On queues with impatience: stability, and the optimality of Earliest Deadline First

Abstract

In this paper, we consider a queue with impatient customers, under general assumptions. We introduce a convenient representation of the system by a stochastic recursive sequence keeping track of the remaining service and patience times of the customers. This description allows us (i) to provide a comprehensive stability condition in the general case, (ii) to give a rigorous proof of the optimality of the Earliest Deadline First (EDF) service discipline in several cases, and (iii) to show that the abandonment probability of the system follows inversely the stochastic ordering of the generic patience time distribution.

Appendix: Technical results

We first have the following result.

Lemma 2

Let $u \in \mathcal{S }^2$. Assume that $\underline{u}$ is such that

$$\begin{aligned} u^1\left(N(u)\right)=\underline{u}^1\left(N(u)\right). \end{aligned}$$

Denote $\alpha$, the permutation of $[[0,N(u)-1 ]]$ such that

$$\begin{aligned} u\left(N(u)-i\right)=\underline{u}\left(N(u)-\alpha (i)\right),\,i \in [[0,N(u)-1 ]]. \end{aligned}$$

Let $\hat{\underline{u}}$ be the transformation of $\underline{u}$ w.r.t. $u$, as defined above. Let $x>0$. Denote $p:=p(u,\,x)=\text{ Card} \mathcal{A }(u,\,x)-1$ (respectively $q:=q(\hat{\underline{u}},\,x)=\text{ Card} \mathcal{A }(\hat{\underline{u}},\,x)-1$) and

$$\begin{aligned} \mathcal{A }(u,\,x)&= \left\{ 0=i_0 < i_1 <...<i_p=\varphi (u,\,x)\right\} ;\\ \mathcal{A }(\hat{\underline{u}},\,x)&= \left\{ 0=k_0<k_1<...<k_q =\varphi \left(\hat{\underline{u}},\,x\right) \right\} . \end{aligned}$$

Then, there exists an injection

$$\begin{aligned} F=\left\{ \begin{array}{ll} \mathcal{A }(u,\,x)&\longrightarrow \mathcal{A }\left(\hat{\underline{u}},\,x\right)\\ i_j&\longmapsto \left\{ \begin{array}{ll} \alpha (i_j)&\text{ if} \alpha (i_j) \in \mathcal{A }(\hat{\underline{u}},\,x),\\ k_m&\text{ if} \alpha (i_j) \not \in \mathcal{A }(\hat{\underline{u}},\,x),\\&\text{ where} u^2\left(N(u)-i_j\right) \ge \underline{u}^2\left(N(u)-k_m\right). \end{array}\right. \end{array}\right. \end{aligned}$$

Proof

We start by showing that for all $\ell \in [[1,p\wedge q ]]$, there exists a bijection

$$\begin{aligned} F_\ell =\left\{ \begin{array}{ll} \mathcal{A }^{i_\ell }(u,\,x)&\longrightarrow \mathcal{A }^{k_\ell }\left(\hat{\underline{u}},\,x\right)\\ i_j&\longmapsto \left\{ \begin{array}{ll} \alpha (i_j)&\text{ if} \alpha (i_j) \in \mathcal{A }^{k_\ell }(\hat{\underline{u}},\,x),\\ k_m&\text{ if} \alpha (i_j) \not \in \mathcal{A }^{k_\ell }(\hat{\underline{u}},\,x),\\&\text{ where} u^2\left(N(u)-i_j\right) \ge \underline{u}^2\left(N(u)-k_m\right). \end{array}\right. \end{array}\right. \end{aligned}$$

For this, we proceed by induction. First, for $\ell =0$ the property clearly holds if $\alpha (0)=0$. If not, as $\underline{u}^2$ is ranked in decreasing order,

$$\begin{aligned} \underline{u}^2\left(N(u)-k_0\right)=\underline{u}^2\left(N(u)\right) \le u^2\left(N(u)\right)=u^2\left(N(u)-k_0\right), \end{aligned}$$

so we can set $F_\ell (i_0)=k_0$.

Assume now that the result holds for $\ell -1$, where $\ell \in [[1,p\wedge q ]]$. There are two possible cases:

• (i) if $k_\ell \le \alpha (i_\ell )$, then

  $$\begin{aligned} \underline{u}^2\left(N(u)-k_\ell \right) \le \underline{u}^2\left(N(u)-\alpha (i_\ell )\right)=u^2\left(N(u)-i_\ell \right), \end{aligned}$$

  and it suffices to set $F_{\ell }(i_m)=F_{\ell -1}(i_m)$ for all $m \le \ell -1$ and $F_{\ell }(i_\ell )=k_\ell$.

