The Benefits of Volume-Conditional Order-Crossing - Singapore ...

This condition is implied by (15). To see this, suppose that ψ is set equal to a value in that range.
The number of patient hedgers in the conditional market when this market clears is then
˜m(1 − λ)
2
> ψ > np − (rL − c)
.
2(rL − c)
The second inequality in (15) simply ensures that the market clears with positive probability, as in
(7). Finally, condition (14) ensures that the interval in (15) in nonempty.
Proof of Proposition 4.2
(i) Suppose first that p ≥ 2κ
λ , so that both impatient hedgers and informed traders participate in
the continuous market if it operates alone, as shown in Proposition 4.1. Let us pick any c ∈ [0, rL]. Let us conjecture that impatient hedgers and informed traders stay in the continuous market after
the conditional market is added, and that patient hedgers start placing orders in the conditional
market. As in the proof of Proposition 3.2, the probability that the conditional market clears under
this conjecture is 1 − ˆm where ˆm = 2ψ
1−λ
number of hedgers in the economy is E ˜m | ˜m ≥ ˆm =
hedgers in the conditional market is
. Also, when the conditional market does clear, the average
1+ ˆm
2
1+ ˆm
2 , so that the average number of patient
(1 − λ). The specialist’s expected profits are therefore
given by (A.6) plus the revenues he generates from the conditional market, that is,
E aλ
˜π MM =
2
+ ˆm
− (p − a)n + (1 − ˆm)1 (1 − λ)c − κ.
2
In equilibrium, this quantity must be equal to zero, and so
a = 2(np + κ) − (1 − ˆm2 )(1 − λ)c
. (A.7)
2n + λ
We need to ensure that all trader types trade in the venue conjectured for the equilibrium. Patient
hedgers do not deviate from trading in the conditional market as c ≤ r L and
a (c≤rL
(rL
2(np + κ) − (1 − ˆm 2 )(1 − λ)2κ
2n + λ
2np + λr L
2n + λ
(p≥ 2κ
λ >rL)
>
(2n + λ)r L
2n + λ
( ˆm≥0)
≥
= r L.
2(np + κ) − (1 − λ)2κ
2n + λ
= 2np + 2λκ
2n + λ
As in the proof of Proposition 3.2, impatient hedgers do not deviate from trading in the continuous
market as long as a ≤ rH and ˆm ≥ a−c
rH−c . The first of these conditions is satisfied as
a (10)
≤ r H(2n + λ) − (1 − ˆm 2 )(1 − λ)c
2n + λ
= r H − (1 − ˆm2 )(1 − λ)c
2n + λ
After replacing a by its expression in (A.7), the second condition reduces to
(1 − λ)c ˆm 2 − (2n + λ)(r H − c) ˆm + 2n(p − c) + (2κ − c) ≤ 0.
29
< r H.