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I'm reading this paper relating to optimal investment with transaction costs where some value function $F(x)$ is optimized. At some boundary $x=u$ it will be optimal to pay a proportional cost $C$ which gives the boundary condition

\begin{equation} F'(u) = -C \end{equation}

The author argues that optimality also implies a boundary condition for the second derivative

\begin{equation} F''(u) = 0 \end{equation}

but I'm struggling to understand why this is the case. Any hints that will help me understand the intuition behind this condition?

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  • $\begingroup$ Does $C$ depend on $u$? $\endgroup$
    – Bob Jansen
    Jul 6, 2020 at 8:56
  • $\begingroup$ $C>0$ is just some constant. $\endgroup$
    – Freelunch
    Jul 6, 2020 at 9:00
  • $\begingroup$ I'm confused, doesn't the second equation follow from $F'$ being constant? $\endgroup$
    – Bob Jansen
    Jul 6, 2020 at 9:11
  • $\begingroup$ It's only constant at $x=u$. Note that $x$ is the variable and $x=u$ is the boundary. $\endgroup$
    – Freelunch
    Jul 6, 2020 at 9:13

1 Answer 1

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I think that they are saying that, at special point $u$, we have:

$$ F'(u) = -\rho. $$

Also, that in a neighborhood from the left, we have:

$$ F'(u- dU) = -\rho $$

for any small positive $dU$.

Then, with Taylor on the left:

$$ F'(u) - F''(u)dU = - \rho $$

Hence: $$ F''(u) = 0 $$

Basically, if the first derivative of a function is constant in a neighborhood of a point, then its second derivative must be null at that point.

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  • $\begingroup$ Why is $F'(u- dU) = -\rho $ true? $\endgroup$
    – Freelunch
    Jul 6, 2020 at 21:06
  • $\begingroup$ Equation (10) it seems. Setting $v$ to $u-dU$. $\endgroup$
    – ir7
    Jul 6, 2020 at 23:36

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