Optimal Structure of Experiential Services: Review and Extensions

Abstract

In many consumer-intensive (B2C) services, delivering memorable customer experiences is often a source of competitive advantage. And yet, there exist few formal guidelines to design the structure of such experiences. In this chapter, we introduce a utility-based model of customer satisfaction when customers are subject to acclimation, satiation, and memory decay. We then review and extend principles for optimizing the structure of an experience to maximize customer satisfaction; specifically, we characterize the optimal sequence of activities, the optimal activity selection, and the optimal information policy about an uncertain outcome. We find that, in general, the optimal experience structure is non-monotone in service levels and makes use of breaks/intermissions to create contrasts and reset satiation levels. However, in many extreme cases, we show that a crescendo design is optimal. We then discuss the implications of our framework for quality management in services, especially as it relates to a potential gap between ex-ante expectation and ex-post satisfaction, and for monetizing customers’ utilities derived while anticipating or recalling the event.

Appendix

Lemma 1:

For any p, q ≥ 1 and x ∈ [0, 1], 1 − x^−p − x^−q + x^−p − q ≥ 0.

Proof:

The derivative of the function 1 − x^−p − x^−q + x^−p − q with respect to p equals x^−p ln x (1 − x^−q) ≥ 0, and similarly for the derivative with respect to q. Hence, for any p, q ≥ 1, 1 − x^−p − x^−q + x^−p − q ≥ 1 − x⁻¹ − x⁻¹ + x⁻² = (1 − x⁻¹)² ≥ 0.∎

Lemma 2:

For any p, q ≥ 1 and x ∈ [0, 1], 1 − x^−p − x^−q + x^−p − q is decreasing in x.

Proof:

The derivative of the function 1 − x^−p − x^−q + x^−p − q with respect to x equals $\left(p+q\right)\ {x}^{-p-q-1}\left(\frac{p}{p+q}\ {x}^q+\frac{q}{p+q}\ {x}^p-1\right)$, and it is negative given that x^q ≤ 1 and x^p ≤ 1.∎

Proof of Proposition 1:

The proof uses an interchange argument. Throughout the proof, since K = 1, we omit the subscript k. Because α = γ = 0, $S\left(\boldsymbol{x},{r}_1,{s}_1\right)={\sum}_{\tau =1}^T{\delta}^{T-\tau}\ v\left({x}_{\tau}-{r}_1\right)-v\left({s}_1\right)$. Suppose that, in the optimal sequence, Activity i starts in time period t and immediately precedes Activity j. In that case, because x_τ = x_i whenever π_τ = i, we obtain

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{r}_1,{s}_1\right)=\sum \limits_{\tau =1}^{t-1}\ {\delta}^{T-\tau}\ v\left({x}_{\tau}-{r}_1\right)+\sum \limits_{\tau =t}^{t+{d}_i-1}\ {\delta}^{T-\tau}\ v\left({x}_i-{r}_1\right)\\ {}+\sum \limits_{\tau =t+{d}_i}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau}\ v\left({x}_j-{r}_1\right)\\ {}+\sum \limits_{\tau =t+{d}_i+{d}_j}^T\ {\delta}^{T-\tau}\ v\left({x}_{\tau}-{r}_1\right)-v\left({s}_1\right).\end{array}} $$

Consider a suboptimal sequence $\tilde{\boldsymbol{x}}$ in which Activities i and j have been permuted. Because x is optimal, we must have that $S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\ge 0$, i.e.,

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\\ {}\kern2em =\sum \limits_{\tau =t}^{t+{d}_i-1}\ {\delta}^{T-\tau}\ v\left({x}_i-{r}_1\right)+\sum \limits_{\tau =t+{d}_i}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau}\ v\left({x}_j-{r}_1\right)\\ {}\kern2em -\sum \limits_{\tau =t}^{t+{d}_j-1}\ {\delta}^{T-\tau}\ v\left({x}_j-{r}_1\right)-\sum \limits_{\tau =t+{d}_j}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau}\ v\left({x}_i-{r}_1\right)\\ {}\kern2em =v\left({x}_i-{r}_1\right)\ \left(\sum \limits_{\tau =t}^{t+{d}_i-1}\ {\delta}^{T-\tau}-\sum \limits_{\tau =t+{d}_j}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau}\right)\\ {}\kern2em +v\left({x}_j-{r}_1\right)\left(\sum \limits_{\tau =t+{d}_i}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau}-\sum \limits_{\tau =t}^{t+{d}_j-1}\ {\delta}^{T-\tau}\right)\\ {}\kern2em =\left(v\left({x}_i-{r}_1\right)\right.\\ {}\kern2em \left.-v\left({x}_j-{r}_1\right)\right){\delta}^{T-t}\frac{1-{\delta}^{-{d}_i}-{\delta}^{-{d}_j}+{\delta}^{-{d}_i-{d}_j}}{1-{\delta}^{-1}}\ge 0,\end{array}} $$

which implies, by Lemma 1, that v(x_i − r₁) ≤ v(x_j − r₁). Because v(x) is increasing, this implies that x_i ≤ x_j. ∎

Proof of Proposition 2:

The proof uses an interchange argument. Throughout the proof, since K = 1, we omit the subscript k. Without loss of generality, we set r₁ = 0. Because α = 0, $S\left(\boldsymbol{x},{r}_1,{s}_1\right)={\sum}_{\tau =1}^T{\delta}^{T-\tau}\ \left(v\left({x}_{\tau}+{s}_{\tau}\right)-v\left({s}_{\tau}\right)\right)$. Suppose that, in the optimal sequence, Activity i starts in time period t and immediately precedes Activity j. In that case, because x_τ = x_i whenever π_τ = i, we obtain

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{r}_1,{s}_1\right)=\sum \limits_{\tau =1}^{t-1}\ {\delta}^{T-\tau}\ \left(v\left({x}_{\tau}+{s}_{\tau}\right)-v\left({s}_{\tau}\right)\right)\\ {}+\sum \limits_{\tau =t}^{t+{d}_i-1}\ {\delta}^{T-\tau}\ \Big(v\left(\left(\tau -t+1\right)\ {x}_i+{s}_t\right)\\ {}-v\left(\left(\tau -t\right)\ {x}_i+{s}_t\right)\Big)\\ {}+\sum \limits_{\tau =t+{d}_i}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau}\Big(v\left(\left(\tau -t-{d}_i+1\right)\ {x}_j+{d}_i\ {x}_i+{s}_t\right)\\ {}-v\left(\left(\tau -t-{d}_i\right)\ {x}_j+{d}_i\ {x}_i+{s}_t\right)\Big)\\ {}+\sum \limits_{\tau =t+{d}_i+{d}_j}^T\ {\delta}^{T-\tau}\ \left(v\left({x}_{\tau}+{s}_{\tau}\right)-v\left({s}_{\tau}\right)\right)\\ {}=\sum \limits_{\tau =1}^{t-1}\ {\delta}^{T-\tau}\ \left(v\left({x}_{\tau}+{s}_{\tau}\right)-v\left({s}_{\tau}\right)\right)-v\left({s}_t\right)\ {\delta}^{T-t}\\ {}+\sum \limits_{\tau =t}^{t+{d}_i-1}\ {\delta}^{T-\tau -1}\ v\left(\left(\tau -t+1\right)\ {x}_i+{s}_t\right)\left(\delta -1\right)\\ {}+\sum \limits_{\tau =t+{d}_i}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau -1}\ v\Big(\left(\tau -t-{d}_i+1\right)\ {x}_j+{d}_i\ {x}_i\\ {}+{s}_t\Big)\left(\delta -1\right)+{\delta}^{T-t-{d}_i-{d}_j}\ v\left({d}_j\ {x}_j+{d}_i\ {x}_i+{s}_t\right)\\ {}+\sum \limits_{\tau =t+{d}_i+{d}_j+1}^T\ {\delta}^{T-\tau}\ \left(v\left({x}_{\tau}+{s}_{\tau}\right)-v\left({s}_{\tau}\right)\right).\end{array}} $$

