Agents are assumed to live for three periods in this model. In periods 1 and 2 they work, in period 3 they retire. Wages are given by \(w_{1},w_{2}\) in periods 1 and 2. All agents are identical conditional on age. There exists a perfect capital market with constant interest rate \(r\) and the price of consumption acts as the numeraire in each period, i.e. it is normalized to one. Let’s call assets at the start of period 1 \(A_{1}\), and we assume that after period 3 all individuals die, and they must have non-negative assets at that point. There is no bequest motive, so everything needs to be consumed by the end of period 3. We assume the following period perferences: \[ U(c_{t},l_{t})=\alpha\ln c_{t}+(1-\alpha)\ln l_{t} \] and point out that \(L-h_{t}=l_{t}\), i.e. \(h\) is hours worked and \(L\) is total time endowment.
\[\begin{align*} \max_{\begin{array}{c} \{c_{t}\}_{t=1}^{3} > 0\\ l_{t}\in(0,L] \end{array}} & U(c_{t},l_{t})\\ \text{subject to } & A_{1}+\sum_{t=1}^{2}\left(\frac{1}{1+r}\right)^{t-1}w_{t}(L-l_{t})=\sum_{t=1}^{3}\left(\frac{1}{1+r}\right)^{t-1}c_{t}\\ & A_{1}\text{ given.} \end{align*}\]
and solve this problem. Provide an expression for \(\lambda\). Show and provide intuition for \(\frac{\partial\lambda}{\partial A_{1}}<0,\frac{\partial\lambda}{\partial w_{t}}<0\).
We disregard the zero lower bound on both \(c_t,l_t\) because of the utility functional form. However, we write explicitly the non-negativity constraint on hours, i.e. the upper bound on \(l_t\). Then the Lagrangian writes as \[\begin{align*} \mathcal{L} & =\alpha\ln c_{1}+(1-\alpha)\ln l_{1}+\beta\left(\alpha\ln c_{2}+(1-\alpha)\ln l_{2}\right)+\beta^{2}\alpha\ln c_{3}\\ & +\lambda\left[A_{1}+\sum_{t=1}^{2}\left(\frac{1}{1+r}\right)^{t-1}w_{t}\left(L-l_{t}\right)-\sum_{t=1}^{3}\left(\frac{1}{1+r}\right)^{t-1}c_{t}\right]\\ & +\mu_{1}\left[L-l_{1}\right]+\mu_{2}\left[L-l_{2}\right] \end{align*}\]
This has the following FOCs: \[\begin{align} \frac{\partial\mathcal{L}}{\partial c_{t}}=0: & \lambda=\frac{\alpha}{c_{1}}=\frac{\beta(1+r)\alpha}{c_{2}}=\frac{\left(\beta(1+r)\right)^{2}\alpha}{c_{3}}\label{eq:foc_c}\\ \frac{\partial\mathcal{L}}{\partial l_{1}}=0: & \frac{1-\alpha}{l_{1}}=\lambda w_{1}+\mu_{1}\label{eq:foc_l1}\\ \frac{\partial\mathcal{L}}{\partial l_{2}}=0: & \frac{\beta(1-\alpha)}{l_{2}}=\lambda\frac{w_{2}}{1+r}+\mu_{2}\label{eq:foc_l2}\\ \frac{\partial\mathcal{L}}{\partial\lambda}=0: & A_{1}+\sum_{t=1}^{2}\left(\frac{1}{1+r}\right)^{t-1}w_{t}\left(L-l_{t}\right)=\sum_{t=1}^{3}\left(\frac{1}{1+r}\right)^{t-1}c_{t}\label{eq:foc_lambda} \end{align}\] We assume an interior solution for this part of the problem, hence \(\mu_{i}=0\). Now plug in for \(c_{t}\) and \(l_{t}\) from (), () and () into (): \[\begin{align} A_{1}+w_{1}L-w_{1}l_{1}+\frac{w_{2}}{1+r}-\frac{w_{2}}{1+r}l_{2} & =c_{1}+\frac{c_{2}}{1+r}+\frac{c_{2}}{\left(1+r\right)^{2}}\nonumber \\ A_{1}+w_{1}L-\frac{1-\alpha}{\lambda}+\frac{w_{2}}{1+r}-\frac{\beta(1-\alpha)}{\lambda} & =\frac{\alpha(1+\beta+\beta^{2})}{\lambda}\nonumber \\ A_{1}+w_{1}L+\frac{w_{2}}{1+r}L & =\frac{1+\beta+\beta^{2}\alpha}{\lambda}\nonumber \\ \lambda & =\frac{1+\beta+\beta^{2}\alpha}{A_{1}+w_{1}L+\frac{w_{2}}{1+r}L}\label{eq:lambda} \end{align}\] We find the required derivatives \[\begin{align*} \frac{\partial\lambda}{\partial A_{1}} & =-\frac{1+\beta+\beta^{2}\alpha}{\left(A_{1}+w_{1}L+\frac{w_{2}}{1+r}L\right)^{2}}<0\\ \frac{\partial\lambda}{\partial w_{t}} & =-\frac{1+\beta+\beta^{2}\alpha}{\left(A_{1}+w_{1}L+\frac{w_{2}}{1+r}L\right)^{2}}\left(\frac{1}{1+r}\right)^{t-1}L<0 \end{align*}\] This means that higher initial wealth \(A_{1}\) and higher period wage \(w_{t}\) causes higher consumption throughout the lifecycle and therefore lower marginal utility of consumption (and, hence, wealth).
