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ERF Training Workshop
Panel Data 5
Raimundo Soto
Instituto de Economía, PUC-Chile
DISCRETE VARIABLE MODELS
• A discrete variable is usually represented by 𝑦𝑖𝑡 = 0,1
• For example,
0 = 𝑑𝑖𝑑 𝑛𝑜𝑡 𝑡𝑎𝑘𝑒 𝑡ℎ𝑒 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
1 = 𝑑𝑖𝑑 𝑡𝑎𝑘𝑒 𝑡ℎ𝑒 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡
• Assume: 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 = 𝑝
• Hence: 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 0 = 1 − 𝑝
2
DISCRETE VARIABLE MODELS
• The expected value of 𝑦𝑖𝑡 is:
𝐸 𝑦 = 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 ∗ 1 + 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 0 ∗ 0
= 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 = 𝑝
• Let us assume now that 𝑝 = 𝐹 𝑥𝑖𝑡, 𝛽
• Then: 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 = 𝐹 𝑥𝑖𝑡, 𝛽 = 𝐸 𝑦|𝑥
3
LINEAR PROBABILITY MODEL
• Let us estimate a linear model of 𝐹 𝑥𝑖𝑡, 𝛽
𝑦𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡
• Naturally, it verifies that
𝐸 𝑌|𝑋 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡
• Problem: forecast 𝑦𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 could lie outside 0,1
4
LINEAR PROBABILITY MODEL
• Furthermore, the model is heteroskedastic and rather
awkward because heteroskedasticity now depends on 𝛽.
• Since 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡 must be either 1 o 0, then
𝜀𝑖𝑡 = −𝛼𝑖 − 𝛽𝑥𝑖𝑡 with probability 1 − 𝐹( 𝑥𝑖𝑡, 𝛼, 𝛽)
𝜀𝑖𝑡 = 1 − 𝛼𝑖 − 𝛽𝑥𝑖𝑡 with probability 𝐹( 𝑥𝑖𝑡, 𝛼, 𝛽)
• Thus, the variance of 𝜀𝑖𝑡 is:
𝑉𝑎𝑟 𝜀𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 1 − 𝛼𝑖 − 𝛽𝑥𝑖𝑡
5
LINEAR PROBABILITY MODEL
• Let’s keep the notion of studying the probability that 𝑦𝑖𝑡 = 1 but
change the specification so that the probability is properly defined
(between 0 and 1)
• The latter obtains if 𝐹 𝑥𝑖𝑡, 𝛽 is a cumulative distribution function
(CDF)
• There are two main families of models:
– Probabilistic models using the logistic CDF, called Logit
– Probabilistic models using the normal CDF, called Probit
• Both types of models are highly non-linear
– Cannot be estimated using Least Squares
– Must be estimated using likelihood functions when such estimator exists
6
NON-LINEAR PROBABILITY MODEL
• Logit Model
Λ 𝑤 =
𝑒 𝑤
1 + 𝑒 𝑤
– Note: variance is 𝜋2
3
• Probit Model
𝜙 𝑤 =
−∞
𝑤
1
2𝜋
𝑒
−𝑤2
2 𝑑𝑤
– Note: 𝜎2 = 1 , it cannot be identified in a two-state model
7
Density Function
8
Normal
Logistic
Cumulative Distribution Function
9
Cumulative Distribution Function
10
Linear Model