Consider a suboptimal sequence $\tilde{\boldsymbol{x}}$ in which Activities i and j have been permuted. Because x is optimal, we must have that $S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\ge 0$. Because the terms associated with activities scheduled before t or after t + d_i + d_j + 1 are identical across both expressions, we must thus have that

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\\ {}\kern2em =\sum \limits_{\tau =t}^{t+{d}_i-1}{\delta}^{T-\tau -1}\ v\left(\left(\tau -t+1\right)\ {x}_i+{s}_t\right)\left(\delta -1\right)\\ {}\kern1em =\sum \limits_{\tau =t+{d}_i}^{t+{d}_i+{d}_j-1}{\delta}^{T-\tau -1}\ v\Big(\left(\tau -t-{d}_i+1\right)\ {x}_j+{d}_i\ {x}_i\\ {}\kern1em +{s}_t\Big)\left(\delta -1\right)\\ {}\kern1em -\sum \limits_{\tau =t}^{t+{d}_j-1}{\delta}^{T-\tau -1}\ v\left(\left(\tau -t+1\right)\ {x}_j+{s}_t\right)\left(\delta -1\right)\\ {}\kern1em -\sum \limits_{\tau =t+{d}_j}^{t+{d}_i+{d}_j-1}\ {\delta}^{T-\tau -1}\ v\Big(\left(\tau -t-{d}_j+1\right)\ {x}_i+{d}_j\ {x}_j\\ {}\kern1em +{s}_t\Big)\left(\delta -1\right)\ge 0,\end{array}} $$

which implies, since v(x) is increasing, that x_i ≤ x_j. ∎

Proof of Proposition 3:

The proof uses an interchange argument. Throughout the proof, since K = 1, we omit the subscript k. Because δ = 1 and γ = 0, $S\left(\boldsymbol{x},{r}_1,{s}_1\right)={\sum}_{\tau =1}^Tv\left({x}_{\tau}-{r}_{\tau}\right)-v\left({s}_1\right)$. Suppose first that Activities i and j are not the last ones. Specifically, suppose that, in the optimal sequence, Activity i starts in time period t and immediately precedes Activity j, and that Activity j precedes Activity l. In that case, because x_τ = x_i whenever π_τ = i, we obtain

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{r}_1,{s}_1\right)=\sum \limits_{\tau =1}^{t-1}\ v\left({x}_{\tau}-{r}_{\tau}\right)+v\left({x}_i-{r}_t\right)+v\left({x}_j-{x}_i\right)+v\left({x}_l-{x}_j\right)\\ {}\kern1em +\sum \limits_{\tau =t+{d}_i+{d}_j+{d}_l}^T\ v\left({x}_{\tau}-{r}_{\tau}\right)-v\left({s}_1\right).\end{array}} $$

Consider a suboptimal sequence $\tilde{\boldsymbol{x}}$ in which Activities i and j have been permuted. Because x is optimal, we must have that $S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\ge 0$. Because the terms associated with activities scheduled before t or after t + d_i + d_j + d_l are identical across both expressions, we obtain that

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\\ {}\kern1em =v\left({x}_i-{r}_t\right)+v\left({x}_j-{x}_i\right)+v\left({x}_l-{x}_j\right)-v\left({x}_j-{r}_t\right)\\ {}\kern1em -v\left({x}_i-{x}_j\right)-v\left({x}_l-{x}_i\right)\ge 0.\end{array}} $$

We next show that this inequality holds if only if x_i ≤ x_j.Suppose first that x_i ≤ x_j. Because v(x) is subadditive, v(x_j − r_t) ≤ v(x_j − x_i) + v(x_i − r_t). Similarly, v(x_l − x_i) ≤ v(x_l − x_j) + v(x_j − x_i). Moreover, because of loss aversion, when x_i ≤ x_j, v(x_j − x_i) ≤  − v(x_i − x_j). Combining these inequalities yields the desired inequality. Conversely, suppose that x_i > x_j. Because v(x) is subadditive, v(x_i − r_t) ≤ v(x_i − x_j) + v(x_j − r_t). Similarly, v(x_l − x_j) ≤ v(x_l − x_i) + v(x_i − x_j). Moreover, because of loss aversion, when.x_i > x_j, v(x_j − x_i) <  − v(x_i − x_j). Combining these inequalities yields the opposite inequality. Hence, $S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\ge 0$implies that x_i ≤ x_j.

Next, suppose that Activities i and j are the last ones. Then,

$$ S\left(\boldsymbol{x},{r}_1,{s}_1\right)=\sum \limits_{\tau =1}^{t-1}\ v\left({x}_{\tau}-{r}_{\tau}\right)+v\left({x}_i-{r}_t\right)+v\left({x}_j-{x}_i\right)-v\left({s}_1\right). $$

Consider a suboptimal sequence $\tilde{\boldsymbol{x}}$ in which Activities i and j have been permuted. Because x is optimal, we must have that

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\\ {}\kern1em =v\left({x}_i-{r}_t\right)+v\left({x}_j-{x}_i\right)-v\left({x}_j-{r}_t\right)-v\left({x}_i-{x}_j\right)\\ {}\kern1em \ge 0.\end{array}} $$

Similar to the argument above for the case where Activities i and j are not the last ones, we can show that this inequality holds if and only if x_i ≤ x_j. Hence, $S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\ge 0$implies that x_i ≤ x_j.∎

Proof of Proposition 4:

The proof uses an interchange argument. Throughout the proof, since K = 1, we omit the subscript k. Because δ = γ = 1, S(x, r₁, s₁) = v(x_T − r_T + s_T) − v(s₁). Suppose first that Activities i and j are not the last ones. Specifically, suppose that, in the optimal sequence, Activity i starts in time period t and immediately precedes Activity j, and that Activity j precedes Activity l. In that case, because x_τ = x_i whenever π_τ = i, we obtain

$$ {\displaystyle \begin{array}{c}{x}_T-{r}_T+{s}_T={x}_T-{r}_t+{s}_t+\sum \limits_{\tau =t}^{T-1}{\left(1-\alpha \right)}^{T-\tau}\ \left({x}_i-{r}_t\right)\\ {}+\sum \limits_{\tau =t+{d}_i}^{T-1}{\left(1-\alpha \right)}^{T-\tau}\ \left({x}_j-{x}_i\right)\\ {}+\sum \limits_{\tau =t+{d}_i+{d}_j}^{T-1}{\left(1-\alpha \right)}^{T-\tau}\ \left({x}_l-{x}_j\right)+\dots.\end{array}} $$

