**Appendix A. Nomenclature**

In this appendix we describe the notation used in this paper.

If *X* is a random variable, we write ||*X*|| for the *p*-norm (E|*X*|*p*)1/*<sup>p</sup>* of *X*. The conditional *p*-norm of *X* on Z*i*, (E|*X*|*p*|Z*i*)1/*p*, is denoted ||*X*||Z*i* . Here we include a summary of notation used for the proofs contained in this appendix as well as all subsequent appendices.

• Time is indexed by *i* for *ti*, *i* = 0, 1, . . . , *m*.



$$\mathcal{H}\_{\rm il} \left( b, \mathbf{S}\_{i\prime}^{\mathbf{j}} \mathcal{N}\_{i\prime} \mathcal{U}\_{i\prime} u \right) = \frac{1}{b-1} \sum\_{k=1 \atop k \neq l}^{b} D\_{i+1} \mathcal{O}\_{i+1}^{\mathbf{k}} \left( b, \mathbf{S}\_{i+1\prime}^{\mathbf{k}} \mathcal{N}\_{i} - I\_{\{u \neq 0\} \prime} l I\_{i} + u \right),$$
 
$$\mathbf{k} = \{ \mathbf{j}, k \}$$


$$\mathfrak{d}\_{\vec{i}} = \frac{1}{b} \sum\_{l=1}^{b} \mathfrak{d}\_{\vec{i}l} \left( b\_{\prime} \mathbf{S}\_{\vec{i}\prime}^{\dagger} \mathcal{N}\_{\vec{i}\prime} \mathcal{U}\_{\vec{i}\prime} \right)^{\dagger}$$


$$\mathcal{U}\_{l} = \{u\_{0\prime}u\_{1\prime}u\_{2\prime}\ldots, u\_{z} : z \in \mathbb{N}\}\,,$$

where *u*0 = 0.


• *hi* **Sj***i*, N*<sup>i</sup>*, *Ui*, *u*is the time-*ti*, state-Z*i* payoff from exercising *u* units with *hi* **Sj***i*, N*<sup>i</sup>*, *Ui*, 0= 0.

• *Hi* **Sj***i* , N*<sup>i</sup>*, *Ui*is the time-*ti*, state-Z*i* true hold value,

$$H\_{\dot{\mathbf{i}}} \left( \mathbf{S}\_{\dot{\mathbf{i}}\prime}^{\dot{\mathbf{j}}} \mathcal{N}\_{\dot{\mathbf{i}}\prime} \mathcal{U}\_{\dot{\mathbf{i}}} \right) = \mathbb{E} \left[ D\_{\dot{\mathbf{i}}+1} B\_{\dot{\mathbf{i}}+1} (\mathbf{S}\_{\dot{\mathbf{i}}+1\prime}^{\mathbf{k}} \mathcal{N}\_{\dot{\mathbf{i}}+1\prime} \mathcal{U}\_{\dot{\mathbf{i}}+1\prime} \mathcal{U}\_{\dot{\mathbf{i}}+1}) \Big| \mathcal{Z}\_{\dot{\mathbf{i}}} \right]$$

• *Bi*(**Sj***i*, N*<sup>i</sup>*, *Ui*) is the time-*ti*, state-Z*i* true option value,

$$B\_{\dot{i}}(\mathbf{S}\_{i\prime}^{\mathbf{j}}\mathcal{N}\_{\dot{i}\prime}^{\prime}\mathcal{U}\_{\dot{i}}) = \max\_{\boldsymbol{\mu}\in\mathcal{U}} \left[ h\_{\dot{i}}\left(\mathbf{S}\_{i\prime}^{\mathbf{j}}\mathcal{N}\_{\dot{i}\prime}\mathcal{U}\_{\dot{i}\prime}\boldsymbol{\mu}\right) + H\_{\dot{i}}\left(\mathbf{S}\_{i\prime}^{\mathbf{j}}\mathcal{N}\_{\dot{i}} - I\_{\{\boldsymbol{\mu}\neq\mathbf{0}\}}, \mathcal{U}\_{\dot{i}} + \boldsymbol{\mu}\right) \right]$$

where *<sup>I</sup>*{*A*} is the indicator function for set *A*.

### **Appendix B. Proofs of Main Results and Lemmas**

### *Appendix B.1. Proofs of Main Results*

In this section we prove the main results of the paper, Theorems 1–5, presented in Section 2.

**Proof of Theorem 1.** Here we prove the more general statement that <sup>E</sup>[*V*<sup>ˆ</sup>*i*(*b*, **Sj***i* , N*<sup>i</sup>*, *Ui*)|Z*i*] ≥ *Bi*(**Sj***i* , N*<sup>i</sup>*, *Ui*) for *i* = 0, 1, ... , *m*. The proof proceeds by backward induction. At expiry the inequality holds trivially since *V* ˆ *<sup>m</sup>*(*b*, **<sup>S</sup>j***<sup>m</sup>*, N*<sup>m</sup>*, *Um*) = *Bm*(**Sj***<sup>m</sup>*, N*<sup>m</sup>*, *Um*) so that <sup>E</sup>[*V*<sup>ˆ</sup>*m*(*b*, **<sup>S</sup>j***<sup>m</sup>*, N*<sup>m</sup>*, *Um*)|Z*m*] ≥ *Bm*(**Sj***<sup>m</sup>*, N*<sup>m</sup>*, *Um*). We now assume the inductive hypothesis, <sup>E</sup>[*V*<sup>ˆ</sup>*i*+<sup>1</sup>(*b*, **<sup>S</sup>j***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>)|Z*<sup>i</sup>*+<sup>1</sup>] ≥ *Bi*+<sup>1</sup>(**Sj***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>), and proceed to the time-*ti* case. We have,

E -*V*ˆ*i b*, **Sj***i* , N*<sup>i</sup>*, *Ui* |Z*i* = E max *<sup>u</sup>*∈U*i hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>k***i*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* |Z*i* ≥ max *<sup>u</sup>*∈U*i hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + E -*Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>k***i*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* |Z*i* = max *<sup>u</sup>*∈U*i hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + E -*Di*+1E -*<sup>V</sup>*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>k***i*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* |Z*<sup>i</sup>*+<sup>1</sup> |Z*i* ≥ max *<sup>u</sup>*∈U*i hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + E -*Di*+<sup>1</sup>*Bi*+<sup>1</sup> **Sk***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup> |Z*i* = max *<sup>u</sup>*∈U*i hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + *Hi* **Sj***i* , N*<sup>i</sup>*, *Ui* = *Bi* **Sj***i* , N*<sup>i</sup>*, *Ui* ,

where the first equality comes from the definition of the high estimator, the first inequality comes from the conditional Jensen's inequality and note that N*<sup>i</sup>*+<sup>1</sup> = N*i* − *<sup>I</sup>*{*u*∗=<sup>0</sup>} and *Ui*+<sup>1</sup> = *ui* + *u*<sup>∗</sup> where *u*<sup>∗</sup> is the value-maximizing volume choice, the second equality uses the tower law and the fact that *Di*+<sup>1</sup> is Z*i*-measurable, and the second inequality invokes the inductive hypothesis.