i.e. functions \(h_{t}^{*}(w_{1},w_{2},A_{1})\) and \(h_{t}^{F}(w_{t},\lambda)\).
From the focs for leisure () and (), we have \[\begin{align*} l_{1} & =\frac{1-\alpha}{\lambda w_{1}}\\ l_{2} & =\frac{\beta(1-\alpha)(1+r)}{\lambda w_{2}} \end{align*}\] Hence the frischian labor supplies are \[\begin{align} h_{1}^{F}(w_{1},\lambda)=L-l_{1} & =L-\frac{1-\alpha}{\lambda w_{1}}\nonumber \\ h_{2}^{F}(w_{2},\lambda)=L-l_{2} & =L-\frac{\beta(1-\alpha)(1+r)}{\lambda w_{2}}\label{eq:frisch-LS} \end{align}\] and we obtain the marshallian ones by substituting our expression for \(\lambda\) from () \[\begin{align} h_{1}^{*}(w_{1},w_{2},A_{1})= & L-\frac{(1-\alpha)}{w_{1}}\left(\frac{1+\beta+\beta^{2}\alpha}{A_{1}+w_{1}L+\frac{w_{2}}{1+r}L}\right)^{-1}\nonumber \\ h_{2}^{*}(w_{1},w_{2},A_{1})= & L-\frac{\beta(1-\alpha)(1+r)}{w_{2}}\left(\frac{1+\beta+\beta^{2}\alpha}{A_{1}+w_{1}L+\frac{w_{2}}{1+r}L}\right)^{-1}\label{eq:marsh-LS} \end{align}\]
Take the following parameter values and evaluate your optimal policy functions for consumption, leisure and assets: \[\alpha=0.3,\beta=0.9,L=8700,A_{1}=1000,r=0.05,w_{1}=5,w_{2}=10\]
# define a model
m1 <- list(A1=1000,
r=0.05,
w=c(5,10),
L=8700,
alpha=0.3,
beta=0.9)
# define our model solution
lambda <- function(m){
r = (1+m$beta+m$alpha*m$beta^2)/(m$A1+m$w[1]*m$L+m$w[2]*m$L/(1+m$r))
return(r)
}
c1 <- function(m,lamb){
m$alpha / lamb
}
c2 <- function(m,lamb){
((1+m$r)*m$beta*m$alpha)/lamb
}
c3 <- function(m,lamb){
(((1+m$r)^2)*(m$beta^2)*m$alpha)/lamb
}
l1 <- function(m,lamb){(1-m$alpha)/(m$w[1] * lamb)}
l2 <- function(m,lamb){m$beta*(1+m$r)*(1-m$alpha)/(m$w[2] * lamb)}
hours <- function(m,leisure){
m$L - leisure
}
# print to a table
lambda1 = lambda(m1)
df = data.frame(period = 1:3,cons = c(c1(m1,lambda1),
c2(m1,lambda1),
c3(m1,lambda1)),
leisure = c(l1(m1,lambda1),
l2(m1,lambda1),
m1$L))
df$hours = m1$L - df$leisure
df## period cons leisure hours
## 1 1 17828.81 8320.112 379.888
## 2 2 16848.23 3931.253 4768.747
## 3 3 15921.57 8700.000 0.000equation \[\Delta\ln h_{2}=\sigma\Delta\ln w_{2}+u_{2}\] using OLS and he claims to be estimating the Frisch elasticity of labor supply. What’s the value of the estimate \(\hat{\sigma}\)? What’s the estimate’s standard error? (Hint: no statistics software needed to answer this question.)
There is no variation in this model as everybody is the same. Hence, \(u_{2}=0\). Then,
\(\sigma=\frac{\Delta\ln h_{2}}{\Delta\ln w_{2}}=\) 3.6499642
under the numerical values from question 4. How would those results change if \(A_{1}=20000\)? Why? For the rest of the problem, use \(A_{1}=1000\). Then calculate the Hicksian elasticity of labor supply in period 1 (i.e. keep discounted lifetime utility constant).
