The name is absent



_____s__-'

for p = 1, ..., n. Analogously, the Lorenz curve of x2 + y2 = (x21 + y12 ... x2n + yn2 ) is
defined by

Lx2+y2 (p) =


∑i=1 (x? + y?)
Σ (*? + y?)

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

pp

Lx1+y1 (p) > Lx2+y2 (p) ' ( ( (xi + yi ) > ( ( (xi + yi ) .
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

(16)


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),

Z?A(Z) + R(Z)

Z?A(Z)

Z?A(Z)


n

∑ L(Xi)Xi + K'(yi)yi
i=1


nn

( (L'(Xi)Xi + K'(yi)yi) ( (L(Xi) + K(yi))


n
Σ
i=1


L(Xi)Xi - L(Xi) 2

™2        Xi

Xi


K'(yi)yi - K(yi)
y?


yi? .


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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