**Proof of Theorem 2.** As with the proof of the bias of the high estimator we prove the more general statement that <sup>E</sup>[*v*<sup>ˆ</sup>*i*(*b*, **Sj***i* , N*<sup>i</sup>*, *Ui*)|Z*i*] ≤ *Bi*(**Sj***i* , N*<sup>i</sup>*, *Ui*) for *i* = 0, 1, ... , *m* by backward induction. Again at expiry the inequality holds trivially since *<sup>v</sup>*<sup>ˆ</sup>*m*(*b*, **<sup>S</sup>j***<sup>m</sup>*, N*<sup>m</sup>*, *Um*)= *Bm*(**Sj***<sup>m</sup>*, N*<sup>m</sup>*, *Um*). We now assume the inductive hypothesis, <sup>E</sup>[*v*<sup>ˆ</sup>*i*+<sup>1</sup>(*b*, **<sup>S</sup>j***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>)|Z*<sup>i</sup>*+<sup>1</sup>] ≤ *Bi*+<sup>1</sup>(**Sj***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>). We also note that since the *<sup>v</sup>*<sup>ˆ</sup>*il*'s are iid we have that, <sup>E</sup>[*v*<sup>ˆ</sup>*i*|Z*i*] = <sup>E</sup>[*v*<sup>ˆ</sup>*il*|Z*i*]. In what follows we define *u*ˆ∗*l* ∈ U*i* to be the volume choice which maximizes a particular *<sup>v</sup>*<sup>ˆ</sup>*il*. That is,

$$\mathfrak{u}\_l^\* = \arg\max\_{u \in \mathcal{U}\_l} \left[ \mathfrak{g}\_{il}(b\_\prime \mathbf{S}\_{i\prime}^\dagger \mathcal{N}\_{i\prime} \mathcal{U}\_{i\prime} u) \right]. \tag{A1}$$

Note that *g*<sup>ˆ</sup>*il*(*b*, **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*) is conditionally independent of *<sup>v</sup>*<sup>ˆ</sup>*i*+1,*<sup>l</sup>*(*b*, **<sup>S</sup>l***i*+1, N*<sup>i</sup>*+1, *Ui*+1, *u*) given Z*i* and subsequently *u*ˆ∗*l* is also independent of *<sup>v</sup>*<sup>ˆ</sup>*i*+1,*<sup>l</sup>* given Z*i* since it is a function of *g*<sup>ˆ</sup>*il*.

Now,

E *<sup>v</sup>*<sup>ˆ</sup>*il b*, **Sj***i* , N*<sup>i</sup>*, *Ui* |Z*i* = E -*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*<sup>i</sup>*, *Ui <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =<sup>0</sup>}|Z*i* + E *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*1 + *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *<sup>u</sup>*1 *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*1}|Z*i* + ... + E *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *uz* + *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> i* + 1, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *uz <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*uz*}|Z*i* = E -*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*<sup>i</sup>*, *Ui* |Z*i* P (*u*<sup>ˆ</sup><sup>∗</sup>*l* = <sup>0</sup>|Z*i*) + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*1 P (*u*<sup>ˆ</sup><sup>∗</sup>*l* = *<sup>u</sup>*1|Z*i*) + E -*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *u*1 |Z*i* P (*u*<sup>ˆ</sup><sup>∗</sup>*l* = *<sup>u</sup>*1|Z*i*) + ... + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *uz* P (*u*<sup>ˆ</sup><sup>∗</sup>*l* = *uz*|Z*i*) + E -*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *uz* |Z*i* P (*u*<sup>ˆ</sup><sup>∗</sup>*l* = *uz*|Z*i*) = E -*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*<sup>i</sup>*, *Ui* |Z*i p*0 + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*1 *p*1 + E -*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *u*1 |Z*i p*1 + ... + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *uz pz* + E -*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *uz* |Z*i pz*

where in the second equality we have used the conditional independence of *g*<sup>ˆ</sup>*il* and *<sup>v</sup>*<sup>ˆ</sup>*i*+1,*l*. Here *p*0 = <sup>P</sup>(*u*<sup>ˆ</sup><sup>∗</sup>*l* = <sup>0</sup>|Z*i*) and *pj* = <sup>P</sup>(*u*<sup>ˆ</sup><sup>∗</sup>*l* = *uj*|Z*i*) for 1 ≤ *j* ≤ *z* and *p*0 + ... + *pz* = 1. Thus, using the tower law, we have,

E *v*ˆ*i b*, **Sj***i* , N*<sup>i</sup>*, *Ui* |Z*i* = E *<sup>v</sup>*<sup>ˆ</sup>*il b*, **Sj***i* , N*<sup>i</sup>*, *Ui* |Z*i* = E -*Di*+1E *<sup>v</sup>*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*<sup>i</sup>*, *Ui* |Z*<sup>i</sup>*+<sup>1</sup> |Z*i p*0 + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*1 *p*1 + E -*Di*+1E *<sup>v</sup>*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *u*1 |Z*<sup>i</sup>*+<sup>1</sup> |Z*i p*1 + ... + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *uz pz* + E -*Di*+1E *<sup>v</sup>*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>k***i*+1, N*i* − 1, *Ui* + *uz* |Z*<sup>i</sup>*+<sup>1</sup> |Z*i pz* ≤ E -*Di*+<sup>1</sup>*Bi*+<sup>1</sup> **Sk***i*+1, N*<sup>i</sup>*, *Ui* |Z*i p*0 + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*1 *p*1 + E -*Di*+<sup>1</sup>*Bi*+<sup>1</sup> **Sk***i*+1, N*i* − 1, *Ui* + *u*1 |Z*i p*1 + ... + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *uz pz* + E -*Di*+<sup>1</sup>*Bi*+<sup>1</sup> **Sk***i*+1, N*i* − 1, *Ui* + *uz* |Z*i pz* = *Hi* **Sj***i* , N*<sup>i</sup>*, *Ui p*0 + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*1 + *Hi* **Sj***i* , N*i* − 1, *Ui* + *<sup>u</sup>*1 *p*1 + ... + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *uz* + *Hi* **Sj***i* , N*i* − 1, *Ui* + *uz pz* ≤ max *<sup>u</sup>*∈U*i hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + *Hi* **Sj***i* , N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* = *Bi*(**Sj***i* , N*<sup>i</sup>*, *Ui*)

where the first inequality follows from the inductive hypothesis and the remaining steps follow from the definitions for *Bi* and *Hi*.

**Proof of Theorem 3.** At expiry we have that *<sup>v</sup>*<sup>ˆ</sup>*m*(*b*, **<sup>S</sup>j***<sup>m</sup>*, N*<sup>m</sup>*, *Um*) = *<sup>V</sup>*<sup>ˆ</sup>*m*(*b*, **<sup>S</sup>j***<sup>m</sup>*, N*<sup>m</sup>*, *Um*) = *Bm*(**Sj***<sup>m</sup>*, N*<sup>m</sup>*, *Um*) so the relation holds trivially. We now take the inductive hypothesis to be

*v* ˆ *<sup>i</sup>*+<sup>1</sup>(*b*, **<sup>S</sup>j***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>) ≤ *<sup>V</sup>*<sup>ˆ</sup>*i*+<sup>1</sup>(*b*, **<sup>S</sup>j***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>) for *ji*+<sup>1</sup> = 1, . . . , *b*. Using the *g*<sup>ˆ</sup>*il* as defined above we first consider the case where for a given tree,

$$\mathfrak{M}\_{l}^{\*} = \arg\max\_{u \in \mathcal{U}\_{l}} \left[ \mathfrak{g}\_{il}(\mathcal{b}\_{\prime} \mathbf{S}\_{i\prime}^{\dagger} \mathcal{N}\_{i\prime} \mathcal{U}\_{l\prime} u) \right],\tag{A2}$$

is the same for all *l* (i.e., *u*ˆ∗*l* = *<sup>u</sup>*<sup>ˆ</sup>∗, for all *l*).

> Then,

*v* ˆ *i b*, **Sj***i* , N*<sup>i</sup>*, *Ui* = 1*b b*∑*l*=1 *<sup>v</sup>*<sup>ˆ</sup>*il b*, **Sj***i* , N*<sup>i</sup>*, *Ui* = 1 *b b* ∑ *l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ∗ + *Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*<sup>ˆ</sup>∗=<sup>0</sup>}, *Ui* + *u*<sup>ˆ</sup><sup>∗</sup> ≤ 1 *b b* ∑ *l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ∗ + *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*<sup>ˆ</sup>∗=<sup>0</sup>}, *Ui* + *u*<sup>ˆ</sup><sup>∗</sup> = *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ∗ + 1*b b*∑*l*=1 -*Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*<sup>ˆ</sup>∗=<sup>0</sup>}, *Ui* + *u*<sup>ˆ</sup><sup>∗</sup> ≤ max *<sup>u</sup>*∈U*i hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + 1*b b*∑*l*=1 -*Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* = *V* ˆ *i b*, **Sj***i* , N*<sup>i</sup>*, *Ui*

where the first inequality comes from the inductive hypothesis and the remaining relations come from the parameter definitions.

