Download Lecture 27: Binary-Valued Dependent Variables

Lecture 28:
Binary-Valued
Dependent
Variables
(Chapter 19.1–19.2)
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Agenda
• Binary-Valued Dependent Variables
(Chapter 19.1)
• Probit/Logit Models (Chapter 19.2)
• Estimating a Probit/Logit Model
(Chapter 19.2)
• Deriving Probit/Logit (Chapter 19.2)
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28-2
Binary Dependent Variables
• We have worked extensively
with regression models in which Y
is continuous.
• We have predicted the effect of
education and experience on earnings.
• We have predicted the effect of
exogenous changes in price on
quantity demanded.
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28-3
Binary Dependent Variables (cont.)
• However, our methods are inappropriate
when the dependent variable takes on just a
few discrete values.
• For example, we may be interested in the
effect of a brand’s advertising on consumers’
decisions to buy that brand.
• We may want to predict the effect of
Head Start on children’s graduating from
high school.
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28-4
Binary Dependent Variables (cont.)
• Discrete-valued dependent variables
are a special case that comes up
sufficiently frequently to warrant its own
special techniques.
• In this lecture, we will focus on
dependent variables that can take on
only 2 values, 0 or 1 (dummy variables).
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28-5
Binary Dependent Variables (cont.)
• Suppose we were to predict whether
NFL football teams win individual
games, using the reported point spread
from sports gambling authorities.
• For example, if the Packers have a
spread of 6 against the Dolphins, the
gambling authorities expect the Packers
to lose by no more than 6 points.
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28-6
Binary Dependent Variables (cont.)
• Using the techniques we have
developed so far, we might regress
Win
i
D
  0  1Spreadi   i
where i indexes games
• How would we interpret the
coefficients and predicted values from
such a model?
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28-7
Binary Dependent Variables (cont.)
Win
i
D
  0  1Spreadi   i
• DiWin is either 0 or 1. It does not make sense
to say that a 1 point increase in the spread
increases DiWin by 1. DiWin can change only
from 0 to 1 or from 1 to 0.
• Instead of predicting DiWin itself, we predict
the probability that DiWin = 1.
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28-8
Binary Dependent Variables (cont.)
Win
i
D
  0  1Spreadi   i
• It can make sense to say that a 1 point
increase in the spread increases the
probability of winning by 1.
• Our predicted values of DiWin are the
probability of winning.
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28-9
Binary Dependent Variables (cont.)
DiWin  0  1Spreadi  i
• When we use a linear regression model
to estimate probabilities, we call the
model the linear probability model.
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28-10
TABLE 19.1 What Point Spreads Say About
the Probability of Winning in the NFL: I
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28-11
Binary Dependent Variables
• Note that the table reports White Robust
Estimated Standard Errors.
• The Linear Probability Model
disturbances are heteroskedastic.
• Heteroskedasticity is the only violation
of the Gauss–Markov assumptions
inherent in using dummy variables as Y.
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28-12
Binary Dependent Variables (cont.)
• The linear probability model works
fine mathematically.
• However, it faces a serious drawback
in interpretation.
• If the point spread is 21 points, the
team’s predicted probability of
winning is:
0.5 - 0.025 • 21 = -0.025
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28-13
Binary Dependent Variables (cont.)
• If X = 21, E(Y | X ) = -0.025
• We predict that the team has a -2.5%
probability of victory.
• If X = -21, we predict that the team has
a 102.5% probability of victory.
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28-14
Figure 19.1 For Some X-Values, E(D|Xi) > 1
For Some Other Values E(D|Xi) < 0
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28-15
Binary Dependent Variables
• Linear regression methods predict
values between -∞ and +∞.
• Probabilities must fall between 0 and 1.
• The linear probability model cannot
guarantee sensible predictions.
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28-16
Binary Dependent Variables (cont.)
• Intuitively, if the linear probability model
predicts a -2.5% chance of victory, we
expect the team to have a very small
probability of winning.
• Similarly, if we predict that the team has
a 102.5% chance of victory, we expect
the team to have a very high probability
of winning.
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28-17
Binary Dependent Variables (cont.)
• We need a procedure to translate
our linear regression results into
true probabilities.
• We need a function that takes a value
from -∞ to +∞ and returns a value from
0 to 1.
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28-18
Binary Dependent Variables (cont.)
• We want a translator such that:



