**Optimization Problem 4**

Code in PSG Text format with parameters from Set 1 for α = 0.75.

```
minimize
vector_c*cvar_dev(vector_a, matrix_s)
value:
cvar_risk(0.75,matrix_s)
```
The code includes two parts. The first part begins with the keyword "minimize" indicating that the objective function is minimized. It implements Step 1 of the Optimization Problem 4, which minimizes deviation from the mixed-quantile quadrangle (for determining optimal vector *C*<sup>∗</sup> of regression coefficients without intercept). The inner product vector\_c\*cvar\_dev(vector\_a, matrix\_s) calculates the mixed CVaR deviation D(*<sup>Z</sup>*0(*C*)). The function cvar\_dev corresponds to the CVaR deviation from the quantile quadrangle. Vector vector\_c contains weights for CVaR Deviation mix corresponding to Set 1. Vector vector\_a contains confidence levels defined by Set 1. The matrix\_s contains scenarios of the residual of the regression *<sup>Z</sup>*(*<sup>C</sup>*0,*<sup>C</sup>*) = *V* − *C*0 − *CTY*.

The second part of the code begins with the keyword "value." This part implements Step 2 of the Optimization Problem 4, calculating the optimal value of intercept *<sup>C</sup>*<sup>∗</sup>0. The PSG function cvar\_risk(0.75,matrix\_s) calculates *CVaR*α(*<sup>Z</sup>*0(*C*<sup>∗</sup>)) with α = 0.75 at the optimal point *C*<sup>∗</sup>.

### **Appendix C Proof of the Lemma 1**

**Proof.** According to the definition, β*v* = 1. However, while proving this lemma, we consider β*v* a bit smaller than 1, i.e., β*v* = 1 − ε>β*<sup>v</sup>*−1, ε > 0, to avoid division by 0. Then, we will consider limit ε → 0 to finish the proof. Note that for α < 1 and for the considered partition, β*<sup>i</sup>*−<sup>1</sup> < β*i* for all *i* = να, ... , ν.

> δ ≥ 0:

First, let us prove that γ*i* ∈ (β*<sup>i</sup>*−1, β*i*) for β*i* < 1. Consider three functions of σ, δ for σ > 0,

$$f\_1(\delta) = \frac{\delta}{\sigma + \delta'} \,\_f\mathfrak{H}(\delta) = \ln(1 + \frac{\delta}{\sigma}) \,\_f\mathfrak{H}(\delta) = \frac{\delta}{\sigma}$$

When δ = 0, all functions equal to 0 and have equal derivatives. When δ > 0, it is true for derivatives that:

$$f'\_{1}(\delta) < f'\_{2}(\delta) < f'\_{3}(\delta)$$

Hence, when > 0, similar inequalities are valid for functions:

$$\frac{\delta}{\sigma + \delta} < \ln \left( 1 + \frac{\delta}{\sigma} \right) < \frac{\delta}{\sigma}.$$

There exist some δγ such that 0 < δγ < δ and δ σ<sup>+</sup>δγ = ln1 + δσ . If δ = β*i* − β*<sup>i</sup>*−1, σ = 1 − β*<sup>i</sup>*, γ = 1 − σ − δγ, then γ = γ*i* and β*<sup>i</sup>*−<sup>1</sup> <γ<β*<sup>i</sup>*. Therefore, γ*i* ∈ (β*<sup>i</sup>*−1, β*i*).