(OK between 0.3 and 0.7)
Cumulative Distribution Function
11
Logistic gives
more probability
to (y=0)
PARAMETERS AND MARGINAL EFFECTS
• In the linear model, 𝑦𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡, parameters are
simply:
𝜕𝐸 𝑦|𝑥
𝜕𝑥
=
𝜕 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡
𝜕𝑥𝑖𝑡
= 𝛽
• In a discrete variable model:
𝜕𝐸 𝑦|𝑥
𝜕𝑥
=
𝜕𝐹 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡
𝜕𝛽𝑥𝑖𝑡
∗
𝜕𝛽𝑥𝑖𝑡
𝜕𝑥𝑖𝑡
= 𝛽𝑓 𝛼𝑖 + 𝛽𝑥𝑖𝑡 ≠ 𝛽
Where 𝑑(∙) = 𝑑𝐹(∙)
12
PARAMETERS AND MARGINAL EFFECTS
• Note, however, that in the logistic:
𝜕𝐸 𝑦|𝑥
𝜕𝑥
=
𝜕Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡
𝜕(𝛼𝑖+𝛽𝑥𝑖𝑡)
=
𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡
1 + 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡 2
= Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 1 − Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡
• So that:
𝜕𝐸 𝑦|𝑥
𝜕𝑥
=
𝜕Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡
𝜕𝛽𝑥𝑖𝑡
∗
𝜕𝛽𝑥𝑖𝑡
𝜕𝑥𝑖𝑡
= 𝛽Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 1 − Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡
13
PARAMETERS AND MARGINAL EFFECTS
• In standard models (not panel) there is a certain
correspondence among estimators of linear models (L),
probit (𝜙) and logit (Λ):
𝛽Λ ≅ 𝛽 𝜙 ∗ 1.6
𝛽𝐿 ≅ 𝛽 𝜙 ∗ 0.4
Except the constant
𝛽𝐿 ≅ 𝛽 𝜙 ∗ 0.4 + 0.5
14
MULTI-STATE MODELS
• So far, we have specified the discrete variable as 𝑦𝑖𝑡 = 0,1
• We could have three (o more) states
• For example,
0 = 𝑡𝑟𝑎𝑣𝑒𝑙 𝑡𝑜 𝐶𝑎𝑖𝑟𝑜 𝑏𝑦 𝑏𝑢𝑠
1 = 𝑡𝑟𝑎𝑣𝑒𝑙 𝑡𝑜 𝐶𝑎𝑖𝑟𝑜 𝑏𝑦 𝑡𝑟𝑎𝑖𝑛
2 = 𝑡𝑟𝑎𝑣𝑒𝑙 𝑡𝑜 𝐶𝑎𝑖𝑟𝑜 𝑏𝑦 𝑝𝑙𝑎𝑛𝑒
• Problems are similar (slightly more complicated)
15
GENERAL PLAN
• It seems we have 4 cases:
16
Fixed
Effects
Random
Effects
Logit
Probit
GENERAL PLAN
• It seems we have 4 cases:
17
Fixed
Effects
Random
Effects
Logit Almost OK Bias
Probit Bias Ok
LOGIT MODEL
• Assume T=2 and consider the (log) likelihood function of a
sample of size N:
log 𝐿 = −
𝑖=1
𝑁
𝑡=1
2
log 1 + 𝑒𝑥𝑝 𝛼𝑖 + 𝛽𝑥𝑖𝑡 +
𝑖=1
𝑁
𝑡=1
2
𝑦𝑖𝑡 𝑒𝑥𝑝 𝛼𝑖 + 𝛽𝑥𝑖𝑡
• Suppose that 𝑥𝑖𝑡 = 0 if 𝑡 = 0 and 𝑥𝑖𝑡 = 1 if 𝑡 = 1. Then the
first derivatives are:
18
LOGIT MODEL
𝜕 log 𝐿
𝜕𝛽
=
𝑖=1
𝑁
𝑡=1
2
𝑦𝑖𝑡 −
𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡
1 + 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡
𝑥𝑖𝑡
𝜕 log 𝐿
𝜕𝛼𝑖
=
𝑖=1
𝑁
𝑡=1
2
𝑦𝑖𝑡 −
𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡
1 + 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡
• Solving:
𝛼𝑖 = ∞ if 𝑦𝑖1 + 𝑦𝑖2 = 2
𝛼𝑖 = −∞ if 𝑦𝑖1 + 𝑦𝑖2 = 0
𝛼𝑖 = −
𝛽
2
if 𝑦𝑖1 + 𝑦𝑖2 = 1
19
LOGIT MODEL
• The estimator of 𝛼𝑖 does not exist if 𝑡=1
𝑇
𝑦𝑖𝑡 = 0 o
𝑡=1
𝑇
𝑦𝑖𝑡 = 𝑇. Obviously!