Consider a suboptimal sequence $\tilde{\boldsymbol{x}}$ in which Activities i and j have been permuted. Because x is optimal, we must have that $S\left(\boldsymbol{x},{r}_1,{s}_1\right)-S\left(\tilde{\boldsymbol{x}},{r}_1,{s}_1\right)\ge 0$. Because the function v(x) is increasing, this implies that ${x}_T-{r}_T+{s}_T\ge {\tilde{x}}_T-{\tilde{r}}_T+{\tilde{s}}_T$, in which ${\tilde{r}}_T$ and ${\tilde{s}}_T$ are the reference point and satiation level in period T corresponding to sequence $\tilde{\boldsymbol{x}}$. Because the terms associated with activities scheduled before t or after t + d_i + d_j + d_l are identical across both expressions, as well as r_t and x_l, we must thus have that

$$ {\displaystyle \begin{array}{c}\left({x}_T-{r}_T+{s}_T\right)-\left({\tilde{x}}_T-{\tilde{r}}_T+{\tilde{s}}_T\right)\\ {}=\left({x}_i-{x}_j\right)\\ {}\times \Big(\sum \limits_{\tau =t}^{T-1}{\left(1-\alpha \right)}^{T-\tau}-\sum \limits_{\tau =t+{d}_i}^{T-1}{\left(1-\alpha \right)}^{T-\tau}\\ {}-\sum \limits_{\tau =t+{d}_j}^{T-1}{\left(1-\alpha \right)}^{T-\tau}+\sum \limits_{\tau =t+{d}_i+{d}_j}^{T-1}{\left(1-\alpha \right)}^{T-\tau}\Big)\ge 0.\end{array}} $$

After expanding the series, we obtain that

$$ {\displaystyle \begin{array}{c}\left({x}_i-{x}_j\right)\times \frac{{\left(1-\alpha \right)}^{T-t+1}}{\alpha}\\ {}\kern2em \times \left(-1+{\left(1-\alpha \right)}^{-{d}_i}+{\left(1-\alpha \right)}^{-{d}_j}-{\left(1-\alpha \right)}^{-{d}_i-{d}_j}\right)\ge 0\end{array}} $$

Using Lemma 1, we obtain that the second term in parentheses is always negative, which implies that x_i ≤ x_j.

When Activities i and j are the last two activities, we obtain, using a similar logic, that

$$ {\displaystyle \begin{array}{l}\left({x}_i-{x}_j\right)\times \left(\sum \limits_{\tau =t}^{T-1}{\left(1-\alpha \right)}^{T-\tau}-\sum \limits_{\tau =t+{d}_i}^{T-1}{\left(1-\alpha \right)}^{T-\tau}-\sum \limits_{\tau =t+{d}_j}^{T-1}{\left(1-\alpha \right)}^{T-\tau}\right)\\ {}\kern5em \ge 0.\end{array}} $$

After expanding the series using the fact that T − t = d_i + d_j, we obtain

$$ {\displaystyle \begin{array}{c}\left({x}_i-{x}_j\right)\times \frac{{\left(1-\alpha \right)}^{d_i+{d}_j+1}}{\alpha}\\ {}\kern2em \times \left(-1+{\left(1-\alpha \right)}^{-{d}_i}+{\left(1-\alpha \right)}^{-{d}_j}-{\left(1-\alpha \right)}^{-{d}_i-{d}_j}\right)\ge 0\end{array}} $$

Using Lemma 1, we obtain that the second term in parentheses is always negative, which implies that x_i ≤ x_j. ∎

Proof of Proposition 5:

The proof uses an interchange argument. Because v_k(x) = w_k x for all k, $S\left(\boldsymbol{x},{\boldsymbol{r}}_1,{\boldsymbol{s}}_1\right)={\sum}_{\tau =1}^T{\sum}_{k=1}^K{w}_k\ {\delta}^{T-\tau}\ \left({x}_{k,\tau}-{r}_{k,\tau}\right)$, i.e., the terms in s_τ cancel each other. Suppose first that Activities i and j are not the last ones. Specifically, suppose that, in the optimal sequence, Activity i starts in time period t and immediately precedes Activity j, and that Activity j precedes Activity l. In that case, because x_{k, τ} = x_{k, i} whenever π_τ = i, we obtain

$$ {\displaystyle \begin{array}{ll}S\left(\boldsymbol{x},{\boldsymbol{r}}_1,{\boldsymbol{s}}_1\right)=& \sum \limits_{\tau =1}^{t-1}\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}\ \left({x}_{k,\tau}-{r}_{k,\tau}\right)\kern1em +\sum \limits_{\tau =t}^T\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t}\ \left({x}_{k,i}-{r}_{k,t}\right)\kern1em \\ {}& +\sum \limits_{\tau =t+{d}_i}^T\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_i}\ \left({x}_{k,j}-{x}_{k,i}\right)\\ {}& +\sum \limits_{\tau =t+{d}_i+{d}_j}^T\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_i-{d}_j}\ \left({x}_{k,l}-{x}_{k,j}\right)\\ {}& +\dots.\end{array}} $$

Consider a suboptimal sequence $\tilde{\boldsymbol{x}}$ in which Activities i and j have been permuted. Because x is optimal, we must have that $S\left(\boldsymbol{x},{\boldsymbol{r}}_1,{\boldsymbol{s}}_1\right)-S\left(\tilde{\boldsymbol{x}},{\boldsymbol{r}}_1,{\boldsymbol{s}}_1\right)\ge 0$. Because the terms associated with activities scheduled before t or after t + d_i + d_j + d_l are identical across both expressions, as well as r_{k, t} and x_{k, l}, we thus obtain that

$$ {\displaystyle \begin{array}{c}S\left(\boldsymbol{x},{\boldsymbol{r}}_1,{\boldsymbol{s}}_1\right)-S\left(\tilde{\boldsymbol{x}},{\boldsymbol{r}}_1,{\boldsymbol{s}}_1\right)\\ {}\kern2em =\sum \limits_{\tau =t}^T\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t}\ \left({x}_{k,i}-{x}_{k,j}\right)\\ {}\kern2em +\sum \limits_{\tau =t+{d}_i}^T\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_i}\ \left({x}_{k,j}-{x}_{k,i}\right)\\ {}\kern2em +\sum \limits_{\tau =t+{d}_j}^T\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_j}\ \left({x}_{k,j}-{x}_{k,i}\right)\\ {}\kern2em +\sum \limits_{\tau =t+{d}_i+{d}_j}^T\sum \limits_{k=1}^K{w}_k\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_i-{d}_j}\ \left({x}_{k,i}-{x}_{k,j}\right)\\ {}\kern2em \ge 0.\end{array}} $$

Equivalently,

$$ {\displaystyle \begin{array}{l}\left(\sum \limits_{k=1}^K{w}_k\ \left({x}_{k,i}-{x}_{k,j}\right)\right)\\ {}\kern3em \times \Big(\sum \limits_{\tau =t}^T{\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t}-\sum \limits_{\tau =t+{d}_i}^T{\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_i}\\ {}\kern3em -\sum \limits_{\tau =t+{d}_j}^T\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_j}\\ {}\kern3em +\sum \limits_{\tau =t+{d}_i+{d}_j}^T{\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_i-{d}_j}\Big)\ge 0.\end{array}} $$

