Appendix
We first need some definitions and preliminary lemmas on multivariate stochastic orders.
Definition A1
Let \({\mathbf{X}}{ = (}X_{1} ,X_{2} ,\ldots,X_{n} )\) and \({\mathbf{Y}}{ = (}Y_{1} ,Y_{2} ,\ldots,Y_{n} )\) be two \(n\)-dimensional random vectors such that
$$ P({\mathbf{X}} \in U) \le P({\mathbf{Y}} \in U),\,{\text{for all upper sets}}\,U \subseteq {\mathbf{R}}^{n} . $$
Then, \({\mathbf{X}}\) is said to be smaller than \({\mathbf{Y}}\) in the usual stochastic order, denoted by \({\mathbf{X}} \le_{st} {\mathbf{Y}}\).
Lemma A1
(Shaked and Shanthikumar 2007) Let \({\mathbf{X}}{ = (}X_{1} ,X_{2} ,\ldots,X_{n} )\) and \({\mathbf{Y}}{ = (}Y_{1} ,Y_{2} ,\ldots,Y_{n} )\) be two \(n\)-dimensional random vectors. If
$$ X_{1} \le_{st} Y_{1} $$
and, for \(i = 2,3,\ldots,n\)
\((X_{i} |X_{1} = x_{1} ,X_{2} = x_{2} ,\ldots,X_{i - 1} = x_{i - 1} ) \le_{st} (Y_{i} |Y_{1} = y_{1} ,Y_{2} = y_{2} ,\ldots,Y_{i - 1} = y_{i - 1} )\) whenever \(x_{j} \le y_{j}\), \(j = 1,2,\ldots,i - 1\), then \({\mathbf{X}} \le_{st} {\mathbf{Y}}\).
Lemma A2
(Shaked and Shanthikumar 2007) Let \({\mathbf{X}}{ = (}X_{1} ,X_{2} ,\ldots,X_{n} )\) and \({\mathbf{Y}}{ = (}Y_{1} ,Y_{2} ,\ldots,Y_{n} )\) be two \(n\) -dimensional random vectors. If \({\mathbf{X}} \le_{st} {\mathbf{Y}}\) and \({\mathbf{g}}\) : \({\mathbf{R}}^{n} \to {\mathbf{R}}^{k}\) is any \(k\)-dimensional increasing (decreasing) function, for any integer \(k\), then the \(k\)-dimensional vectors \({\mathbf{g}}({\mathbf{X}})\) and \({\mathbf{g}}({\mathbf{Y}})\) satisfy \({\mathbf{g}}({\mathbf{X}}) \le_{st} ( \ge_{st} ){\mathbf{g}}({\mathbf{Y}})\) .
Theorem A1
Suppose that the distribution
\(G(x)\)
is DFR, i.e.,
\(\lambda_{L} (x)\)
is decreasing in
\(x\)
. Then,
$$ L_{1} \le_{st} L_{2} \le_{st} L_{3} \le_{st} \ldots. $$
Proof
First, we show \(L_{1} <_{st} L_{2}\). Observe that
$$ P(L_{1} > w) = \overline{G}(w) = \exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (x){\rm d}x } \right\}, $$
whereas
$$ P(L_{2} > w|L_{1} = w_{1} ) = \overline{G}(w|qw_{1} ) = \exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (qw_{1} + x){\rm d}x}\right\} , $$
and \(P(L_{2} > w) = E[P(L_{2} > w|L_{1} )]\). When \(\lambda_{L} (x)\) is decreasing,
$$ P(L_{1} > w) \le P(L_{2} > w|L_{1} = w_{1} ),\,{\text{for }}\,{\text{all }}\,{\text{realization }}\,{\text{of}}\,w_{1} > 0,\,{\text{for all}}\,w > 0. $$
(A1)
This implies that \(P(L_{1} > w) \le E[P(L_{2} > w|L_{1} )] = P(L_{2} > w)\), for all \(w > 0\).
