You can find the source code for this file in the class repository. The direct link is here.

# Model

Let’s start with studying static labor supply. We will consider the decision of the agent under the following rule:

$\max_{c,h} \frac{c^{1+\eta}}{1+\eta} - \beta \frac{h^{1+\gamma}}{1+\gamma}\\ \text{s.t. } c = \rho \cdot w\cdot h -r + \mu - \beta_0 \cdot 1[h>0] \\$ The individual takes his wage $$w$$ as given, he chooses hours of work $$h$$ and consumption $$c$$ subject to a given non labor income $$\mu$$ as well as a tax regime defined by $$\rho,r$$. $$\beta_0$$ is a fixed cost associated with working. We ignore participation here, see the homework for including it.

We note already that the non labor income can control for dynamic labor supply since we can have $$\mu= b_t - (1+r)b_{t+1}$$. This is part of a larger maximization problem where the agents choose optimaly $$b_t$$ over time. We will get there next time.

## Finding solution

The first order conditions give us $$w(wh +r - \mu)^\eta = \beta h^\gamma$$. There is no closed-form but we can very quickly find an interior solution by using Newton maximization on the function $$f(x) = w(wh +r - \mu)^\eta - \beta h^\gamma$$. We iterate on

$x \leftarrow x - f(x)/f'(x).$

# function which updates choice of hours using Newton step
# R here is total unearned income (including taxes when not working and all)
ff.newt <- function(x,w,R,eta,gamma,beta) {
f0 = w*(w*x + R)^eta - beta*x^gamma
f1 =  eta*w^2 * (w*x + R)^(eta-1) - gamma * beta *x^(gamma-1)
x  = x - f0/f1
x  = ifelse(w*x + R<=0, -R/w + 0.0001,x) # make sure we do not step out of bounds for next iteration
x  = ifelse(x<0, 0.0001,x)
x
}

## Simulating data

We are going to simulate a data set where agents will choose participation as well as the number of hours if they decide to work. To do that we will solve for the interior solution under a given tax rate and compare this to the option of no-work.

p  = list(eta=-1.5,gamma = 0.8,beta=1,nu=0.5) # define preferences
tx = list(rho=1,r=0) # define a simple tax
N=50000
simdata = data.table(i=1:N,X=rnorm(N))
simdata <- simdata[,lw := X     + rnorm(N)*0.2];      # add a wage which depends on X
simdata <- simdata[,mu := exp(0.3*X + rnorm(N)*0.2)]; # add non-labor income that also depends on X

# we then solve for the choice of hours and consumption
simdata <- simdata[, h := pmax(-mu+tx$r ,0)/exp(lw)+1] # starting value # for loop for newton method (30 should be enough, it is fast) for (i in 1:30) { simdata[, h := ff.newt(h,tx$rho*exp(lw),mu-tx$r,p$eta,p$gamma,p$beta) ]
}

# attach consumption, value of working
simdata <- simdata[, c  := tx$rho*exp(lw)*h + mu]; simdata <- simdata[, u1 := c^(1+p$eta)/(1+p$eta) - p$beta * h^(1+p$gamma)/(1+p$gamma) ];

At this point we can regress $$\log(w)$$ on $$\log(c)$$ and $$\log(h)$$ and find precisely the parameters of labor supply:

pander(summary(simdata[,lm(lw ~ log(c) + log(h))]))
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.186e-14 1.464e-16 -149.3 0
log(c) 1.5 7.615e-17 1.97e+16 0
log(h) 0.8 1.429e-16 5.6e+15 0
Fitting linear model: lw ~ log(c) + log(h)
Observations Residual Std. Error $$R^2$$ Adjusted $$R^2$$
50000 8.672e-15 1 1

The regression still works, among each individual who chooses to work, the FOC is still satisfied.