After expanding the series, we obtain

$$ {\displaystyle \begin{array}{l}\left(\sum \limits_{k=1}^K{w}_k\ \left({x}_{k,i}-{x}_{k,j}\right)\right)\times \frac{1}{\delta -\left(1-\alpha \right)}\\ {}\kern3em \times \left(-{\left(1-\alpha \right)}^{T-t+1}+{\left(1-\alpha \right)}^{T-t-{d}_i+1}\right)\\ {}\kern3em +{\left(1-\alpha \right)}^{T-t-{d}_j+1}-{\left(1-\alpha \right)}^{T-t-{d}_i-{d}_j+1}+{\delta}^{T-t+1}\\ {}\kern3em -{\delta}^{T-t-{d}_i+1}-{\delta}^{T-t-{d}_j+1}+{\delta}^{T-t-{d}_i-{d}_j+1}\Big)\ge 0.\end{array}} $$

Using Lemmas 1 and 2, that the last two terms are nonnegative if and only if $t\le T+1-\frac{\ln \left(\frac{1-{\delta}^{-{d}_i}-{\delta}^{-{d}_j}+{\delta}^{-{d}_i-{d}_j}}{1-{\left(1-\alpha \right)}^{-{d}_i}-{\left(1-\alpha \right)}^{-{d}_j}+{\left(1-\alpha \right)}^{-{d}_i-{d}_j}}\right)}{\ln \left(\frac{1-\alpha}{\delta}\right)}$.

When Activities i and j are the last two activities, we obtain, using a similar logic, that

$$ {\displaystyle \begin{array}{l}\left(\sum \limits_{k=1}^K{w}_k\ \left({x}_{k,i}-{x}_{k,j}\right)\right)\\ {}\kern3em \times \Big(\sum \limits_{\tau =t}^T{\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t}-\sum \limits_{\tau =t+{d}_i}^T{\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_i}\\ {}\kern3em -\sum \limits_{\tau =t+{d}_j}^T\ {\delta}^{T-\tau}{\left(1-\alpha \right)}^{\tau -t-{d}_j}\Big)\ge 0.\end{array}} $$

After expanding the series, it can be checked that the second term in parentheses is nonnegative if and only if $t\ge T+1-\frac{\ln \left(\frac{1-{\delta}^{-{d}_i}-{\delta}^{-{d}_j}}{1-{\left(1-\alpha \right)}^{-{d}_i}-{\left(1-\alpha \right)}^{-{d}_j}}\right)}{\ln \left(\frac{1-\alpha}{\delta}\right)}$. Because, for any p, q ≥ 1, the function 1 − x^−p − x^−q is increasing in x, $1-{\delta}^{-{d}_i}-{\delta}^{-{d}_j}\ge 1-{\left(1-\alpha \right)}^{-{d}_i}-{\left(1-\alpha \right)}^{-{d}_j}$ if and only if δ ≥ 1 − α. Hence, the term $\frac{\ln \left(\frac{1-{\delta}^{-{d}_i}-{\delta}^{-{d}_j}}{1-{\left(1-\alpha \right)}^{-{d}_i}-{\left(1-\alpha \right)}^{-{d}_j}}\right)}{\ln \left(\frac{1-\alpha}{\delta}\right)}$ is always negative, and therefore, $t<T+1-\frac{\ln \left(\frac{1-{\delta}^{-{d}_i}-{\delta}^{-{d}_j}}{1-{\left(1-\alpha \right)}^{-{d}_i}-{\left(1-\alpha \right)}^{-{d}_j}}\right)}{\ln \left(\frac{1-\alpha}{\delta}\right)}$ for all Activities i and j. As a result, we must have that ${\sum}_{k=1}^K{w}_k\ \left({x}_{k,i}-{x}_{k,j}\right)\le 0$ if Activities i and j are the last ones and if Activity i precedes Activity j in the optimal sequence.∎

Proof of Proposition 6:

We first show by induction that ${x}_{T-k}={r}_{T-k}+{\left({v}^{\hbox{'}}\right)}^{-1}\left(\frac{\lambda \left(1+ k\alpha \right)}{\delta^k}\right)$. Consider the Lagrangean function $L\left(\left({x}_1,\dots, {x}_T\right),{r}_1,0\right)=S\left(\left({x}_1,\dots, {x}_T\right),{r}_1,0\right)-\lambda\ \left(\sum \limits_{t=1}^T{x}_t-B\right)$. Because the second-order optimality condition associated with x_T, i.e., $\frac{\partial^2L\left(\left({x}_1,\dots, {x}_T\right),{r}_1,0\right)}{\partial {x}_T^2}={v}^{{\prime\prime}}\left({x}_T-{r}_T\right)<0$, is always satisfied by concavity of v(x), every stationary point defines a global maximum. Because x_T = r_T + (v^’)⁻¹(λ) satisfies the first-order optimality condition associated with x_T, i.e., $\frac{\partial L\left(\left({x}_1,\dots, {x}_T\right),{r}_1,0\right)}{\partial {x}_T}={v}^{\prime}\left({x}_T-{r}_T\right)-\lambda =0$, it is optimal to set ${x}_T^{\ast}={r}_T+{\left({v}^{\hbox{'}}\right)}^{-1}\left(\lambda \right)$.

Fix k > 0 and suppose that it is optimal to set ${x}_{T-l}^{\ast}={r}_{T-l}+{\left({v}^{\hbox{'}}\right)}^{-1}\left(\frac{\lambda \left(1+ l\alpha \right)}{\delta^l}\right)$ for l = 0, …, k − 1. Using the induction hypothesis, we obtain that

$$ {\displaystyle \begin{array}{l}\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-k},{x}_{T-k+1}^{\ast},\dots, {x}_T^{\ast}\right),{r}_1,0\right)}{\partial {x}_{T-k}}\\ {}\kern5em ={\delta}^k\ {v}^{\prime}\left({x}_{T-k}-{r}_{T-k}\right)\\ {}\kern5em -\alpha \sum \limits_{l=0}^{k-1}{\delta}^l\ {\left(1-\alpha \right)}^{k-1-l}{v}^{\prime}\left({x}_{T-l}^{\ast}-{r}_{T-l}\right)-\lambda \\ {}\kern5em ={\delta}^k\ {v}^{\prime}\left({x}_{T-k}-{r}_{T-k}\right)-\alpha \sum \limits_{l=0}^{k-1}{\left(1-\alpha \right)}^{k-1-l}\left(\lambda \left(1+l\ \alpha \right)\right)\\ {}\kern5em -\lambda ={\delta}^k\ {v}^{\prime}\left({x}_{T-k}-{r}_{T-k}\right)-\lambda \left(1+k\ \alpha \right),\end{array}} $$

because

$$ {\displaystyle \begin{array}{l}\sum \limits_{l=0}^{k-1}{\left(1-\alpha \right)}^{-l}\left(1+l\ \alpha \right)\\ {}\kern5em =\frac{1-{\left(1-\alpha \right)}^{-k}}{1-{\left(1-\alpha \right)}^{-1}}\\ {}\kern5em +\alpha {\left(1-\alpha \right)}^{-1}\frac{1-k\ {\left(1-\alpha \right)}^{-\left(k-1\right)}+\left(k-1\right){\left(1-\alpha \right)}^{-k}}{{\left(1-{\left(1-\alpha \right)}^{-1}\right)}^2}\\ {}\kern5em =\frac{1-\alpha}{\alpha}\Big(-1+{\left(1-\alpha \right)}^{-k}+1-k\ {\left(1-\alpha \right)}^{-\left(k-1\right)}\\ {}\kern5em +\left(k-1\right){\left(1-\alpha \right)}^{-k}\Big)\\ {}\kern5em =\frac{1-\alpha}{\alpha}\ k\ {\left(1-\alpha \right)}^{-k}\ \left(-\left(1-\alpha \right)+1\ \right)=k\ {\left(1-\alpha \right)}^{1-k}.\end{array}} $$