Next consider the case where the low estimator gives two different estimated optimal exercise amounts, *u*ˆ1, *u*ˆ2, across all *l* branches where *u*ˆ1 = *u*ˆ2. That is *u*ˆ∗*l* = *u*ˆ1 or *u*ˆ∗*l* = *u*ˆ2 for all *l* = 1, . . . , *b*. As above we take *u*ˆ∗*l* to be the optimal exercise amount determined by the *l*-th leave one out estimator, then,

*v* ˆ *i b*, **Sj***i* , N*<sup>i</sup>*, *Ui* = 1*b* ∑*bl*=<sup>1</sup> *<sup>v</sup>*<sup>ˆ</sup>*il b*, **Sj***i* , N*<sup>i</sup>*, *Ui* = 1 *b* ∑*bl*=<sup>1</sup> *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ∗*l* + *Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =<sup>0</sup>}, *Ui* + *u*ˆ∗*l* = 1 *b* ∑*bl*=<sup>1</sup> *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ1 + *Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ1=<sup>0</sup>}, *Ui* + *u*ˆ1 *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ1} + *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ2 + *Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ2=<sup>0</sup>}, *Ui* + *u*ˆ2 *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ2} = 1*b* <sup>∑</sup>*bl*=<sup>1</sup> *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ1} × 1*b* <sup>∑</sup>*bl*=<sup>1</sup>*hi***Sj***i* ,N*<sup>i</sup>*,*Ui*,*u*ˆ1+*Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup>*<sup>b</sup>*,**Sl***i*+1,N*<sup>i</sup>*−*I*{*u*ˆ1=<sup>0</sup>},*Ui*+*u*ˆ1*<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ1} 1 *b* <sup>∑</sup>*bl*=<sup>1</sup> *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ1} + 1*b* <sup>∑</sup>*bl*=<sup>1</sup> *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ2} × 1*b* <sup>∑</sup>*bl*=<sup>1</sup>*hi***Sj***i* ,N*<sup>i</sup>*,*Ui*,*u*ˆ2+*Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup>*<sup>b</sup>*,**Sl***i*+1,N*<sup>i</sup>*−*I*{*u*ˆ2=<sup>0</sup>},*Ui*+*u*ˆ2*<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ2} 1 *b* <sup>∑</sup>*bl*=<sup>1</sup> *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ2} = *p* × 1 *b* <sup>∑</sup>*bl*=<sup>1</sup>*hi***Sj***i* ,N*<sup>i</sup>*,*Ui*,*u*ˆ1+*Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup>*<sup>b</sup>*,**Sl***i*+1,N*<sup>i</sup>*−*I*{*u*ˆ1=<sup>0</sup>},*Ui*+*u*ˆ1*<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ1} 1 *b* <sup>∑</sup>*bl*=<sup>1</sup> *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ1} + (1 − *p*) × 1 *b* <sup>∑</sup>*bl*=<sup>1</sup>*hi***Sj***i* ,N*<sup>i</sup>*,*Ui*,*u*ˆ2+*Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup>*<sup>b</sup>*,**Sl***i*+1,N*<sup>i</sup>*−*I*{*u*ˆ2=<sup>0</sup>},*Ui*+*u*ˆ2*<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ2} 1 *b* <sup>∑</sup>*bl*=<sup>1</sup> *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ2} ,

where *p* = 1*b* ∑*bl*=<sup>1</sup> *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*ˆ1}.

Without loss of generality, suppose that *u*ˆ∗*l* = *u*ˆ1 for *l* = 1, ... , *k* and *u*ˆ∗*l* = *u*ˆ2 for *l* = *k* + 1, ... , *b*. Then the above ratios become

$$\frac{\sum\_{l=1}^{k} \left[ h\_{\hat{i}} \left( \mathbf{S}\_{\hat{i}'}^{\hat{l}} \mathcal{N}\_{\hat{i}'} \mathcal{U}\_{\hat{i}} \mathcal{U}^{\hat{1}} \right) + D\_{\hat{i}+1} \mathfrak{P}\_{\hat{i}+1} \left( b \, \mathcal{S}\_{\hat{i}+1'}^{\mathbf{I}} \mathcal{N}\_{\hat{i}} - I\_{\{\hat{\mathcal{U}}^{\hat{1}} \neq \mathbf{0}\}} \, \mathcal{U}\_{\hat{i}} + \mathfrak{i}^{\mathbf{1}} \right) \right]}{k} \tag{A3}$$

and

$$\frac{\sum\_{l=k+1}^{b} \left[ h\_i \left( \mathbf{S}\_{i'}^{\dagger} \mathcal{N}\_{i'} \, \mathcal{U}\_{i'} \, \hat{\mathbf{n}}^2 \right) + D\_{i+1} \vartheta\_{i+1} \left( b, \mathbf{S}\_{i+1'}^{\dagger} \mathcal{N}\_i - I\_{\{\hat{\mathbf{n}}^2 \neq 0\} \prime} \, \mathcal{U}\_i + \hat{\mathbf{n}}^2 \right) \right]}{b-k} \,\tag{A4}$$

respectively. Now for any *i*∗ ≤ *k* < *j*∗ ≤ *b* we have

$$\left\| \, \S\_{ii^\*} \left( b\_{\prime} \mathbf{S}\_{i^{\prime}}^{\dagger} \mathcal{N}\_{i \prime} \, \mathcal{U}\_{i \prime} \mathcal{U}^{1} \right) \right\| > \left\| \, \S\_{ij^\*} \left( b\_{\prime} \mathbf{S}\_{i \prime}^{\dagger} \mathcal{N}\_{i \prime} \, \mathcal{U}\_{i \prime} \mathcal{U}^{1} \right) \right\| $$

which from the definition of *g*<sup>ˆ</sup>*il*, implies that

$$D\_{i+1} \mathfrak{d}\_{i+1} \left( b\_\prime \mathbf{S}\_{i+1\prime}^{\mathbf{i}^\*} \mathcal{N}\_{i\prime} \mathcal{U}\_i + \mathfrak{u}^1 \right) \le D\_{i+1} \mathfrak{d}\_{i+1} \left( b\_\prime \mathbf{S}\_{i+1\prime}^{\mathbf{i}^\*} \mathcal{N}\_{i\prime} \mathcal{U}\_i + \mathfrak{u}^1 \right).$$

Therefore,

$$\max\_{1 \le a \le k} \left[ D\_{i+1} \mathfrak{d}\_{i+1} \left( b, \mathbf{S}\_{i+1}^{\mathbf{a}}, \mathcal{N}\_{i\prime} \mathcal{U}\_{i\prime} + \mathfrak{d}^{1} \right) \right] \le \min\_{k+1 \le a \le b} \left[ D\_{i+1} \mathfrak{d}\_{i+1} \left( b, \mathbf{S}\_{i+1\prime}^{\mathbf{a}}, \mathcal{N}\_{i\prime} \mathcal{U}\_{i\prime} + \mathfrak{d}^{1} \right) \right].$$