# Heterogeneity in $$\beta$$

Finally we want to add heterogeneity in the $$\beta$$ parameter.

simdata <- simdata[,betai := exp(p$beta*X+rnorm(N)*0.1)] simdata <- simdata[, h := pmax(-mu+tx$r ,0)/exp(lw)+1]
for (i in 1:30) {
simdata <- simdata[, h := ff.newt(h,tx$rho*exp(lw),mu-tx$r,p$eta,p$gamma,betai) ]
}

# attach consumption
simdata <- simdata[, c  := exp(lw)*h + mu];

# let's check that the FOC holds
sfit = summary(simdata[,lm(lw ~ log(c) + log(h) + log(betai))])
expect_equivalent(sfit$r.squared,1) expect_equivalent(coef(sfit)["log(c)",1],-p$eta)
expect_equivalent(coef(sfit)["log(h)",1],p$gamma) # omitting betai doesn't work sfit2 = summary(simdata[,lm(lw ~ log(c) + log(h))]) expect_false(coef(sfit2)["log(c)",1]==-p$eta)

# finally we can try a natural regression since we don't observe betai directly:
sfit3 = summary(simdata[,lm(lw ~ log(c) + log(h) + X)])
pander(sfit3)
Estimate Std. Error t value Pr(>|t|)
(Intercept) -0.07801 0.001801 -43.32 0
log(c) 1.272 0.003112 408.8 0
log(h) 0.5756 0.001981 290.6 0
X 0.9375 0.001467 639 0
Fitting linear model: lw ~ log(c) + log(h) + X
Observations Residual Std. Error $$R^2$$ Adjusted $$R^2$$
50000 0.08742 0.9926 0.9926

## Simple case of $$\eta=0$$

p  = list(eta=0,gamma = 0.8,beta=1,beta0=0,nu=0.5) # define preferences
tx = list(rho=1,r=0) # define a simple tax
N=500
simdata = data.table(i=1:N,X=rnorm(N))
simdata <- simdata[,lw := X     + rnorm(N)*0.2];      # add a wage which depends on X
simdata <- simdata[,mu := exp(0.3*X + rnorm(N)*0.2)]; # add non-labor income that also depends on X
simdata <- simdata[,eps := rnorm(N)*0.1]
simdata <- simdata[,betai := exp(p$nu*X+eps)] simdata <- simdata[, h := (tx$rho*exp(lw)/betai)^(1/p$gamma)] sfit3 = summary(simdata[,lm(log(h) ~ lw + X)]) pander(sfit3) Estimate Std. Error t value Pr(>|t|) (Intercept) 0.01516 0.005392 2.811 0.005138 lw 1.243 0.02812 44.22 2.084e-174 X -0.6236 0.02877 -21.67 7.916e-74 Fitting linear model: log(h) ~ lw + X Observations Residual Std. Error $$R^2$$ Adjusted $$R^2$$ 500 0.1205 0.9672 0.9671 sfit4 = summary(simdata[,lm(lw ~ log(h) + X)]) pander(sfit4) Estimate Std. Error t value Pr(>|t|) (Intercept) -0.01141 0.003869 -2.95 0.003327 log(h) 0.6413 0.0145 44.22 2.084e-174 X 0.6033 0.009897 60.96 8.046e-233 Fitting linear model: lw ~ log(h) + X Observations Residual Std. Error $$R^2$$ Adjusted $$R^2$$ 500 0.08651 0.9925 0.9925 Then we can construct a counter-factual revenue p2 = list(eta=0,gamma = 1/sfit3$coefficients["lw","Estimate"],beta=1,beta0=0)
tx2 = tx
tx2$rho = 0.9 simdata <- simdata[, h2 := (tx2$rho*exp(lw)/betai)^(1/p2$gamma)] simdata[, list(totearnings =mean(exp(lw+h)), R1=mean((1-tx$rho)*exp(lw+h)),R2=mean((1-tx2$rho)*exp(lw+h2)) ,R3=mean((1-tx2$rho)*exp(lw+h)) )]
##    totearnings R1       R2       R3
## 1:    138.3258  0 6.018509 13.83258

## Heterogeneity in $$\beta$$ revisited

Finally we want to add heterogeneity in the $$\beta$$ parameter.