The closer to -∞ is the value from our linear
regression model, the closer to 0 is our
predicted probability.
The closer to +∞ is the value from our
linear regression model, the closer to 1 is
our predicted probability.
No predicted probabilities are less than 0
or greater than 1.
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28-19
Figure 19.2 A Graph of Probability of
Success and X
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28-20
Binary Dependent Variables
• How can we construct such a translator?
• How can we estimate it?
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28-21
Probit/Logit Models (Chapter 19.2)
• In common practice, econometricians use
TWO such “translators”:


probit
logit
• The differences between the two models
are subtle.
• For present purposes there is no practical
difference between the two models.
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28-22
Probit/Logit Models (cont.)
• Notice that the slope varies dramatically.
• When the team is very very likely or very very
unlikely to win, a small change in the point
spread has very little impact.
• When the team’s chance of victory
is 50/50, a small change in the point spread
can lead to a large change in probabilities.
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28-23
Figure 19.2 A Graph of Probability of
Success and X
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28-24
Probit/Logit Models
•
Both the Probit and Logit models have
the same basic structure.
1. Estimate a latent variable Z using a linear
model. Z ranges from negative infinity to
positive infinity.
2. Use a non-linear function to transform Z
into a predicted Y value between 0 and 1.
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28-25
Probit/Logit Model (cont.)
• Suppose there is some unobserved
continuous variable Z that can take on
values from negative infinity to infinity.
• The higher E(Z) is, the more probable it
is that a team will win, or a student will
graduate, or a consumer will purchase a
particular brand.
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28-26
Probit/Logit Model (cont.)
• We call an unobserved variable, Z, that
we use for intermediate calculations, a
latent variable.
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28-27
Probit/Logit Model (cont.)
• Z is a linear function of the explanators:
Z   0  1 X1i   2 X 2i  ...   K X Ki   i
• Our goal is to estimate these  ’s.
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28-28
Probit/Logit Model (cont.)
• We will focus particularly on E(Z):
E(Z )  0  1 X1i  2 X2i  ... K X Ki
• It is convenient to consider the E(Z)
separately from its stochastic
component.
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28-29
Probit/Logit (cont.)
• The predicted probability of Y is a
non-linear function of E(Z).
• The probit model uses the standard
normal cumulative density function.
• The logit model uses the logistic
cumulative density function.
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28-30
Probit/Logit (cont.)
• The logistic cumulative density
function is computationally much more
tractable than the standard normal, but
modern computers can calculate probits
quite easily.
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28-31
Probit/Logit Model (cont.)
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28-32
Probit/Logit (cont.)
• To predict the Prob(Y ) for a given
X value, begin by calculating the
fitted Z value from the predicted linear
coefficients.
• For example, if there is only one
explanator X:
E(Z )  Zˆi  0  1 Xˆ i
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28-33
Probit/Logit Model (cont.)
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28-34
Probit/Logit Model (cont.)
• Then use the nonlinear function to
translate the fitted Z value into
a Prob(Y ):
ˆ
Prob(Y )  F (Z )
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28-35
Probit/Logit Model (cont.)
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28-36
Estimating a Probit/Logit Model
(Chapter 19.2)
• In practice, how do we implement a
probit or logit model?
• Either model is estimated using a
statistical method called the method of
maximum likelihood.
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28-37
Estimating a Probit/Logit Model (cont.)
• You must specify three elements:
1. The dummy outcome variable (whether
the NFL team actually won game i)
2. The explanator/s (the NFL team’s point
spread for game i)
3. Which nonlinear function F(•) you wish
to use (you specify F when you tell the
computer whether to use logit or probit)
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28-38
Estimating a Probit/Logit Model (cont.)
• The computer then calculates the  ’s
that assigns the highest probability to the
outcomes that were observed.
• The computer can calculate the  ’s for you.
You must know how to interpret them.
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28-39
Estimating a Probit/Logit Model (cont.)
• For example, let us estimate the probability of
winning an NFL game using the logit model.
• We could just as easily have used the
probit model.
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28-40
TABLE 19.3 What Point Spreads Say About
the Probability of Winning in the NFL: III
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28-41
Estimating a Probit/Logit Model
• The estimated slope of the point spread
is -0.1098
• A 1-point increase in the point spread
decreases E(Z ) by 0.1098 units.
• How do we interpret the slope dZ/dX ?
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28-42
Estimating a Probit/Logit Model (cont.)
• In a linear regression, we look to
coefficients for three elements:
1. Statistical significance: You can still
read statistical significance from the
slope dZ/dX. The z-statistic reported for
probit or logit is analogous to OLS’s
t-statistic.
2. Sign: If dZ/dX is positive, then
dProb(Y)/dX is also positive.
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28-43
Estimating a Probit/Logit Model (cont.)

