# Deriving the tax incidence formula -- what am I doing wrong?

I'm trying to derive the tax incidence for an excise $$t$$ so that producers receive $$p$$ and consumers pay $$q=p+t$$ per unit. The formula I am seeing everywhere is some version of

$$\frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\epsilon_D}{\epsilon_S - \epsilon_D},$$

but I keep ending up (derivation below) with:

$$\frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\epsilon_D}{(q/p) \epsilon_S - \epsilon_D}.$$

I use definitions of elasticity $$\epsilon_D = \frac{q}{D}\frac{\partial D}{\partial q}$$ and $$\epsilon_S = \frac{p}{S}\frac{\partial S}{\partial p}$$. I can see that I would get the expected result if I used $$\epsilon_D = \frac{p}{D}\frac{\partial D}{\partial q}$$ instead, but surely this would not be sound, since the consumer adjusts their demand based on $$q$$, not $$p$$?

What is my mistake here?

## My derivation

Starting from the equilibrium condition

$$D(q) = S(p),$$

I differentiate by $$t$$ to get

$$\frac{\partial D}{\partial q} \frac{\mathrm{d} q}{\mathrm{d} t} + \frac{\partial D}{\partial q} \frac{\partial q}{\partial p} \frac{\mathrm{d} p}{\mathrm{d} t} = \frac{\partial S}{\partial p} \frac{\mathrm{d} p}{\mathrm{d} t},$$

Which rearranges to

$$\frac{\mathrm{d} p}{\mathrm{d} t} = \frac{ \frac{\partial D}{\partial q} }{ \frac{\partial S}{\partial p} - \frac{\partial D}{\partial q} },$$

Noting that $$\frac{\partial q}{\partial p} = \frac{\partial q}{\partial t} = 1$$.

Now I apply the definitions of the elasticities $$\epsilon_D = \frac{q}{D}\frac{\partial D}{\partial q}$$ and $$\epsilon_S = \frac{p}{S}\frac{\partial S}{\partial p}$$, giving

$$\frac{\mathrm{d} p}{\mathrm{d} t} = \frac{ \frac{D}{q}\epsilon_D }{ \frac{S}{p} \epsilon_S - \frac{D}{q} \epsilon_D } = \frac{\epsilon_D}{\frac{q}{p} \epsilon_S - \epsilon_D},$$

where I simplify by dividing through with $$\frac{D}{q}$$ (recalling that $$S = D$$).

• Looking at these slides it seems that the tax incidence is the derivative of price with respect to tax evaluated for a tax $t = 0$. In this case $q = p$ so the ratio becomes 1.
– tdm
Commented Dec 6, 2023 at 15:30