Recovering a Value Function from Its Convex Dual
Summary
The document asks how derivatives of a convex value function relate to its dual, defined using the slope of the original function as the dual variable. The answer derives the first derivative of the dual by applying the chain rule: terms involving the change in the state cancel because the dual variable equals the value function’s slope. Differentiating again gives the negative reciprocal of the original function’s second derivative, assuming the inverse relationship between state and slope is well defined.
This identity is useful in dual methods that transform a nonlinear Hamilton–Jacobi–Bellman equation into a linear equation. However, the response only addresses the derivative identities. It does not answer the question of reconstructing the original value function from a solved dual function, nor does it discuss boundary conditions, convexity requirements, or cases where the mapping is not invertible.
Key ideas
- The dual variable is the derivative of the original value function with respect to its state.
- The first derivative of the dual function equals the negative of the state variable.
- The second derivative of the dual is the negative reciprocal of the original function’s curvature when the slope mapping is invertible.
- The response derives these identities but leaves the value function reconstruction question unresolved.
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# Recover value function from its dual
# Recover value function from its dual
I take as my reference Optimal Investment by Rogers (2013). Denote the agent's value function, which is convex, by $V(w)$, where $w$ is the state variable. Define the dual variable $z = V_w$, where the subscript denotes differentiation. Also define the dual value function $$ J(z) = V(w) - w z\ . $$ I have two questions. Firstly, Rogers claims that the following identities hold: $$ J_z = -w\ , \quad J_{zz} = -1/V_{ww}\ . $$ The first identity is clear enough, but how does one derive the second?
Secondly, Rogers demonstrates how this function is helpful in transforming the non-linear HJB equation for $V$ into a linear differential equation for $J$. Fair enough. What I don't understand is how to recover the original value function once we have solved for $J(z)$. If I substitute the available information into the above equation I end up with the differential equation $$ J(V_w) = V - w V_w\ , $$ which Mathematica says can be solved as $$ V(w) = C w + J(C)\ , $$ where $C$ is a constant of integration. But $V$ will not in general be a linear function: where I am going wrong? Or more precisely, how can $V$ be recovered from $J$?
## Answer by Attack68 (score 1)
https://quant.stackexchange.com/a/80147
Draft attempt..
$$ d_z J = d_z V - w - z d_z w $$ where $$ d_z V = \partial_z V + \partial_w V d_z w$$ noting $\partial_z V = 0$ and $\partial_w V \equiv z$ $$ \implies d_z V = z d_z w $$ and $$ \implies d_z J = -w $$ Now, $$ d^2_{zz} J = - d_z w$$ If, $$ d_z w = \frac{1}{d_w z} = \frac{1}{d^2_{ww} V }$$ then, $$ d^2_{zz} J = \frac{-1}{d^2_{ww} V} $$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.