A Partition-Based Optimization Approach for Level Set Approximation: Probabilistic Branch and Bound

Abstract

We present a partition-based random search optimization algorithm, called probabilistic branch and bound (PBnB), to approximate a level set that achieves a user-defined target. Complex systems are often modeled with computer simulations, both deterministic and stochastic, and a decision-maker may desire a set of near-optimal solutions, such as solutions with performance metrics in the best 10% overall, instead of a single estimate of a global optimum. Our approach is valid for black-box, ill-structured, noisy function evaluations, involving both integer and real-valued decision variables. PBnB iteratively maintains, prunes, or branches subregions of a bounded solution space based on an updated confidence interval of a target quantile. Finite-time probability bounds are derived on the maximum volume of incorrectly maintained or incorrectly pruned regions. Thus, the user has a statistical quantification of the output. For example, with probability greater than 0.9, the final maintained subregion is inside the target level set with the volume of incorrectly maintained points less than 2% of the volume of the initial set. Numerical results on noisy and non-noisy test functions demonstrate the performance of the PBnB algorithm. Tests on a sphere function, with spherical level sets, allow a comparison between the theoretical bounds and numerical results. PBnB has been applied to several application areas including: weather impacts on air traffic flow management; policy decisions on screening and treatment budget allocation for hepatitis C; combining portable ultrasound machines with reserved MRI usage for orthopedic care; and optimizing water distribution networks using simulation.

Appendix

Proof of Theorem 1

Proof

We consider the iterative effect on δ _k as subregions are pruned or maintained. We use the superscript k to denote the iteration that subregions are pruned $\{\sigma ^k_i:{P}_i = 1\}$ or maintained $\{\sigma ^k_i:{M}_i = 1\}$. By (6.15) in the algorithm, we have

$$\displaystyle \begin{aligned} \delta_k&=\frac{\delta_{k-1} v\left(\widetilde{\varSigma}^C_{k-1}\right) - \sum_{i:{M}_i = 1}v\left(\sigma^{k-1}_i\right)} {v\left(\widetilde{\varSigma}^C_{k-1}\right) -\sum_{i:{P}_i = 1}v\left(\sigma^{k-1}_i\right) - \sum_{i:{M}^{k-1}_i = 1} v\left(\sigma^{k-1}_i\right)} \end{aligned} $$

and removing the pruned and maintained subregions from $\widetilde {\varSigma }^C_{k-1}$ yields the next current set of subregions $\widetilde {\varSigma }^C_{k}$, used in the denominator, then

$$\displaystyle \begin{aligned} &=\frac{\delta_{k-1} v\left(\widetilde{\varSigma}^C_{k-1}\right) - \sum_{i:{M}_i = 1}v\left(\sigma^{k-1}_i\right)}{v\left(\widetilde{\varSigma}^C_{k}\right)} \end{aligned} $$

and invoking (6.15) in the algorithm again to replace δ _k−1 with its equivalence in terms of δ _k−2 (assuming that the maintained regions are in the level set and pruned regions are out of the level set), we have

$$\displaystyle \begin{aligned} &=\frac{\frac{\delta_{k-2}v\left(\widetilde{\varSigma}^C_{k-2}\right)- \sum_{i:{M}_i = 1}v\left(\sigma^{k-2}_i\right)}{v\left(\widetilde{\varSigma}^C_{k-1}\right)} v\left(\widetilde{\varSigma}^C_{k-1}\right) - \sum_{i:{M}_i = 1}v\left(\sigma^{k-1}_i\right)}{v\left(\widetilde{\varSigma}^C_{k}\right)}\\ &=\frac{\delta_{k-2} v\left(\widetilde{\varSigma}^C_{k-2}\right) - \sum_{l=k-2}^{k-1}\sum_{i:{M}_i = 1}v\left(\sigma^{l}_i\right)}{v\left(\widetilde{\varSigma}^C_{k}\right)}\\ &\quad \vdots \\ &=\frac{\delta_{1} v\left(\widetilde{\varSigma}^C_{1}\right) - \sum_{l=1}^{k-1}\sum_{i:{M}_i = 1}v\left(\sigma^{l}_i\right)}{v\left(\widetilde{\varSigma}^C_{k}\right)} \end{aligned} $$