• (ii) if $k_\ell > \alpha (i_\ell )$, we have

  $$\begin{aligned} x \wedge \underline{u}^2\left(N(u)-\alpha (i_\ell )\right)&= x \wedge u^2\left(N(u)-i_\ell \right)\\&> \sum _{j=1}^{\ell -1} u^1\left(N(u)-i_j\right)\\&= \sum _{j=1}^{\ell -1} \hat{\underline{u}}^1\left(N(u)-k_j\right)\\&\ge \sum _{j=1}^{\ell -2} \hat{\underline{u}}^1\left(N(u)-k_j\right). \end{aligned}$$

  Therefore, $\alpha (i_\ell ) \in \mathcal{A }^{k_{\ell -1}}\left(\underline{u},\,x\right),$ so $\alpha (i_\ell )=F_{\ell -1}(i_j)$ for some $j\in [[0,\ell -1 ]]$. We have

  $$\begin{aligned} x \wedge \underline{u}^2\left(N(u)-\alpha (i_j)\right)&= x \wedge u^2\left(N(u)-i_j\right) \nonumber \\&\ge x \wedge \underline{u}^2\left(N(u)-F_{\ell -1}(i_j)\right) \nonumber \\&= x \wedge \underline{u}^2\left(N(u)-\alpha (i_\ell )\right)\nonumber \\&= x \wedge u^2\left(N(u)-i_\ell \right)\nonumber \\&> \sum _{j=1}^{\ell -1} u^1\left(N(u)-i_j\right)\nonumber \\&= \sum _{j=1}^{\ell -1} \hat{\underline{u}}^1\left(N(u)-k_j\right). \end{aligned}$$

  (45)

  Furthermore, as $i_\ell \not \in \mathcal{A }^{i_{\ell -1}}\left(u,\,x\right),$ it is clear by the very construction of $F_{\ell -1}$ that $\alpha (i_j) \not \in \mathcal{A }^{k_{\ell -1}}\left(\underline{u},\,x\right)$, so from (45), $\alpha (i_j)>k_{\ell -1}$. But by definition of $\underline{u}$ and $k_\ell$,

  $$\begin{aligned} k_\ell =\min \biggl \{q \in [[k_{\ell -1}+1,N(u)-1 ]]; x \wedge \underline{u}^2\left(N(u)-q\right) > \sum _{j=1}^{\ell -1} \hat{\underline{u}}^1\left(N(u)-k_j\right)\biggl \}. \end{aligned}$$

  Therefore, we deduce from (45) that

  $$\begin{aligned} \alpha (i_j)\ge k_\ell . \end{aligned}$$

  (46)

  There are three subcases:

  • (a) If $k_\ell =\alpha (i_j)$, we can set

    $$\begin{aligned} \left\{ \begin{array}{ll} F_{\ell }(i_\ell )&=\alpha (i_\ell );\\ F_{\ell }(i_j)&=k_\ell ;\\ F_{\ell }(i_m)&=F_{\ell -1}\left(i_m\right),\, \text{ for} \text{ all} m \in [[0,\ell -1 ]]\setminus \{j\}. \end{array} \right. \end{aligned}$$

  • (b) If $k_\ell =\alpha (i_p)$ for some $p \in \mathcal{A }^{i_{\ell -1}}\left(u,\,x\right),\,p\ne j$, in view of (46) we have

    $$\begin{aligned} \underline{u}^2\left(N(u)-F_{\ell -1}(i_p)\right)&\le u^2\left(N(u)-i_p\right)\\&= \underline{u}^2\left(N(u)-k_\ell \right)\\&\le \underline{u}^2\left(N(u)-\alpha (i_j)\right)\\&= u^2\left(N(u)-i_j\right). \end{aligned}$$

    Therefore, we can set

    $$\begin{aligned} \left\{ \begin{array}{ll} F_{\ell }(i_\ell )&=\alpha (i_\ell );\\ F_{\ell }(i_p)&=k_\ell ;\\ F_{\ell }(i_j)&=F_{\ell -1}\left(i_p\right);\\ F_{\ell }(i_m)&=F_{\ell -1}\left(i_m\right),\, \text{ for} \text{ all} m \in [[0,\ell -1 ]]\setminus \{j,p\}. \end{array} \right. \end{aligned}$$