This implies that Equation (A3)

$$\begin{split} &\frac{1}{k} \sum\_{l=1}^{k} \left[ h\_{i} \left( \mathbf{S}\_{i'}^{\dagger} \mathcal{N}\_{i'} \mathcal{U}\_{i}, \hat{\mathfrak{u}}^{1} \right) + D\_{i+1} \mathfrak{v}\_{i+1} \left( b\_{\prime} \mathbf{S}\_{i+1'}^{\dagger} \mathcal{N}\_{i} - I\_{\{\mathfrak{u}^{1} \neq 0\}}, l \mathcal{U}\_{i} + \hat{\mathfrak{u}}^{1} \right) \right] \\ &\leq \frac{1}{b} \sum\_{l=1}^{b} \left[ h\_{\hat{\mathfrak{u}}} \left( \mathbf{S}\_{i'}^{\dagger} \mathcal{N}\_{i'} \mathcal{U}\_{i}, \hat{\mathfrak{u}}^{1} \right) + D\_{i+1} \mathfrak{v}\_{i+1} \left( b\_{\prime} \mathbf{S}\_{i+1'}^{1} \mathcal{N}\_{i} - I\_{\{\mathfrak{u}^{1} \neq 0\}}, l \mathcal{U}\_{i} + \hat{\mathfrak{u}}^{1} \right) \right], \end{split}$$

and similarly for Equation (A4)

$$\begin{split} &\frac{1}{b-k}\sum\_{l=k+1}^{b}\left[h\_{i}\left(\mathbf{S}\_{i'}^{\dagger}\mathcal{N}\_{i'}\mathcal{U}\_{i},\hat{\mu}^{2}\right)+D\_{i+1}\mathfrak{v}\_{i+1}\left(b,\mathbf{S}\_{i+1}^{\mathbf{1}},\mathcal{N}\_{i}-\operatorname{I}\_{\{\hat{\mu}^{2}\neq 0\}},\mathcal{U}\_{i}+\hat{\mu}^{2}\right)\right] \\ &\leq\frac{1}{b}\sum\_{l=1}^{b}\left[h\_{i}\left(\mathbf{S}\_{i'}^{\dagger}\mathcal{N}\_{i'}\mathcal{U}\_{i},\hat{\mu}^{2}\right)+D\_{i+1}\mathfrak{v}\_{i+1}\left(b,\mathbf{S}\_{i+1'}^{\dagger}\mathcal{N}\_{i}-\operatorname{I}\_{\{\hat{\mu}^{2}\neq 0\}},\mathcal{U}\_{i}+\hat{\mu}^{2}\right)\right]. \end{split}$$

Therefore

*v* ˆ *i b*, **Sj***i* , N*<sup>i</sup>*, *Ui* ≤ *p* × 1*b b*∑*l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ1 + *Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ1=<sup>0</sup>}, *Ui* + *u*ˆ1 + (1 − *p*) × 1*b b*∑*l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ2 + *Di*+1*v*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ2=<sup>0</sup>}, *Ui* + *u*ˆ2 ≤ *p* × 1 *b b* ∑ *l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ1 + *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ1=<sup>0</sup>}, *Ui* + *u*ˆ1 + (1 − *p*) × 1*b b*∑*l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ2 + *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ2=<sup>0</sup>}, *Ui* + *u*ˆ2 ≤ max 1*b b*∑*l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ1 + *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ1=<sup>0</sup>}, *Ui* + *u*ˆ1 , 1 *b b* ∑ *l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u*ˆ2 + *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*ˆ2=<sup>0</sup>}, *Ui* + *u*ˆ2 ≤ max *<sup>u</sup>*∈U*i* 1*b b*∑*l*=1 *hi* **Sj***i* , N*<sup>i</sup>*, *Ui*, *u* + *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>l***i*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* = *V* ˆ *<sup>i</sup>*(**Sj***i* , N*<sup>i</sup>*, *Ui*),

where the second inequality comes from the inductive hypothesis, the third inequality is an application of Jensen's inequality, the fourth inequality comes from maximizing over a larger set, and the final equality is the definition of the high-biased estimator.

For the cases where the low estimator gives *z*<sup>∗</sup> distinct estimated optimal exercise amounts, *u* ˆ1 , ... , *u* ˆ*z*<sup>∗</sup> , across all *z* branches, *z*<sup>∗</sup> = 3, ... , *z*, arguments similar to those given above (for 2 distinct estimated optimal exercise amounts) show that,

$$
\left(\boldsymbol{\upsigma}\_{i}(\mathbf{S}\_{i\prime}^{\mathbf{j}}\mathcal{N}\_{i\prime}\mathcal{U}\_{i})\right)\leq\hat{\mathcal{V}}\_{i}(\mathbf{S}\_{i\prime}^{\mathbf{j}}\mathcal{N}\_{i\prime}\mathcal{U}\_{i})\,.
$$

Since we restrict the number of volume choices to be finite, the theorem is proven.

Prior to proving Theorems 4 and 5 we first state the following preliminary result.

**Lemma A1.** *If* ||*hi*(**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *u*)|| < ∞ *for all ti, for some p* ≥ 1*, then the following are true for all* 0 ≤ *ti* ≤ *tk* ≤ *tm:*

$$\|\|B\_k(\mathbf{S}\_{k'}\mathcal{N}\_{k'}\mathcal{U}\_k)\|\|\_{\mathcal{Z}\_i} < \infty \tag{A5}$$

$$\sup\_{b} \left\| \left| \hat{\mathcal{V}}\_{k} (b, \mathbf{S}\_{k}, \mathcal{N}\_{k}, \mathcal{U}\_{k}) \right| \right\|\_{\mathcal{Z}\_{i}} < \infty \tag{A6}$$

$$\sup\_{b} \left\| \left\| \left\| \left\| b\_{k}(b\_{\prime} \mathbf{S}\_{k\prime} \mathcal{N}\_{k\prime} \mathcal{U}\_{k}) \right\| \right\|\_{\mathcal{Z}\_{i}} < \infty \right\| \tag{A7}$$

The proof of this lemma can be found in the appendix. A second preliminary result is

**Lemma A2.** *Let a*1,..., *an*, *b*1,..., *bn*, *c*1,..., *cn be real numbers. Then,*

$$A\_n \equiv \left| \max \left( a\_1 + b\_1, \dots, a\_n + b\_n \right) - \max \left( a\_1 + c\_1, \dots, a\_n + c\_n \right) \right| \le 2 \sum\_{i=1}^{n+1} \left| b\_i - c\_i \right| \equiv B\_n. \tag{A8}$$

Its proof can also be found in the appendix. We now prove Theorems 4 and 5.

**Proof of Theorem 4.** Here we take *R* = 1 and state that if the convergence holds for a single realization of the forest then it will hold for the mean of any number of realizations due to the independence of each repeated valuation. Here we prove by backward induction the more general statement *<sup>V</sup>*<sup>ˆ</sup>*i*(*b*, **S***i*, N*<sup>i</sup>*, *Ui*) − *Bi*(**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*)Z*i* → 0 for any generic node in a given tree and for all *i* = 0, ... , *m*. At expiry the relation holds trivially since at *ti* = *tm* we have that *V* ˆ *<sup>m</sup>*(*b*, **S***<sup>m</sup>*, N*<sup>m</sup>*, *Um*) = *Bm*(**<sup>S</sup>***<sup>m</sup>*, N*<sup>m</sup>*, *Um*). The inductive hypothesis is taken to be *<sup>V</sup>*<sup>ˆ</sup>*i*+<sup>1</sup>(*b*, **S***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>) − *Bi*+<sup>1</sup>(**<sup>S</sup>***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>)Z*<sup>i</sup>*+<sup>1</sup> → 0.