and by the initial setting of δ ₁ and Σ˜1C,

$$\displaystyle \begin{aligned} &=\frac{\delta v(S) - \sum_{l=1}^{k-1}\sum_{i:{M}_i = 1}v\left(\sigma^{l}_i\right)}{v\left(\widetilde{\varSigma}^C_{k}\right)} \end{aligned} $$

and $\sum _{l=1}^{k-1}\sum _{i:{M}_i = 1}v\left (\sigma ^{l}_i\right )=v\left (\widetilde {\varSigma }^M_{k}\right )$ since it denotes the volume of all maintained subregions at the end of the k − 1 iteration, therefore,

$$\displaystyle \begin{aligned} &=\frac{\delta v(S) - v\left(\widetilde{\varSigma}^M_{k}\right)}{v\left(\widetilde{\varSigma}^C_{k}\right)}. \end{aligned} $$

(6.44)

Based on the definition of quantile, and when X is uniformly sampled on S, we have

$$\displaystyle \begin{aligned} y(\delta, S)&=\mathop{\arg\min}\limits_{y\in \{f(x):x \in S\}} \left\{P(f(X)\leq y | X \in S)\geq \delta \right\} \\ &=\mathop{\arg\min}\limits_{y\in \{f(x):x \in S\}} \left\{\frac{v(\{x\in S: f(x)\leq y\})}{v(S)}\geq \delta \right\} \end{aligned} $$

and subtracting v(Σ˜kM)−𝜖 kM+𝜖 kP v(S) from both sides and multiplying v(S) v(Σ˜kC) on both sides,

$$\displaystyle \begin{aligned} &=\mathop{\arg\min}\limits_{y\in \{f(x):x \in S\}} \left\{\frac{v\left(\{x\in S: f(x)\leq y\}\right)-v\left(\widetilde{\varSigma}^M_{k}\right)+ \epsilon^M_k-\epsilon^P_k}{v\left(\widetilde{\varSigma}^C_k\right)}\right.\\ &\quad \qquad \qquad \qquad \geq \left. \frac{\delta v(S)-v\left(\widetilde{\varSigma}^M_{k}\right)+ \epsilon^M_k-\epsilon^P_k}{v\left(\widetilde{\varSigma}^C_k\right)} \right\} \end{aligned} $$

and by (6.44), also $v\left (\{x\in \widetilde {\varSigma }^P_k:f(x)<y\}\right )=\epsilon ^P_k$ and $v\left (\{x\in \widetilde {\varSigma }^M_k:f(x)<y\}\right )= v\left (\widetilde {\varSigma }^M_{k}\right )-\epsilon ^M_k$,

$$\displaystyle \begin{aligned} &=\mathop{\arg\min}\limits_{y\in \{f(x):x \in S\}} \left\{\frac{v\left(\left\{x\in S\setminus \left\{\widetilde{\varSigma}^P_k \cup \widetilde{\varSigma}^M_{k}\right\}: f(x)\leq y\right\}\right)}{v\left(\widetilde{\varSigma}^C_k\right)}\right.\\ &\left.\quad \qquad \qquad \qquad \geq \delta_k + \frac{\epsilon^M_k}{v\left(\widetilde{\varSigma}^C_k\right)}- \frac{\epsilon^P_k}{v\left(\widetilde{\varSigma}^C_k\right)}\right\} \end{aligned} $$

and since Σ˜kC = S ∖{Σ˜kP ∪Σ˜kM}, and X is uniformly distributed in Σ˜kC and Σ˜kC ⊂ S,

$$\displaystyle \begin{aligned} &=\mathop{\arg\min}\limits_{y\in \{f(x):x \in \widetilde{\varSigma}^C_k\}} \left\{P\left(f(X)\leq y \bigg| X \in \widetilde{\varSigma}^C_k\right)\geq \delta_k + \frac{\epsilon^M_k}{v\left(\widetilde{\varSigma}^C_k\right)}- \frac{\epsilon^P_k}{v\left(\widetilde{\varSigma}^C_k\right)}\right\}\\ &= y\left(\delta_k + \frac{\epsilon^M_k}{v\left(\widetilde{\varSigma}^C_k\right)}- \frac{\epsilon^P_k}{v\left(\widetilde{\varSigma}^C_k\right)}, \widetilde{\varSigma}^C_k\right). \end{aligned} $$