The z-statistic on the point spread is -7.22,
well exceeding the 5% critical value of 1.96.
The point spread is a statistically significant
explanator of winning NFL games.

The sign of the coefficient is negative.
A higher point spread predicts a lower
chance of winning.
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28-44
Estimating a Probit/Logit Model (cont.)
3. Magnitude: the magnitude of dZ/dX
has no particular interpretation.
We care about the magnitude of
dProb(Y)/dX.

From the computer output for a probit or
logit estimation, you can interpret the
statistical significance and sign of each
coefficient directly. Assessing magnitude
is trickier.
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28-45
Estimating a Probit/Logit Model (cont.)
• Problems in Interpreting Magnitude:
1. The estimated coefficient relates X to Z.
We care about the relationship between
X and Prob(Y = 1).
2. The effect of X on Prob(Y = 1) varies
depending on Z.
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28-46
Estimating a Probit/Logit Model (cont.)
• There are two basic approaches to
assessing the magnitude of the
estimated coefficient.
• One approach is to predict Prob(Y ) for
different values of X, to see how the
probability changes as X changes.
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28-47
Estimating a Probit/Logit Model (cont.)
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28-48
Estimating a Probit/Logit Model (cont.)
• Note Well: the effect of a 1-unit
change in X varies greatly, depending
on the initial value of E(Z ).
• E(Z ) depends on the values of
all explanators.
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28-49
Estimating a Probit/Logit Model (cont.)
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28-50
Estimating a Probit/Logit Model (cont.)
• For example, let’s consider the effect of 1
point change in the point spread, when
we start 1 standard deviation above the
mean, at SPREAD = 5.88 points.
• Note: In this example, there is only one
explanator, SPREAD. If we had other
explanators, we would have to specify
their values for this calculation, as well.
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28-51
Estimating a Probit/Logit Model (cont.)
• Step One: Calculate the E(Z ) values
for X = 5.88 and X = 6.88, using the
fitted values.
• Step Two: Plug the E(Z ) values into the
formula for the logistic density function.
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28-52
Estimating a Probit/Logit Model (cont.)
Z (5.88)  0  0.1098 5.88  0.6456
Z (6.88)  0  0.1098 6.88  0.7554
ˆ)
exp(
Z
For the logit, F ( Zˆ ) 
1  exp( Zˆ )
F (0.7554)  F (0.6456)  3.20  3.44  0.024.
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28-53
Estimating a Probit/Logit Model (cont.)
• Changing the point spread from 5.88 to 6.88
predicts a 2.4 percentage point decrease in
the team’s chance of victory.
• Note that changing the point spread from
8.88 to 9.88 predicts only a 2.1 percentage
point decrease.
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28-54
Estimating a Probit/Logit Model (cont.)
• The other approach is to use calculus.
dProb(Y ) dProb(Y ) dZˆ dF ˆ


1
dX 1
X 1 dZˆ
dZˆ
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28-55
Estimating a Probit/Logit Model (cont.)
dProb(Y ) dProb(Y ) dZˆ dF ˆ


