Finding the optimal tax

Question

I have a problem regarding the last two questions of this exercises because I can not understand how to set up my model in order to find the optimal tax the optimal quantity to be as there was not a tax.

Please share your ideas with me. Thanks.

Answer 1

Prove that $\Pi_j(x_j)$ is strictly concave in $x_j$.

$\Pi_j(x_j) = G(x_j) + F\left(\frac{x_j}{y}\right) =G(x_j) + F\left(R_j(x_j)\right) $

Differentiating it we get

$\Pi_j'(x_j) =G'(x_j) + F'\left(R_j(x_j)\right)R_j'(x_j) $

Differentating $\Pi_j'(x_j)$, we get

$\Pi_j''(x_j) =G''(x_j) + F''\left(R_j(x_j)\right)(R_j'(x_j))^2 + F'\left(R_j(x_j)\right)R_j''(x_j) < 0 $

Therefore, $\Pi_j$ is concave.

In a symmetric Nash equilibrium, determine the number of hours devoted to academic work. Call it $x^{**}$.

Symmetric Nash equilibrium where everyone choose $x^{**}$ satisfy :

$G'(x^{**}) + F'\left(1\right)R_j'(x^{**}) = 0 $

Determine if $x^{**}$ is greater than, less than, or equal to $x^*$?

Note that $x^*$ satisfy :

$G'(x^{*}) = 0$

and $x^{**}$ satisfy :

$G'(x^{**}) + F'\left(1\right)R_j'(x^{**}) = 0$

Combining the two, we get

$F'\left(1\right)R_j'(x^{**}) = G'(x^{*}) - G'(x^{**})$

Since $F'\left(1\right) > 0$ and $R_j'(x^{**}) > 0$ holds, we obtain

$G'(x^{*}) - G'(x^{**}) > 0$

Since $G$ is concave, we get $x^* < x^{**}$.

Compare the students' welfare at $x^*$ and $x^{**}$.

$\Pi_j(x^{**})=G(x^{**}) + F(1) = G(x^{**}) < G(x^{*})$

Therefore, students are worse off at $x^{**}$ compared to $x^{*}$.

Find the tax $\theta^* > 0$ on academic work, that ensures that $x^{**}= x^{*}$.

Symmetric Nash equilibrium where everyone choose $x^{**}$ in the presence of tax $\theta$ satisfy :

$G'(x^{**}) + F'\left(1\right)R_j'(x^{**}) - \theta = 0 $

To determine the sign of $\partial x^{**}/\partial \theta$, we differentiate the above condition with respect to $\theta$ :

$\displaystyle G''(x^{**})\frac{\partial x^{**}}{\partial \theta} + F'\left(1\right)R_j''(x^{**})\frac{\partial x^{**}}{\partial \theta} - 1 = 0 $

This yields

$\displaystyle \frac{\partial x^{**}}{\partial \theta} = \frac{1}{G''(x^{**})+F'\left(1\right)R_j''(x^{**})} < 0 $

Finally, we want to find $\theta^*$ that ensures $x^{**} = x^*$. To do so, we'll use

$G'(x^{**}) + F'\left(1\right)R_j'(x^{**}) - \theta = 0 $

and substitute $x^{**} = x^*$ in above to get

$\theta^* = F'\left(1\right)R_j'(x^{*})$