Since $0 \leq \epsilon ^M_k \leq \frac {\epsilon \widetilde {\varSigma }^M_k}{v(S)}$ and $0 \leq \epsilon ^P_k \leq \frac {\epsilon \widetilde {\varSigma }^P_k}{v(S)}$, an upper bound of $y\left (\delta _k + \frac {\epsilon ^M_k}{v\left (\widetilde {\varSigma }^C_k\right )}- \frac {\epsilon ^P_k}{v\left (\widetilde {\varSigma }^C_k\right )}, \widetilde {\varSigma }^C_k\right )$ can be achieved when $\epsilon ^P_k=0$ and $\epsilon ^M_k=\frac {\epsilon v\left (\widetilde {\varSigma }^M_{k}\right )}{v(S)}$, yielding

$$\displaystyle \begin{aligned} y(\delta, S) \leq y\left(\delta_k + \frac{\epsilon v\left(\widetilde{\varSigma}^M_{k}\right)}{v(S)v\left(\widetilde{\varSigma}^C_k\right)}, \widetilde{\varSigma}^C_k\right) = y\left(\delta_{ku}, \widetilde{\varSigma}^C_k\right).\end{aligned} $$

(6.45)

Similarly, we have a lower bound when $\epsilon ^M_k=0$ and $\epsilon ^P_k=\frac {\epsilon v(\widetilde {\varSigma }^P_{k})}{v(S)}$, yielding

$$\displaystyle \begin{aligned} y(\delta, S) \geq y\left(\delta_k - \frac{\epsilon v\left(\widetilde{\varSigma}^P_{k}\right)}{v(S)v\left(\widetilde{\varSigma}^C_k\right)}, \widetilde{\varSigma}^C_k\right) = y\left(\delta_{kl}, \widetilde{\varSigma}^C_k\right).\end{aligned} $$

(6.46)

Note, if $\epsilon ^M_k=0$ and $\epsilon ^P_k=0$, that is, there is no error in pruning and maintaining, then $y(\delta , S)=y\left (\delta _k, \widetilde {\varSigma }^C_{k}\right )$.

At the beginning of any iteration k, the current set $\widetilde {\varSigma }^C_k$ is uniformly sampled for N _k = c _k samples. Since the samples are independent and uniformly distributed in the current set $\widetilde {\varSigma }^C_{k}$, each sample acts like a Bernoulli trial and falls in a δ _kl or δ _ku level set with δ _kl or δ _ku probability, respectively. Therefore, using properties of a binomial distribution, we can build a 1 − α _k quantile confidence interval as f(z _(r)) ≤ y(δ, S) ≤ f(z _(s)) (Conover 1999) with $y\left (\delta _{kl}, \widetilde {\varSigma }^C_{k}\right )$ and $y\left (\delta _{ku}, \widetilde {\varSigma }^C_{k}\right )$ based on (6.45) and (6.46), where f(z _(r)) and f(z _(s)) are the rth and sth order samples that have the following binomial properties

$$\displaystyle \begin{aligned} & P\left(f(z_{(r)})>y\left(\delta_{kl}, \widetilde{\varSigma}^C_{k}\right)\right) \leq \sum^{r-1}_{i=0}{{N_k}\choose{i}}(\delta_{kl})^i\left(1-\delta_{kl}\right)^{N_k-i}{} \end{aligned} $$

(6.47)

$$\displaystyle \begin{aligned} & P\left(f(z_{(s)})\geq y\left(\delta_{ku}, \widetilde{\varSigma}^C_{k}\right)\right) \geq \sum^{s-1}_{i=0}{{N_k}\choose{i}}(\delta_{ku})^i(1-\delta_{ku})^{N_k-i}.{} \end{aligned} $$

(6.48)

The 1 − α _k confidence interval can be approximated by two one-sided intervals. We split α _k into two halves, and allocate one half to each probability bound. Therefore, find the maximum r for which (6.47) is less than or equal to $\frac {\alpha _k}{2}$ and the minimum s for which (6.48) is greater than or equal to $1-\frac {\alpha _k}{2}$, that is

$$\displaystyle \begin{aligned} \max r: \sum^{r-1}_{i=0}{{N_k}\choose{i}}(\delta_{kl})^i(1-\delta_{kl})^{N_k-i} \leq \frac{\alpha_k}{2}\ \text{ and } {} \end{aligned} $$

(6.49)

$$\displaystyle \begin{aligned} \min s: \sum^{s-1}_{i=0}{{N_k}\choose{i}}(\delta_{ku})^i(1-\delta_{ku})^{N_k-i} \geq 1-\frac{\alpha_k}{2} . {} \end{aligned} $$