Further, we show how to calculate the integral 1−<sup>ε</sup> α *CVaR*β(*X*)*d*β as a sum of integrals over the partition:

$$\overline{\mathcal{R}}\_{\mathfrak{a}}(X) = \lim\_{\varepsilon \to 0} \frac{1}{1 - \alpha} \int\_{\mathfrak{a}}^{1 - \varepsilon} CVaR\_{\beta}(X) d\beta = \lim\_{\beta\_v \to 1} \frac{1}{1 - \alpha} \sum\_{i = \nu\_{\mathfrak{a}}}^{\nu} \int\_{\beta\_{i - 1}}^{\beta\_i} CVaR\_{\beta}(X) d\beta.$$

Let us denote *<sup>C</sup>*β*i* = *CVaR*β*i*(*X*) and *Vi* = *VaR*γ*i*(*X*). Note that *VaR*γ(*X*) is a singleton for every γ ∈ (β*<sup>i</sup>*−1, β*i*) and it equals *Vi*, because γ*i* ∈ (β*<sup>i</sup>*−1, β*i*).

Below, we use value *Vi* for the closed interval [β*<sup>i</sup>*−1, β*i*] while calculating the integral over this interval because the value of the integral does not depend on the finite values of *VaR*γ(*X*) at the boundary points β*<sup>i</sup>*−1, β*<sup>i</sup>*.

Using the definition of CVaR (Rockafellar and Uryasev (2002), Proposition 8 CVaR for scenario models) we write:

$$\begin{array}{lcl} \frac{\mathbb{M}\_i}{\mathbb{M}\_{i-1}} \operatorname{CVaR}\_{\beta}(X) d\beta &=& \int\_{\beta\_{i-1}}^{\mathbb{M}\_i} \frac{1}{1-\beta} \Big( \mathbb{C}\_{\beta\_i}(1-\beta\_i) + V\_i(\beta\_i-\beta) \Big) d\beta \\ &=& \mathbb{C}\_{\beta\_i}(1-\beta\_i) \int\_{\beta\_{i-1}}^{\beta\_i} \frac{1}{1-\beta} d\beta + V\_i \int\_{\beta\_{i-1}}^{\beta\_i} \frac{\beta\_i-\beta}{1-\beta} d\beta \\ &=& \mathbb{C}\_{\beta\_i}(1-\beta\_i) \ln \Big( \frac{1-\beta\_{i-1}}{1-\beta\_i} \Big) + V\_i \Big[ (\beta\_i-\beta\_{i-1}) - (1-\beta\_i) \ln \Big( \frac{1-\beta\_{i-1}}{1-\beta\_i} \Big) \Big] \end{array} \tag{A7}$$

Let us make the transformation of Equation (A7) using the expression for γ*i* in the Set 1 definition:

$$\begin{split} \int\_{\beta\_{i-1}}^{\beta\_i} \mathsf{CVA}\_{\beta}(X) d\beta &= (\beta\_i - \beta\_{i-1}) \Big[ V\_i + \frac{1}{\beta\_i - \beta\_{i-1}} \ln \Big( \frac{1 - \beta\_{i-1}}{1 - \beta\_i} \Big) (\mathbb{C}\_{\beta\_i} - V\_i)(1 - \beta\_i) \Big] \\ &= (\beta\_i - \beta\_{i-1}) \Big[ V\_i + \frac{1}{1 - \gamma\_i} (\mathbb{C}\_{\beta\_i} - V\_i)(1 - \beta\_i) \Big] = (\beta\_i - \beta\_{i-1}) \frac{1}{1 - \gamma\_i} \Big[ \mathbb{C}\_{\beta\_i}(1 - \beta\_i) + V\_i(\beta\_i - \gamma\_i) \Big] \\ &= (\beta\_i - \beta\_{i-1}) \mathsf{CVA}\_{\beta\_i}(X) \end{split}$$

The last equality is valid because γ*i* ∈ (β*<sup>i</sup>*−1, β*i*).