• The estimator of 𝛽 is inconsistent with fixed T if 𝑛 → ∞
because of the incidental parameter problem (Neyman and
Scott, 1948): as N grows, so thus the number of parameters
to be estimated (one 𝛼𝑖 for each i)
• In fact, the estimator of 𝛽 is inconsistent, 𝑝𝑙𝑖𝑚 𝛽 = 2𝛽
(Hsiao, pp. 160-161)
20
LOGIT MODEL
• However, there is a Logit estimator that is consistent, called
conditional logit
• It uses only the information from units that switch states,
i.e., when a unit from 0 → 1 or from 1 → 0
• Therefore, the estimator is conditional on observing a
change in state, which allows identifying first 𝛽 and later
𝛼𝑖
• Note that this estimator eliminates:
– All units that do not change state (always 0 or 1)
– All variables that do not change in time. 
21
LOGIT MODEL
• The conditional Logit model allows estimating 𝛽 using
Newton-Raphson algorithm iterative (o similar techniques,
e.g. BHHH). The same algorithm produces the variance of
the estimators, 𝑣𝑎𝑟 𝛽 .
• Scores (1st derivatives) 𝑠 𝛽 =
𝜕ℒ 𝑐 𝛽
𝜕𝛽
= 𝑖 1(0 < 𝑦𝑖+ <
22
PROBIT MODEL
23
• Following the linear probability model, we assume that
individual effects are random and distribute with Normal
distribution
𝑦𝑖𝑡 ≠ 0 ↔ 𝛽𝑥𝑖𝑡 + 𝛼𝑖 + 𝜀𝑖𝑡 > 0
• Let 𝜈𝑖𝑡 = 𝛼𝑖 + 𝜀𝑖𝑡
• Then, 𝜈𝑖𝑡 ∼ 𝑁 0, 𝜎𝜈
2
• And: 𝐹 𝑦, 𝑧 =
𝜙 𝑧 𝑠𝑖 𝑦 ≠ 0
1 − 𝜙 𝑧 𝑦 = 0
PROBIT MODEL
24
• The likelihood of each panel is
𝑃𝑟 𝑦𝑖1, 𝑦𝑖2, … , 𝑦𝑖𝑛|𝑥𝑖1, 𝑥𝑖2, … , 𝑥𝑖𝑛 =
−∞
∞
𝑒
− 𝜈𝑖𝑡
2
2𝜎2
2𝜋𝜎2
𝑡=1
𝑛
𝐹 𝑦𝑖𝑡, 𝛼𝑖 + 𝛽𝑥𝑖𝑡 𝑑𝜈𝑖
• This integral can be represented by
=
−∞
∞
𝑔 𝑦𝑖𝑡, 𝑥𝑖𝑡, 𝜈𝑖 𝑑𝜈𝑖
• And approximated using Gauss-Hermite quadrature
methods (e.g., using 𝑒−𝑥2
):
PROBIT MODEL
25
• Quadrature’s notion:
−∞
∞
𝑒−𝑥2
ℎ 𝑥 𝑑𝑥 ≈
𝑚=1
𝑀
𝜔 𝑚
∗ℎ 𝑎 𝑚
∗
where 𝜔 𝑚 are weights, ℎ 𝑎 𝑚
∗
are quadrature ordinates and M are
quadrature points
– The idea is properly approximate that ℎ(𝑥) using a polynomial of
order M (this is the key issue)
• The above integral de can be written as :
−∞
∞
𝑓 𝑥 𝑑𝑥 ≈
𝑚=1
𝑀
𝜔 𝑚
∗ 𝑒 𝑎 𝑚
∗ 2
ℎ 𝑎 𝑚
∗
PROBIT MODEL
26
• The likelihood function of each panel (unit) is
approximated using:
𝑙𝑖 = 2 𝜎𝑖
𝑚=1
𝑀
𝑤 𝑚
∗ 𝑒 𝑎 𝑚
∗ 2
𝑔 𝑦𝑖𝑡, 𝑥𝑖𝑡, 2 𝜎𝑖 𝑎 𝑚
∗ + 𝜀𝑖
• The likelihood function of the complete sample (all units)
is approximated using:
𝐿 ≈
𝑖=1
𝑛
𝑤𝑖 log 2 𝜎𝑖
𝑚=1
𝑀
𝑤 𝑚
∗
𝑒 𝑎 𝑚
∗ 2 𝑒− 2 𝜎 𝑖 𝑎 𝑚
∗ + 𝜀 𝑖
2𝜋𝜎2
𝑡=1
𝑛
𝐹 𝑦𝑖𝑡, 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 2 𝜎𝑖 𝑎 𝑚
∗
+ 𝜀𝑖
Likelihood Estimation
• The likelihood function is “the probability that a sample of
observations of size n is a realization of a particular
distribution 𝑓 𝑦𝑖𝑡, 𝑥𝑖𝑡|𝜃 "
• Let’s call it ℒ 𝑛 𝜃|𝑦𝑖𝑡, 𝑥𝑖𝑡
27
Example of Likelihood Estimation
• Consider “bicycle accidents in the campus” in a
given year. Suppose we record the following
sample
{2,0,3,4,1,3,0,2,3,4,3,5}
• What do you think is the model that generated this
sample?