(6.50)

Combining (6.47)–(6.50), as in Conover (1999), we have

$$\displaystyle \begin{aligned} P\left(f(z^k_{(r)}) \leq y\left(\delta_{kl}, \widetilde{\varSigma}^C_k\right)\leq y\left(\delta_{ku}, \widetilde{\varSigma}^C_k\right) \leq f\left(z^k_{(s)}\right)\right)\geq 1-\alpha_k. \end{aligned} $$

(6.51)

When there is no noise, $0 \leq \epsilon ^P_k \leq \frac {\epsilon v\left (\widetilde {\varSigma }^P_{k}\right )}{v(S)}$ and $0 \leq \epsilon ^M_k \leq \frac {\epsilon v\left (\widetilde {\varSigma }^M_{k}\right )}{v(S)}$, the 1 − α _k confidence interval of y(δ, S) is given by [f(z _(r)), f(z _(s))] based on (6.45), (6.46), and (6.51), that is

$$\displaystyle \begin{aligned} P\left(f(z_{(r)})\leq y(\delta, S) \leq f(z_{(s)})\right) \geq 1-\alpha_k.\end{aligned} $$

□

Proof of Theorem 2

Proof

We note that the event $v\left (L(\delta , S)\cap \hat {\sigma }^k_p\right )\leq D^P_k\epsilon _k$ is equivalent to the event $v\left (\left \{x: f(x) \leq y(\delta , S), x\in \hat {\sigma }^k_p\right \}\right )\leq D^P_k\epsilon _k$ by the definition of L(δ, S), and therefore, the probability of that event, that is, that the volume of the incorrectly pruned region is less than or equal to $D^P_k\epsilon _k$, can be expressed as

$$\displaystyle \begin{aligned} P\left(v\left(L(\delta , S)\cap \hat{\sigma}^k_p\right)\leq D^P_k\epsilon_k \big| A_k\right)&=P\left(v\left(\{x: f(x) \leq y(\delta , S), x\in \hat{\sigma}^k_p\}\right)\right.\\ &\quad \qquad \left.\leq D^P_k\epsilon_k \big| A_k\right). \end{aligned} $$

(6.52)

Now, consider the probability expression of quantile in (6.4) from the main article, and let $\delta _p=\frac {D^P_k\epsilon _k}{v(\hat {\sigma }^k_p)}$. We first prove the theorem under the special case that y(δ, S) is continuous in δ and $y(\delta _p,\hat {\sigma }^k_p)$ is continuous in δ _p, which implies that $v\left (\left \{x: f(x) = y, x\in \hat {\sigma }^k_p\right \}\right )=0, \forall y$ and that (6.4) holds at equality. When X is a uniform sample in $\hat {\sigma }^k_p$, we have

then multiplying $v(\hat {\sigma }^k_p)$ on both sides, we have

$$\displaystyle \begin{aligned} & \qquad v\left(\left\{x: f(x) < y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right) = D^P_k\epsilon_k. \end{aligned} $$

Hence, we have

$$\displaystyle \begin{aligned} D^P_k\epsilon_k &= v\left(\left\{x: f(x) < y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right) \\ &=v\left(\left\{x: f(x) \leq y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right)\notag\\ &\quad - v\left(\left\{x: f(x) = y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right) \end{aligned} $$

and in the special case that $v\left (\{x: f(x) = y, x\in \hat {\sigma }^k_p\}\right )=0, \forall y$, we have

$$\displaystyle \begin{aligned} &\qquad = v\left(\{x: f(x) \leq y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\}\right). {} \end{aligned} $$

(6.53)

We substitute the expression for $D^P_k\epsilon _k$ from (6.53) into the probability expression in (6.52), yielding

and from the properties of level sets, if $y(\delta ,S) \leq y\left (\delta _p , \hat {\sigma }^k_p\right )$, then $\{x: f(x) \leq y(\delta , S), x\in \hat {\sigma }^k_p\} \subseteq \{x: f(x) \leq y\left (\delta _p , \hat {\sigma }^k_p\right ), x\in \hat {\sigma }^k_p\}$, therefore,

$$\displaystyle \begin{aligned} &\quad = P\left(\left. y(\delta, S) \leq y\left(\delta_p , \hat{\sigma}^k_p\right) \right| A_k\right) \end{aligned} $$

and in the special case that $v\left (\left \{x: f(x) = y, x\in \hat {\sigma }^k_p\right \}\right )=0, \forall y$, we have that