Now,

*V*<sup>ˆ</sup>*i*(*b*, **S***i*, N*<sup>i</sup>*, *Ui*) − *Bi*(**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*)<sup>Z</sup>*i* = max *<sup>u</sup>*∈U*i hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *u*) + 1*b b*∑*j*=1 *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>***ji*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* − max *<sup>u</sup>*∈U*i hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *u*) + *Hi* **S***i*, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *<sup>u</sup>*<sup>Z</sup>*i* = max *<sup>u</sup>*∈U*i hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *u*) + 1*b b*∑*j*=1 *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>***ji*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* − max *<sup>u</sup>*∈U*i hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *u*) + *Hi b*, **S***ji*, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* Z*i* ≤ 2 *z*∑*k*=0 1*b b*∑*j*=1 *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>***ji*+1, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk* − *Hi* **S***i*, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk* Z*i* ≤ 2 *z* ∑ *k*=0 1*b b*∑*j*=1 *Di*+1*V*<sup>ˆ</sup>*i*+<sup>1</sup> *b*, **<sup>S</sup>***ji*+1, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk* − *Hi* **S***i*, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk*Z*i* ≤ 2 *z* ∑ *k*=0 1*b b*∑*j*=1 *Di*+<sup>1</sup> -*<sup>V</sup>*<sup>ˆ</sup>*i*+<sup>1</sup> *i*, **<sup>S</sup>***ji*+1, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk* − *Bi*+<sup>1</sup> **S***ji*+1, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk*<sup>Z</sup>*i* + 2 *z* ∑ *k*=0 *Di*+<sup>1</sup>*Bi*+<sup>1</sup> **S***ji*+1, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk* − *Hi* **S***i*, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk*<sup>Z</sup>*i* = 2 *z* ∑ *k*=0 (*Ek* + *Ck*)

where the first equality comes from the definitions of the estimator and the true value. The third step comes as a result of Lemma A2, the fourth step comes from a generalization of the triangle inequality. In the final step we rewrite the expression for convience in what follows.

First we deal with the *Ck*'s. Given Z*i* we have that *Di*+<sup>1</sup>*Bi*+<sup>1</sup>(**S***ji*+1, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk*) for *j* = 1, ... , *b* and *k* = 0, ... , *z* are iid with means of *Hi*(**<sup>S</sup>***i*, N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>}, *Ui* + *uk*) and finite *p*-norms. Then by Theorem I.4.1 of Gut (1988) we have that all *Ck*'s in the above expression go to zero.

Next we consider the *Ek*'s. Here we have, by the properties of *p*-norms and the fact that the terms being averaged are iid, that

$$E\_k \le \left\| \left\| \dot{\mathcal{V}}\_{i+1} \left( b\_\prime \mathbf{S}\_{i+1}, \mathcal{N}\_i - I\_{\{u\_k \ne 0\}\prime} \, \mathrm{l}I\_i + \mathfrak{u}\_k \right) - B\_{i+1} \left( \mathbf{S}\_{i+1\prime} \mathcal{N}\_i - I\_{\{u\_k \ne 0\}\prime} \, \mathrm{l}I\_i + \mathfrak{u}\_k \right) \right\|\_{\mathcal{Z}\_i}$$

since *Ek* is bounded by the *p*-norm of any one of the terms being averaged. By the inductive hypothesis

$$\left\| \left\| \dot{\mathcal{V}}\_{i+1} \left( b\_{\prime} \mathbf{S}\_{i+1\prime} \mathcal{N}\_{i+1\prime} \mathcal{U}\_{i+1} \right) - B\_{i+1} \left( \mathbf{S}\_{i+1\prime} \mathcal{N}\_{i+1\prime} \mathcal{U}\_{i+1} \right) \right\|\_{\mathcal{Z}\_{i+1}} \to 0\_{\star}$$

where N*<sup>i</sup>*+<sup>1</sup> = N*i* − *<sup>I</sup>*{*uk*=<sup>0</sup>} and *Ui*+<sup>1</sup> = *Ui* + *uk*.

Also by a standard condition for uniform integrability (see Gut (1988) p. 178) we have that

$$\left\| \left\| \hat{V}\_{i+1} \left( b\_{\prime} \mathbf{S}\_{i+1 \prime \prime} \mathcal{N}\_{i+1 \prime} \mathcal{U}\_{i+1} \right) - B\_{i+1} \left( \mathbf{S}\_{i+1 \prime} \mathcal{N}\_{i+1 \prime} \mathcal{U}\_{i+1} \right) \right\|\_{\mathcal{Z}\_i} \to 0,\tag{A9}$$

provided

$$\sup\_{b} \mathbb{E}\left[ \left| \mathcal{V}\_{i+1} \left( b, \mathbf{S}\_{i+1}, \mathcal{N}\_{i+1}, \mathcal{U}\_{i+1} \right) - B\_{i+1} \left( \mathbf{S}\_{i+1}, \mathcal{N}\_{i+1}, \mathcal{U}\_{i+1} \right) \right|^{p+c} \left| \mathcal{Z}\_{i} \right] < \infty$$

for some . From Lemma A1 we know that

$$\sup\_{b} \mathbb{E}\left[ \left| \hat{\mathcal{V}}\_{i+1} \left( b\_{\prime} \mathbf{S}\_{i+1}, \mathcal{N}\_{i+1}, \mathcal{U}\_{i+1} \right) \right|^{p+c} \left| \mathcal{Z}\_{i} \right] < \infty \right]$$

and that

$$\mathbb{E}\left[|B\_{i+1}\left(\mathbf{S}\_{i+1},\mathcal{N}\_{i+1},\mathcal{U}\_{i+1}\right)|^{p+\mathfrak{c}}\mid\mathcal{Z}\_{i}\right] < \infty.$$

Thus (A9) holds for each *k* = 0, . . . , *z* and hence the result is proven.

**Proof of Theorem 5.** As with the proof of Theorem 4 we proceed by backward induction. Again at expiry the relation holds trivially since *<sup>v</sup>*<sup>ˆ</sup>*m*(*b*, **S***<sup>m</sup>*, N*<sup>m</sup>*, *Um*) = *Bm*(**<sup>S</sup>***<sup>m</sup>*, N*<sup>i</sup>*, *Ui*). The inductive hypothesis is taken to be *<sup>v</sup>*<sup>ˆ</sup>*i*+<sup>1</sup>(*b*, **S***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>) − *Bi*+<sup>1</sup>(**<sup>S</sup>***i*+1, N*<sup>i</sup>*+1, *Ui*+<sup>1</sup>)Z*<sup>i</sup>*+<sup>1</sup> → 0.

Let *g*<sup>ˆ</sup>*il*(*b*, **Sj***i*, N*<sup>i</sup>*, *Ui*, *u*) be as defined at the start of Appendix A and note that, with probability one,

$$\begin{split} &h\_{i}\left(\mathbf{S}\_{i'}^{\mathbf{j}}\mathcal{N}\_{i'}\mathcal{U}\_{i'}\boldsymbol{u}^{1}\right) + H\_{i}\left(\mathbf{S}\_{i'}^{\mathbf{j}}\mathcal{N}\_{i} - I\_{\{\boldsymbol{u}^{1}\neq 0\}\prime}\mathcal{U}\_{i} + \boldsymbol{u}^{1}\right) \\ &\neq h\_{i}\left(\mathbf{S}\_{i'}^{\mathbf{j}}\mathcal{N}\_{i'}\mathcal{U}\_{i}\boldsymbol{u}^{2}\right) + H\_{i}\left(\mathbf{S}\_{i'}^{\mathbf{j}}\mathcal{N}\_{i} - I\_{\{\boldsymbol{u}^{2}\neq 0\}\prime}\mathcal{U}\_{i} + \boldsymbol{u}^{2}\right), \end{split}$$

for all *u*1, *u*2 ∈ U*i*, *u*1 = *u*2.