  • (c) Finally, in the other cases $\alpha (i_j) \not \in \mathcal{A }^{k_{\ell }}(\underline{u},\,x)$ and $\underline{u}^2\left(N(u)-k_\ell \right)\le u^2\left(N(u)-i_j\right)$, so we can set again

    $$\begin{aligned} \left\{ \begin{array}{ll} F_{\ell }(i_\ell )&=\alpha (i_\ell );\\ F_{\ell }(i_j)&=k_\ell ;\\ F_{\ell }(i_m)&=F_{\ell -1}\left(i_m\right),\, \text{ for} \text{ all} m \in [[0,\ell -1 ]]\setminus \{j\}. \end{array} \right. \end{aligned}$$

It follows from the existence of $F_p$, that $p\le q$. Indeed, if we assume that some index $i_{q+1}$ belongs to $\mathcal{A }\left(u,\,x\right)$, then

$$\begin{aligned} x \wedge \underline{u}^2\left(N(u)-\alpha \left(i_{q+1}\right)\right)&= x \wedge u^2\left(N(u)-i_{q+1}\right)\\&> \sum _{j=1}^{q} u^1\left(N(u)-i_j\right)\nonumber \\&= \sum _{j=1}^{q} \hat{\underline{u}}^1\left(N(u)-k_j\right). \end{aligned}$$

Therefore, $\alpha \left(i_{q+1}\right) \in \mathcal{A }(\underline{u},\,x)$, say $\alpha \left(i_{q+1}\right)=F_{\ell }(i_{j}),$ where $i_j \in \mathcal{A }^{i_\ell }(u,\,x)$. So by the very construction of the mapping $F_\ell ,\,\alpha (i_j) \not \in \mathcal{A }^{k_\ell }(\underline{u},\,x)$, whereas

$$\begin{aligned} x \wedge \underline{u}^2\left(N(u)-\alpha \left(i_j\right)\right)&= x \wedge u^2\left(N(u)-i_j\right)\\&\ge x \wedge \underline{u}^2\left(N(u)-F_\ell (i_j)\right)\\&= x \wedge \underline{u}^2\left(N(u)-\alpha \left(i_{q+1}\right)\right)\\&= x \wedge u^2\left(N(u)-i_{q+1}\right)\\&> \sum _{j=1}^{q} \hat{\underline{u}}^1\left(N(u)-k_j\right). \end{aligned}$$

Therefore, there exists an index

$$\begin{aligned} k_{q+1}=\min \biggl \{q \not \in \mathcal{A }^{k_\ell }(\underline{u},\,x);\,x \wedge \underline{u}^2\left(N(u)-\alpha \left(i_j\right)\right)>\sum _{j=1}^{q} \hat{\underline{u}}^1\left(N(u)-k_j\right)\biggl \}, \end{aligned}$$

a contradiction. This shows that $q \ge p.$

The proof of the Lemma is thus completed, by setting $F \equiv F_{p}$, which is injective by construction. $\square$

Lemma 3

Let $u$ and $v \in \mathcal{S }^2$ such that $u^2$ and $v^2$ are ordered in decreasing order, and such that

$$\begin{aligned} u&\prec _2 v \text{ in} \mathcal{S }^2;\\ v^1\left(N(v)\right)&= u^1\left(N(u)\right).\nonumber \end{aligned}$$

(47)

Let $x>0$. Denote $q:=q(u,\,x)=\text{ Card} \mathcal{A }(u,\,x)-1$ (respectively $r:=r(\hat{v},\,x)=\text{ Card} \mathcal{A }(\hat{v},\,x)-1$) and

$$\begin{aligned} \mathcal{A }(u,\,x)&= \left\{ 0=k_0 < k_1 <...<k_q=\varphi (u,\,x)\right\} ;\\ \mathcal{A }(\hat{v},\,x)&= \left\{ 0=l_0<l_1<...<l_r =\varphi (\hat{v},\,x) \right\} , \end{aligned}$$

where $\hat{v}$ is the modified version of $v$ with respect to $u$ as defined above. Then, $q \le r$ and for any $\ell \in [[0,q ]]$,

$$\begin{aligned} N(u)-k_\ell \le N(v)-l_\ell . \end{aligned}$$

(48)