 Taking into account that lim ε→0 γν = 1, *CVaR*1−<sup>ε</sup>(*X*) = *CVaR*1(*X*), and *VaR*1−<sup>ε</sup>(*X*) = *VaR*1(*X*), we obtain:

$$\overline{\mathcal{R}}\_{\alpha}(X) = \lim\_{\varepsilon \to 0} \sum\_{i=\nu\_{\alpha}}^{\nu} \frac{\beta\_{i} - \beta\_{i-1}}{1 - \alpha} \text{CVaR}\_{\mathcal{V}\_{i}}(X) = \sum\_{i=\nu\_{\alpha}}^{\nu} p\_{i} \text{CVaR}\_{\mathcal{V}\_{i}}(X)$$

Deviation is calculated as follows:

$$\overline{\mathcal{D}}\_{\alpha}(X) = \sum\_{i=\nu\_{\alpha}}^{\nu} p\_i \text{CVaR}\_{\mathcal{V}\_i}(X) - E[X]$$

By the definition of CVaR (Rockafellar and Uryasev (2002)):

$$C VaR\_{\alpha}(X) = \sum\_{i=\nu\_{\alpha}}^{\nu} p\_i VaR\_{\mathcal{Y}\_i}(X) = \overline{\mathcal{S}}\_{\alpha}(X).$$

Lemma 1 is proved. - *JRFM* **2019**, *12*, 107

### **Appendix D Proof of Lemma 2**

**Proof.** Similar to proof of Lemma 1, we consider β*v* as a bit smaller than 1, i.e., β*v* = 1 − ε>β*<sup>v</sup>*−1, ε > 0, to avoid division by 0. Then, we consider limit ε → 0 to finish the proof. We denote *<sup>C</sup>*β*i* = *CVaR*β*i*(*X*) and *Vi* = *VaR*γ(*X*) for any γ ∈ (β*<sup>i</sup>*−1, β*i*) because *VaR*γ(*X*) does not change on this interval.

Additionally we denote δ*i* = β*i* − β*<sup>i</sup>*−1, *i* = να, ... , ν, and σ*i* = 1 − β*<sup>i</sup>*, *i* = να − 1, ... , ν. Note that all δ*i* > 0, all σ*i* > 0, and δ*i* = δ*j* = δ for all *i*, *j* < ν.

$$\overline{\mathcal{R}}\_{a}(X) = \lim\_{\varepsilon \to 0} \frac{1}{1 - \alpha} \int\_{a}^{1 - \varepsilon} CVaR\_{\beta}(X) d\beta = \lim\_{\beta\_{\mathbb{P}} \to 1} \frac{1}{1 - \alpha} \sum\_{i = \nu\_{a}}^{\nu} \int\_{\beta\_{i - 1}}^{\beta\_{i}} CVaR\_{\beta}(X) d\beta. \tag{A8}$$

Equation (A7) in Lemma 1 is valid for any interval [θ, β*i*] such that β*<sup>i</sup>*−<sup>1</sup> ≤ θ ≤ β*<sup>i</sup>*, therefore:

$$\int\_{0}^{\beta\_{i}} \mathbb{C}VaR\_{\beta}(X)d\beta = \mathbb{C}\_{\beta\_{i}}(1-\beta\_{i})\ln\left(\frac{1-\theta}{1-\beta\_{i}}\right) + V\_{i}\left[\left(\beta\_{i}-\theta\right)-\left(1-\beta\_{i}\right)\ln\left(\frac{1-\theta}{1-\beta\_{i}}\right)\right] \tag{A9}$$

Let us express *Vi* from *CVaR*β*<sup>i</sup>*−<sup>1</sup> (*X*) using the definition of CVaR from Rockafellar and Uryasev (2002) (Equation (25)), then insert this expression into Equation (A9):

$$\begin{array}{c} \mathsf{C}\_{\beta\_{i-1}} = \frac{1}{1 - \beta\_{i-1}} \Big( \mathsf{C}\_{\beta\_i} (1 - \beta\_i) + V\_i (\beta\_i - \beta\_{i-1}) \Big) = \frac{1}{\sigma\_{i-1}} \Big( \mathsf{C}\_{\beta\_i} \sigma\_i + V\_i \delta\_i \Big), \\\ V\_i = \frac{1}{\delta\_i} \Big( \mathsf{C}\_{\beta\_{i-1}} \sigma\_{i-1} - \mathsf{C}\_{\beta\_i} \sigma\_i \Big) \end{array}$$