• What do you think is the distribution that
generated this sample?
28
Example of Likelihood Estimation
• OK, let’s try Poisson (although I think it is Normal)
• Poisson distribution of each observation:
• When observations are independent, the joint probability
or likelihood function is the product of the marginals
29
Example of Likelihood Estimation
• We want to pick 𝜃 so as to make this probability
(likelihood) to be a maximum. There are two ways:
– Trying different values (most often used)
– Use calculus
• Our likelihood function is really ugly (non linear)
• But we can use logs to make it much nicer
30
Example of Likelihood Estimation
• Taking logs:
• To get the optimal 𝜃: derive twice, equalize to zero,
make sure second derivative is negative, and
obtain 𝜃 from first derivative:
• 1st derivative: −12 + 30 ∗
1
𝜃
• 2nd derivative: −30 ∗
1
𝜃2 which is negative
• Therefore 𝜃 = 2.5
31
Example of Likelihood Estimation
• Of all Poisson distributions, the one that best
describes the data is that with parameter 2.5
• What about my normal?
– Well I can fit the normal to the data and find 𝜇, 𝜎2
• Is you model better than mine?
– No way!!!
32
Likelihood Estimation
• The likelihood function is “the probability that a sample of
observations of size n is a realization of a particular
distribution 𝑓 𝑦𝑖𝑡, 𝑥𝑖𝑡|𝜃 "
ℒ 𝑛 𝜃|𝑦𝑖𝑡, 𝑥𝑖𝑡
• The joint distribution is the product of the conditional
density times the marginal density:
𝑓 𝑦𝑖𝑡, 𝑥𝑖𝑡|𝜃 = 𝑓 𝑦𝑖𝑡|𝑥𝑖𝑡, 𝜃 𝑓 𝑥𝑖𝑡|𝜃
33
Likelihood Estimation
• A statistic is sufficient with regards to a model and its
unknown parameters if "no other statistic can be calculated
using the same sample that could bring additional
information vis-à-vis the true value of the parameters”.
• Usually, a sufficient statistic is a simple function of the data,
e.g., the sum of the observations.
• In our case, a sufficient statistic of 𝑓 𝑥𝑖𝑡|𝜃 is the sum of
the observations (the units that change state, because in
the others 𝛼𝑖 is undefined)
34
Likelihood Estimation
• Therefore, the conditional log likelihood function is:
ℒ 𝑐 𝛽 =
𝑖
1 0 < 𝑦𝑖+ < 𝑇
𝑡
𝑦𝑖𝑡 𝑥𝑖𝑡
′
𝛽 − log
𝑧 𝑦 𝑖+
𝑒 𝑡 𝑧 𝑡 𝑥 𝑖𝑡´𝛽
Where the 𝑧𝑡 represents all possible cases where there is a
change in state.
