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probit: marginal effects [Jan. 6th, 2008|11:29 am]
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I've got already the next question: In a Probit model for whether or not a household has any expenditure on tobacco would the following interpretation for the marginal effect (value = -0.1)of an independent variable "log of total expenditures" be correct? If total expenditures increase 1% the probability of that there are expendiutres on tobacco decreases 10%?

Thanks again!!

[User Picture]From: warhol
2008-01-06 07:58 pm (UTC)
It means that a 10% increase in total expenditures reduces the probability of spending on tobacco by one percentage point.
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From: pantaa_rheii
2008-01-07 01:00 pm (UTC)
are you sure? i just found this:

"Again, it might be useful to calculate the "elasticities" or effects on the probabilities for a one unit change or a one standard deviation change in each X from the sample mean, holding the other X constant at the sample means. For age, a one year increase reduces the probability of liberalism by .0024, a standard deviation increase in age reduces it by .0340. For education, a one year increase reduces the probability of liberalism by .0049; a one standard deviation unit increase reduces the probability by .0165."
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[User Picture]From: warhol
2008-01-07 01:39 pm (UTC)
The logit and probit are different models, and they are interpreted differently.

If you have indeed estimated a probit, and if you have indeed calculated the marginal effects of the variables (not the straight coefficients), and the marginal effect of "log x" is -0.1, then the interpretation is that a 1 unit increase in log(x), which represents a 100% increase in x, reduces P[y=1] by 0.1; that is, a 100% increase in x reduces the probability by 10 percentage points.
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From: pantaa_rheii
2008-01-07 02:26 pm (UTC)
thanks again!

so is this how I always (e.g. in ols as well) have to interpret a coefficient if the dependet variable is in the log-form -- a 100% increase of x leads to xxx increase of xxx?

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[User Picture]From: warhol
2008-01-07 03:35 pm (UTC)
Yes. A change in the (natural) logarithm is always interpreted as (approximately) a percentage change: ∆ln(z) ≈ ∆z/z.
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