Before proceeding, we stop to make three claims:

$$1. \qquad \left\| \begin{array}{l} 1 \\ \end{array} \right\| \stackrel{1}{b} \sum\_{l=1}^{b} D\_{i+1} \psi\_{i+1} \left( b, \mathbf{S}\_{i+1}^{1}, \mathcal{N}\_{i} - I\_{\{u \neq 0\}}, \mathcal{U}\_{i} + u \right) - H\_{i} \left( \mathbf{S}\_{i'}^{j}, \mathcal{N}\_{i} - I\_{\{u \neq 0\}}, \mathcal{U}\_{i} + u \right) \right\|\_{\mathcal{Z}\_{i}} \to 0$$

$$\text{2.} \qquad \left\| \left\| \left\| \left\| \left( b, \mathbf{S}\_{i'}^{\mathbf{j}}, \mathcal{N}\_{i\boldsymbol{\nu}} \boldsymbol{\Lambda} \mathbf{I}\_{i\boldsymbol{\nu}} \boldsymbol{\Lambda} \right) - \left[ h\_{i} \left( \mathbf{S}\_{i'\boldsymbol{\nu}}^{\mathbf{j}} \mathcal{N}\_{i\boldsymbol{\nu}} \boldsymbol{\Lambda} \mathbf{I}\_{i\boldsymbol{\nu}} \boldsymbol{\Lambda} \right) - H\_{i} \left( \mathbf{S}\_{i'}^{\mathbf{j}} \mathcal{N}\_{i\boldsymbol{\nu}} - I\_{\{u \neq 0\}}, l\boldsymbol{\Lambda}\_{i} + u \right) \right] \right\| \right\|\_{\mathcal{Z}\_{i}} \to 0$$

$$\mathcal{B}. \qquad \left\| I\_{\{\mathbb{A}\_l^\* = \mathsf{u}\}} - I\_{\{\mathsf{u}^\* = \mathsf{u}\}} \right\|\Big|\_{\mathcal{Z}\_i} \to 0;$$

for all *u* ∈ U*i* and where

$$\begin{split} \mathcal{U}\_{l}^{\*} &= \arg\max\_{u \in \mathcal{U}\_{l}} \left[ \mathcal{G}\_{il}(\boldsymbol{b}, \mathbf{S}\_{\boldsymbol{i}'}^{\mathrm{j}}, \mathcal{N}\_{i'} \boldsymbol{\mathsf{U}}\_{i} \boldsymbol{u}) \right] \quad \text{and} \\ \boldsymbol{u}^{\*} &= \arg\max\_{u \in \mathcal{U}\_{l}} \left[ \boldsymbol{h}\_{l} \left( \mathbf{S}\_{\boldsymbol{i}'}^{\mathrm{j}}, \mathcal{N}\_{i'} \boldsymbol{\mathsf{U}}\_{i'} \boldsymbol{u} \right) - \boldsymbol{H}\_{i} \left( \mathbf{S}\_{\boldsymbol{i}'}^{\mathrm{j}}, \mathcal{N}\_{i} - \boldsymbol{I}\_{\{u \neq 0\}}, \mathcal{U}\_{i} + \boldsymbol{u} \right) \right] \end{split}$$

The proof of item (i) is the same as the proof of the corresponding step in Theorem 4. Since the estimators in (i) and (ii) differ only in the omission of one term in *g*<sup>ˆ</sup>*il*, similar arguments prove that (ii) also holds.

Now for (iii), if *u*ˆ∗*l* = *u*<sup>∗</sup> then the result holds trivially. Now suppose that *u*ˆ∗*l* = *u* = *u*<sup>∗</sup> for some *u* ∈ U*i*. Then,

$$\begin{split} & \left\| \left\| I\_{\{\boldsymbol{\uptheta}\_{l}^{\*} = \boldsymbol{u} \}} - I\_{\{\boldsymbol{u}^{\*} = \boldsymbol{u} \}} \right\| \right\|\_{\mathcal{Z}\_{i}} = \left\| I\_{\{\boldsymbol{d}\_{l}^{\*} = \boldsymbol{u} \}} \right\|\_{\mathcal{Z}\_{i}} = \left[ \mathbf{P} \left( \hat{\boldsymbol{u}}\_{l}^{\*} = \boldsymbol{u} \,|\, \mathcal{Z}\_{i} \right) \right]^{\frac{1}{p}} \\ & = \left[ \mathbf{P} \left( \mathcal{S}\_{il} (\boldsymbol{b}, \mathbf{S}\_{i'}^{\dagger} \mathcal{N}\_{i'} \mathcal{U}\_{i} \mathcal{U}\_{i} \boldsymbol{u}) \geq h\_{i} (\mathbf{S}\_{i'}^{\dagger} \mathcal{N}\_{i'} \mathcal{U}\_{i} \boldsymbol{u}) + H\_{i} (\mathbf{S}\_{i'}^{\dagger} \mathcal{N}\_{i} - I\_{\{\boldsymbol{u} \neq 0\}} \mathcal{U}\_{i} + \mathbf{u}) \right) \right]^{1/p} \\ & \to 0. \end{split} $$

Since (ii) holds and convergence in *p*-norm implies convergence in probability. Thus (iii) is proven. Now proceeding from the definition of the low estimator and the true option value for all *u* ∈ U*i*,

*<sup>v</sup>*<sup>ˆ</sup>*i* (*b*, **S***i*, N*<sup>i</sup>*, *Ui*) − *Bi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*)Z*i* = 1*b* ∑*bl*=<sup>1</sup> *<sup>v</sup>*<sup>ˆ</sup>*il* (*b*, **S***i*, N*<sup>i</sup>*, *Ui*) − *Bi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*)<sup>Z</sup>*i* = 1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*<sup>i</sup>*, *Ui <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =<sup>0</sup>} + *ahi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *<sup>u</sup>*1) + *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − 1, *Ui* + *<sup>u</sup>*1 *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*1} + ... + *ahi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *uz*) <sup>+</sup>*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − 1, *Ui* + *uz <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*uz*} − *Bi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*) Z*i* ≤ 1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*<sup>i</sup>*, *Ui <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =<sup>0</sup>} − *Hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*) *<sup>I</sup>*{*u*∗=<sup>0</sup>}<sup>Z</sup>*i* + 1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − 1, *Ui* + *u*1 *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*1} − *Hi* (**<sup>S</sup>***i*, N*i* − 1, *Ui* + *<sup>u</sup>*1) *<sup>I</sup>*{*u*∗=*u*1}<sup>Z</sup>*i* + ... + 1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − 1, *Ui* + *uz <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*uz*} − *Hi* (**<sup>S</sup>***i*, N*i* − 1, *Ui* + *uz*) *<sup>I</sup>*{*u*∗=*uz*}<sup>Z</sup>*i* + *hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *<sup>u</sup>*1) *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*u*1} − *hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *<sup>u</sup>*1) *<sup>I</sup>*{*u*∗=*uz*}<sup>Z</sup>*i* + ... + *hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *uz*) *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*uz*} − *hi* (**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *uz*) *<sup>I</sup>*{*u*∗=*uz*}<sup>Z</sup>*i*(A10)

where the inequality in the third step is due to a generalization of the triangle inequality.

The immediate consequence of claim (iii) above is that all terms in Equation (A10) with the form

$$\left\| h\_i(\mathbf{S}\_{i\prime}\mathcal{N}\_{i\prime}\mathcal{U}\_{i\prime}\mathbf{u})I\_{\{\mathcal{U}\_l^\*=\mathbf{u}\}} - h\_i(\mathbf{S}\_{i\prime}\mathcal{N}\_{i\prime}\mathcal{U}\_{i\prime}\mathbf{u})I\_{\{\mathbf{u}^\*=\mathbf{u}\}} \right\|\_{\mathcal{Z}\_i} \to 0$$

for all *u* ∈ U*i*. Thus

$$\sum\_{k=0}^{z} \left\| h\_i(\mathbf{S}\_{i\prime}, \mathcal{N}\_{i\prime} \mathcal{U}\_{i\prime} \boldsymbol{u}\_k) \boldsymbol{I}\_{\{\mathcal{U}\_l^\* = \boldsymbol{u}\_k\}} - h\_i(\mathbf{S}\_{i\prime} \mathcal{N}\_{i\prime} \mathcal{U}\_{i\prime} \boldsymbol{u}\_k) \boldsymbol{I}\_{\{\boldsymbol{u}^\* = \boldsymbol{u}\_k\}} \right\|\_{\mathcal{Z}\_i} \to 0.$$

It remains to show that the remaining terms in (A10) converge in the *p*-norm. Taking one of these terms, that is, fix a *u* ∈ U*i*, we now show this converges in the *p*-norm to zero.