By substituting *Vi* into Equation (A9), we obtain:

$$\int\_{0}^{\wp\_{i}} \mathrm{C}VaR\_{\beta}(X)d\beta = \mathbb{C}\_{\beta \dot{\gamma}}\sigma\_{i} \ln\left(\frac{1-\theta}{\sigma\_{i}}\right) + \frac{1}{\delta\_{l}} \Big(\mathbb{C}\_{\beta \dot{\gamma}-1}\sigma\_{i-1} - \mathbb{C}\_{\beta \dot{\gamma}}\sigma\_{i}\Big) \Big[ (\beta\_{i} - \theta) - \sigma\_{i} \ln\left(\frac{1-\theta}{\sigma\_{i}}\right) \Big] \tag{A10}$$

The last equation contains two CVaRs, *<sup>C</sup>*β*i* and *<sup>C</sup>*β*<sup>i</sup>*−<sup>1</sup> . Let us express the coefficients of these CVaRs. *<sup>C</sup>*β*i*in Equation (A10) has the following coefficient:

$$q\_{1i} = \sigma\_i \left[ \ln \left( \frac{1-\theta}{\sigma\_i} \right) \left( 1 + \frac{\sigma\_i}{\delta\_i} \right) - \frac{\beta\_i - \theta}{\delta\_i} \right] \tag{A11}$$

and *<sup>C</sup>*β*<sup>i</sup>*−<sup>1</sup> has the coefficient:

$$q\_{2i-1} = \frac{\sigma\_{i-1}}{\delta\_i} \left[ (\beta\_i - \theta) - \sigma\_i \ln \left( \frac{1-\theta}{\sigma\_i} \right) \right] \tag{A12}$$

Coefficient *q*2*i*−<sup>1</sup> ≥ 0 because the value in square brackets is the same as in Equations (A7) and (A9), and is a result of the integration of a non-negative function over the interval [θ, β*i*]. The coefficient *q*1*i* ≥ 0 because it is an integral of the non-negative function σ*i* δ*i* δ*i*−β*i*+β 1−β over the same interval.

When summing up integrals to obtain 1 1−<sup>α</sup> 1−<sup>ε</sup> α *CVaR*β(*X*)*d*β, every *<sup>C</sup>*β*i* , aside from *<sup>C</sup>*βνα−<sup>1</sup> and *<sup>C</sup>*βν , enters the sum two times with coefficients depending on ν, α, β*<sup>i</sup>*−1, β*<sup>i</sup>*, and β*<sup>i</sup>*+1. Once in Equation (A11) for *i*, and the second time in Equation (A12) for *i* + 1. All coefficients are non-negative.

Let us explain this in more detail.

If *i* is such that να < *i* < ν, then βνα < β*i* < βν and θ = β*<sup>i</sup>*−<sup>1</sup> for *i* in Equation (A11) and θ = β*i* for *i* + 1 in Equation (A12). Then, the coefficient for *<sup>C</sup>*β*i* in Equation (A8) equals:

$$q\_{i} = \frac{1}{1 - \alpha} (q\_{1i} + q\_{2i}) = \frac{1}{1 - \alpha} (\sigma\_{i} \Big[ \ln \left( \frac{\sigma\_{i-1}}{\sigma\_{i}} \right) \left( \frac{\sigma\_{i-1}}{\delta\_{i}} \right) - 1 \Big] + \frac{\sigma\_{i}}{\delta\_{i+1}} \Big[ \delta\_{i+1} - \sigma\_{i+1} \ln \left( \frac{\sigma\_{i}}{\sigma\_{i+1}} \right) \Big])$$