p  = list(eta=-1.5,gamma = 0.8,beta=1,beta0=0,nu=0.5) # define preferences
tx = list(rho=1,r=0) # define a simple tax
N=5000
simdata = data.table(i=1:N,X=rnorm(N))
simdata <- simdata[,lw := X     + rnorm(N)*0.2];      # add a wage which depends on X
simdata <- simdata[,mu := exp(0.3*X + rnorm(N)*0.2)]; # add non-labor income that also depends on X
simdata <- simdata[,eps := rnorm(N)*0.1]
simdata <- simdata[,betai := exp(p$nu*X+eps)] simdata <- simdata[, h := (tx$rho*exp(lw)/betai)^(1/p$gamma)] simdata <- simdata[, betai := exp(p$nu*X+rnorm(N)*0.1)]
simdata <- simdata[, h := pmax(-mu+tx$r ,0)/exp(lw)+1] for (i in 1:30) { simdata <- simdata[, h := ff.newt(h,tx$rho*exp(lw),mu-tx$r,p$eta,p$gamma,betai) ] } # attach consumption simdata <- simdata[, c := exp(lw)*h + mu]; # let's check that the FOC holds sfit = summary(simdata[,lm(lw ~ log(c) + log(h) + log(betai))]) expect_equivalent(sfit$r.squared,1)
expect_equivalent(coef(sfit)["log(c)",1],-p$eta) expect_equivalent(coef(sfit)["log(h)",1],p$gamma)

sfit2 = summary(simdata[,lm(lw ~ log(c) + log(h))])
expect_false(coef(sfit2)["log(c)",1]==-peta) sfit3 = summary(simdata[,lm(lw ~ log(c) + log(h) + X)]) pander(sfit2) Estimate Std. Error t value Pr(>|t|) (Intercept) -0.5546 0.007959 -69.69 0 log(c) 2.397 0.008055 297.6 0 log(h) 0.5769 0.01228 47 0 Fitting linear model: lw ~ log(c) + log(h) Observations Residual Std. Error $$R^2$$ Adjusted $$R^2$$ 5000 0.1807 0.9683 0.9683 Next we have established the following condition in the class: $E \left[ \log(w) - \gamma \log h - \eta \log c + \nu x | W,R,X \right] = 0$ This suggests an IV strategy where we instrument using $$W,R,X$$. We can use either hours or consumption as a dependent variable and instrument the other. We use $$h$$ and implement using 2SLS (identical to IV in this case). We then first regress $$c$$ on the three variables to get the predicted component, that we then use in a regression. fit_c = simdata[, lm( log(c) ~ lw + log(mu) + X ) ] simdata[, c_iv := predict(fit_c)] sfit_iv2 = simdata[, lm( log(h) ~ c_iv + lw + X )] pander(sfit_iv2) Fitting linear model: log(h) ~ c_iv + lw + X Estimate Std. Error t value Pr(>|t|) (Intercept) -0.002048 0.007381 -0.2774 0.7815 c_iv -1.868 0.01715 -108.9 0 lw 1.259 0.01067 117.9 0 X -0.636 0.007769 -81.87 0 Where we find as regression coefficients $$1/\gamma$$ and $$\eta/\gamma$$. Note of course that the inference out of the second regression is no valid (need to use IB formula, not only the second stage). fit_c = simdata[, lm( log(c) ~ lw + log(mu) + X ) ] simdata[, c_iv := predict(fit_c)] sfit_iv2 = simdata[, lm( log(h) ~ c_iv + lw + X )] pander(sfit_iv2) Fitting linear model: log(h) ~ c_iv + lw + X Estimate Std. Error t value Pr(>|t|) (Intercept) -0.002048 0.007381 -0.2774 0.7815 c_iv -1.868 0.01715 -108.9 0 lw 1.259 0.01067 117.9 0 X -0.636 0.007769 -81.87 0 # Heckman selection We cover here a simple version of the Heckman selection. In the labor supply homework, you will see a version of this closer to the structure of the model presented at the begining of this page. We start with the following model: \begin{align} y_i^* & = \gamma x_i + \epsilon_i \\ d_i & = 1[ \delta_1 x_i + \delta_2 z_i + u_i \geq 0] \\ y_i & = d_i \cdot y_i^* \end{align} And only $$(y_i, d_i, x_i, z_i)$$ are observed. We assume that $$(\epsilon_i,u_i)$$ are jointly normal, mean zero and independent of $$x_i$$. N=5000 p = list(gamma = 1.0, delta1 = 0.8, delta2 = 0.8, alpha = 1.0, sigma_v = 1.0, sigma_u = 1.0) X = 2*runif(N)-1 Z = 2*runif(N)-1 U = psigma_u * rnorm(N)
D = p$delta1 * X + p$delta2 * Z + U >= 0
E = p$alpha * U + p$sigma_v * rnorm(N)
Ys = p\$gamma * X + E