35

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ERF Training Workshop Panel Data 5

  • 1. ERF Training Workshop Panel Data 5 Raimundo Soto Instituto de Economía, PUC-Chile
  • 2. DISCRETE VARIABLE MODELS • A discrete variable is usually represented by 𝑦𝑖𝑡 = 0,1 • For example, 0 = 𝑑𝑖𝑑 𝑛𝑜𝑡 𝑡𝑎𝑘𝑒 𝑡ℎ𝑒 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 1 = 𝑑𝑖𝑑 𝑡𝑎𝑘𝑒 𝑡ℎ𝑒 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 • Assume: 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 = 𝑝 • Hence: 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 0 = 1 − 𝑝 2
  • 3. DISCRETE VARIABLE MODELS • The expected value of 𝑦𝑖𝑡 is: 𝐸 𝑦 = 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 ∗ 1 + 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 0 ∗ 0 = 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 = 𝑝 • Let us assume now that 𝑝 = 𝐹 𝑥𝑖𝑡, 𝛽 • Then: 𝑃𝑟𝑜𝑏 𝑦𝑖𝑡 = 1 = 𝐹 𝑥𝑖𝑡, 𝛽 = 𝐸 𝑦|𝑥 3
  • 4. LINEAR PROBABILITY MODEL • Let us estimate a linear model of 𝐹 𝑥𝑖𝑡, 𝛽 𝑦𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡 • Naturally, it verifies that 𝐸 𝑌|𝑋 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 • Problem: forecast 𝑦𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 could lie outside 0,1 4
  • 5. LINEAR PROBABILITY MODEL • Furthermore, the model is heteroskedastic and rather awkward because heteroskedasticity now depends on 𝛽. • Since 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡 must be either 1 o 0, then 𝜀𝑖𝑡 = −𝛼𝑖 − 𝛽𝑥𝑖𝑡 with probability 1 − 𝐹( 𝑥𝑖𝑡, 𝛼, 𝛽) 𝜀𝑖𝑡 = 1 − 𝛼𝑖 − 𝛽𝑥𝑖𝑡 with probability 𝐹( 𝑥𝑖𝑡, 𝛼, 𝛽) • Thus, the variance of 𝜀𝑖𝑡 is: 𝑉𝑎𝑟 𝜀𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 1 − 𝛼𝑖 − 𝛽𝑥𝑖𝑡 5
  • 6. LINEAR PROBABILITY MODEL • Let’s keep the notion of studying the probability that 𝑦𝑖𝑡 = 1 but change the specification so that the probability is properly defined (between 0 and 1) • The latter obtains if 𝐹 𝑥𝑖𝑡, 𝛽 is a cumulative distribution function (CDF) • There are two main families of models: – Probabilistic models using the logistic CDF, called Logit – Probabilistic models using the normal CDF, called Probit • Both types of models are highly non-linear – Cannot be estimated using Least Squares – Must be estimated using likelihood functions when such estimator exists 6
  • 7. NON-LINEAR PROBABILITY MODEL • Logit Model Λ 𝑤 = 𝑒 𝑤 1 + 𝑒 𝑤 – Note: variance is 𝜋2 3 • Probit Model 𝜙 𝑤 = −∞ 𝑤 1 2𝜋 𝑒 −𝑤2 2 𝑑𝑤 – Note: 𝜎2 = 1 , it cannot be identified in a two-state model 7
  • 10. Cumulative Distribution Function 10 Linear Model (OK between 0.3 and 0.7)
