4 ms·
Maybe I am wrong. But how can dU(AW)/dW = A dU(AW)/dW be true? If you set z = dU(AW)/dW and substitute, then z = Az. This can only be true in trivial cases. Al
by ckuehne 16y ago
Maybe I am wrong. But how can dU(AW)/dW = A dU(AW)/dW be true? If you set z = dU(AW)/dW and substitute, then z = Az. This can only be true in trivial cases.
Also if you apply ln(P) as utility function then the partial derivate for W yields the same result to U(P) = U(AW) as to U(P) = U(W), namely 1/W. So in that case your inequality does not hold.
- yummyfajitas 16y agoI think the Liebniz notation is confusing you. Therefore: [U(G(W))]' = U'(G(W))G'(W) [ln(AW)] ' = ln'(AW) [AW]' = (1/(AW)) A = 1/W If G(W) = AW, then G'(W)=A. So in that case [U(P)=ln(P)] your inequality does not hold. The inequality I gave is a criteria, not a theorem. I didn't claim it was always satisfied, I just claimed it was possible. If it is satisfied (e.g., U(P)=sqrt(P)), a NIT creates an incentive for work. If equality holds (U(P)=ln(P)), a NIT has no marginal effect on work. If the reverse inequality holds (U(P)=1-1/P), a NIT creates a disincentive for work.
- ckuehne 16y agoActually I think the Leibniz notation is confusing you a little (or at least you are using it very sloppy): you use dU(AW)/dW to denote the derivative of the chained functions as well as the derivative of the outer function applied to the inner function, all in the same equation. As I said: substitute the aforementioned term with z and your equation reads z = Az.