Because $\frac{\partial}{\partial {x}_{T-k}}\left(\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-k},{x}_{T-k+1}^{\ast},\dots, {x}_T^{\ast}\right),{r}_1,0\right)}{\partial {x}_{T-k}}\right)={\delta}^k\ {v}^{{\prime\prime}}\left({x}_{T-k}-{r}_{T-k}\right)<0$ and because ${x}_{T-k}={r}_{T-k}+{\left({v}^{\hbox{'}}\right)}^{-1}\left(\frac{\lambda \left(1+ k\alpha \right)}{\delta^k}\right)$ solves $\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-k},{x}_{T-k+1}^{\ast},\dots, {x}_T^{\ast}\right),{r}_1,0\right)}{\partial {x}_{T-k}}=0$, it is optimal to set ${x}_{T-k}^{\ast}={r}_{T-k}+{\left({v}^{\hbox{'}}\right)}^{-1}\left(\frac{\lambda \left(1+ k\alpha \right)}{\delta^k}\right)$. This completes the induction step.

Because v^″(x) < 0, ${x}_{T-k}^{\ast}-{r}_{T-k}={\left({v}^{\hbox{'}}\right)}^{-1}\left(\frac{\lambda \left(1+ k\alpha \right)}{\delta^k}\right)$ is decreasing in k.

Finally, suppose that v(x) = x^β for some 0 < β < 1 when x ≥ 0. In that case, ${x}_{T-k}^{\ast}\ge {x}_{T-k-1}^{\ast}$ if and only if ${r}_{T-k}+{\left(\frac{\lambda \left(1+ k\alpha \right)}{\beta\ {\delta}^k}\right)}^{\frac{1}{\beta -1}}\ge {x}_{T-k-1}^{\ast}$, i.e., if and only if ${\left(\frac{\lambda \left(1+ k\alpha \right)}{\beta\ {\delta}^k}\right)}^{\frac{1}{\beta -1}}\ge \left(1-\alpha \right)\ {\left(\frac{\lambda \left(1+\left(k+1\right)\alpha \right)}{\beta\ {\delta}^{k+1}}\right)}^{\frac{1}{\beta -1}}$, i.e., if and only if $\frac{\left(\frac{\lambda \left(1+ k\alpha \right)}{\beta\ {\delta}^k}\right)}{\left(\frac{\lambda \left(1+\left(k+1\right)\alpha \right)}{\beta\ {\delta}^{k+1}}\right)}\le {\left(1-\alpha \right)}^{\beta -1}$, i.e., if and only if $\left(\frac{\delta\ \left(1+ k\alpha \right)}{\left(1+\left(k+1\right)\alpha \right)}\right)\le {\left(1-\alpha \right)}^{\beta -1}$. The left-hand side is increasing in k, whereas the right-hand side is constant, so there is at most one crossing. Because the left-hand side is equal to $\left(\frac{\delta}{\left(1+\alpha \right)}\right)$ when k = 0 and to δ when k → ∞, and that both values are smaller than (1 − α)^β − 1, we conclude that ${x}_{T-k}^{\ast}\ge {x}_{T-k-1}^{\ast}$ for all k.∎

Proof of Proposition 7:

Without loss of generality, we set r₁ = 0. Consider the Lagrangean function $L\left(\left({x}_1,\dots, {x}_T\right),0,{s}_1\right)=S\left(\left({x}_1,\dots, {x}_T\right),0,{s}_1\right)-\lambda\ \left(\sum \limits_{t=1}^T{x}_t-B\right)$. Because the second-order optimality condition associated with x_T, i.e., $\frac{\partial^2L\left(\left({x}_1,\dots, {x}_T\right),0,{s}_1\right)}{\partial {x}_T^2}={v}^{{\prime\prime}}\left({x}_T+{s}_T\right)<0$, is always satisfied by concavity of v(x), every stationary point defines a global maximum. Because x_T =  − s_T + (v^’)⁻¹(λ) satisfies the first-order optimality condition associated with x_T, i.e., $\frac{\partial L\left(\left({x}_1,\dots, {x}_T\right),0,{s}_1\right)}{\partial {x}_T}={v}^{\prime}\left({x}_T+{s}_T\right)-\lambda =0$, it is optimal to set ${x}_T^{\ast}=-{s}_T+{\left({v}^{\hbox{'}}\right)}^{-1}\left(\lambda \right)$.

We next show by induction that ${v}^{\prime}\left({x}_{T-k}^{\ast}+{s}_{T-k}\right)-\left(\gamma /\delta \right)\ {v}^{\hbox{'}}\left(\gamma\ \left({x}_{T-k}^{\ast}+{s}_{T-k}\right)\right)=\left(1-\gamma \right)\lambda /{\delta}^k$ for all k = 1, …, T − 1. Because

$$ {\displaystyle \begin{array}{l}\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-1},{x}_T^{\ast}\right),0,{s}_1\right)}{\partial {x}_{T-1}}\\ {}\kern5em =\delta\ {v}^{\prime}\left({x}_{T-1}+{s}_{T-1}\right)+\gamma\ \left({v}^{\prime}\left({x}_T^{\ast}+{s}_T\right)-{v}^{\hbox{'}}\left({s}_T\right)\right)-\lambda \\ {}\kern5em =\delta\ {v}^{\prime}\left({x}_{T-1}+{s}_{T-1}\right)+\gamma\ \left(\lambda -{v}^{\prime}\left(\gamma\ \left({x}_{T-1}+{s}_{T-1}\right)\right)\right)\\ {}\kern5em -\lambda, \end{array}} $$

we obtain that

$$ {\displaystyle \begin{array}{l}\frac{\partial}{\partial {x}_{T-1}}\left(\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-1},{x}_T^{\ast}\right),0,{s}_1\right)}{\partial {x}_{T-1}}\right)\\ {}\kern6em =\delta\ {v}^{{\prime\prime}}\left({x}_{T-1}+{s}_{T-1}\right)-{\gamma}^2\ {v}^{{\prime\prime}}\left(\gamma\ \left({x}_{T-1}+{s}_{T-1}\right)\right)<0,\end{array}} $$

by assumption, and it is thus optimal to set ${x}_{T-1}^{\ast}$ such that $\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-1},{x}_T^{\ast}\right),0,{s}_1\right)}{\partial {x}_{T-1}}=\delta\ {v}^{\prime}\left({x}_{T-1}+{s}_{T-1}\right)+\gamma\ \left(\lambda -{v}^{\prime}\left(\gamma\ \left({x}_{T-1}+{s}_{T-1}\right)\right)\right)-\lambda =0$.