  • 11. Cumulative Distribution Function 11 Logistic gives more probability to (y=0)
  • 12. PARAMETERS AND MARGINAL EFFECTS • In the linear model, 𝑦𝑖𝑡 = 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡, parameters are simply: 𝜕𝐸 𝑦|𝑥 𝜕𝑥 = 𝜕 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡 𝜕𝑥𝑖𝑡 = 𝛽 • In a discrete variable model: 𝜕𝐸 𝑦|𝑥 𝜕𝑥 = 𝜕𝐹 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡 𝜕𝛽𝑥𝑖𝑡 ∗ 𝜕𝛽𝑥𝑖𝑡 𝜕𝑥𝑖𝑡 = 𝛽𝑓 𝛼𝑖 + 𝛽𝑥𝑖𝑡 ≠ 𝛽 Where 𝑑(∙) = 𝑑𝐹(∙) 12
  • 13. PARAMETERS AND MARGINAL EFFECTS • Note, however, that in the logistic: 𝜕𝐸 𝑦|𝑥 𝜕𝑥 = 𝜕Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡 𝜕(𝛼𝑖+𝛽𝑥𝑖𝑡) = 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡 1 + 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡 2 = Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 1 − Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 • So that: 𝜕𝐸 𝑦|𝑥 𝜕𝑥 = 𝜕Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝜀𝑖𝑡 𝜕𝛽𝑥𝑖𝑡 ∗ 𝜕𝛽𝑥𝑖𝑡 𝜕𝑥𝑖𝑡 = 𝛽Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 1 − Λ 𝛼𝑖 + 𝛽𝑥𝑖𝑡 13
  • 14. PARAMETERS AND MARGINAL EFFECTS • In standard models (not panel) there is a certain correspondence among estimators of linear models (L), probit (𝜙) and logit (Λ): 𝛽Λ ≅ 𝛽 𝜙 ∗ 1.6 𝛽𝐿 ≅ 𝛽 𝜙 ∗ 0.4 Except the constant 𝛽𝐿 ≅ 𝛽 𝜙 ∗ 0.4 + 0.5 14
  • 15. MULTI-STATE MODELS • So far, we have specified the discrete variable as 𝑦𝑖𝑡 = 0,1 • We could have three (o more) states • For example, 0 = 𝑡𝑟𝑎𝑣𝑒𝑙 𝑡𝑜 𝐶𝑎𝑖𝑟𝑜 𝑏𝑦 𝑏𝑢𝑠 1 = 𝑡𝑟𝑎𝑣𝑒𝑙 𝑡𝑜 𝐶𝑎𝑖𝑟𝑜 𝑏𝑦 𝑡𝑟𝑎𝑖𝑛 2 = 𝑡𝑟𝑎𝑣𝑒𝑙 𝑡𝑜 𝐶𝑎𝑖𝑟𝑜 𝑏𝑦 𝑝𝑙𝑎𝑛𝑒 • Problems are similar (slightly more complicated) 15
  • 16. GENERAL PLAN • It seems we have 4 cases: 16 Fixed Effects Random Effects Logit Probit
  • 17. GENERAL PLAN • It seems we have 4 cases: 17 Fixed Effects Random Effects Logit Almost OK Bias Probit Bias Ok
  • 18. LOGIT MODEL • Assume T=2 and consider the (log) likelihood function of a sample of size N: log 𝐿 = − 𝑖=1 𝑁 𝑡=1 2 log 1 + 𝑒𝑥𝑝 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 𝑖=1 𝑁 𝑡=1 2 𝑦𝑖𝑡 𝑒𝑥𝑝 𝛼𝑖 + 𝛽𝑥𝑖𝑡 • Suppose that 𝑥𝑖𝑡 = 0 if 𝑡 = 0 and 𝑥𝑖𝑡 = 1 if 𝑡 = 1. Then the first derivatives are: 18
  • 19. LOGIT MODEL 𝜕 log 𝐿 𝜕𝛽 = 𝑖=1 𝑁 𝑡=1 2 𝑦𝑖𝑡 − 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡 1 + 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡 𝑥𝑖𝑡 𝜕 log 𝐿 𝜕𝛼𝑖 = 𝑖=1 𝑁 𝑡=1 2 𝑦𝑖𝑡 − 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡 1 + 𝑒 𝛼 𝑖+𝛽𝑥 𝑖𝑡 • Solving: 𝛼𝑖 = ∞ if 𝑦𝑖1 + 𝑦𝑖2 = 2 𝛼𝑖 = −∞ if 𝑦𝑖1 + 𝑦𝑖2 = 0 𝛼𝑖 = − 𝛽 2 if 𝑦𝑖1 + 𝑦𝑖2 = 1 19