If *i* = να < ν, then θ = α in Equation (A11) and θ = β*i* in Equation (A12). Then, the coefficient for *<sup>C</sup>*β*i* in Equation (A8) equals:

$$\begin{array}{c} q\_{i} = \frac{1}{1 - a} (q\_{1i} + q\_{2i}) = \frac{1}{1 - a} \Big( \sigma\_{i} \Big[ \ln \left( \frac{1 - a}{\sigma\_{i}} \right) \left( \frac{\sigma\_{i}}{\delta\_{i}} \right) - \frac{\delta\_{i}}{\delta\_{i}} \Big] + \frac{\sigma\_{i}}{\delta\_{i+1}} \Big[ \delta\_{i+1} - \sigma\_{i+1} \ln \left( \frac{\sigma\_{i}}{\sigma\_{i+1}} \right) \Big] \Big) \\\ = \frac{\sigma\_{i}}{1 - a} \Big[ 1 - \frac{\delta\_{a}}{\delta\_{i}} + \frac{\sigma\_{i} - 1}{\delta\_{i}} \ln \left( \frac{1 - a}{\sigma\_{i}} \right) - \frac{\sigma\_{i+1}}{\delta\_{i+1}} \ln \left( \frac{\sigma\_{i}}{\sigma\_{i+1}} \right) \Big] \end{array}$$

If *i* = να − 1, then *<sup>C</sup>*β*i*enters in the sum only in Equation (A12) with θ = α. Then:

$$q\_i = \frac{1}{1-\alpha} q\_{2i} = \frac{\sigma\_i}{1-\alpha} \left[ \frac{\delta\_\alpha}{\delta\_{i+1}} - \frac{\sigma\_{i+1}}{\delta\_{i+1}} \ln \left( \frac{1-\alpha}{\sigma\_{i+1}} \right) \right]$$

If *i* = ν>να, then *<sup>C</sup>*β*i*enters in the sum only in Equation (A11) with θ = β*<sup>i</sup>*−1. Then:

$$q\_i = \frac{1}{1-\alpha} q\_{1i} = \frac{\sigma\_i}{1-\alpha} \left[ \ln \left( \frac{\sigma\_{i-1}}{\sigma\_i} \right) \left( \frac{\sigma\_{i-1}}{\delta\_i} \right) - 1 \right]$$

Also, in the case when *i* = ν = να, *<sup>C</sup>*β*i*enters in the sum only in Equation (A11) with θ = α. Then:

$$\eta\_{i} = \frac{1}{1-\alpha} q\_{1i} = \frac{\sigma\_{i}}{1-\alpha} \left[ \ln \left( \frac{1-\alpha}{\sigma\_{i}} \right) \left( 1 + \frac{\sigma\_{i}}{\delta\_{i}} \right) - \frac{\delta\_{\alpha}}{\delta\_{i}} \right]$$

Then, the risk in the CVaR quadrangle equals:

$$\overline{\mathcal{R}}\_{\alpha}(X) = \lim\_{\iota \to 0} \sum\_{i=\iota\_{\alpha}-1}^{\nu} q\_i C V a R\_{\beta\_i}(X).$$

Deviation is calculated as follows: Dα(*X*) = Rα(*X*) − *<sup>E</sup>*[*X*].

Taking into account that all β*i* are fixed for *i* < ν, the limit operation affects only coefficients for *i* = ν. Namely, lim ε→0 βν = 1, limε→0 σν = 0, limε→0 σν ln(σν) = 0, limε→0 δν = δ. Then, the limit values of coefficients *qi* are equal to:

*q*ν = 0 in both cases, when ν>να and ν = να. *qi* = 1 1−<sup>α</sup> × σ*i* δ <sup>σ</sup>*i*−<sup>1</sup> ln σ*i*−1 σ*i* for *i* such that να < *i* = ν − 1. *qi* = 1 1−<sup>α</sup> × σ*i* δ <sup>σ</sup>*i*−<sup>1</sup> ln σ*i*−1 σ*i* + <sup>σ</sup>*i*+1 ln <sup>σ</sup>*i*+1 σ*i* for *i* such that να < *i* < ν − 1. *qi* = 1 1−<sup>α</sup> × σ*i* δ δ − δα + σ*i*−1 ln 1−ασ*i* + <sup>σ</sup>*i*+1 ln <sup>σ</sup>*i*+1 σ*i* for *i* = να < ν − 1. *qi* = 1 1−<sup>α</sup> × σ*i* δ δα + <sup>σ</sup>*i*+1 ln <sup>σ</sup>*i*+1 1−<sup>α</sup> for *i* = να − 1 < ν − 1.

