Answer to Question #157848 in Microeconomics for ved

Question

Let Y be a production set. We say that the technology is

additive if y in Y and y

in Y implies y + y

in Y . We say that the technology

is divisible if y in Y and 0< t<  1 implies that ty is in Y . Show that if a

technology is both additive and divisible, then Y must be convex and exhibit

constant returns to scale.

Answer 1

a) PROVE CONVEXITY: In simple words a set is convex if, given any two points within the set, it contains the whole line segment that joins them.

In mathematical terms, the line segment joining two points is the weighted average of the 2 points with the weights adding up to 1. Say the points are y and y" and the corresponding weights are a and b.

Then the expression for the line segment is (ay + by") where [a + b = 1].

^ Easy intuition: If a = 0, b = 1 we are at one extreme which is y". If a = 1 and b = 0 we are at another extreme which is y. Intermediate combinations represent different bundles of y and y".

RTP: (ay + by") belongs to Y || given [a + b = 1] & [y and y" belong to Y]

Observe that if [a + b = 1] then [0 <= a <= 1] - same for b. Because y and y" both belong to Y, we can use the divisibility criteria to say ay and by" both belong to Y.

Next we can use the additivity criteria to say if ay and by" belong to Y, the (ay + by") also belongs to Y.

Hence Y is convex.

b) PROVE CRS: The production set Y exhibits constant returns to scale (CRS) if y ∈ Y implies αy ∈ Y for any scalar α ≥ 0

Continuing the proof above, y and y" belongs to Y implies (y + y") belongs to y {additivity criteria}. But y and y" need not be unique; if y = y", (y + y") = 2y also belongs to Y. Thus with 2y and y both in Y, (2y + y) = 3y also belongs in Y.

In general with cy and dy both in Y, (c + d)*y belongs in y; we have seen so far that c and d can be non-negative integers; so knowing the closure property of addition in this set we can say (c + d) will also be a non-negative integer.

If we incorporate the divisibility criteria then (c + d) can be any real number i.e. for any scalar alpha => 0, you can express alpha = (c + d) s.t. cy and dy are both in Y and so (c + d)*y will also be in Y.

Thus CRS is proved.