The name is absent

_____s__-'

for p = 1, ..., n. Analogously, the Lorenz curve of x² + y² = (x²₁ + y₁² ≤ ... ≤ x²_n + y_n² ) is
defined by

∑i=₁ (x? + y?)
Σ (*? + y?)

for p = 1, ..., n. Then, from hypothesis, for p = 0, n,

pp

Lx¹+y^{1 (p)} > Lx²+y^{2 (p)} '⇔ ( ( (^xi ^{+ y}i ) > ( ( (^xi ^{+ y}i ) ^.
t=1 t=1

Finally, given 0 < p < n,

p  ppppp

∑ (x¹ + y1) = ∑4 + ∑y1 >∑χ? + ∑y? = Σ (x? + y?)

t=1 t=1 t=1 t=1 t=1 t=1

for X² is LD by x¹, y² is LD by y¹ and the hypothesis of the lemma. ■

Proof. [Proof of Lemma 1] We have

R(kZ ) = ( L(kxi) + K (kyi).
i=1

If we take the derivative of (16) with respect to k and evaluate it in k = 1 we obtain

R(Z )Z = ∑ L(Xi)Xi + K '(yi)yi*
i=1

On the other hand, from (7),

Example 1

Consider the quasi-progressive dual tax schedule T (X, y) = L(X) + K(y)where L(X) and
K(y) are unidimensional tax schedules, on the labor income and on the capital income re-
spectively, that are applied to the following income distribution z = {(2, 1), (3, 4), (6, 5)} in
the way defined in the table below:

33

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