 hold for the subsequence $\left\{ {\tau }_j^+\right\}$.

Now examine the same two ratios for the subsequence $\left\{ {\tau }_j^- \right\}$, where $\lim _j \tau _j^{-(1)} = -\infty$. This implies that $\tau _j^{-(1)}$ is negative in the tail of $\left\{ {\tau }_j^-\right\}$. Therefore,

$$\begin{aligned}&\lim _j\frac{p\big ({\tau }_j^-|\mathbf {S}\big )}{p\bigg ({\tau }_j^- - \text {sign}\big (\tau _j^{-(1)}\big )y\mathbf {e}_1|\mathbf {S}\bigg )} = \lim _j \frac{p\big ({\tau }_j^- - y\mathbf {e}_1|\mathbf {S}\big )}{p\big ({\tau }_j^- + y\mathbf {e}_1|\mathbf {S}\big )}\\&\text {and} \\&\lim _j \frac{p\bigg ({\tau }_j^- + \text {sign}\big (\tau _j^{-(1)}\big )y\mathbf {e}_1\bigg )}{p\big ({\tau }_j^-|\mathbf {S}\big )} = \lim _j \frac{p\big ({\tau }_j^- - y\mathbf {e}_1|\mathbf {S}\big )}{p\big ({\tau }_j^-|\mathbf {S}\big )}. \end{aligned}$$

This gives the following for the first ratio after similar cancelation to that used to establish that the limit holds for $\left\{ {\tau }_j^+\right\}$. The limit of the ratio can be written as

$$\begin{aligned}&\lim _j e^{-\frac{y}{5}\sum _{i\in G}S_i + 2y\tau _j^{-(1)} + y^2+ \frac{y}{2}\tau _j^{-(1)} -40y + \frac{1}{4}y^2}\\&\quad = \lim _j e^{-\frac{y}{5}\sum _{i\in G}S_i + \frac{5}{2}y\tau _j^{-(1)} - 40y + \frac{5}{4}y^2}\\&\quad \le e^{\frac{5}{4}(2.25)^2}\lim _j e^{\frac{5}{2}y\tau _j^{-(1)}}, \end{aligned}$$

which tends to 0 as j approaches $\infty$. After similar cancelation as has been used throughout the rest of the proof, the second limit can be written as

$$\begin{aligned}&\lim _j e^{-\frac{y}{5}\sum _{i \in G}S_i + 2y\tau _j^{-(1)} - y^2 + \frac{y}{2}\tau _j^{-(1)} - 40y - \frac{1}{4}y^2}\\&\quad \le \lim _j e^{5.625\tau _j^{-(1)}}, \end{aligned}$$

which tends to 0 as j tends to $\infty$. This establishes that $({\tau }_t)_{t\ge 0}$ satisfies the conditions provided by Fort et al. (2003), and as a result, the chain satisfies the specified drift condition for an appropriate choice of s for the small set based on an appropriate choice of d that is presented by Rosenthal (1995). $\square$