  • 20. LOGIT MODEL • The estimator of 𝛼𝑖 does not exist if 𝑡=1 𝑇 𝑦𝑖𝑡 = 0 o 𝑡=1 𝑇 𝑦𝑖𝑡 = 𝑇. Obviously! • The estimator of 𝛽 is inconsistent with fixed T if 𝑛 → ∞ because of the incidental parameter problem (Neyman and Scott, 1948): as N grows, so thus the number of parameters to be estimated (one 𝛼𝑖 for each i) • In fact, the estimator of 𝛽 is inconsistent, 𝑝𝑙𝑖𝑚 𝛽 = 2𝛽 (Hsiao, pp. 160-161) 20
  • 21. LOGIT MODEL • However, there is a Logit estimator that is consistent, called conditional logit • It uses only the information from units that switch states, i.e., when a unit from 0 → 1 or from 1 → 0 • Therefore, the estimator is conditional on observing a change in state, which allows identifying first 𝛽 and later 𝛼𝑖 • Note that this estimator eliminates: – All units that do not change state (always 0 or 1) – All variables that do not change in time.  21
  • 22. LOGIT MODEL • The conditional Logit model allows estimating 𝛽 using Newton-Raphson algorithm iterative (o similar techniques, e.g. BHHH). The same algorithm produces the variance of the estimators, 𝑣𝑎𝑟 𝛽 . • Scores (1st derivatives) 𝑠 𝛽 = 𝜕ℒ 𝑐 𝛽 𝜕𝛽 = 𝑖 1(0 < 𝑦𝑖+ < 22
  • 23. PROBIT MODEL 23 • Following the linear probability model, we assume that individual effects are random and distribute with Normal distribution 𝑦𝑖𝑡 ≠ 0 ↔ 𝛽𝑥𝑖𝑡 + 𝛼𝑖 + 𝜀𝑖𝑡 > 0 • Let 𝜈𝑖𝑡 = 𝛼𝑖 + 𝜀𝑖𝑡 • Then, 𝜈𝑖𝑡 ∼ 𝑁 0, 𝜎𝜈 2 • And: 𝐹 𝑦, 𝑧 = 𝜙 𝑧 𝑠𝑖 𝑦 ≠ 0 1 − 𝜙 𝑧 𝑦 = 0
  • 24. PROBIT MODEL 24 • The likelihood of each panel is 𝑃𝑟 𝑦𝑖1, 𝑦𝑖2, … , 𝑦𝑖𝑛|𝑥𝑖1, 𝑥𝑖2, … , 𝑥𝑖𝑛 = −∞ ∞ 𝑒 − 𝜈𝑖𝑡 2 2𝜎2 2𝜋𝜎2 𝑡=1 𝑛 𝐹 𝑦𝑖𝑡, 𝛼𝑖 + 𝛽𝑥𝑖𝑡 𝑑𝜈𝑖 • This integral can be represented by = −∞ ∞ 𝑔 𝑦𝑖𝑡, 𝑥𝑖𝑡, 𝜈𝑖 𝑑𝜈𝑖 • And approximated using Gauss-Hermite quadrature methods (e.g., using 𝑒−𝑥2 ):
  • 25. PROBIT MODEL 25 • Quadrature’s notion: −∞ ∞ 𝑒−𝑥2 ℎ 𝑥 𝑑𝑥 ≈ 𝑚=1 𝑀 𝜔 𝑚 ∗ℎ 𝑎 𝑚 ∗ where 𝜔 𝑚 are weights, ℎ 𝑎 𝑚 ∗ are quadrature ordinates and M are quadrature points – The idea is properly approximate that ℎ(𝑥) using a polynomial of order M (this is the key issue) • The above integral de can be written as : −∞ ∞ 𝑓 𝑥 𝑑𝑥 ≈ 𝑚=1 𝑀 𝜔 𝑚 ∗ 𝑒 𝑎 𝑚 ∗ 2 ℎ 𝑎 𝑚 ∗
  • 26. PROBIT MODEL 26 • The likelihood function of each panel (unit) is approximated using: 𝑙𝑖 = 2 𝜎𝑖 𝑚=1 𝑀 𝑤 𝑚 ∗ 𝑒 𝑎 𝑚 ∗ 2 𝑔 𝑦𝑖𝑡, 𝑥𝑖𝑡, 2 𝜎𝑖 𝑎 𝑚 ∗ + 𝜀𝑖 • The likelihood function of the complete sample (all units) is approximated using: 𝐿 ≈ 𝑖=1 𝑛 𝑤𝑖 log 2 𝜎𝑖 𝑚=1 𝑀 𝑤 𝑚 ∗ 𝑒 𝑎 𝑚 ∗ 2 𝑒− 2 𝜎 𝑖 𝑎 𝑚 ∗ + 𝜀 𝑖 2𝜋𝜎2 𝑡=1 𝑛 𝐹 𝑦𝑖𝑡, 𝛼𝑖 + 𝛽𝑥𝑖𝑡 + 2 𝜎𝑖 𝑎 𝑚 ∗ + 𝜀𝑖
  • 27. Likelihood Estimation • The likelihood function is “the probability that a sample of observations of size n is a realization of a particular distribution 𝑓 𝑦𝑖𝑡, 𝑥𝑖𝑡|𝜃 " • Let’s call it ℒ 𝑛 𝜃|𝑦𝑖𝑡, 𝑥𝑖𝑡 27
