Dynare timing and redefinition

Stochastic simulations
1

Hi Prof, I am new to dynare and I have a puzzle on timing and redefining:

In usual stochastic growth model, suppose I have production function as y=z k^alpha, then in dynare timing it is y=z k(-1)^alpha.
define: m=z alpha k(-1)^(alpha-1), then I focus on m(+1).
Also define: n=z(+1) alpha k^(alpha-1).
My questions are :

  1. Are m(+1) and n the same? conceptually and numerically?
  2. How do I interpret the above? Does m refers to realized value, and n refers to expected value? If NOT, how should I construct the expected & realized values in this case?

Many thanks.

2

Without knowing how m(+1) is used, it is impossible to tell. But yes, n is the expected value, while m is the realization. You may want to have a look at Expected value of a power

3

Hi Prof., Thanks for your reply!
I am using it in an Euler equation. The original formula is
1/c(t)=beta E_{t} [1/c(t+1) (alpha z(t+1) k(t+1)^ (alpha-1) +1 -delta].
If I put into dynare, i need to change k(t+1) to k(t).

So I wonder if I define m as: m= z alpha k(-1)^(alpha-1)
Then is it correct to write the above Euler equation as:
1/c=beta 1/c(+1) (m(+1)+1-delta) ?
Or should I ONLY define n as: n= z(+1) alpha k^(alpha-1)
and write 1/c=beta 1/c(+1) (n+1-delta)?

It seems that I check it computationally, the diff is around 10^-16. Both seem fine.

Thanks.

4

At first order, both are equivalent and you do not need to worry.At second order, your definition with n would theoretically be different, because it contains the conditional expectations.

5

Hi Prof. Thanks for your reply.
Sorry but I dont think i understand the 1st order 2nd order thing. I assume you are referring to order of perturbation. But for my understanding, whatever order I do( or if I do second order), in dynare, m refers to realized element, BUT m(+1) still contains an expectation? so still equivalent to n?

Or, i wonder whether you have any example for the 1st order and 2nd order you mentioned, for cases m(+1) and n are different?

Many thanks

6

n= z(+1) alpha k^(alpha-1)
means
n= E_t[z_{t+1} \alpha k_t^{\alpha-1}]
When you write
1/c=\beta 1/c(+1) (n+1-\delta)
you get
1/c_t=E_t[\beta 1/c_{t+1} (E_t[z_{t+1} \alpha k_t^{\alpha-1}]+1-\delta)]
Because of Jensen’s Inequality, this is not equal to
1/c_t=E_t[\beta 1/c_{t+1}(z_{t+1} \alpha k_t^{\alpha-1}+1-\delta)]
i.e. you cannot pull the second E_t out.

7

Ohh! Got this! Thanks Prof.!

But following to that, the same problem should also occur if I use
m=z alpha k(-1)^(alpha-1)

and put m(+1) in replace of n in the Euler equation. Because in this case, m(+1) is also an expectation term, which cannot be pulled out?

And further, if I do want to show what my original Euler equation means, under the re-definition of marginal product of capital (say, to save place), how should I do this to avoid the Jensen Inequality? Seems no alternative but type the marginal product out?

Thanks and sorry for keeping disturbing you.

8

No, in Dynare the conditional expectation is always around the full equation by definition. In the definition of the marginal product

both sides are contained in the information set a time t. Thus, the conditional expectation E_t drops out and does not shop up when you shift m one period into the future.

9

Hi Prof., thanks for your reply.
So it implies I should always use m and put m(+1) in my Euler equation. for n, it would cause trouble? Hence I’d better stick to m and m(+1).

Thanks again for your help and patience.

10

Yes, for the first part. For the second one, again if you are doing a first order approximation then Jensen’s Inequality will not matter because everything is linear. Problems therefore will only arise at order>1.

11

That’s great! I am very clear now. Thanks for your help a lot!