1
ˆ
ˆ
dX 1
X 1 dZ
dZ
dF
Unfortunately,
varies, depending
dZˆ
on Zˆ . However, a sample value can
be calculated for a representative Zˆ
value. Typically, we use the Zˆ
calculated at the mean values for each X .
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28-56
Estimating a Probit/Logit Model (cont.)
• Some econometrics software packages can
calculate such “pseudo-slopes” for you.
• In STATA, the command is “dprobit.”
• EViews does NOT have this function.
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28-57
Checking Understanding
• The following table reports a probit on
the probability of holding interestbearing assets, as a function of total
financial assets (LNFINAST) and
dummy variables for having a pension
(PENSION) or IRA (IRAS).
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28-58
Checking Understanding (cont.)
• What can you determine about the
effects of each explanator, based
directly on the table?
• Suggest a follow-up calculation to give a
clearer understanding of the impact on Y
of having a pension (the dummy
variable PENSION).
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28-59
TABLE 19.2 Probit Estimates of The Probability
of Holding Interest-Bearing Assets
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28-60
Checking Understanding
• We can directly see that all three
explanators are statistically significant
(using the z-statistics).
• Also, all three explanators have positive
coefficients. Increasing total financial
assets, having a pension, and having an
IRA all increase the probability of
holding interest-bearing assets.
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28-61
Checking Understanding (cont.)
• To assess the magnitude of the
coefficient on PENSION, we need to
conduct a follow-up calculation.
• A reasonable calculation would be to
predict Prob(Y ) when PENSION = 0
and when PENSION = 1, holding the
other explanators fixed at their
sample means.
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28-62
Deriving Probit/Logit (Chapter 19.2)
• Where do the Logit and Probit
estimators come from?
• How does the latent variable Z
determine whether Y = 1 or Y = 0?
• What role do the i ’s play?
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28-63
Deriving Probit/Logit (cont.)
We assume Yi acts "as if"
determined by latent variable Z.
Z i   0  1 X 1i  ..   K X Ki   i
Yi  1 if Z i  0
Yi  0 if Z i  0
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28-64
Deriving Probit/Logit (cont.)
• Note: the assumption that the
breakpoint falls at 0 is arbitrary.
• 0 can adjust for whichever breakpoint
you might choose to set.
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28-65
Deriving Probit/Logit (cont.)
• We assume we know the distribution
of i.
• In the probit model, we assume i is
distributed by the standard normal.
• In the logit model, we assume i is
distributed by the logistic.
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28-66
Deriving Probit/Logit (cont.)
• The key to Probit/Logit: since we know
the distribution of i , we can calculate
the probability that a given observation
receives a shock i that pushes Z into
the Z > 0 or Z < 0 region.
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28-67
Deriving Probit/Logit (cont.)
1. Calculate E(Zi )  0  1 X1i  ... K X Ki
2. Determine the regions of i such that
E(Zi )  i  0 or E(Zi )  i  0
3. Using the distribution of i , calculate
the probability of drawing an i from
each region.
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28-68
Deriving Probit/Logit (cont.)
For example, suppose E ( Z i )  1
If  i  -1, then E ( Z i )   i  0
If E ( Z i )   i  0, then Yi  1
For the standard normal distribution,
what is the Prob( i  -1)?
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28-69
Deriving Probit/Logit (cont.)
•
For the standard normal distribution,
Prob(i > -1) ≈ 0.83
•
If Zi = 1, we predict there is an 83%
chance that Y = 1.
•
For another example, suppose we are
estimating a probit and E(Zi) = -2. For
what values of i will Zi > 0 (so Y = 1)?
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28-70
Deriving Probit/Logit (cont.)
•
Suppose we are estimating a probit
and E(Zi) = -2.
•
If i > 2, Zi > 0 (so Y = 1).
•
For the standard normal distribution,
Prob(i) > 2 ≈ 0.025. We predict a
2.5% chance that Y = 1.
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28-71
Deriving Probit/Logit (cont.)
More generally, suppose  i has a
cumulative density function F
That is, Prob( i  a)  F (a)
Prob( i  a)  1- F (a)
If F is symmetric, 1 – F (a)  F (-a)
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28-72
Deriving Probit/Logit (cont.)
More generally, Prob(Yi  1)
 Prob( i   E ( Z i ))
 Prob( i   ˆ0  ˆ1 X 1i  ..ˆK X Ki )
 1  Prob( i   ˆ0  ˆ1 X 1i  ..ˆK X Ki )
 1  F ( ˆ  ˆ X  ..ˆ X )
0
1
1i
K
Ki
 F ( ˆ0  ˆ1 X 1i  ..ˆK X Ki )
(for a symmetric distribution)
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28-73
Review
• Frequently econometricians wish to
estimate the probability that a discrete
event occurs.
• The Linear Probability Model:
estimating a probability by using a
linear model (e.g. OLS) with a dummy
variable for Y.
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28-74
Review (cont.)
• Problems with the Linear Probability
Model:
– OLS disturbances are heteroskedastic.
– OLS predictions range from - ∞ to + ∞.
A probability needs to range from 0 to 1.
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28-75
Review (cont.)
• Solution: Probit or Logit
• Assume a latent variable, Z, mediates
between the explanators and the
dummy variable Y.
• The higher Z is, the higher the
probability that Y = 1.
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28-76
Review
• To predict the Prob(Y ) for a given
X value, begin by calculating the
fitted Z value from the predicted
linear coefficients.
• For example, if there is only one
explanator X :
E(Z )  Zˆi  0  1 Xˆ i
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28-77
Review (cont.)
• Then use the nonlinear function to
translate the fitted Z value into
a Prob(Y ):
ˆ
Prob(Y )  F (Z )
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28-78
Review (cont.)
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28-79
Review (cont.)
• The estimated coefficients relate X to Z.
• In interpreting the coefficients, look for:
1. Statistical significance: You can still
read statistical significance from the slope
dZ/dX. The z-statistic reported for probit or
logit is analogous to OLS’s t-statistic.
2. Sign: If dZ/dX is positive, then
dProb(Y )/dX is also positive.
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28-80
Review (cont.)
3. Magnitude: the magnitude of dZ/dX
has no particular interpretation.
We care about the magnitude of
dProb(Y)/dX.