  • 28. Example of Likelihood Estimation • Consider “bicycle accidents in the campus” in a given year. Suppose we record the following sample {2,0,3,4,1,3,0,2,3,4,3,5} • What do you think is the model that generated this sample? • What do you think is the distribution that generated this sample? 28
  • 29. Example of Likelihood Estimation • OK, let’s try Poisson (although I think it is Normal) • Poisson distribution of each observation: • When observations are independent, the joint probability or likelihood function is the product of the marginals 29
  • 30. Example of Likelihood Estimation • We want to pick 𝜃 so as to make this probability (likelihood) to be a maximum. There are two ways: – Trying different values (most often used) – Use calculus • Our likelihood function is really ugly (non linear) • But we can use logs to make it much nicer 30
  • 31. Example of Likelihood Estimation • Taking logs: • To get the optimal 𝜃: derive twice, equalize to zero, make sure second derivative is negative, and obtain 𝜃 from first derivative: • 1st derivative: −12 + 30 ∗ 1 𝜃 • 2nd derivative: −30 ∗ 1 𝜃2 which is negative • Therefore 𝜃 = 2.5 31
  • 32. Example of Likelihood Estimation • Of all Poisson distributions, the one that best describes the data is that with parameter 2.5 • What about my normal? – Well I can fit the normal to the data and find 𝜇, 𝜎2 • Is you model better than mine? – No way!!! 32
  • 33. Likelihood Estimation • The likelihood function is “the probability that a sample of observations of size n is a realization of a particular distribution 𝑓 𝑦𝑖𝑡, 𝑥𝑖𝑡|𝜃 " ℒ 𝑛 𝜃|𝑦𝑖𝑡, 𝑥𝑖𝑡 • The joint distribution is the product of the conditional density times the marginal density: 𝑓 𝑦𝑖𝑡, 𝑥𝑖𝑡|𝜃 = 𝑓 𝑦𝑖𝑡|𝑥𝑖𝑡, 𝜃 𝑓 𝑥𝑖𝑡|𝜃 33
  • 34. Likelihood Estimation • A statistic is sufficient with regards to a model and its unknown parameters if "no other statistic can be calculated using the same sample that could bring additional information vis-à-vis the true value of the parameters”. • Usually, a sufficient statistic is a simple function of the data, e.g., the sum of the observations. • In our case, a sufficient statistic of 𝑓 𝑥𝑖𝑡|𝜃 is the sum of the observations (the units that change state, because in the others 𝛼𝑖 is undefined) 34
  • 35. Likelihood Estimation • Therefore, the conditional log likelihood function is: ℒ 𝑐 𝛽 = 𝑖 1 0 < 𝑦𝑖+ < 𝑇 𝑡 𝑦𝑖𝑡 𝑥𝑖𝑡 ′ 𝛽 − log 𝑧 𝑦 𝑖+ 𝑒 𝑡 𝑧 𝑡 𝑥 𝑖𝑡´𝛽 Where the 𝑧𝑡 represents all possible cases where there is a change in state. 35