Fix k > 0 and suppose that it is optimal to set ${x}_{T-l}^{\ast}$ such that ${v}^{\prime}\left({x}_{T-l}^{\ast}+{s}_{T-l}\right)-\left(\gamma /\delta \right)\ {v}^{\hbox{'}}\left(\gamma\ \left({x}_{T-l}^{\ast}+{s}_{T-l}\right)\right)=\left(1-\gamma \right)\lambda /{\delta}^l$ for all l = 1, …, k. Using the induction hypothesis, we obtain that

$$ {\displaystyle \begin{array}{l}\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-k},{x}_{T-k+1}^{\ast},\dots, {x}_T^{\ast}\right),{r}_1,{s}_1\right)}{\partial {x}_{T-k}}\\ {}\kern5em ={\delta}^k\ {v}^{\prime}\left({x}_{T-k}+{s}_{T-k}\right)\\ {}\kern5em +\sum \limits_{l=0}^{k-1}{\delta}^l\ {\gamma}^{k-l}\ \left({v}^{\prime}\left({x}_{T-l}^{\ast}+{s}_{T-l}\right)-{v}^{\hbox{'}}\left({s}_{T-l}\right)\right)-\lambda \\ {}\kern5em ={\delta}^k\ {v}^{\prime}\left({x}_{T-k}+{s}_{T-k}\right)-{\delta}^{k-1}\ \gamma\ {v}^{\prime}\left({s}_{T-k+1}\right)\\ {}\kern5em +\sum \limits_{l=1}^{k-1}{\delta}^l\ {\gamma}^{k-l}\ \left({v}^{\prime}\left({x}_{T-l}^{\ast}+{s}_{T-l}\right)-\left(\frac{\gamma}{\delta}\right)\ {v}^{\prime}\left({s}_{T-l+1}\right)\right)\\ {}\kern5em +{\gamma}^k{v}^{\prime}\left({x}_T^{\ast}+{s}_T\right)-\lambda \\ {}\kern5em ={\delta}^k\ {v}^{\prime}\left({x}_{T-k}+{s}_{T-k}\right)-{\delta}^{k-1}\ \gamma\ {v}^{\prime}\left({s}_{T-k+1}\right)\\ {}\kern5em +\sum \limits_{l=1}^{k-1}\ {\gamma}^{k-l}\lambda\ \left(1-\gamma \right)+{\gamma}^k\lambda -\lambda \\ {}\kern5em ={\delta}^k\ {v}^{\prime}\left({x}_{T-k}+{s}_{T-k}\right)-{\delta}^{k-1}\ \gamma\ {v}^{\prime}\left({s}_{T-k+1}\right)+\gamma\ \lambda -\lambda.\end{array}} $$

Hence,

$$ {\displaystyle \begin{array}{l}\frac{\partial}{\partial {x}_{T-k}}\left(\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-k},{x}_{T-k+1}^{\ast},\dots, {x}_T^{\ast}\right),{r}_1,{s}_1\right)}{\partial {x}_{T-1}}\right)\\ {}\kern7em ={\delta}^k\ {v}^{{\prime\prime}}\left({x}_{T-k}+{s}_{T-k}\right)-{\delta}^{k-1}\ {\gamma}^2\ v{\hbox{'}}^{\hbox{'}}\left(\gamma\ \left({x}_{T-k}+{s}_{T-k}\right)\right)\\ {}\kern7em <0,\end{array}} $$

and it therefore is optimal to set ${x}_{T-k}^{\ast}$ such that $\frac{\partial L\left(\left({x}_1,\dots, {x}_{T-k},{x}_{T-k+1}^{\ast},\dots, {x}_T^{\ast}\right),{r}_1,{s}_1\right)}{\partial {x}_{T-k}}=0$. This completes the induction step.

Because ${v}^{\prime}\left({x}_{T-k}^{\ast}+{s}_{T-k}\right)-\left(\frac{\gamma}{\delta}\right)\ {v}^{\hbox{'}}\left(\gamma\ \left({x}_{T-k}^{\ast}+{s}_{T-k}\right)\right)=\frac{\left(1-\gamma \right)\lambda}{\delta^k}<\frac{\left(1-\gamma \right)\lambda}{\delta^{k+1}}={v}^{\prime}\left({x}_{T-k-1}^{\ast}+{s}_{T-k-1}\right)-\left(\frac{\gamma}{\delta}\right)\ {v}^{\hbox{'}}\left(\gamma\ \left({x}_{T-k-1}^{\ast}+{s}_{T-k-1}\right)\right)$ and because the function ${v}^{\prime}(x)-\left(\frac{\gamma}{\delta}\right)\ {v}^{\hbox{'}}\left(\gamma\ x\right)$ is decreasing by assumption, we obtain that ${x}_{T-k}^{\ast}+{s}_{T-k}\ge {x}_{T-k-1}^{\ast}+{s}_{T-k-1}$ for any k = 1, …, T − 1. Moreover, because $\left(1-\gamma \right){v}^{\prime}\left({x}_T^{\ast}+{s}_T\right)=\left(1-\gamma \right)\lambda =\delta\ {v}^{\hbox{'}}\left({x}_{T-1}^{\ast}+{s}_{T-1}\right)-\gamma\ {v}^{\hbox{'}}\left(\gamma\ \left({x}_{T-1}^{\ast}+{s}_{T-1}\right)\right)<\left(\delta -\gamma \right)\ {v}^{\hbox{'}}\left({x}_{T-1}^{\ast}+{s}_{T-1}\right)\le \left(1-\gamma \right)\ {v}^{\hbox{'}}\left({x}_{T-1}^{\ast}+{s}_{T-1}\right)$, since v^’(x) > 0, v^’’(x) > 0, and δ ≤ 1; therefore, ${x}_T^{\ast}+{s}_T>{x}_{T-1}^{\ast}+{s}_{T-1}$.

