# Comparative statics of a monopoly

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?

• Hi! Can you please type your calculations into Mathjax and write separate lines separately so that the question is readable? May 27, 2021 at 5:49
• Also, what is t'? And do I understand correctly that you just didn't type in your final few lines of calculations? May 27, 2021 at 5:50
• 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? May 27, 2021 at 11:39

$$\pi = PQ(P) - C(Q(P), \mu) - tQ(P)$$ 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).