data = data.table(y = Ys * D, ys=Ys, d = D, x=X,z=Z,e=E,u=U)
data[ d==FALSE, y:=NA]

While $$E[ \epsilon_i | x_i ]=0$$ we can’t use that for identification since what we observe is $$(y_i, x_i)$$ and note $$(y^*_i, x_i)$$, so we can’t construct $$E[ y^*_i x_i ]$$.

We then need to rely on something else. We are going to look at $$E[ y^*_i x_i | d_i \geq 0 ]$$ which is then equal to $$E[y_i x_i | d_i \geq 0 ]$$ which is observed!

fit1 = lm(y ~ x,data)
summary(fit1)
##
## Call:
## lm(formula = y ~ x, data = data)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -3.5073 -0.9004 -0.0255  0.8355  4.5879
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)  0.72108    0.02636   27.36   <2e-16 ***
## x            0.53792    0.04583   11.74   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.252 on 2453 degrees of freedom
##   (2545 observations deleted due to missingness)
## Multiple R-squared:  0.05318,    Adjusted R-squared:  0.05279
## F-statistic: 137.8 on 1 and 2453 DF,  p-value: < 2.2e-16

We we compare the regression coefficient 0.5379224 to the true parameter 1. We can look at the distribution of the error.

#ggplot(data, aes(x = x, y= e)) + geom_smooth() +
#  theme_bw() + geom_hline(yintercept=0,linetype=2)
ggplot(data[d==TRUE], aes(x = x, y= e)) + geom_smooth() +
theme_bw() + geom_hline(yintercept=0,linetype=2)
## geom_smooth() using method = 'gam' and formula 'y ~ s(x, bs = "cs")' We then look at

\begin{align} E[y_i | x_i, z_i, d_i=1 ] & = E[y^*_i | x_i, z_i, d_i=1 ] \\ & = E[\gamma x_i +\epsilon_i | x_i, z_i, d_i=1 ] \\ & = \gamma x_i + E[\epsilon_i | x_i, z_i, d_i=1 ] \\ \end{align} $$(\epsilon_i,u_i)$$ are jointly normal so we know we can write $$\epsilon_i = \alpha u_i + v_i$$ where $$v_i$$ is normal, mean zero and independent of $$u_i$$. We can then write the conditional expectation as follows:

\begin{align} E[\epsilon_i | x_i, z_i, d_i =1 ] & =E[\epsilon_i | x_i, z_i, \delta_1 x_i + \delta_2 z_i + u_i \geq 0 ] \\ & =E[\alpha u_i + v_i | x_i, z_i, \delta_1 x_i + \delta_2 z_i + u_i \geq 0 ] \\ & = \alpha E[u_i | x_i, z_i, \delta_1 x_i + \delta_2 z_i + u_i \geq 0 ] \\ & \quad \quad \quad \quad + E[v_i | x_i, z_i, \delta_1 x_i + \delta_2 z_i + u_i \geq 0 ] \\ & = \alpha E[u_i | x_i, z_i, \delta_1 x_i + \delta_2 z_i + u_i \geq 0 ] + 0 \\ & = \alpha E[u_i | u_i \geq -\delta_2 z_i -\delta_1 x_i ] \\ \end{align} Finally, we note that $$u_i$$ is a normal distribution with some variance $$\sigma^2_u$$. And we want to compute the mean of that distribution conditional on being lower than some value.