Proof

We first show by induction that (48) holds for any $\ell \in [[0,q\wedge r ]]$. It is clear in view of (47) that $N(u) \le N(v)$, so (48) holds for $\ell =0$. Suppose that it holds true for $\ell -1$ (with $\ell \in [[1,q\wedge r ]]$). We have

$$\begin{aligned} x\wedge v^2\Bigl (N(v)-\bigl (N(v)-N(u)+k_j\bigl )\Bigl )&= x\wedge v^2\left(N(u)-k_j\right)\nonumber \\&\ge x\wedge u^2\left( N(u)-k_j\right)\nonumber \\&> \sum _{j=0}^{\ell -1} u^1\left(N(u)-k_j\right)\nonumber \\&= \sum _{j=0}^{\ell -1} \hat{v}^1\left(N(v)-l_j\right). \end{aligned}$$

(49)

But in view of the recurrence assumption,

$$\begin{aligned} N(v)-N(u)+k_\ell > N(v)-\left(N(u)-k_{\ell -1}\right) \ge l_{\ell -1}. \end{aligned}$$

Therefore, by the very definition of $l_\ell$, we deduce from (49) that

$$\begin{aligned} N(v)-N(u)+k_\ell \le l_\ell , \end{aligned}$$

which completes the recurrence.

It remains to show that $q \le r$. Assume that this does not hold, i.e., some index $k_{r+1} \in \mathcal{A }(u,\,x)$. This implies that

$$\begin{aligned} x\wedge v^2\left(N(u)-k_{r+1}\right)&\ge x\wedge u^2\left(N(u)-k_{r+1}\right)\\&> \sum _{j=1}^r u^1\left(N(u)-k_j\right)\\&= \sum _{j=1}^r \hat{v}^1\left(N(v)-k_j\right). \end{aligned}$$

As the term on the right-hand side is larger that the latter sum up to any $\ell \le r$, this implies for some $\ell \le r,\, N(u)-k_{r+1}=N(v)-l_\ell .$ But in view of the previous result, this implies that

$$\begin{aligned} N(u)-k_\ell \le N(v)-l_\ell = N(u)-k_{r+1}, \end{aligned}$$

a contradiction. Thus $q \le r$. $\square$

Lemma 4

Let $p \le q$ and $u \in \mathbf{R }^p,\,v\in \mathbf{R }^q$. Assume that there exists an injection $\gamma :\,[[1,p ]]\rightarrow [[1,q ]]$ such that

$$\begin{aligned} u(i) \le v\left(\gamma \left(i\right)\right),\,i\in [[1,p ]]. \end{aligned}$$

Thus,

$$\begin{aligned} \underline{u}(i) \le \underline{v}(i),\,i\in [[1,p ]]. \end{aligned}$$

Proof

Denote $\delta$ and $\varepsilon$ the respective permutations of $[[1,p ]]$ and $[[1,q ]]$ such that

$$\begin{aligned} \underline{u}=u\left(\delta (.)\right) \text{ and} \underline{v}=v\left(\varepsilon (.)\right). \end{aligned}$$

We proceed by induction. First,

$$\begin{aligned} \underline{u}(1)=u(\delta (j)) \le v\left(\gamma \circ \delta (j)\right) \le \underline{v}(1). \end{aligned}$$

Assume that the property holds until $j$. First, if

$$\begin{aligned} \gamma \circ \delta (j+1)\not \in \varepsilon \left([[1,j ]]\right):=\big \{\varepsilon (1),\,\varepsilon (2),\,...,\,\varepsilon (j)\big \}, \end{aligned}$$

then $\gamma \circ \delta (j+1)=\varepsilon (\ell )$ for some $\ell >j$, so in particular

$$\begin{aligned} \underline{u}(j+1)=u\left(\delta (j+1)\right)\le v\left(\gamma \circ \delta (j+1)\right)=v\left(\varepsilon (\ell )\right) \le \underline{v}(j+1). \end{aligned}$$

If now $\gamma \circ \delta (j+1)\in \varepsilon \left([[1,j ]]\right)$, as $\gamma \circ \delta$ is injective, there exist $k \le j$ and $m > j$ such that

$$\begin{aligned} \gamma \circ \delta (k) =\varepsilon (m). \end{aligned}$$

So we have that

$$\begin{aligned} \underline{u}(j+1) \le \underline{u}(k) = u\left(\delta (k)\right) \le v\left(\gamma \circ \delta (k)\right) = v\left(\varepsilon (m)\right) \le \underline{v}(j+1), \end{aligned}$$

which completes the proof. $\square$

This finally leads to the following result.

