Appendix 1: Proofs
Proposition 1
From (7), (xp + xmx)dp — xmxdp + xm(p — p)dx = xpdp. Hence, using (3) to
eliminate xp , we have
dp
dpx
xm x
(18)
(19)
S — (x — xx)xm — xsp
dp _ -(p — p)xm
dxx S — (x — xx)xm — xsp
Proposition 2
Differentiating (6) w.r.t. px , we then obtain 15
|
dP |
=dp+ |
1 |
—+— |
xx dp p')CX)2 xp dp. |
(20) | |
|
By assumption, |
∈p |
p dx = pττ< 0, |
(21) | |||
|
6P |
p dxs = ——> 0. |
(22) |
15We use brackets with superscript 2, (x)2 to define x square function. The superscript 2
without brackets represents variables for the type-2 household.
27