Ansatz y ecuación HJB

Question

Supongamos que tenemos una ecuación HJB de la forma $$ \frac{\partial v}{\partial t}+\frac{1}{2}\sigma^{2}\frac{\partial^{2}v}{\partial s^{2}}+max_{\delta^{a}}\left\{ \lambda^{a}(\delta^{a})\left[v(t,s,x+s+\delta^{a},q-1)-v(t,s,x,q)\right]\right\}+max_{\delta^{b}}\left\{ \lambda^{b}(\delta^{b})\left[v(t,s,x-s+\delta^{b},q+1)-v(t,s,x,q)\right]\right\} $$ con condición de terminal $$ v(T,s,x,q)=-e^{-\gamma(x+qs)} $$ Buscaremos una solución de la forma $$ v(t,s,x,q)=-e^{-\gamma\left(x+\theta(t,s,q)\right)}=f(x,\theta(t,s,q)) $$ por sustitución directa en la ecuación de HJB y aplicación de la regla de la cadena obtenemos $$ \frac{\partial f(x,\theta(t,s,q))}{\partial\theta(t,s,q)}\frac{\partial\theta(t,s,q)}{\partial t}+\frac{1}{2}\sigma^{2}\left[\frac{\partial f(x,\theta(t,s,q))}{\partial\theta(t,s,q)}\frac{\partial^{2}\theta(t,s,q)}{\partial s^{2}}+\frac{\partial^{2}f(x,\theta(t,s,q))}{\partial\theta(t,s,q)^{2}}\left(\frac{\partial\theta(t,s,q)}{\partial s}\right)^{2}\right]+max_{\delta^{a}}\left\{ \lambda^{a}(\delta^{a})\left[f(x+s+\delta^{a},\theta(t,s,q-1))-f(x,\theta(t,s,q))\right]\right\} +max_{\delta^{b}}\left\{ \lambda^{b}(\delta^{b})\left[f(x-s+\delta^{b},\theta(t,s,q+1))-f(x,\theta(t,s,q))\right]\right\} $$ tomando derivadas de $f$ $$ \gamma f(x,\theta(t,s,q))\frac{\partial\theta(t,s,q)}{\partial t}+\frac{1}{2}\sigma^{2}\left[\gamma f(x,\theta(t,s,q))\frac{\partial^{2}\theta(t,s,q)}{\partial s^{2}}-\gamma^{2}f(x,\theta(t,s,q))\left(\frac{\partial\theta(t,s,q)}{\partial s}\right)^{2}\right]+max_{\delta^{a}}\left\{ \lambda^{a}(\delta^{a})\left[f(x+s+\delta^{a},\theta(t,s,q-1))-f(x,\theta(t,s,q))\right]\right\}+max_{\delta^{b}}\left\{ \lambda^{b}(\delta^{b})\left[f(x-s+\delta^{b},\theta(t,s,q+1))-f(x,\theta(t,s,q))\right]\right\} $$ dividiendo por $\gamma f(x,\theta(t,s,q))$ $$ \frac{\partial\theta(t,s,q)}{\partial t}+\frac{1}{2}\sigma^{2}\left[\frac{\partial^{2}\theta(t,s,q)}{\partial s^{2}}-\gamma\left(\frac{\partial\theta(t,s,q)}{\partial s}\right)^{2}\right]+max_{\delta^{a}}\left\{ \frac{\lambda^{a}(\delta^{a})}{\gamma}\left[e^{-\gamma\left(s+\delta^{a}+\theta(t,s,q-1)-\theta(t,s,q)\right)}-1\right]\right\}+max_{\delta^{b}}\left\{ \frac{\lambda^{b}(\delta^{b})}{\gamma}\left[e^{\gamma\left(s-\delta^{b}-\theta(t,s,q+1)+\theta(t,s,q)\right)}-1\right]\right\} $$ ¿Es esto correcto? Se afirma que con este ansatz deberíamos tener en cambio $$ \frac{\partial\theta(t,s,q)}{\partial t}+\frac{1}{2}\sigma^{2}\left[\frac{\partial^{2}\theta(t,s,q)}{\partial s^{2}}-\gamma\left(\frac{\partial\theta(t,s,q)}{\partial s}\right)^{2}\right]+max_{\delta^{a}}\left\{ \frac{\lambda^{a}(\delta^{a})}{\gamma}\left[1-e^{-\gamma\left(s+\delta^{a}+\theta(t,s,q-1)-\theta(t,s,q)\right)}\right]\right\} +max_{\delta^{b}}\left\{ \frac{\lambda^{b}(\delta^{b})}{\gamma}\left[1-e^{\gamma\left(s-\delta^{b}-\theta(t,s,q+1)+\theta(t,s,q)\right)}\right]\right\} $$ No estoy muy seguro de por qué los signos son diferentes, pero creo que me estoy perdiendo algo realmente trivial.

Answer 1

Los derivados tenían un signo erróneo (debía ser (-f) en lugar de f, por ejemplo)