If the number of atoms used in calculating Rα(*X*) is 3 or 2, then the coefficients have values:

 $q\_{i} = \frac{1}{1 - \alpha} \times \frac{\sigma\_{i}}{\delta}$  $\left[\delta - \delta\_{\alpha} + \sigma\_{i - 1} \ln\left(\frac{1 - \alpha}{\sigma\_{i}}\right)\right]$ for  $i = \nu\_{\alpha} = \nu - 1$ .  $q\_{i} = \frac{1}{1 - \alpha} \times \frac{\sigma\_{i}}{\delta}$  $\left[\delta\_{\alpha}\right] = 1 \text{ for } i = \nu\_{\alpha} - 1 = \nu - 1.$ 

It can be shown that <sup>ν</sup>*i*=να−<sup>1</sup> *qi* = 1 by sequentially summing up coefficients. By recalling that σ*i* = <sup>δ</sup>(*v* − *i*), we can rewrite the equations for coefficients *qi* using δ and *j* = *v* − *i*. Lemma 2 is proved. -

### **Appendix E Proof of the Lemma 3**

**Proof.** Distribution of random value *X* defines the partition of the interval [0, 1]: δ = 1/<sup>ν</sup>, β*i* = *i*δ, for *i* = 0, 1, ... , ν.

CDF *FX*(*x*) = *prob*{*<sup>X</sup>* ≤ *x*} is a non-decreasing and right-continuous function and it is constant for every right-open interval [β*<sup>i</sup>*−1, β*i*), *i* = 1, ... , ν. Therefore, from the definitions of *VaR*<sup>−</sup>γ (*X*) and *VaR*+γ (*X*), we have:

$$VaR^-\_\gamma(X) = VaR^+\_\gamma(X) = VaR\_\gamma(X) = const \text{ for } \gamma \in (\beta\_{i-1}, \beta\_i).$$

$$VaR^+\_{\beta\_i}(X) = VaR\_\gamma(X) \text{ for } \gamma \in (\beta\_{i\prime}\beta\_{i+1}) \text{ and } i = 0, \dots, \nu - 1.$$

$$VaR^-\_{\beta\_i}(X) = VaR\_\gamma(X) \text{ for } \gamma \in (\beta\_{i-1}\beta\_i) \text{ and } i = 1, \dots, \nu.$$

For *i* = 0 we have β*i* = 0 and *VaR*<sup>−</sup>0(*X*) = −∞.

Thus, according to the definition of statistic for the mixed-quantile quadrangle, for Set 2:

$$\mathcal{S}\_{\alpha}^{II}(X) = \sum\_{i=\nu\_{\alpha}-1}^{\nu-1} q\_i \, VaR\_{\beta\_i}(X)$$

where *VaR*β*i*(*X*) are intervals (that may have zero lengths for some *i*), and *qi* and να are defined above for the Set 2. Therefore, <sup>S</sup>*II*α (*X*) is also an interval and it has a non-zero length if not all *VaR*<sup>−</sup>β*i*(*X*), *i* = να − 1, ... , ν − 1 are equal.

If να − 1 = 0, then <sup>S</sup>*II*α (*X*) is a left-open interval with the lower bound −∞.

