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tel-00703797, version 2 - 7 Jun 2012
ceiling function. Indeed, we have
4.4. Focus on the dependence structure
P (W = k) = qP (0 = k) + (1 − q)P (⌈Y ⌉ = k) = qδk,0 + (1 − q)(1 − δk,0)e −θ(k−1) (1 − e −θ ).
Furthermore, P (W ≤ x) = q + (1 − q)P (⌊Y ⌋ < ⌊x⌋) = q + (1 − q)(1 − e −ν(⌊x⌋) ) for all x ∈ R+.
In particular for x = k ∈ N, we have FW (k) = q + (1 − q)(1 − e −θk ).
Let Wi = Ii⌈Yi⌉ with Ii’s and Yi be conditionnaly independent exponential random variables,
i.e. Yi|Θ = θ ∼ E(θ). We have
and
P (Wi ≤ x) = 1 − (1 − q)LΘ(⌊x⌋),
P (W1 ≤ x, W2 ≤ y) = 1 − (1 − q)(LΘ(⌊x⌋) + LΘ(⌊y⌋)) + (1 − q) 2 LΘ(⌊x⌋ + ⌊y⌋).
Let DW1,W2 be the distribution function of the pair (FW1 (W1), FW2 (W2)). Similarly to the
previous subsection, we get
0
DW1,W2 (u, v) =
P (W1 ≤ lu, W2 ≤ lv)
if u < q or v < q
otherwise
.
And we get for u, v ≥ q,
DW1,W2 (u, v) = 1 − (1 − q)(LΘ(⌊lu⌋) + LΘ(⌊lv⌋)) + (1 − q) 2 LΘ(⌊lu⌋ + ⌊lv⌋).
There is also a jump when u or v equal q. Indeed, we have DW1,W2 (q, v) = q −q(1−q)LΘ(⌊lv⌋)
and DW1,W2 (u, v) ≥ q2 for u, v ≥ q.
Differences between DZ1,Z2
and DW1,W2
Let u ∈ Im(FW1 ) and v ∈ Im(FW2 ), i.e. there exist n, m ∈ N, such that u = FW1 (n) and
v = FW2 (m). Firstly, we have
u = FW1 (n) = FZ1 (n + 1) and v = FW2 (m) = FZ2 (m + 1).
Since we have lu = n and lv = m, we have
DZ1,Z2 (u, v) = 2 − (1 − q)(LΘ(n) + LΘ(m)) − 1 + (1 − q) 2 LΘ(n + m) = DW1,W2 (u, v).
Now, we focus on the maximal difference between these two functions, after deriving an
elementary lemma.
Lemma 4.4.5. Let f be a completely monotone function, c > 0 and p ∈ [0, 1]. The function
fc,p : x ↦→ f(x) − pf(x + c) is also completely monotone.
Proof. As f being completely monotone, (−1) n f (n) is decreasing. Thus, we have
(−1) n f (n) (x) ≥ (−1) n f (n) (x + c) ≥ p(−1) n f (n) (x + c),
as x < x + c. Hence, (−1) n f (n)
c,p (x) ≥ 0, i.e. fc,p is completely monotone.
Proposition 4.4.6. Let ∆i,j be the difference between functions DW1,W2 and DZ1,Z2 at the
point u = FW1 (i) and v = FW1 (i). The maximal stairstep is ∆∞,1 = ∆1,∞.
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