# Assumptions of Ramsey Cass Model

Field of Economics became mathematical and model based. I can grasp the appeal of the model but fail to understand why do we introduce certain assumptions.

(Actually I am trying to relax the assumptions of the model, Ramsey Cass Kooperman (RCK) model, just as an exercise but before I can relax them I need to know why they were significant in the first place)

So, in the context of RCK model Why does the assumption of

1.Compactness,

2.twice continuously differentiable function, and

1. homogeneity (homothetic) are taken.

Why are the relevant?

I know this that homogeneity produce constant returns to scale with Cobb Douglas production function and lt also implies homothetic. Why do we need homothetic functions?

• Hi: I don't know the other ones but twice differentiable implies that, if the hessian of the function is psd, then the function is convex. The convexity of a function implies that the optimization of the function will lead to a global maximum ( minimum ) which is helpful when optimizing it. Also, if a function is convex, then the algorithms for optimizing it are more straightforward. – mark leeds Jul 24 '18 at 13:25
• just read it again and what I said is related to the concavity of profuction function because a function is convex if and only if the negative of it is concave. So, if you have a concave function, you can multiply by -1 to make it convex. – mark leeds Jul 24 '18 at 19:36
• your welcome: I'd be interested in reasons for compactness ( I think it has to do with implying boundedness of the function but you may need other conditions also ) and constant elasticity of substitution so, if you get any answers for those outside of this thread and can post them here, it's appreciated. – mark leeds Jul 24 '18 at 23:02
• I'll definitely post them here! – Elina Gilbert Jul 25 '18 at 5:51
• Τhis post asks for a compact introduction to almost all the underlying mathematical theory of core micro- and macro-economics, and their logic (for the supplier side). And then, the "what happens if they are relaxed" sends us to explore Space... this is too broad to be answerable, please try to considerably narrow it down. Your 2nd post about the Euler things was much more suitable than this one, so you can do it. – Alecos Papadopoulos Aug 21 '18 at 9:51

In the presentation that the OP linked to, "compactness" is not assumed, but derived as a result from more primitive properties that are assumed, one of which is the "twice differentiability" of the functions involved.

The presentation deals with the vary basic deterministic growth model in discrete time. The per capita production function in each period is $g(k_t)$ for which it is assumed that it is continuous and differentiable $g'>0,\;\; g''<0$. The law of motion of capital is then

$$k_{t+1} = g(k_t) + (1-\delta)k_t -c_t$$

with $0<\delta<1$ being the depreciation rate. The sequence starts at some arbitrary, finite initial value.

The presentation wants to show the existence and uniqueness of the solution. To do that, in page 6 it first proves that the "choice set" is bounded -namely that the set from which the economic agents may choose the optimal levels of consumption and capital is not the whole $\mathbb R_+$.

This is proved by considering the maximal sequence of capital levels, irrespective of utility optimization considerations. If the maximal sequence is bounded, it means that the optimal sequence can only be chosen from a bounded set of values.

The maximum capital sequence obtains if we set consumption equal to zero: what happens when we consume nothing and we use all our capital stock and all the new output as capital only. We have then the sequence

$$k^{max}_{t+1} = g(k^{max}_{t}) + (1-\delta)k^{max}_{t}$$

Now that consumption is set to zero, the sequence of capital is governed by a non-linear first-order difference equation. A sufficient (although not necessary) condition for the maximal sequence to be bounded is that this difference equation is asymptotically globally stable, namely, if starting from any initial point we converge to a fixed point (it is not necessary because we could have bounded oscillations also and obtain a bounded sequence).

The fixed point of the maximal capital sequence satisfies

$$k^{max}_*: k^{max}_{t+1}=k^{max}_{t} \implies g(k^{max}_*) = \delta k^{max}_*$$

By standard theory, global stability of the fixed point will obtain if the derivative of the difference equation, evaluated at the fixed point, is less than unity in absolute value,

$$\left |\frac {\partial k^{max}_{t+1}(k^{max}_*)}{\partial k^{max}_{t}}\right| = g'(k^{max}_*)+1-\delta <1 \implies g'(k^{max}_*) < \delta$$

At the fixed point, it holds that

$$\frac {g(k^{max}_*)}{k^{max}_*} = \delta$$

For fixed point stability we want

$$g'(k^{max}_*) < \delta$$

So fixed point stability requires

$$g'(k^{max}_*)<\frac {g(k^{max}_*)}{k^{max}_*}$$

or that "the marginal product will be less than the average product". Under the assumption that $g''<0$ everywhere, this will also hold for the fixed point. Hence, we obtain fixed point stability, and so also convergence and so boundedness of the maximal capital sequence. This tells us why we assume 2nd-order differentiability: by having the 2nd derivative and making assumptions about it, we can easily determine whether the equilibrium points are unique and things like that. Note also that differentiability of any degree implies continuity.

If the maximal capital sequence is bounded, so will be any other capital sequence, and hence the choice set for capital is bounded. If it is for capital it is also for consumption. Then, the full capital equation expressed as an inequality

$$k_{t+1} \leq g(k_t) + (1-\delta)k_t -c_t$$

defines a set (a unidimensional set, an interval on the real line) that is a) bounded below at zero (by the assumption that capital cannot be negative b) bounded above as proven c) it can attain its boundary points. So it is closed and bounded and therefore compact.

More over, we assume that there are no "holes" in it, that there are not "forbidden" values in this set. This makes the set convex.

Given these properties (incl. continuity of the production function), by standard theorems we are assured that a maximum, i.e. a solution exists, and it is unique, which is what the presentation wanted to prove at this point.