Regarding the choice for k(bar) in the exercise 1, I think the answer should be k(bar) = max { k(0) , k* } where k(0) is the initial value of per capita capital stock and k* is the solution to the equation f(k*) = (n + delta)k*.

The reasoning is as follows – Consider the path of pure accumulation defined by c(t) = 0 for all t. Then from equation (12) we get, an equation whereby k(t+1) is expressed in terms of k(t) and other parameters of the model (This equation is obtained by setting the LHS of (12) as 0 and then bringing k(t+1) to the LHS). Then we can say that k(t+1) > ( ( < or =) (n + delta)k(t). So suppose k(0) k*. In that case k will keep on falling till it reaches k*. So maximum value of k = k(0). In the last case when k(0) = k*, the value of k will remain unchanged at this value and so maximum value = k(0) = k*.

Is this answer & reasoning correct ?

With regards,

Bikramaditya Datta

I went through both solutions. They are correct, though, in my opinion, you should search for shorter proofs. However, as I said, there is nothing wrong with the solutions.

Hint for the natural choice for a bound. Suppose bar{k} is the bound you established in your proof. Since k_0 is exogenously specified, is there any reason to suppose that k_0 < bar{k}?

ddg

]]>Dear Bikramaditya:

I will check your solution and reply.

ddg

]]>Done.

]]>I am attaching the partial solution of question 1 and the complete solution of question 2 – can you please take a look at them and comment –

Proof of Exercise 1

With regards,

Bikramaditya Datta

]]>In the last line of page 9 where you have written there exists a number k(subscript 0 and superscript t) with – I think that the orders of t and 0 should be interchanged that is, it should read there exists a number k(subscript t and superscript 0).

With regards,

Bikramaditya Datta ]]>

Regarding the last comment – I have found the relevant theorem – Walter Rudin, theorem 3.24 (Page 60).

With regards,

Bikramaditya Datta ]]>

In the proof of proposition 1, in the 3rd line, when you say that the series is absolutely convergent , the result you have stated – is it the same as what is known as Abel’s Convergence test – and are you applying that test to the absolute value of U in the proof ?

With regards,

Bikramaditya Datta ]]>

Dear Bikramaditya:

Since U is really a redefinition of u and the latter is differentiable by assumption, one need not prove that U is differentiable.

ddg

]]>In exercise 2 for proving U to be concave do I also have to first proof the differentiability of U (for we are using differentiability to construct the Hessian) ?

With regards,

Bikramaditya Datta ]]>