We can look at that.

\begin{align} E[u_i | u_i \geq a ] & = \int_a^{\infty} u f_{u>a}(u) du \\ & = \int_a^{\infty} u \frac{1}{1-\Phi(\frac{a}{\sigma_u})} \frac{1}{\sigma_a} \phi ( \frac{u}{\sigma_u}) du \\ & = \frac{1}{1-\Phi(\frac{a}{\sigma_u})} \int_a^{-\infty} u \frac{1}{\sigma_u \sqrt{2 \pi }} \exp ( - \frac{u^2}{2\sigma_u^2}) du \\ & = \frac{1}{1-\Phi(\frac{a}{\sigma_u})} \Big[ \frac{-\sigma_u}{ \sqrt{2 \pi}} \exp ( - \frac{u^2}{2\sigma_u^2}) \Big]^{\infty}_a \\ & = \sigma_u \frac{ \phi(\frac{a}{\sigma_u})}{1-\Phi(\frac{a}{\sigma_u})} \\ \end{align} And so we we have that

$\alpha E[u_i | u_i \geq -\delta_2 z_i -\delta_1 x_i] = \alpha \sigma_u \frac{ \phi(\frac{-\delta_2 z_i -\delta_1 x_i}{\sigma_u})}{1-\Phi(\frac{-\delta_2 z_i -\delta_1 x_i}{\sigma_u})}$

Which means that we need a value for $$\frac{\delta_1}{\sigma_u}$$ and $$\frac{\delta_2}{\sigma_u}$$. Luckily the participation decision given by $$\delta_2 z_i + \delta_1 x_i + u_i \geq 0$$ and so the probit regression of $$d_i$$ on $$z_i$$ and $$x_i$$ gives precisely these coefficients!

Indeed

fit2 = glm(d ~ x + z, family = binomial('probit'), data)
summary(fit2)
##
## Call:
## glm(formula = d ~ x + z, family = binomial("probit"), data = data)
##
## Deviance Residuals:
##     Min       1Q   Median       3Q      Max
## -2.3203  -0.9748  -0.4359   0.9818   2.3760
##
## Coefficients:
##              Estimate Std. Error z value Pr(>|z|)
## (Intercept) -0.001577   0.019017  -0.083    0.934
## x            0.735774   0.033998  21.642   <2e-16 ***
## z            0.825689   0.033776  24.446   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance: 6929.9  on 4999  degrees of freedom
## Residual deviance: 5849.5  on 4997  degrees of freedom
## AIC: 5855.5
##
## Number of Fisher Scoring iterations: 3

We can then use this to reconstruct our inverse mills ratio. Note that we don’t actually now the coefficient in the front. We don’t need it though.

data[, ai := predict(fit2)]
data[, m := dnorm(ai)/pnorm(ai)]

ggplot(data[d==TRUE], aes(x = x, y= e)) + geom_smooth(se=FALSE) +
geom_smooth(aes(y=m),color='red',se=FALSE) + theme_bw() #+ geom_point()
## geom_smooth() using method = 'gam' and formula 'y ~ s(x, bs = "cs")'
## geom_smooth() using method = 'gam' and formula 'y ~ s(x, bs = "cs")' summary(lm(y ~ x + m,data))
##
## Call:
## lm(formula = y ~ x + m, data = data)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -3.3161 -0.8642 -0.0360  0.8090  4.4850
##
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)  0.10138    0.07329   1.383    0.167
## x            0.85439    0.05709  14.966   <2e-16 ***
## m            0.83924    0.09284   9.040   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 1.232 on 2452 degrees of freedom
##   (2545 observations deleted due to missingness)
## Multiple R-squared:  0.08372,    Adjusted R-squared:  0.08297
## F-statistic:   112 on 2 and 2452 DF,  p-value: < 2.2e-16

We can see that the regression coefficient is close to it’s true value!