Observe that
$$ P(L_{3} > w|L_{1} = w_{1} ,L_{2} = w_{2} ) = \overline{G}(w|q(qw_{1} + w_{2} )) =\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q(qw_{1} + w_{2} ) + x){\rm d}x}\right\} . $$
Thus, \(P(L_{3} > w|L_{1} = w_{1} ,L_{2} ) =\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{2} w_{1} + qL_{2} + x){\rm d}x}\right\} \ge\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (qL_{2} + x){\rm d}x }\right\}\), and
$$\begin{aligned}& P(L_{3} > w|L_{1} = w_{1} ) = E_{{(L_{2} |L_{1} = w_{1} )}} [P(L_{3} > w|L_{1} = w_{1} ,L_{2} )] \\ &\qquad\ge E_{{(L_{2} |L_{1} = w_{1} )}} \left[\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (qL_{2} + x){\rm d}x}\right\} \right ].\end{aligned} $$
On the other hand, \(P(L_{2} > w) = E\left[\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (qL_{1} + x){\rm d}x }\right\} \right]\). As \(L_{1} \le_{st} (L_{2} |L_{1} = w_{1} )\), for all realization of \(w_{1} > 0\), due to Lemma 1-(i),(ii),
$$ P(L_{3} > w|L_{1} = w_{1} ) \ge E_{{(L_{2} |L_{1} = w_{1} )}} \left[\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (qL_{2} + x){\rm d}x}\right\} \right] $$
\(\ge E\left[\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (qL_{1} + x){\rm d}x}\right\} \right] = P(L_{2} > w)\), for all realization of \(w_{1} > 0\), for all \(w > 0\),
which implies that \(P(L_{3} > w) \ge P(L_{2} > w)\), for all \(w > 0\).
In general, for \(n = 3,4,\ldots\),
$$ \begin{aligned} & P(L_{n} > w|L_{1} = w_{1} ,L_{2} = w_{2} ,\ldots,L_{n - 1} = w_{n - 1} ) \\ &\quad = \exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{n - 1} w_{1} + q^{n - 2} w_{2} + \cdots + qw_{n - 1} + x){\rm d}x}\right\}\end{aligned} $$
and
$$ \begin{aligned} & P(L_{n + 1} > w|L_{1} = w_{1} ,L_{2} = w_{2} ,\ldots,L_{n} = w_{n} ) \\ &\quad =\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{n} w_{1} + q^{n - 1} w_{2} + \cdots + qw_{n} + x){\rm d}x}\right\} . \end{aligned} $$
Thus,
$$ P(L_{n} > w) = E_{{(L_{1} ,L_{2} \ldots,L_{n - 1} )}} \left[\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{n - 1} L_{1} + q^{n - 2} L_{2} + \cdots + qL_{n - 1} + x){\rm d}x}\right\} \right ] $$
and
$$ \begin{aligned} & P(L_{n + 1} > w|L_{1} \\ & \quad = w_{1} ) = E_{{(L_{2} ,L_{3} \ldots,L_{n} |L_{1} = w_{1} )}} [P(L_{n + 1} > w|L_{1} = w_{1} ,L_{2} ,L_{3} ,\ldots,L_{n} )] \\ &\quad \ge E_{{(L_{2} ,L_{3} \ldots,L_{n} |L_{1} = w_{1} )}} \left[\exp\left \{ - \int\limits_{0}^{w} {\lambda_{L} (q^{n - 1} L_{2} + \cdots + q^{2} L_{n - 1} + qL_{n} + x){\rm d}x}\right\} \right]. \\ \end{aligned} $$
Now we stochastically compare \((L_{1} ,L_{2} ,\ldots,L_{n - 1} )\) with \((L_{2} ,L_{3} ,\ldots,L_{n} |L_{1} = w_{1} )\). From (A1),
$$ L_{1} \le_{st} (L_{2} |L_{1} = w_{1} ). $$
Also,
$$ \begin{aligned} P(L_{i} > w|L_{1} & = w_{1} ,L_{2} = w_{2} ,\ldots,L_{i - 1} = w_{i - 1} ) \\ & =\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{i - 1} w_{1} + q^{i - 2} w_{2} + \cdots + qw_{i - 1} + x){\rm d}x}\right\} , \\ \end{aligned} $$
whereas
$$ \begin{aligned} P(V_{i} & > w|L_{1} = w_{1} ,V_{1} = v_{1} ,V_{2} = v_{2} ,\ldots,V_{i - 1} = v_{i - 1} ) \\ & =\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{i} w_{1} + q^{i - 1} v_{1} + \cdots + qv_{i - 1} + x){\rm d}x}\right\} , \\ \end{aligned} $$