Let να − 1 > 0. For the Set 1, we defined confidence levels as internal points in intervals γ*i* ∈ (β*<sup>i</sup>*−1, β*i*). Using these definitions of γ*i*, we can express <sup>S</sup>*II*α (*X*) as the interval:

$$S\_a^{II}(X) = \left[\sum\_{i=\nu\_a-1}^{\nu-1} q\_i VaR\_{\mathcal{Y}\_i}(X), \sum\_{i=\nu\_a-1}^{\nu-1} q\_i VaR\_{\mathcal{Y}\_i+1}(X)\right]$$

To simplify notations, let us denote: *Vi* = *VaR*γ*i*(*X*); *C*γ = *CVaR*γ(*X*); and *L*, *U* are bounds for <sup>S</sup>*II*α (*X*) such that <sup>S</sup>*II*α (*X*) = [*L*, *<sup>U</sup>*].

Statistic Sα(*X*) of the CVaR quadrangle is defined in Lemma 1 (see Equation (1)) as Sα(*X*) = <sup>ν</sup>*i*=να *piVi*.

Therefore, we wish to prove that:

$$L = \sum\_{i=\nu\_a - 1}^{\nu - 1} q\_i V\_i \le \overline{\mathbf{S}}\_a(\mathbf{X}) \le \sum\_{i=\nu\_a - 1}^{\nu - 1} q\_i V\_{i+1} = \mathbf{U}$$

Let us prove the right inequality for the upper bound. According to Lemmas 1 and 2, it is valid that:

$$\overline{\mathcal{R}}\_{\alpha}(X) = \sum\_{i=\nu\_{\alpha}}^{\nu} p\_i \mathbb{C}\_{\mathbb{Y}\_i} = \sum\_{i=\nu\_{\alpha}-1}^{\nu-1} q\_i \mathbb{C}\_{\beta\_i}$$

Let *di* = *qi*−<sup>1</sup> − *pi*. Then, from the last equality:

$$\sum\_{i=\nu\_a}^{\nu} p\_i \mathbb{C}\_{\mathcal{V}\_i} = \sum\_{i=\nu\_a}^{\nu} q\_{i-1} \mathbb{C}\_{\beta\_{i-1}} = \sum\_{i=\nu\_a}^{\nu} p\_i \mathbb{C}\_{\beta\_{i-1}} + \sum\_{i=\nu\_a}^{\nu} d\_i \mathbb{C}\_{\beta\_{i-1}}$$

$$\sum\_{i=\nu\_a}^{\nu} d\_i \mathbb{C}\_{\beta\_{i-1}} = \sum\_{i=\nu\_a}^{\nu} p\_i \Big(\mathbb{C}\_{\gamma i} - \mathbb{C}\_{\beta\_{i-1}}\Big) \ge 0$$

The last inequality is valid because *pi* > 0, γ*i* > β*<sup>i</sup>*−1, and *<sup>C</sup>*γ*i* ≥ *<sup>C</sup>*β*<sup>i</sup>*−<sup>1</sup> .

Because *<sup>C</sup>*β*<sup>i</sup>*−<sup>1</sup> may have arbitrary but ordered values (*C*β*<sup>i</sup>*−<sup>1</sup> ≤ *<sup>C</sup>*β*i* due to ordered β*i*), *i* = να, ... , ν, the last inequality is valid for any ordered sequence of values *zi*−1 ≤ *zi*. Therefore, it is valid for VaRs *Vi* that:

$$\sum\_{i=\nu\_a}^{\nu} d\_i \mathbf{V}\_i = \sum\_{i=\nu\_a}^{\nu} (q\_{i-1} - p\_i) \mathbf{V}\_i \ge 0$$

This leads to the inequality for the upper bound:

$$M = \sum\_{i=\nu\_{\alpha}}^{\nu} q\_{i-1} V\_i \ge \sum\_{i=\nu\_{\alpha}}^{\nu} p\_i V\_i = \overline{\mathcal{S}}\_{\alpha}(X).$$

Let us prove the inequality for the lower bound *L*:

$$L = \sum\_{i=\nu\_a}^{\nu} q\_{i-1} V\_{i-1} = \sum\_{i=\nu\_a}^{\nu} (p\_i + d\_i)(V\_i - \Delta V\_i) = \overline{\mathcal{S}}\_a(\mathcal{X}) - \sum\_{i=\nu\_a}^{\nu} p\_i \Delta V\_i + \sum\_{i=\nu\_a}^{\nu} d\_i V\_{i-1} \tag{A13}$$

where Δ*Vi* = *Vi* − *Vi*−1.