$$\displaystyle \begin{aligned} &\quad = P\left(\left. y(\delta, S) < y\left(\delta_p , \hat{\sigma}^k_p\right) \right| A_k\right), \end{aligned} $$

and by the condition A _k and the pruned assumption, we have $y(\delta , S) \leq y\left (\delta _{ku}, \widetilde {\varSigma }_k^C\right ) \leq f(z_{(s)}) < f(x_{(p),(1)})$, where x _(p),(1) is the best sample out of $D^P_k N^p_k$ independent samples in $\hat {\sigma }^k_p$, therefore,

$$\displaystyle \begin{aligned} &\quad \geq P\left(\left.f(x_{(p),(1)}) \leq y\left(\delta_p , \hat{\sigma}^k_p\right) \right| A_k\right)\\ &\quad =1-P\left(\left.f(x_{(p),(1)}) > y\left(\delta_p , \hat{\sigma}^k_p\right) \right| A_k\right), \end{aligned} $$

and since each of the DkPNkp independent uniform samples X in σ̂pk satisfies

$$\displaystyle \begin{aligned} &\quad \geq 1-\left(1-\delta_p\right)^{D^P_k N^p_k}. {} \end{aligned} $$

(6.54)

Since $N^p_k =\left \lceil \frac {\ln {\alpha _k}}{\ln {(1-\frac {\epsilon _k}{v(\sigma _i)}})}\right \rceil$ in Step 4, and $\delta _p=\frac {D^P_k \epsilon _k}{v(\hat {\sigma }_p)}=\frac {D^P_k \epsilon _k}{D^P_k v(\sigma _i)}=\frac {\epsilon _k}{v(\sigma _i)}$, where σ _i is a subregion pruned at the kth iteration, and $D^P_k \geq 1$, we know $N^p_k \geq \frac {\ln {\alpha _k}}{\ln \left (1-\frac {\epsilon _k}{v(\sigma _i)}\right )}=\frac {\ln {\alpha _k}}{\ln {(1-\delta _p)}} \Rightarrow \ln {\left (1-\delta _p\right )^{D^P_k N^p_k}}\leq \ln {\alpha _k} \Rightarrow \left (1-\delta _p\right )^{D^P_k N^p_k}\leq \alpha _k$. Multiplying − 1 on both sides and adding one to both sides, the inequality becomes $1-\left (1-\delta _p\right )^{D^P_k N^p_k}\geq 1-\alpha _k$, hence

$$\displaystyle \begin{aligned} & P\left(v\left(\{x: f(x) \leq y(\delta , S), x\in \hat{\sigma}^k_p\}\right)\leq D^P_k\epsilon_k |A_k\right) \\ &\quad \geq 1-\alpha_k. \end{aligned} $$

(6.55)

This and (6.52) yield the theorem statement in (6.21) in the special case.

Now, in the more general case where y(δ, S) and $y(\delta _p,\hat {\sigma }_p^k)$ may have discontinuities, the $v\left (\left \{x: f(x) = y, x\in \hat {\sigma }^k_p\right \}\right )$ may be positive for some y. The flow of the proof is the same, however, the possibility of discontinuities changes equalities to inequalities while accounting for $v\left (\left \{x: f(x) = y, x\in \hat {\sigma }^k_p\right \}\right )$, as follows.

When X is a uniform sample in $\hat {\sigma }^k_p$, the probability expression of quantile in (6.4) with $\delta _p=\frac {D^P_k\epsilon _k}{v\left (\hat {\sigma }^k_p\right )}$ now can be expressed as

$$\displaystyle \begin{aligned} & P\left(f(X)<y\left(\delta_p,\hat{\sigma}^k_p\right)\right)=\frac{v\left(\left\{x: f(x) < y(\delta_p , \hat{\sigma}^k_p), x\in \hat{\sigma}^k_p\right\}\right)}{v(\hat{\sigma}^k_p)}\leq \delta_p =\frac{D^P_k\epsilon_k}{v(\hat{\sigma}^k_p)}, \end{aligned} $$

then multiplying $v(\hat {\sigma }^k_p)$ on both sides, we have

$$\displaystyle \begin{aligned} & v\left(\left\{x: f(x) < y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right)\leq D^P_k\epsilon_k. \end{aligned} $$

