Law and Economics course, going over Shavell 2011 paper, specifically proposition 4.

The first claim is that ?= 1 and assumes t= 0 and if ?

To prove W?? will be higher when ?= 1 if y

which gives this equation to prove the claim.

Y

but im unsure what the equation means..

VOL 101 NO. 3 CORRECTIVE TAXATION VERSUS LIABILITY 275 x'(py) reflects y, it is excessive; x'(py) > x'(y) because p x(py) > x(y). From p b(x'(Xpy)) - x(Xpy)y. declines with the expected harm y, it is excess Hence, Wyz is higher at A = 1 than at A 0 and let x(y) = x'(t + Xpy). Note first that x(y) crosses x"(y) the tax and liability compare. once at ye = t/(1 - Ap). In particular, at a cross- PROPOSITION 3: Liability is superior to the ing point, b'(x) = y = t + Xpy, implying that ye is as stated. Now to show that A = 1 is opti- tax if and only if the probability of suit p is above mal, I demonstrate that if A p), where x(y) =x(1-T+ (x+ 6)py). With respect to the first claim, note that This will be so if (t - T) + (X + 6)pyc = yes implying that T = opy . By picking o suffi- WL 0. is first-best at p = 1. Hence, the asserted threshold p" must exist. Regarding the second To prove that Wyz will be higher, consider y y.). > b(x (pE(y))) - x(pE (y))E(y). Hence I claim that y b(x (py)) - x (py)y grand of Wy is higher for y if enough probability mass is contained in this *(y) > x(y). Now the right side of the inequal- neighborhood, the tax will be superior to liability. ity is equivalent to opy py)ylf(y)dy. = [x x"(t + Apy) [b'(x(t + Xpy)) - y)If(y)dy. This is positive at t = 0 since x"(Xpy) 0. The claim that " of harm paid by an injurer is 1. Also, the optimal tax " is positive; II. Model with Expected Harm and t"