Dynare timing and redefinition

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.

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

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.


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.

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

n= z(+1) alpha k^(alpha-1)
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.

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.

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.

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.

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.

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