The Frisch elasticities are given by
\[\begin{align*} \frac{\partial h_{1}^{F}(w_{1},\lambda)}{\partial w_{1}}\frac{w_{1}}{h_{1}^{F}} & =\frac{1-\alpha}{L\lambda w_{1}-1+\alpha}\\ \frac{\partial h_{2}^{F}(w_{2},\lambda)}{\partial w_{2}}\frac{w_{2}}{h_{2}^{F}} & =\frac{\beta(1-\alpha)(1+r)}{L\lambda w_{2}-\beta(1-\alpha)(1+r)} \end{align*}\]
and evaluates to
frisch_e1 <- function(m,lamb){
(1-m$alpha)/(lamb*m$w[1]*m$L - 1 + m$alpha)
}
frisch_e2 <- function(m,lamb){
m$beta*(1+m$r)*(1-m$alpha) / (lamb*m$w[2]*m$L -(m$beta*(1+m$r)*(1-m$alpha)))
}\(\varepsilon_{f,1}=\) 21.9014863 and \(\varepsilon_{f,2}=\) 0.8243786
Let’s see what happens to our current leisure function () when we plug in this new model:
m2 <- m1
m2$A1 <- 20000
lambda1_1 = lambda(m2)we get a first period leisure of 9561.36, which is larger than total time available: TRUE. So this is not an admissible solution.
We have a corner solution where at least \(l_1 = L\). Doing the same calculation for \(l_2\) we get 4517.74, which is fine. Although this is not entirely correct, because \(\lambda\) is different if we don’t work in the first period. The budget constraint becomes
\[\begin{align*} A_{1}+\frac{w_{2}}{1+r}L&=\frac{\alpha+\beta+\alpha\beta^{2}}{\lambda}\\\lambda&=\frac{\alpha+\beta+\alpha\beta^{2}}{A_{1}+\frac{w_{2}}{1+r}L} \end{align*}\]
hence we define a new \(\lambda\) as
lambda_2 <- function(m){
r = (m$alpha+m$beta+m$alpha*(m$beta^2))/(m$A1+m$w[2]*m$L/(1+m$r))
return(r)
}
lambda2 = lambda_2(m2)
df2 = data.frame(period = 1:3,cons = c(c1(m2,lambda2),
c2(m2,lambda2),
c3(m2,lambda2)),
leisure = c(m2$L,
l2(m2,lambda2),
m2$L))
df2$hours = m2$L - df2$leisure
df2## period cons leisure hours
## 1 1 21384.02 8700.000 0.000
## 2 2 20207.90 4715.177 3984.823
## 3 3 19096.47 8700.000 0.000And with that we get a new Frisch elasticity for period 2 of 1.18.
Going back to \(A_{1}=1000\), the Hicksian elasticity is derived either from the slutzky equation, or from a complete solution of the dual of the above maximization problem. The Slutzky equation tells us in this case that the compensated response (holding lifetime utility V fixed) is equal to the substitution effect minus the income effect:
\[\begin{align*} \frac{\partial h_{1}^{H}(w_{1},V)}{\partial w_{1}} =\frac{\partial h_{1}^{*}(w_{1},w_{2},A)}{\partial w_{1}}-\frac{\partial h_{1}^{*}(w_{1},w_{2},A)}{\partial A_{1}}h_{1}^{*}(w_{1},w_{2},A) \end{align*}\]
and then the hicksian elasticity is defined as
\[ \varepsilon_{1,H}=\frac{\partial h_{1}^{H}(w_{1},V)}{\partial w_{1}} \frac{w_1}{h_{1}^*(w_{1},w_{2})} \]
We get the required partial derivatives \[\begin{align*} \frac{\partial h_{1}^{*}(w_{1},w_{2},A)}{\partial w_{1}}&=\frac{(1-\alpha)A_{1}}{w_{1}^{2}(1+\beta+\beta^{2}\alpha)}+\frac{(1-\alpha)w_{2}L}{w_{1}^{2}(1+\beta+\beta^{2}\alpha)(1+r)}\\ \frac{\partial h_{1}^{*}(w_{1},w_{2},A)}{\partial A_{1}}&=-\frac{(1-\alpha)}{w_{1}(1+\beta+\beta^{2}\alpha)} \end{align*}\]
and compute the elasticity as
h1_w1 = ((1-m1$alpha)*m1$A1 ) / (m1$w[1]^2 *(1+m1$beta+(m1$beta^2)*m1$alpha)) +
((1-m1$alpha)*m1$w[2]*m1$L ) / (m1$w[1]^2 *(1+m1$beta+(m1$beta^2)*m1$alpha)*(1+m1$r))
h1_A1 = - (1-m1$alpha ) / (m1$w[1] *(1+m1$beta+(m1$beta^2)*m1$alpha))
deriva = h1_w1 - h1_A1 * df[1,]$hours
hicks = deriva * m1$w[1] / df[1,]$hourswhich yields result \(\varepsilon_{1,H}=\) 14.747.