Finally, suppose that v(x) = x^β for some lnδ/ ln γ < β < 1 when x ≥ 0. In that case, ${x}_{T-k}^{\ast}=-{s}_{T-k}+{\left(\frac{\delta^{k-1}\ \beta\ \left(\delta -{\gamma}^{\beta}\right)}{\lambda\ \left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}$ for k = 1, …, T − 1 and ${x}_T^{\ast}=-{s}_T+{\left(\frac{\beta}{\lambda}\right)}^{\frac{1}{1-\beta}}$. Suppose that T ≥ 3. Then, ${x}_T^{\ast}\ge {x}_{T-1}^{\ast}$ if and only if ${\left(\frac{\beta}{\lambda}\right)}^{\frac{1}{1-\beta}}\ge {x}_{T-1}^{\ast}+{s}_T={x}_{T-1}^{\ast}+\gamma \left({x}_{T-1}^{\ast}+{s}_{T-1}\right)=\left(1+\gamma \right)\left({x}_{T-1}^{\ast}+{s}_{T-1}\right)-{s}_{T-1}=\left(1+\gamma \right)\left({x}_{T-1}^{\ast}+{s}_{T-1}\right)-\gamma \left({x}_{T-2}^{\ast}+{s}_{T-2}\right)=\left(1+\gamma \right){\left(\frac{\beta\ \left(\delta -{\gamma}^{\beta}\right)}{\lambda\ \left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}-\gamma {\left(\frac{\delta\ \beta\ \left(\delta -{\gamma}^{\beta}\right)}{\lambda\ \left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}$. Hence, ${x}_T^{\ast}\ge {x}_{T-1}^{\ast}$ if and only if $1\ge \left(\left(1+\gamma \right)-{\gamma \delta}^{\frac{1}{1-\beta}}\right){\left(\frac{\left(\delta -{\gamma}^{\beta}\right)}{\left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}$, which is always true since $\left(\left(1+\gamma \right)-{\gamma \delta}^{\frac{1}{1-\beta}}\right){\left(\frac{\left(\delta -{\gamma}^{\beta}\right)}{\left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}\le \left(\left(1+\gamma \right)-\gamma \right){\left(\frac{\left(1-{\gamma}^{\beta}\right)}{\left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}\le 1$. Similarly, suppose that T − k ≥ 3. Then, ${x}_{T-k}^{\ast}\ge {x}_{T-k-1}^{\ast}$ if and only if ${\left(\frac{\delta^{k-1}\ \beta\ \left(\delta -{\gamma}^{\beta}\right)}{\lambda\ \left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}\ge {x}_{T-k-1}^{\ast}+{s}_{T-k}={x}_{T-k-1}^{\ast}+\gamma \left({x}_{T-k-1}^{\ast}+{s}_{T-k-1}\right)=\left(1+\gamma \right)\left({x}_{T-k-1}^{\ast}+{s}_{T-k-1}\right)-{s}_{T-k-1}=\left(1+\gamma \right)\left({x}_{T-k-1}^{\ast}+{s}_{T-k-1}\right)-\gamma \left({x}_{T-k-2}^{\ast}+{s}_{T-k-2}\right)=\left(1+\gamma \right){\left(\frac{\delta^k\ \beta\ \left(\delta -{\gamma}^{\beta}\right)}{\lambda\ \left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}-\gamma {\left(\frac{\delta^{k+1}\ \beta\ \left(\delta -{\gamma}^{\beta}\right)}{\lambda\ \left(1-\gamma \right)}\right)}^{\frac{1}{1-\beta}}$. Hence, ${x}_{T-k}^{\ast}\ge {x}_{T-k-1}^{\ast}$ if and only if $1\ge {\delta}^{\frac{1}{1-\beta}}\left(\left(1+\gamma \right)-{\gamma \delta}^{\frac{1}{1-\beta}}\right)$, which is always true.∎

Proof of Proposition 8:

The proof proceeds by backward induction by showing that ${W}_t\left({r}_t\right)=\sqrt{\left(1+{\sum}_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)\ {r}_t\ \left(1-{r}_t\right)}$ for all t. To initialize the induction step, we have ${W}_T\left({r}_T\right)={\max}_{{\overline{x}}_T\ge {r}_T\ge {\underline{x}}_T}\sqrt{\frac{r_T-{\underline{x}}_T}{{\overline{x}}_T-{\underline{x}}_T}{\left({\overline{x}}_T-{r}_T\right)}^2+\frac{{\overline{x}}_T-{r}_T}{{\overline{x}}_T-{\underline{x}}_T}{\left({r}_T-{\underline{x}}_T\right)}^2}=\sqrt{\left({\overline{x}}_T-{r}_T\right)\left({r}_T-{\underline{x}}_T\right)}$. Because the objective function is increasing in ${\overline{x}}_T$ and decreasing in ${\underline{x}}_T$, it is optimal to set ${\overline{x}}_T=1$ and ${\underline{x}}_T=0$. Hence, ${W}_T\left({r}_T\right)=\sqrt{\ {r}_T\ \left(1-{r}_T\right)}$. Fix t < T and suppose that ${W}_{t+1}\left({r}_{t+1}\right)=\sqrt{\left(1+{\sum}_{\tau =1}^{T-t-1}{\delta}^{-2\tau}\right)\ {r}_{t+1}\ \left(1-{r}_{t+1}\right)}$. In that case,

$$ {\displaystyle \begin{array}{c}{W}_t\left({r}_t\right)={\max}_{{\overline{x}}_t\ge {r}_t\ge {\underline{x}}_t}\sqrt{\frac{r_t-{\underline{x}}_t}{{\overline{x}}_t-{\underline{x}}_t}{\left({\overline{x}}_t-{r}_t\right)}^2+\frac{{\overline{x}}_t-{r}_t}{{\overline{x}}_t-{\underline{x}}_t}{\left({r}_t-{\underline{x}}_t\right)}^2}\\ {}\kern2em +{\delta}^{-1}\frac{r_t-{\underline{x}}_t}{{\overline{x}}_t-{\underline{x}}_t}\ {W}_{t+1}\left({\overline{x}}_t\right)+{\delta}^{-1}\frac{{\overline{x}}_t-{r}_t}{{\overline{x}}_t-{\underline{x}}_t}{W}_{t+1}\left({\underline{x}}_t\right)\\ {}\kern2em ={\max}_{{\overline{x}}_t\ge {r}_t\ge {\underline{x}}_t}\sqrt{\left({\overline{x}}_t-{r}_t\right)\left({r}_t-{\underline{x}}_t\right)}\\ {}\kern2em +\frac{r_t-{\underline{x}}_t}{{\overline{x}}_t-{\underline{x}}_t}\ \sqrt{\left(\sum \limits_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)\ {\overline{x}}_t\ \left(1-{\overline{x}}_t\right)}\\ {}\kern2em +\frac{{\overline{x}}_t-{r}_t}{{\overline{x}}_t-{\underline{x}}_t}\sqrt{\left(\sum \limits_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)\ {\underline{x}}_t\ \left(1-{\underline{x}}_t\right)}.\end{array}} $$

Taking the first-order optimality conditions yields that ${\overline{x}}_t=\frac{1}{2}+\sqrt{{\left({r}_t-\frac{1}{2}\right)}^2+\frac{1}{1+\left({\sum}_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)}{r}_t\left(1-{r}_t\right)}$ and ${\underline{x}}_t=\frac{1}{2}-\sqrt{{\left({r}_t-\frac{1}{2}\right)}^2+\frac{1}{1+\left({\sum}_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)}{r}_t\left(1-{r}_t\right)}.$Substituting these values into the objective function yields:

$$ {\displaystyle \begin{array}{l}{W}_t\left({r}_t\right)\\ {}=\sqrt{\frac{1}{1+\left({\sum}_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)}{r}_t\left(1-{r}_t\right)}\\ {}+\sqrt{\left(\sum \limits_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)\ \left(\frac{1}{4}-{\left({r}_t-\frac{1}{2}\right)}^2-\frac{1}{1+\left({\sum}_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)}{r}_t\left(1-{r}_t\right)\right)}\\ {}=\sqrt{\left(1+\left(\sum \limits_{\tau =1}^{T-t}{\delta}^{-2\tau}\right)\right){r}_t\left(1-{r}_t\right)},\end{array}} $$

completing the induction step. To see that this solution is indeed optimal, note that W_t(r_t) corresponds to the optimal solution of the problem of allocation a total variance r_t(1 − r_t) across the T − t + 1 remaining periods so as to maximize ${\sum}_{\tau =t}^T{\delta}^{t-\tau}\ \sqrt{v_{\tau}}$ subject to ${\sum}_{\tau =t}^T{v}_{\tau}\le {r}_t\left(1-{r}_t\right)$ (Ely et al. 2015, online appendix). ∎

