Causation

Parameter Constraints and the Concavity of v(t) = βt^α

For the function v(t)=βtαv(t) = \beta t^\alpha to represent convex preferences, it must be concave in tt. This is confirmed by checking its second derivative, v(t)=α(α1)βtα2v''(t) = \alpha(\alpha-1)\beta t^{\alpha-2}. This derivative is negative because while α\alpha, β\beta, and tt are all positive, the term (α1)(\alpha-1) is negative due to the constraint $0 < \alpha < 1$. The product of positive terms and one negative term results in a negative value, proving the function's concavity.

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Updated 2026-05-02

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