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Consider a profit maximising monopolist with linear demand Q(P*) and total production cost C(Q(P*)) who faces a per unit tax t. Suppose the slope of marginal cost is decreasing in some parameter, μ. Let P* denote the monopolist’ profit maximising choice of price. Being careful to explain your method and interpret your result, determine the comparative static:

∂²P*/∂μ∂t

I got to P*= price Q = Q(P*) t=t(Q(P*)) C=C(Q(P*))

πmax = P*(Q(P*))-C(Q(P*))-t(Q(P*))

dπ/dP* = Q'(P*)[-C'(Q(P*))-t'(Q(P*))+P*]+Q(P*)=0 dπ/dP* = μ + Q'(P*)[-t'(Q(P*))+P*]+Q(P*)

Equilibrum: PQ'(P)=t'(Q(P*))Q(P))-μ-Q(P*)

P* = t'(Q(P*))- μ/(Q'(P*)) - Q(P*)/Q'(P*)

This is where I got to, when I solved the comparative static partial differentiation I got an answer of 0, I don't believe this is correct, can anyone help me solve this? Thanks!

If 0 is somehow the correct answer, what does it mean?

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    $\begingroup$ Hi! Can you please type your calculations into Mathjax and write separate lines separately so that the question is readable? $\endgroup$
    – Giskard
    May 27 at 5:49
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    $\begingroup$ Also, what is t'? And do I understand correctly that you just didn't type in your final few lines of calculations? $\endgroup$
    – Giskard
    May 27 at 5:50
  • $\begingroup$ So total tax is tQ so it comes out as t(Q(P)) so t(Q(P*)) differentiates to t'(Q(P*)) * Q'(P*) if I am not mistaken? $\endgroup$
    – Javid
    May 27 at 11:39
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The profit function is given by:

$$ \pi = PQ(P) - C(Q(P), \mu) - tQ(P) $$ Assume demand is linear, s $$ p = \alpha - \beta Q \to Q = b(\alpha - P), $$ where $b = 1/\beta$. So: $$ Q_P = -b, $$ where I use subscripts to denote the partial derivatives.

The first order condition for profit maximisation gives: $$ \begin{align*} &Q + P Q_P - C_Q Q_P - t Q_P = 0,\\ \iff &Q - b P + b C_Q + bt = 0 \end{align*} $$ then differentiating this with respect to $t$ gives: $$ \begin{align*} &(Q_P - b + b C_{QQ} Q_P)\frac{\partial P}{\partial t} + b = 0,\\ \to &(-2b - b^2 C_{QQ})\frac{\partial P}{\partial t} = -b,\\ \to &\frac{\partial P}{\partial t} = \frac{1}{2 + b C_{QQ}} \end{align*} $$ Differentiating the first order condition with respect to $\mu$ gives: $$ (Q_P - b + b C_{QQ} Q_P)\frac{\partial P}{\partial \mu} + b C_{Q\mu} = 0,\\ \to (-2b - b^2 C_{QQ})\frac{\partial P}{\partial \mu} = - b C_{Q\mu},\\ \to \frac{\partial P}{\partial \mu} = \frac{C_{Q,\mu}}{2 + b C_{QQ}} $$

Then differentiating $\frac{\partial P}{\partial t}$ once more with respect to $\mu$ gives: $$ \frac{\partial^2 P}{\partial t \partial \mu} = -\frac{1}{(2 + bC_{QQ})^2}b\left(C_{QQQ}Q_P \frac{\partial P}{\partial \mu}+ C_{QQ\mu}\right),\\ = -\frac{1}{(2 + b C_{QQ})^2}b \left(-b C_{QQQ} \frac{C_{Q,\mu}}{2 + b C_{QQ}} + C_{QQ\mu} \right) $$ Where the last line uses the expression for $\frac{\partial P}{\partial \mu}$ from above.

Now assume that costs take the form: $$ C(Q, \mu) = \delta + \eta Q + \frac{\gamma(\mu)\, Q^2}{2} $$ So the slope of the marginal costs depends on $\mu$. Then $C_{QQ} = \gamma(\mu)$, $C_{QQQ} = 0$ and $C_{QQ\mu} = \gamma_\mu$ , so: $$ \frac{\partial^2 P}{\partial t \partial \mu} = -\frac{b \gamma_\mu}{(2 + b \gamma)^2}. $$ We have that $\gamma_\mu < 0$ by assumption (slope decreases in $\mu$) then we see that the sign of $\frac{\partial^2 P}{\partial t \partial \mu}$ is positive (at least if I didn't make any mistakes).

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