From the computer output for a probit or
logit estimation, you can interpret the
statistical significance and sign of each
coefficient directly. Assessing magnitude
is trickier.
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28-81
Review (cont.)
• Problems in Interpreting Magnitude:
1. The estimated coefficient relates X to Z.
We care about the relationship between
X and Prob(Y = 1).
2. The effect of X on Prob(Y = 1) varies
depending on Z.
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28-82
Review (cont.)
• There are two basic approaches to
assessing the magnitude of the
estimated coefficient.
• One approach is to predict Prob(Y ) for
different values of X, to see how the
probability changes as X changes.
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28-83
Review (cont.)
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28-84
Review (cont.)
• The other approach is to use calculus.
dProb(Y ) dProb(Y ) dZˆ dF ˆ


1
dX 1
X 1 dZˆ
dZˆ
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28-85
Review (cont.)
dProb(Y ) dProb(Y ) dZˆ dF ˆ


1
dX 1
X 1 dZˆ
dZˆ
dF
Unfortunately,
varies, depending
dZˆ
on Zˆ . However, a sample value can
be calculated for a representative Zˆ
value. Typically, we use the Zˆ
calculated at the mean values for each X .
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Review (cont.)
We assume Yi acts "as if"
determined by latent variable Z.
Zi  0  1 X1i  .. K X Ki   i
Yi  1 if Z i  0
Yi  0 if Zi  0
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Review (cont.)
• We assume we know the distribution
of i.
• In the probit model, we assume i is
distributed by the standard normal.
• In the logit model, we assume i is
distributed by the logistic.
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Review (cont.)
1. Calculate E(Zi )  0  1 X1i  ... K XKi
2. Determine the regions of i such that
E(Zi )   i  0 or E(Zi )  0
3. Using the distribution of i , calculate
the probability of drawing an i from
each region.
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Review (cont.)
 i has a cumulative density function F
That is, Prob( i  a)  F (a)
Prob( i  a)  1- F (a)
If F is symmetric, 1 – F (a)  F (-a )
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Review (cont.)
More generally, Prob(Yi  1)
 Prob( i   E ( Z i ))
 Prob( i   ˆ0  ˆ1 X 1i  ..ˆK X Ki )
 1  Prob( i   ˆ0  ˆ1 X 1i  ..ˆK X Ki )
 1  F ( ˆ  ˆ X  ..ˆ X )
0
1
1i
K
Ki
 F ( ˆ0  ˆ1 X 1i  ..ˆK X Ki )
(for a symmetric distribution)
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