Exercise for Capital Income Tax

1. Derive $l^{\ast }$ when $\delta >0$, show that $l^{\ast }$ depends on ${\tau }_r$ if $\delta >0$

Answer: recall the definition of $\delta$ is the capital depreciation rate,

Now, the asset-accumulation equation of $\dot{K}$ becomes:

$$\begin{array}{rl}\dot{K}& =I=wl+rK-{\tau }_{r}rK-T-C-\delta K\\ & =[r-{\tau }_{r}r-\delta ]K+wl-C-T\end{array}$$

We should then redo the Hamiltonian:

$$\begin{array}{rl}\frac{\delta H}{\delta C}& =\frac{1}{C}-\lambda =0⟹\lambda =\frac{1}{C}\\ \frac{\delta H}{\delta l}& =-\frac{\beta }{L-l}+\lambda w=0⟹\lambda w=\frac{\beta }{L-l}\\ \frac{\delta H}{\delta K}& =\lambda (r-{\tau }_{r}r-\delta )=\rho \lambda -\dot{\lambda }\\ ⟹-& \frac{\dot{\lambda }}{\lambda }=(1-{\tau }_r)r-\delta -\rho =\frac{\dot{C}}{C}\end{array}$$

We assume the labor-income tax = 0, so we won’t have ${\tau }_r$ in the equation 2.

Use the 1, 3 equation:

We could get:

$$\frac{\dot{C}}{C}=(1-{\tau }_r)r-\delta -\rho$$

Also recall the equation we use when finding the $l^{\ast }$:

$$l^s=l^d:l=L-\frac{\beta C}{(1-\alpha )Y}l$$

Since it is at the steady state, that is $\dot{C}=0$,

Government Side

$$G=T+{\tau }_{r}rK$$

Comparative Analysis

Always remember that we would have following steps for comparative analysis:

1. Capital Market: What does $\dot{K}$ change?
2. Labour Market: $l^d$?
3. Output Market: What does $Y$change? And lead to the decrease in $l^s$
4. Capital Market: $K^d$ shift?

$$\begin{array}{rl}\dot{K}& =I=\underset{Y}{\underbrace{wl+rK}}-\underset{G=\gamma Y}{\underbrace{{\tau }_{r}rK+T}}-C-\delta K=0\\ 0& =Y-{\tau }_{r}rK-C-\delta K\\ C& =Y-{\tau }_{r}rK-\delta K\\ \frac{C}{Y}& =(1-\gamma )\frac{Y}{Y}-\frac{\delta K}{Y}\end{array}$$

That is to say, we have to derive $\frac{K}{Y}$, the capital-output ratio.

At long-run,

$$\begin{array}{rlr}{K}^d:& r=\alpha \frac{Y}{K}⟹\frac{K}{Y}=\frac{\alpha }{r}& \\ K^s:& \dot{C}=0⟹(1-{\tau }_r)r=\rho +\delta ⟹r=\frac{\rho +\delta }{1-{\tau }_r}& \end{array}$$

So that when $K^d=K^s$, we could get: $\frac{K}{Y}=\frac{\alpha (1-{\tau }_r)}{(\rho +\delta ))}$

That is, $\frac{C}{Y}=(1-\gamma )-\frac{\delta \alpha (1-{\tau }_r)}{\rho +\delta }$, so that ${\tau }_r\uparrow ⟹\frac{C}{Y}\uparrow$

So we plug it in the original equation, the we could get:

$$l^{\ast }=\frac{L}{1+\frac{\beta }{1-\alpha }[(1-\gamma )-\frac{\delta \alpha (1-{\tau }_r)}{\rho +\delta }]}$$

So that the ${\tau }_r\uparrow ⟹l^{\ast }\downarrow$

The negative effect of ${\tau }_r$ on $l^d$ is bigger than the positive effect of ${\tau }_r$ on $l^s$,

The overall effect of ${\tau }_r$ on $l^{\ast }$ depends on $\frac{C}{Y}$

• The negative effect of ${\tau }_r$ on $l^{\ast }$ via $l^d$ is captured by the decrease in $Y$.

• The positive effect of ${\tau }_r$ on $l^{\ast }$ via $l^s$ is captured by the decrease in $C$.

• If $\delta =0$, the percent decrease in $Y=$ the percent decrease in $C$ because $I^{\ast }=0$ in Long-run.

• If $\delta >0$, the percent decrease in $Y>$ the percent decrease in $C$ because $I^{\ast }>0$

Important:

\begin{align} \dot{K} &= I - \delta K= wl + rK -\tau_{r}rK - T - C - \delta K \ &= [r-\tau_{r}r -\delta]K + wl -C - T \end{align}

What happens when $\tau_{r}$ and $\gamma$ change simultaneously?

\begin{align} G & = T + \tau_{r}rK \ \gamma Y & = \tau_{r}rK \ \gamma & = \tau_{r} \frac{rK}{Y} = \alpha \tau_{r} \end{align}

- If $\delta = 0$, $l^* = \frac{L}{1+\frac{\beta}{1-\alpha}(1-\gamma)}= \frac{L}{1+\frac{\beta}{1-\alpha}(1-\alpha \tau_{r})}=l^*(\underbrace{ \tau_{r} }_{ + })$ - If $\delta> 0$, $l^* = \frac{L}{1+\frac{\beta}{1-\alpha}[1-\gamma-\frac{\delta\alpha(1-\tau_{r})}{\delta+\rho}]}= \frac{L}{1+\frac{\beta}{1-\alpha}[1-\alpha \tau_{r}-\frac{\delta\alpha(1-\tau_{r})}{\delta+\rho}]}=l^*(\underbrace{ \tau_{r} }_{ + })$ In both cases, $l^*$ is increasing in $\tau_{r}$, when $\gamma$ and $\tau_{r}$ change simultaneously. > Do it yourself!