Binary Choice/Censored Data Flashcards
(12 cards)
What major flaw does the linear probability model have?
E(yi|xi)=P(yi=1|xi)=xi’β;
Mean E(yi|xi)=P(yi=1|xi)=xi’β
Variance: Var(yi|xi)=P(|xi)(1-P(|xi))=xi’β(1-xi’β)
It assumes the conditional probability function to be linear. This does not restrict P(y=1|x1,…,xk) to lie between 0 and 1. Thus you can get probabilities lower than 0 and higher than 1. Therefore we need probit/logit regression
E(yi|xi)=F(xi’β0): then probit:
F is the standard normal distribution: F(t)=φ(t)
E(yi|xi)=F(xi’β0): then logit:
F is the standard logistic distribution with the location parameter 0 and scale parameter 1: F(t)=exp(t)/(1+exp(t))=1/(1+exp(-t))
Coefficients in the probit/logit models cannot be interpreted in the same way as in the linear regression models, because:
probit/logit only reveal the signs of the effect, not the magnitude. However, when we look at marginal effects, interpretation is possible. This marginal effect is reported either at the sample mean of xi or the average marginal effect is taken
Marginal effect (continuous variables):
∂E(yi|xi1,…,xip)/∂xij=∂P(yi=1|xi1,…,xip)/∂xij
Marginal effect (discrete variables):
P(yi=1|xi1,…,xij+1,…,xip)-P(yi=0|xi1,…,xij,…,xip)
Marginal effect of variable xij:
pj(x)=∂P(yi=1|xi1,…,xip)/∂xij=∂F(x’β)/∂xij=f(x’β)βj
Measures of fit:
1: Percentage correctly predicted (PCP)=ΣI(yi=yi^)/n where yi^=I(F(x’βn^)>0.5)
2: Pseudo-R^2=1-ln(Ln(βn^)/ln(Ln((1,0,…,0)’)
Censored data:
Some values are not observable. Only the lower bound is known. Transformation to censoring from below at 0 (yi* is the latent/uncensored variable, yi is observable):
yi=max(a,yi) <–> yi-a=max(0,yi-a)
yi=min(a,yi) <–> -yi=max(-a,-yi)
Assume yi=max(0,yi*)
Tobit model
In this model, the data is censored. We estimate the model with maximum likelihood:
1: (xi,yi) random sample
2: yi=xi’β+εi; yi=max(0,yi)
3: εi|xi~N(0,σ^2)
4: (xi,εi*) with E(xixi’)>0 has full rank
5: εi=yi-xi’β
Tobit model likelihood contribution for yi=0:
P(yi=0|xi)=1-φ(xi’β/σ)
Tobit model likelihood contribution for yi>0:
f(yi|xi)=(1/σ)φ((yi-xi’β)/σ)
