Download Price indices – comparing the arithmetic and geometric mean

1
A Note on Geometric vs Arithmetic Mean
Price index can be calculated as a geometric or an arithmetic mean. What is
a price index? The (minimum) cost to achieve a ’basket’of goods.
Geometric mean
n
PG =
i=1
1
pin
This may be transformed into
n
1X
ln PG =
ln pi
n i=1
n
Interpretation of PG =
i=1
1
pin is that I spend a constant (and in this case equal)
fraction of my total expenditure on each good i ( piExi = n1 ), where E is total
expenditure, xi is quantity of good i, n is the number of goods. Expenditure
n
shares constant =) utility is Cobb-Douglas u(x1 ; :::; xn ) =
Since
i=1
1
xin .
1
xi
pj
pi xi
= then
=
(foc)
E
n
xj
pi
Why is PG the true price index for C-D preferences? It solves the minimum
expenditure (dual consumer problem) to buy u units of utility (see below for
derivation). i.e.
n
e(p; u) =
i=1
1
pin
The composition of the ’basket’ is ‡exible in the sense that I can substitute
n
between goods subject to being on the utility surface
i=1
1
xin = u. The ’basket’
is de…ned by a utility level e.g. u.
Pn
Arithmetic mean (Laspeyre’
s Index) PA = n1 i=1 pi
P
n
Interpretation of PA = n1 i=1 pi is that in my ’basket’ includes 1 unit of
each of n goods. Preferences in this case are Leontief, so the composition of the
basket is …xed, and the ’basket’ may be thought of interchangeably as a …xed
set of goods, or de…ned by the utility level u:
1.1
Simple example
Two goods, x1 and x2 with prices p1 and p2 . Initially p1 = p2 = 1
Arithmetic mean My basket contains 1 unit of each good. Arithmetic
mean is
PA =
1
1
p1 + p2 = 1
2
2
1
Geometric mean My expenditure share on each good is 12 . Since p1 =
1
1
p2 = 1 and u = 1 then e(p1 ; p2 ; u) = 2p12 p22 = 2 and x1 = x2 = 1: Geometric
mean is
1
1
= p12 p22 or
1
1
ln PG =
ln p1 + ln p2
2
2
Suppose both prices increase by 50% (p1 = p2 = 1:5). Then both price
indices also increase by 50%. i.e. PA = PG = 1:5. For the geometric mean,
since pp12 doesn’t change then xx21 doesn’t change. I am buying the same bundle
of goods and it costs 50% more.
Alternatively suppose that p1 increases by 50% and p2 doesn’t change ) pp12
increases to 1:5. Then
PG
1
1
PG = 1 2 1:5 2 = 1:225
1
1
PA = :1 + 1:5 = 1:25
2
2
In the case of the arithmetic mean I continue to consume 1 unit of each good
(as I alway will regardless of (relative) prices).
For the geometric mean, Hicksian (compensated) demand curves with u = 1
are (see below for derivation)
x1 (p1 ; p2 ; u)
=
p2
p1
x2 (p1 ; p2 ; u)
=
p1
p2
p1
p2
=
Since
1
2
1
2
1:5 then x1 =
1
2
1
1:5
1
= 0:816 and x2 = (1:5) 2 = 1:225
There is substitution away from the more expensive good 1 toward the relatively
less expensive good 2. i.e. x1 falls from 1 to 0:816 and x2 increases from 1 to
1:225 . The geometric mean allows for substitution, while the arithmetic mean
does not.
Comment: One of the critisms of CPI based on arithmetric mean is that
it overstates in‡ation because it does not account for substitution away from
relatively more expensive goods. On the other hand the geometric mean allows
substitution (subject to elasticity of substition 1) - (e.g. for the price changes
above PA = 1:25 > PG = 1:225 ).
!1
n
X
As a further extension, when utility is CES, u(x1 ; :::; xn ) =
xj
; then
i=1
the price index is
PG =
n
X
i=1
p1j
"
!11"
where " =
1
1
2
is the elasticity of substitution
This analysis provides a link between volumes and prices!
1.1.1
Why is PG the true price index for C-D Utility?
Solve expenditure minimization problem
min
n
X
n
pi xi s.t. u
=
i=1
foc pi xi
1
i=1
= pj xj for all i; j , xi =
n
Substituting into u
xin
=
n
1
i=1
xin gives u =
i=1
pj
xj
pi
pj
xj
pi
1
n
=
pj xj
n
i=1
xj (p; u)
1 n n1
u p Hicksian Demand
=
pj i=1 i
n
X
Sub xj (p; u) into
pj xj gives
i=j
n
e(p; u) = nu
De…ne units of utility so that u
=
i=1
pi
1
then
n
n
e(p; u) =
1
n
i=1
3
1
pin
1
pin
Expenditure function - 2 good case
1
min E
f oc p1
p1 x1
x1
u
x2 (p1 ; p2 ; u)
1
= p1 x1 + p2 x2 s.t. u = x12 x22
1 u
=
, p1 x1 = u; p2 x2 = u
2 x1
2
2
p2 x2
= p2 x2 () x1 =
p1
p2 x2
=
p1
1
2
1
2
= x1 x2 =
=
p2 x2
p1
1
2
1
x22
1
2
p1
p2
u Hicksian
1
p2 2
x1 (p1 ; p2 ; u) =
u by symmetry
p1
Set u = 1 then
x2
=
1
2
p1
p2
1
x1
Sub x1 (p1 ; p2 ; u) and x2 (p1 ; p2 ; u) into E
p2 2
=
p1
= p1 x1 + p2 x2
1
e(p1 ; p2 ; u)
Davies, Spring 2010
4
=
1
2p12 p22 u expenditure function