where \(V_{j} = L_{j + 1}\), \(j = 1,2,\ldots\). Then, we have
$$ \begin{aligned} &\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{i - 1} w_{1} + q^{i - 2} w_{2} + \cdots + qw_{i - 1} + x){\rm d}x}\right\} \\ &\quad \le\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{i} w_{1} + q^{i - 1} v_{1} + \cdots + qv_{i - 1} + x){\rm d}x}\right\} , \end{aligned} $$
for all \(w > 0\), whenever \(w_{j} \le v_{j}\), \(j = 1,2,\ldots,i - 1\), which implies that
$$\begin{aligned} & (L_{i} |L_{1} = w_{1} ,L_{2} = w_{2} ,\ldots,L_{i - 1} = w_{i - 1} ) \\ &\quad \le_{st} (V_{i} |L_{1} = w_{1} ,V_{1} = v_{1} ,V_{2} = v_{2} ,\ldots,V_{i - 1} = v_{i - 1} ), \end{aligned} $$
(A2)
whenever \(w_{j} \le v_{j}\), \(j = 1,2,\ldots,i - 1\). Inequality (A2) holds for all \(i = 2,\ldots,n - 1\). Thus, by Lemma A1, \((L_{1} ,L_{2} ,\ldots,L_{n - 1} ) \le_{st} (V_{1} ,V_{2} ,\ldots,V_{n - 1} |L_{1} = w_{1} )\) and thus
$$ (L_{1} ,L_{2} ,\ldots,L_{n - 1} ) \le_{st} (L_{2} ,L_{3} ,\ldots,L_{n} |L_{1} = w_{1} ). $$
(A3)
Observe that
$$ h(w_{1} ,w_{2} ,\ldots,w_{n - 1} ) =\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{n - 1} w_{1} + q^{n - 2} w_{2} + \ldots + qw_{n - 1} + x){\rm d}x}\right\} $$
is an increasing function of \((w_{1} ,w_{2} ,\ldots,w_{n - 1} )\). Then, from (A3) and Lemma A3,
$$ \begin{aligned} P(L_{n} > w) & = E_{{(L_{1} ,L_{2} \ldots,L_{n - 1} )}} \left[ {\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{n - 1} L_{1} + q^{n - 2} L_{2} + \ldots + qL_{n - 1} + x){\rm d}x}\right\} } \right] \\ & \le E_{{(L_{2} ,L_{3} \ldots,L_{n} |L_{1} = w_{1} )}} \left[\exp \left\{ - \int\limits_{0}^{w} {\lambda_{L} (q^{n - 1} L_{2} + \cdots + q^{2} L_{n - 1} + qL_{n} + x){\rm d}x}\right\} \right] \\ &\le P(L_{n + 1} > w|L_{1} = w_{1} ), \\ \end{aligned} $$
for all realization of \(w_{1} > 0\), for all \(w > 0\). This again implies that
$$ P(L_{n} > w) \le P(L_{n + 1} > w),\,{\text{ for }}\,{\text{all}}\,w > 0. $$
Therefore, we have shown that \(P(L_{n} > w) \le P(L_{n + 1} > w)\), for all \(n = 1,2,\ldots\)
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Theorem A2
Suppose that the distribution
\(G(x)\)
is DFR, i.e.,
\(\lambda_{L} (x)\)
is decreasing in
\(x\)
. Then,
$$ Z_{1} \le_{st} Z_{2} \le_{st} Z_{3} \le_{st} \ldots. $$
Proof
It is obvious that \(Z_{1} \le_{st} Z_{2}\). In the proof of Theorem A1, we have shown that
$$ (L_{1} ,L_{2} , \ldots ,L_{n} ) \le_{st} (L_{2} ,L_{3} , \ldots ,L_{n + 1} |L_{1} = w_{1} ). $$
Furthermore,
$$ h(w_{1} ,w_{2} ,\ldots,w_{n - 1} ) = q^{n - 1} w_{1} + q^{n - 2} w_{2} + \cdots + q^{{}} w_{n - 1} $$
is an increasing function of \((w_{1} ,w_{2} ,\ldots,w_{n - 1} )\). Then, from Lemma A2,
$$ \begin{aligned}&q^{n-1}L_1+ q^{n-2}L_2+\cdots+qL_{n-1}\le_{st} (q^{n-1}L_2+q^{n-2}L_3+\cdots+qL_n|L_1=w_1 )\\ &\quad\le_{st} (q^{n}w_1+q^{n-1}L_2+q^{n-2}L_3+\cdots+qL_n|L_1=w_1 )\end{aligned}$$
which implies that \(Z_{n} \le_{st} (Z_{n + 1} |L_{1} = w_{1} )\), for all realizations of \(w_{1}\). Therefore, we have
$$ Z_{n} \le_{st} Z_{n + 1} ,\,n = 1,2,\ldots $$
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