Proof of Proposition 9:

The proof proceeds by backward induction. In period T, ${W}_T\left({r}_T\right)={\max}_{{\overline{x}}_T\ge {r}_T\ge {\underline{x}}_T}\frac{r_T-{\underline{x}}_T}{{\overline{x}}_T-{\underline{x}}_T}\ \left({\overline{x}}_T-{r}_T\right)+\frac{{\overline{x}}_T-{r}_T}{{\overline{x}}_T-{\underline{x}}_T}\left({r}_T-{\underline{x}}_T\right)$. Because the objective function is increasing in ${\overline{x}}_T$ and decreasing in ${\underline{x}}_T$, it is optimal to set ${\overline{x}}_T=1$ and ${\underline{x}}_T=0$. Hence, W_T(r_T) = 2r_T(1 − r_T). Therefore, in period T − 1,

$$ {\displaystyle \begin{array}{c}{W}_{T-1}\left({r}_{T-1}\right)={\max}_{{\overline{x}}_{T-1}\ge {r}_{T-1}\ge {\underline{x}}_{T-1}}\frac{r_{T-1}-{\underline{x}}_{T-1}}{{\overline{x}}_{T-1}-{\underline{x}}_{T-1}}\ \left({\overline{x}}_{T-1}-{r}_{T-1}\right)\\ {}+\frac{{\overline{x}}_{T-1}-{r}_{T-1}}{{\overline{x}}_{T-1}-{\underline{x}}_{T-1}}\left({r}_{T-1}-{\underline{x}}_{T-1}\right)\\ {}+{\delta}^{-1}\frac{r_{T-1}-{\underline{x}}_{T-1}}{{\overline{x}}_{T-1}-{\underline{x}}_{T-1}}\ {W}_T\left({\overline{x}}_{T-1}\right)\\ {}+{\delta}^{-1}\frac{{\overline{x}}_{T-1}-{r}_{T-1}}{{\overline{x}}_{T-1}-{\underline{x}}_{T-1}}{W}_T\left({\underline{x}}_{T-1}\right)\\ {}={\max}_{{\overline{x}}_{T-1}\ge {r}_{T-1}\ge {\underline{x}}_{T-1}}2\ \frac{r_{T-1}-{\underline{x}}_{T-1}}{{\overline{x}}_{T-1}-{\underline{x}}_{T-1}}\ \left({\overline{x}}_{T-1}-{r}_{T\_1}\right)\\ {}+2\ {\delta}^{-1}\Big(\frac{r_{T-1}-{\underline{x}}_{T-1}}{{\overline{x}}_{T-1}-{\underline{x}}_{T-1}}{\overline{x}}_{T-1}\ \left(1-{\overline{x}}_{T-1}\right)\\ {}+\frac{{\overline{x}}_{T-1}-{r}_{T-1}}{{\overline{x}}_{T-1}-{\underline{x}}_{T-1}}{\underline{x}}_{T-1}\left(1-{\underline{x}}_{T-1}\right)\Big).\end{array}} $$

Taking the first-order optimality conditions with respect to ${\underline{x}}_{T-1}$ and ${\overline{x}}_{T-1}$ (and ignoring the suboptimal non-informative solutions) yields the following solution: ${\underline{x}}_{T-1}={r}_{T-1}-\delta /4$ and ${\overline{x}}_{T-1}={r}_{T-1}+\delta /4$ if ${r}_{T-1}\in \left[\frac{\delta}{4},1-\frac{\delta}{4}\right]$, ${\underline{x}}_{T-1}=0$ and ${\overline{x}}_{T-1}=\sqrt{\delta\ {r}_{T-1}}$ if ${r}_{T-1}\in \left[0,\frac{\delta}{4}\right]$, and ${\underline{x}}_{T-1}=1-\sqrt{\delta\ \left(1-{r}_{T-1}\right)}$ and ${\overline{x}}_{T-1}=1$ if ${r}_{T-1}\in \left[1-\frac{\delta}{4},1\right].$ Hence,

$$ {\displaystyle \begin{array}{l}{W}_{T-1}\left({r}_{T-1}\right)\\ {}=\left\{\begin{array}{cc}2{r}_{T-1}\left(1+\delta \right)-4{r}_{T-1}\sqrt{\delta\ {r}_{T-1}}& \mathrm{if}\ {r}_{T-1}\in \left[0,\frac{\delta}{4}\right]\\ {}\kern0.5em {\delta}^{-1}2{r}_{T-1}\left(1-{r}_{T-1}\right)+\frac{\delta}{8}& \mathrm{if}\ {r}_{T-1}\in \left[\frac{\delta}{4},1-\frac{\delta}{4}\right]\\ {}2\left(1-{r}_{T-1}\right)\left(1+\delta \right)-4\left(1-{r}_{T-1}\right)\sqrt{\delta\ \left(1-{r}_{T-1}\right)}& \mathrm{if}\ {r}_{T-1}\in \left[1-\frac{\delta}{4},1\right].\end{array}\right.\end{array}} $$

Consider next period T − 2:

$$ {\displaystyle \begin{array}{c}{W}_{T-2}\left({r}_{T-2}\right)={\max}_{{\overline{x}}_{T-2}\ge {r}_{T-2}\ge {\underline{x}}_{T-2}}\frac{r_{T-2}-{\underline{x}}_{T-2}}{{\overline{x}}_{T-2}-{\underline{x}}_{T-2}}\ \left({\overline{x}}_{T-2}-{r}_{T-2}\right)\\ {}+\frac{{\overline{x}}_{T-2}-{r}_{T-2}}{{\overline{x}}_{T-2}-{\underline{x}}_{T-2}}\left({r}_{T-2}-{\underline{x}}_{T-2}\right)\\ {}+{\delta}^{-1}\frac{r_{T-2}-{\underline{x}}_{T-2}}{{\overline{x}}_{T-2}-{\underline{x}}_{T-2}}\ {W}_{T-1}\left({\overline{x}}_{T-2}\right)\\ {}+{\delta}^{-1}\frac{{\overline{x}}_{T-2}-{r}_{T-2}}{{\overline{x}}_{T-2}-{\underline{x}}_{T-2}}{W}_{T-1}\left({\underline{x}}_{T-2}\right).\end{array}} $$

When ${r}_{T-2}\in \left[\frac{\delta +{\delta}^2}{4},1-\frac{\delta +{\delta}^2}{4}\right]$, the function is maximized at ${\underline{x}}_{T-2}={r}_{T-2}-{\delta}^2/4$ and ${\overline{x}}_{T-2}={r}_{T-2}+{\delta}^2/4$. ∎