Hence, we have

$$\displaystyle \begin{aligned} & D^P_k\epsilon_k \geq v\left(\left\{x: f(x) < y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right) \\ & \qquad =v\left(\left\{x: f(x) \leq y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right) \\ & \qquad \qquad - v\left(\left\{x: f(x) = y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right). {} \end{aligned} $$

(6.56)

We substitute the expression for $D^P_k\epsilon _k$ from (6.56) into the probability expression in (6.52), and due to the possibility of discontinuities, this is a stricter event, yielding an inequality in the probability as follows:

$$\displaystyle \begin{aligned} & P\left(v\left(\{x: f(x) \leq y(\delta , S), x\in \hat{\sigma}^k_p\}\right)\leq D^P_k\epsilon_k |A_k\right) \\ &\quad \geq P\left(\left.v\left(\left\{x: f(x) \leq y(\delta , S), x\in \hat{\sigma}^k_p\right\}\right) \right.\right.\\ &\quad \leq v\left(\left\{x: f(x) \leq y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right) \\ &\qquad \left.\left. - v\left(\left\{x: f(x) = y\left(\delta_p , \hat{\sigma}^k_p\right), x\in \hat{\sigma}^k_p\right\}\right) \right| A_k\right) \end{aligned} $$

and since $=v\left (\{x: f(x) < y\left (\delta _p , \hat {\sigma }^k_p\right ), x\in \hat {\sigma }^k_p\}\right )$

and now comparing the level sets associated with y(δ, S) and $y\left (\delta _p , \hat {\sigma }^k_p\right )$, we see that, if $y(\delta , S)<y\left (\delta _p , \hat {\sigma }^k_p\right )$, then even in the presence of discontinuities, $\{x: f(x) \leq y(\delta , S), x\in \hat {\sigma }^k_p\} \subseteq \{x: f(x) < y\left (\delta _p , \hat {\sigma }^k_p\right ), x\in \hat {\sigma }^k_p\}$, so we have the following

$$\displaystyle \begin{aligned} &\quad \geq P\left(\left. y(\delta, S) < y\left(\delta_p , \hat{\sigma}^k_p\right) \right| A_k\right), \end{aligned} $$

and by the condition A _k and the pruned assumption, we have $y(\delta , S) \leq y(\delta _{ku}, \widetilde {\varSigma }_k^C) \leq f(z_{(s)}) < f(x_{(p),(1)})$, where x _(p),(1) is the best sample out of $D^P_k N^p_k$ independent samples in $\hat {\sigma }^k_p$, therefore,

$$\displaystyle \begin{aligned} &\quad \geq P\left(\left.f(x_{(p),(1)}) \leq y\left(\delta_p , \hat{\sigma}^k_p\right) \right| A_k\right)\\ &\quad =1-P\left(\left.f(x_{(p),(1)}) > y\left(\delta_p , \hat{\sigma}^k_p\right) \right| A_k\right), \end{aligned} $$

and since each of the ${D^P_k N^p_k}$ independent uniform samples X in ${\hat {\sigma }^k_p}$ satisfies

$$\displaystyle \begin{aligned} P\left(f(X) > y\left(\delta_p , \hat{\sigma}^k_p\right) \right)=1-P\left(f(X) \leq y\left(\delta_p , \hat{\sigma}^k_p\right) \right) \leq 1- \delta_p, \end{aligned} $$

we have

$$\displaystyle \begin{aligned} &\quad \geq 1-\left(1-\delta_p\right)^{D^P_k N^p_k} \end{aligned} $$

(6.57)

which is the same inequality as in (6.54).

As in the special case, since $N^p_k =\left \lceil \frac {\ln {\alpha _k}}{\ln {(1-\frac {\epsilon _k}{v(\sigma _i)}})}\right \rceil$ in Step 4, and $\delta _p=\frac {D^P_k \epsilon _k}{v(\hat {\sigma }_p)}=\frac {D^P_k \epsilon _k}{D^P_k v(\sigma _i)}=\frac {\epsilon _k}{v(\sigma _i)}$, where σ _i is a subregion pruned at the kth iteration, and $D^P_k \geq 1$, we have that

$$\displaystyle \begin{aligned} & P(v(\{x: f(x) \leq y(\delta , S), x\in \hat{\sigma}^k_p\})\leq D^P_k\epsilon_k |A_k) \\ &\quad \geq 1-\alpha_k\end{aligned} $$

(6.58)

which yields the theorem statement in (6.21) in the general case, too. □