Comment by yummyfajitas | Hacker News Reader

Comment by yummyfajitas | Hacker News Reader

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First, if the utility function was "downward sloping" nobody would ever work.

Aggh, I miswrote. I meant concave down, which is actually what I used in my calculation. Thanks for the correction. In defense of my readers, they probably glossed over my incorrect definitions, and didn't notice any problem because my calculations didn't match my definition.

But I will defend my calculus: dU(AW)/dW = A dU(AW)/dW follows from the chain rule. Another way to do your calculation: d(ln(AW))/dW = A x 1/(AW) = 1/W.

Also, the NIT I described has been proposed (that's how it was described in my intro to econ class), and it does not always create an incentive to work. The relevant criteria is A dU(AW)/dW > dU(W)/dW. Since dU(AW)/dW decreases with A, there the marginal utility of work can be negative with respect to NIT.

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.

I 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.

Actually 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.

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