1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*<sup>ˆ</sup>∗=*<sup>u</sup>*} − *Hi* **S***i*, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*<sup>∗</sup>*l* <sup>=</sup>*<sup>u</sup>*}<sup>Z</sup>*i* = 1*b* ∑*bl*=<sup>1</sup> -*Di*+1*v*<sup>ˆ</sup>*i*+*i*,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*<sup>u</sup>*} − [*Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*∗=*<sup>u</sup>*} + 1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*∗=*<sup>u</sup>*} − *Hi* **S***i*, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*∗=*<sup>u</sup>*}<sup>Z</sup>*i* ≤ 1*b* ∑*bl*=<sup>1</sup> -*Di*+1*v*<sup>ˆ</sup>*i*+*i*,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*<sup>u</sup>*} − *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*∗=*<sup>u</sup>*} + 1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*∗=*<sup>u</sup>*} − *Hi* **S***i*, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u <sup>I</sup>*{*u*∗=*<sup>u</sup>*}<sup>Z</sup>*i* ≤ *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *<sup>u</sup>*<sup>Z</sup>*i* · *<sup>I</sup>*{*u*<sup>ˆ</sup><sup>∗</sup>*l* =*<sup>u</sup>*} − *<sup>I</sup>*{*u*∗=*<sup>u</sup>*}<sup>Z</sup>*i* + *<sup>I</sup>*{*u*∗=*<sup>u</sup>*}<sup>Z</sup>*i* · 1*b* ∑*bl*=<sup>1</sup> *Di*+1*v*<sup>ˆ</sup>*i*+1,*<sup>l</sup> b*, **<sup>S</sup>***li*+1, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* − *Hi* **S***i*, N*i* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Ui* + *u* Z*i* ,

where the first step comes from adding and subtracting the same term, the second comes from applying the triangle inequality and the third step comes from factoring out common terms.

Now by (iii),

$$\left\| \left| I\_{\{\hat{\mu}^\*\_l = u\}} - I\_{\{\hat{\mu}^\* = u\}} \right| \right\|\_{\mathcal{Z}\_i} \to 0\_{\prime\prime} $$

by (i),

$$\left\| \frac{1}{b} \sum\_{l=1}^{b} D\_{i+1} \hat{v}\_{i+1,l} \left( b\_\star \mathbf{S}\_{i+1}^l, \mathcal{N}\_i - I\_{\{u \neq 0\}}, lI\_i + u \right) - H\_i \left( \mathbf{S}\_{i\prime} \mathcal{N}\_i - I\_{\{u \neq 0\}}, lI\_i + u \right) \right\|\_{\mathcal{Z}\_i} \to 0, \quad i \to \infty$$

and we note that

$$\begin{aligned} \left\| \left| I\_{\{\boldsymbol{u}^\* = \boldsymbol{u}\}} \right| \right\|\_{\mathcal{Z}\_i} &< \infty \quad \text{and} \\ \left\| D\_{i+1} \hat{v}\_{i+1,l} \left( b\_{\prime} \mathbf{S}\_{i+1\prime}^l \mathcal{N}\_i - I\_{\{\boldsymbol{u} \neq 0\}}, lI\_i + \boldsymbol{u} \right) \right\|\_{\mathcal{Z}\_i} &< \infty \end{aligned}$$

by (A7).

> Hence we have proven the consistency of the low-biased estimator.

*Appendix B.2. Lemma Proofs*

**Proof of Lemma A1.** If every *hi*(**<sup>S</sup>***i*, N*<sup>i</sup>*, *Ui*, *u*) has finite *p*-th moment, then each ||*hi*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*, *<sup>u</sup>*)||Z*i* is finite. Since the max, discounting, and conditional expectation operators preserve finiteness of moments then it follows that ||*Bk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*)||Z*i*and also ||*Hk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*)||Z*i*must also be finite.

Proceeding to (A6), fix *ti* and proceed by backward induction on *tk* from *tm* to *ti*. At expiry (A6) follows from (A5). Then for *tk* < *tm*,

sup *b V*<sup>ˆ</sup>*k*(*b*, **S***k*, N*<sup>k</sup>*, *Uk*)<sup>Z</sup>*i* = sup*b* max *<sup>u</sup>*∈U*k hk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*, *u*) + *H*ˆ *<sup>k</sup>*(*b*, **S***k*, N*k* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Uk* + *<sup>u</sup>*)<sup>Z</sup>*i* = sup *b* max *<sup>u</sup>*∈U*k*,*<sup>u</sup> hk*(**S***k*, N*<sup>k</sup>*, *Uk*, *u*) + 1*b b*∑*j*=1 *Dk*+1*V*<sup>ˆ</sup>*k*+<sup>1</sup>(*b*, **<sup>S</sup>***jk*+1, N*k* − *<sup>I</sup>*{*u*=<sup>0</sup>}, *Uk* + *u*) Z*i* ≤ *hk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*, <sup>0</sup>)Z*i* + sup*b* 1*b b*∑*j*=1 *Dk*+1*V*<sup>ˆ</sup>*k*+<sup>1</sup>(*b*, **<sup>S</sup>***jk*+1, N*<sup>k</sup>*, *Uk*)Z*i* + *hk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*, *<sup>u</sup>*1)Z*i* + sup*b* 1*b b*∑*j*=1 *Dk*+1*V*<sup>ˆ</sup>*k*+<sup>1</sup>(*b*, **<sup>S</sup>***jk*+1, N*k* − 1, *Uk* + *u*1)Z*i* + ... + *hk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*, *ur*)Z*i* + sup*b* 1*b b*∑*j*=1 *Dk*+1*V*<sup>ˆ</sup>*k*+<sup>1</sup>(*b*, **<sup>S</sup>***jk*+1, N*k* − 1, *Uk* + *ur*)Z*i* ≤ sup *b V*<sup>ˆ</sup>*k*+<sup>1</sup>(*b*, **S***k*+1, N*<sup>k</sup>*, *Uk*)<sup>Z</sup>*i* + *hk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*, *<sup>u</sup>*1)Z*i* + sup*b V*<sup>ˆ</sup>*k*+<sup>1</sup>(*b*, **S***k*+1, N*k* − 1, *Uk* + *<sup>u</sup>*1)<sup>Z</sup>*i* + ... + *hk*(**<sup>S</sup>***k*+1, N*<sup>k</sup>*, *Uk*, *ur*)Z*i* + sup*b V*<sup>ˆ</sup>*k*+<sup>1</sup>(*b*, **S***k*+1, N*k* − 1, *Uk* + *ur*)<sup>Z</sup>*i* ,

where *hk*(**<sup>S</sup>***k*, N*<sup>k</sup>*, *Uk*, 0) = 0. This is the sum of a finite number of terms, each of which is finite. For (iii) the proof is similar to that of (ii).

