Creating a shock

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Creating a shock

Postby aleksey73 » Fri Jul 08, 2011 8:14 am


I have a model forecasting the oil price that I built using add factors. I would like to simulate a shock that my add factors aer 10$ below what I think they should be from 2010 to 2020.

What I did is, for the sample 2010 to 2020, I had 10$ to the value of my add factors, then I put the sample back to 2000 to 2020.
After I solve the model and I was expecting to have a constant 10$ difference between my baseline and actuals values, but i had something going from 8$ to 15$... So I was wondering if there was an other way to solve the model to be sure that we'll get a 10$ difference between the baseline and actual values.

Thank you!

EViews Chris
EViews Developer
Posts: 161
Joined: Wed Sep 17, 2008 10:39 am

Re: Creating a shock

Postby EViews Chris » Mon Jul 11, 2011 2:05 pm

There's a few potential issues here - I'm afraid I don't have enough details on exactly what you are doing to know what is going on.

If your model is a multiple equation model, including an add factor on a single equatiion will not necessarily change that variable by the specified amount because the equation is part of a simultaneous system. Some of the impact of the add factor may be 'absorbed' by other parts of the system so that the particular endogenous variable may not move by the specified amount.

Another issue can come up if the equation in your model has an expression on the left hand side of the equals sign. In this case you will have to be careful in choosing whether the add factor is applied to the intercept or on the endogenous variable of the equation.

As a simple example, if your equation is

log(y) = x + 5

Then adding an intercept (residual) add factor will make the expression:

log(y) = x + 5 + y_a

which means that:

y = exp(x + 5 + y_a)

so that changes from y_a will be inside the exponentitation and will not be one-to-one.

On the other hand, if use you an 'endogenous variable shift' add factor, then the expression will be:

log(y - y_a) = x + 5

so that:

y = exp(x + 5) + y_a

so the add factor is outside the exponentiation.

Hope that helps.


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