Let us calculate the upper estimate of *L*. Let us set *<sup>V</sup>*να−<sup>1</sup> = *V*να , therefore Δ*V*να = 0. This increases the right hand side of Equation (A13). By recalling that *pi* = δ/(1 − α) = 1 <sup>ν</sup>(<sup>1</sup>−<sup>α</sup>), *i* = να + 1, ... , ν, for the Set 1, we have:

$$\sum\_{i=\nu\_a}^{\nu} p\_i \Delta V\_i = \delta(V\_{\nu} - V\_{\nu\_a})/(1 - \alpha)$$

Because <sup>ν</sup>*i*=να *pi* = <sup>ν</sup>*i*=να *qi*−1, then <sup>ν</sup>*i*=να *di* = 0 and <sup>ν</sup>*i*=να *diVi*−<sup>1</sup> = <sup>ν</sup>*i*=να *di*(*Vi*−<sup>1</sup> − *D*) for any *D*. Because <sup>ν</sup>*i*=να *dizi* ≥ 0 for any increasing ordered *zi*, and we can set *<sup>z</sup>*να = 0, *zi* = 1, *i* = να + 1, ... , ν, then:

$$\sum\_{i=\nu\_a}^{\nu} d\_i z\_i = \sum\_{i=\nu\_{a+1}}^{\nu} d\_i = -d\_{\nu\_a} = p\_{\nu\_a} - q\_{\nu\_a - 1} \ge 0$$

Taking into account that *p*να , *q*να−<sup>1</sup> ≥ 0 and *p*να ≤ δ 1−<sup>α</sup> , we have <sup>ν</sup>*i*=να+<sup>1</sup> *di* ≤ δ 1−<sup>α</sup> . Then:

$$\sum\_{i=\nu\_a}^{\nu} d\_i V\_{i-1} = \sum\_{i=\nu\_a}^{\nu} d\_i \big( V\_{i-1} - V\_{\nu\_{a-1}} \big) \leq \sum\_{i=\nu\_a+1}^{\nu} d\_i \big( V\_{\nu-1} - V\_{\nu\_{a-1}} \big) \leq \left( V\_{\nu-1} - V\_{\nu\_{a-1}} \right) \frac{\delta}{1 - \alpha}$$

Let us return to estimation *L* by taking into account that *<sup>V</sup>*να−<sup>1</sup> = *V*να :

$$\begin{array}{c} \mathcal{L} \leq \overline{\mathcal{S}}\_{a}(\mathcal{X}) - \sum\_{i=\nu\_{a}}^{\nu} p\_{i} \Delta V\_{i} + \sum\_{i=\nu\_{a}}^{\nu} d\_{i} V\_{i-1} \leq \overline{\mathcal{S}}\_{a}(\mathcal{X}) - \frac{\delta(V\_{\mathcal{V}} - V\_{\nu\_{a}})}{1 - a} + \frac{\delta\left(V\_{\mathcal{V}} - V\_{\mathcal{V}\_{a-1}}\right)}{1 - a} \\ = \overline{\mathcal{S}}\_{a}(\mathcal{X}) - \frac{\delta(V\_{\mathcal{V}} - V\_{\nu-1})}{1 - a} \leq \overline{\mathcal{S}}\_{a}(\mathcal{X}) \end{array}$$

Therefore, we have proved that *L* ≤ Sα(*X*) ≤ *U*. Lemma 3 is proved. -