Theorem 5

Let $u$ and $v$ be two elements of $\mathcal{S }^2$ such that $v^2$ is ordered in decreasing order. Denote $\hat{v}$ and $\hat{\underline{u}}$, the modified versions of $v$ and $\underline{u}$ with respect to $u$. Assume that

$$\begin{aligned} u&\prec _2 v\, \text{ in} \mathcal{S }^2;\\ u^1\left(N(u)\right)&= \underline{u}^1\left(N(u)\right)=v^1\left(N(v)\right). \end{aligned}$$

Let $x>0$. Then,

$$\begin{aligned} H_2\left( u,\,x\right) \prec _2 H_2\left(\hat{v},\,x\right)\text{ in} \mathcal{S }^2 \end{aligned}$$

and if $H_2\left(u,\,x\right)\ne \mathbf 0$, we have

$$\begin{aligned} \left(H_2\left(u,\,x\right)\right)^1\Bigl (N\left(H_2\left(u,\,x\right) \right)\Bigl )=\left(H_2\left(\hat{v},\,x\right)\right)^1\Bigl (N\left(H_2\left(\hat{v},\,x\right) \right)\Bigl ), \end{aligned}$$

(50)

where the mapping $H_2$ is defined by (16).

Proof

The result holds true whenever $H_2\left(u,\,x\right)=\mathbf 0$, which is the case in particular if $u^2=\mathbf 0 _2$ since $\mathcal{A }(u,\,x)=\{0\}$ by construction.

If not, we consider the sequences $u,\underline{u}$ and $v$ and apply Lemmas 2, 3 and 4. Here again, we denote $\alpha$ the permutation such that

$$\begin{aligned} u\left(N(u)-i\right)=\underline{u}\left(N(u)-\alpha (i)\right) \end{aligned}$$

and $F$, the injection defined in Lemma 2. We denote as well

$$\begin{aligned} \mathcal{A }(u,\,x)&= \left\{ 0=i_0 < i_1 <...<i_p=\varphi (u,\,x)\right\} ;\\ \mathcal{A }(\hat{\underline{u}},\,x)&= \left\{ 0=~_0 < k_1 <...<k_q=\varphi (\underline{u},\,x)\right\} ;\\ \mathcal{A }(\hat{v},\,x)&= \left\{ 0=l_0<l_1<...<l_r =\varphi (\hat{v},\,x) \right\} , \end{aligned}$$

with $p \le q \le r.$

By the very definition of $H_2$, any positive second component of $H_2(u,\,x)$ reads $u^2\left(N(u)-i\right)-x$, where $i\in [[i_p+1,N(u)-1 ]]$ and

$$\begin{aligned} u^2\left(N(u)-i\right)>x > \sum _{j=1}^{p-1} u^1\left(N(u)-i_j\right). \end{aligned}$$

We then necessarily have that

$$\begin{aligned} \sum _{j=1}^{p} \hat{v}^1\left(N(v)-l_j\right)=\sum _{j=1}^{p} u^1\left(N(u)-i_j\right) \ge x, \end{aligned}$$

(51)

because the contrary would imply that

$$\begin{aligned} u^2\left(N(u)-i\right) \wedge x > \sum _{j=1}^{p} u^1\left(N(u)-i_j\right), \end{aligned}$$

so some index $i_{p+1} \in [[i_p+1,i ]]$ would belong to $\mathcal{A }(u,\,x)$, a contradiction. Therefore, (51) implies that $r \le p$, so we have that

$$\begin{aligned} p=q=r \end{aligned}$$

(52)

and in particular, $F$ defines a bijection from $\mathcal{A }(u,\,x)$ to $\mathcal{A }(\hat{\underline{u}},\,x)$. Furthermore, this implies that

$$\begin{aligned} \left(H_2\left(u,\,x\right)\right)^1\Bigl (N\left(H_2\left(u,\,x\right) \right)\Bigl )&= \sum _{j=1}^{p} u^1\left(N(u)-i_j\right)-x\\&= \sum _{j=1}^{p} \hat{v}^1\left(N(v)-l_j\right)-x\\&= \left(H_2\left(\hat{v},\,x\right)\right)^1\Bigl (N\left(H_2\left(\hat{v},\,x\right) \right)\Bigl ), \end{aligned}$$

hence (50) is verified. Now, there are two cases:

• (i) if $\alpha (i) \not \in \mathcal{A }\left(\hat{\underline{u}},\,x\right)$, we have $\alpha (i)> k_p$ and

  $$\begin{aligned} \underline{u}^2\left(N(u)-\alpha (i)\right)=u^2\left(N(u)-i\right)>x> \sum _{j=1}^{p-1} \hat{\underline{u}}^1\left(N(u)-k_j\right), \end{aligned}$$

  so some positive component of $\left(H_2\left(\hat{\underline{u}},\,x\right)\right)^2$ reads

  $$\begin{aligned} \underline{u}^2\left(N(u)-\alpha (i)\right)-x=u^2\left(N(u)-i\right)-x. \end{aligned}$$

  (53)

• (ii) if $\alpha (i) \in \mathcal{A }\left(\hat{\underline{u}},\,x\right),$ then $\alpha \circ F^{-1} \circ \alpha (i) > k_p$ and we have that

  $$\begin{aligned} \underline{u}^2\left(N(u)-\alpha \circ F^{-1} \circ \alpha (i)\right)&= u^2\left(N(u)- F^{-1} \circ \alpha (i)\right)\\&\ge \underline{u}^2\left(N(u)- \alpha (i)\right)\\&= u^2\left(N(u)-i\right)\\&> x >\sum _{j=1}^{p-1} \hat{\underline{u}}^1\left(N(u)-k_j\right), \end{aligned}$$

  thus some positive component of $\left(H_2\left(\hat{\underline{u}},\,x\right)\right)^2$ is given by

  $$\begin{aligned} \underline{u}^2\left(N(u)-\alpha \circ F^{-1} \circ \alpha (i)\right)-x \ge u^2\left(N(u)-i\right)-x. \end{aligned}$$

  (54)

Now, any positive second component of $H_2\left(\hat{\underline{u}},\,x\right)$ reads $\underline{u}^2\left(N(u)-j\right)-x$, where $j \in [[k_p+1,N(u) ]]$. Assume that $N(v)-N(u)+j \in \mathcal{A }(\hat{v},\,x)$ and denote the index $\ell \in [[0,p ]]$ such that $N(v)-N(u)+j=l_\ell$. From Lemma 3, $N(u)-k_\ell \le N(v)-l_\ell ,$ thus $N(u)-k_p \le N(u)-j$, a contradiction. Therefore, $N(v)-N(u)+j \not \in \mathcal{A }(\hat{v},\,x)$. But as

$$\begin{aligned} v^2\left(N(v)-\left(N(v)-N(u)+j\right)\right)&= v^2\left(N(u)-j\right)\\&\ge \underline{u}^2\left(N(u)-j\right)\\&> x>\sum _{j=1}^{p-1} \hat{\underline{u}}^1\left(N(u)-k_j\right), \end{aligned}$$

the index $N(v)-N(u)+j$ is strictly larger than $l_p$. This means that some positive second component of $v$ reads

$$\begin{aligned} v^2\left(N(v)-\left(N(v)-N(u)+j\right)\right)-x \ge \underline{u}^2\left(N(u)-j\right)-x. \end{aligned}$$

(55)

Therefore, applying (53), (54) and (55) to any index $i \in [[i_p+1,N(u)-1 ]]$, and to $j=\alpha (i)$ in the case (i) and to $j=\alpha \circ F^{-1} \circ \alpha (i)$ in the case (ii), we obtain that the number $a$ of positive components of $\left(H_3(u,\,x)\right)^2$ is less than the number $b$ of components of $\left(H_3(\hat{v},\,x)\right)^2$, and that there exists an injection $\gamma :[[1,a ]]\rightarrow [[1,b ]]$ such that for all $i$,

$$\begin{aligned} \left(H_2(u,\,x)\right)^2\left(\gamma (i)\right)\le \left(H_2(\hat{v},\,x)\right)^2\left(i\right). \end{aligned}$$

We conclude with Lemma 4, to obtain that

$$\begin{aligned} H_2(u,\,x) \prec _2 H_2(\hat{v},\,x), \end{aligned}$$

which completes the proof. $\square$