**Proof of Lemma A2.** In order to prove Lemma A2 we proceed by induction by considering the cases for *n* = 1 and *n* = 2. For *n* = 1,

$$A\_1 = |\max(a\_1 + b\_1) - \max(a\_1 + c\_1)| = |b\_1 - c\_1| < B\_1$$

therefore *A*1 ≤ *B*1. Now, for *n* = 2

$$A\_1 = |\max(a\_1 + b\_1, a\_2 + b\_2) - \max(a\_1 + c\_1, a\_2 + c\_2)|$$

Consider the following,

	- (a) *a*1 + *c*1 > *a*2 + *c*2 Then

*A*2 = |*a*1 + *b*1 − *a*1 − *c*1| = |*b*1 − *c*1| ≤ *B*2

> (b) *a*1 + *c*1 < *a*2 + *c*2 Note that conditions (i) and (b) imply that

$$b\_2 - b\_1 < a\_1 - a\_2 < c\_2 - c\_1 \tag{A11}$$

and we have

$$\begin{aligned} A\_2 &= |a\_1 + b\_1 - a\_2 - c\_2| \\ &= |a\_1 + b\_1 - c\_1 + c\_1 - a\_2 - c\_2| \\ &\le |b\_1 - c\_1| + |(a\_1 - a\_2) - (c\_2 - c\_1)| \\ &\le |b\_1 - c\_1| + |(b\_2 - b\_1) - (c\_2 - c\_1)| \\ &= |b\_1 - c\_1| + |b\_2 - c\_2 - (b\_1 - c\_1)| \\ &\le 2|b\_1 - c\_1| + |b\_2 - c\_2| \\ &\le 2|b\_1 - c\_1| + 2|b\_2 - c\_2| \\ &= B\_2 \end{aligned}$$

where the first inequality comes from the triangle inequality, the second comes from Inequality (A11) and the third inequality comes from another application of the triangle inequality.

$$2. \qquad a\_2 + b\_2 > a\_1 + b\_1$$

(a) *a*2 + *c*2 > *a*1 + *c*1 Then

$$A\_2 = |a\_2 + b\_2 - a\_2 - c\_2| = |b\_2 - c\_2| \le B\_2$$

(b) *a*2 + *c*2 < *a*1 + *c*1 Note that conditions (ii) and (b) imply that

$$b\_1 - b\_2 < a\_2 - a\_1 < c\_1 - c\_2 \tag{A12}$$

and we have

$$\begin{aligned} A\_2 &= |a\_2 + b\_2 - a\_1 - c\_1| \\ &= |a\_2 + b\_2 - c\_2 + c\_2 - a\_1 - c\_1| \\ &\le |b\_2 - c\_2| + |(a\_2 - a\_1) - (c\_1 - c\_2)| \\ &\le |b\_2 - c\_2| + |(b\_1 - b\_2) - (c\_1 - c\_2)| \\ &= |b\_2 - c\_2| + |b\_1 - c\_1 - (b\_2 - c\_2)| \\ &\le 2|b\_2 - c\_2| + |b\_1 - c\_1| \\ &\le 2|b\_2 - c\_2| + 2|b\_1 - c\_1| \\ &= B\_2 \end{aligned}$$

where again the first inequality comes from the triangle inequality, the second comes from Inequality (A12) and the third inequality comes from another application of the triangle inequality.

Therefore *A*2 ≤ *B*2.

Now assume that the inductive hypothesis *An* ≤ *Bn* is true. We need to show that *An*+<sup>1</sup> ≤ *Bn*+1. First define *in* and *jn* such that

$$a\_{i\_n} + b\_{i\_n} = \max(a\_1 + b\_{1'} \dots \iota\_n + b\_n)$$

and

$$a\_{i\_n} + c\_{i\_n} = \max(a\_1 + c\_{1'}, \dots, a\_n + c\_n)$$

respectively. Now,

$$\begin{aligned} A\_{n+1} &= \left| \max(a\_1 + b\_1, \dots, a\_n + b\_n, a\_{n+1} + b\_{n+1}) - \max(a\_1 + c\_1, \dots, a\_n + c\_n, a\_{n+1} + c\_{n+1}) \right| \\ &= \left| \max(a\_{i\_n} + b\_{i\_n}, a\_{n+1} + b\_{n+1}) - \max(a\_{j\_n} + c\_{j\_n}, a\_{n+1} + c\_{n+1}) \right| \end{aligned}$$

 Consider the following,

1. *ain* + *bin* > *an*+1 + *bn*+<sup>1</sup>

$$\mathbf{(a)}\qquad a\_{\bar{\jmath}\_n} + c\_{\bar{\jmath}\_n} > a\_{n+1} + c\_{n+1}$$

$$\begin{aligned} A\_{n+1} &= |a\_{i\_n} + b\_{i\_n} - a\_{j\_n} - c\_{j\_n}| \\ &\le 2 \sum\_{i=1}^n |b\_i - c\_i| \\ &\le 2 \sum\_{i=1}^{n+1} |b\_i - c\_i| \\ &= B\_{n+1} \end{aligned}$$

where the first inequality comes from the inductive hypothesis.

(b) *ajn* + *cjn* < *an*+1 + *cn*+1 By the definitions of *in* and *jn* and (b) we have

$$a\_{i\_n} + c\_{i\_n} \le a\_{j\_n} + c\_{j\_n} < a\_{n+1} + c\_{n+1} \dots$$

This combined with (i) gives

$$b\_{n+1} - b\_{i\_n} < a\_{i\_n} - a\_{n+1} < c\_{n+1} - c\_{i\_n} \tag{A13}$$

Then

$$\begin{split} A\_{n+1} &= |a\_{i\_n} + b\_{i\_n} - a\_{n+1} - c\_{n+1}| \\ &= |a\_{i\_n} + b\_{i\_n} - c\_{i\_n} + c\_{i\_n} - a\_{n+1} - c\_{n+1}| \\ &\leq |b\_{i\_n} - c\_{i\_n}| + |(a\_{i\_n} - a\_{n+1}) - (c\_{n+1} - c\_{i\_n})| \\ &\leq |b\_{i\_n} - c\_{i\_n}| + |(b\_{n+1} - b\_{i\_n}) - (c\_{n+1} - c\_{i\_n})| \\ &= |b\_{i\_n} - c\_{i\_n}| + |b\_{n+1} - c\_{n+1} - (b\_{i\_n} - c\_{i\_n})| \\ &\leq 2|b\_{i\_n} - c\_{i\_n}| + |b\_{n+1} - c\_{n+1}| \\ &\leq B\_{n+1} \end{split}$$

where again the first inequality comes from the triangle inequality, the second comes from Inequality (A13) and the third inequality comes from another application of the triangle inequality.

$$2. \qquad a\_{i\_n} + b\_{i\_n} < a\_{n+1} + b\_{n+1}$$

> (a) *ajn* + *cjn* < *an*+1 + *cn*+1

$$\begin{aligned} A\_{n+1} &= |a\_{n+1} + b\_{n+1} - a\_{n+1} - c\_{n+1}| \\ &= |b\_{n+1} - c\_{n+1}| \\ &\le 2 \sum\_{i=1}^{n+1} |b\_i - c\_i| \\ &= B\_{n+1} \end{aligned}$$

(b) *ajn* + *cjn* > *an*+1 + *cn*+1 By the definitions of *in* and *jn* and (ii) we have

$$a\_{j\_n} + b\_{j\_n} \le a\_{i\_n} + b\_{i\_n} < a\_{n+1} + b\_{n+1} \dots$$

This combined with (b) gives

$$b\_{j\_n} - b\_{n+1} < a\_{n+1} - a\_{j\_n} < c\_{j\_n} - c\_{n+1} \tag{A14}$$

Then

$$\begin{split} A\_{n+1} &= |a\_{n+1} + b\_{n+1} - a\_{j\_n} - c\_{j\_n}| \\ &= |a\_{n+1} + b\_{n+1} - c\_{n+1} + c\_{n+1} - a\_{j\_n} - c\_{j\_n}| \\ &\leq |b\_{n+1} - c\_{n+1}| + |(a\_{n+1} - a\_{j\_n}) - (c\_{j\_n} - c\_{n+1})| \\ &\leq |b\_{n+1} - c\_{n+1}| + |(b\_{j\_n} - b\_{n+1}) - (c\_{j\_n} - c\_{n+1})| \\ &= |b\_{n+1} - c\_{n+1}| + |b\_{j\_n} - c\_{j\_n} - (b\_{n+1} - c\_{n+1})| \\ &\leq 2|b\_{n+1} - c\_{n+1}| + |b\_{j\_n} - c\_{j\_n}| \\ &\leq B\_{n+1} \end{split}$$

where again the first inequality comes from the triangle inequality, the second comes from Inequality (A14) and the third inequality comes from another application of the triangle inequality.

Therefore *An*+<sup>1</sup> ≤ *Bn*+<sup>1</sup> and the Lemma is proven.
