Timeline for Differentiability of value function
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2015 at 16:44 | comment | added | avk255 | Great. Many thanks for your help. :) | |
| Dec 2, 2015 at 16:44 | vote | accept | avk255 | ||
| Dec 2, 2015 at 16:22 | comment | added | Bjørn Kjos-Hanssen | By substitution and the chain rule from ordinary calculus | |
| Dec 2, 2015 at 16:17 | comment | added | avk255 | Awesome. Thanks. Sorry to bother you with one more question then. If you need the derivative of $\int_0^t [W_s < -x] ds$ how does that translate to requiring the derivative of $\int_0^t [ W_s < x ] ds$? | |
| Dec 2, 2015 at 8:42 | comment | added | Bjørn Kjos-Hanssen | @Aditya you're right, I fixed the sign now. | |
| Dec 2, 2015 at 8:41 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
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| Dec 2, 2015 at 3:37 | comment | added | avk255 | Also, in going from line 1 to 2, should it not be $W_s < -x$ instead of $x$? | |
| Dec 2, 2015 at 3:05 | comment | added | avk255 | Yes, $V''$ won't exist at $0$, certainly. But when $x > 0$ and when $x < 0$, it will exist and I can use the DEs I mentioned, I think. And then, I can get the two constants using continuity and differentiability at $0$. | |
| Dec 2, 2015 at 3:03 | comment | added | Bjørn Kjos-Hanssen | Not sure $V''$ exists | |
| Dec 2, 2015 at 2:59 | comment | added | avk255 | The values computed using the two approaches don't match. What am I doing wrong? | |
| Dec 2, 2015 at 2:51 | comment | added | avk255 | Thanks a lot. I can sort of understand your argument. But here's a related question then. Assuming it is differentiable, I can compute the functional form explicitly as, when $x > 0$, $3 V(x) = 1 + 1/2 V''(x)$ and a similar DE when $x \le 0$. Now, I can solve these, kill one constant by conditions at $\infty$ and $-\infty$. For the remaining two constants I use continuity and differentiability at $0$. But here's an alternative way of getting $V(0)$. If $r = 3$ throughout then $V(0) = 1/3$. If $r = 7$, then $V(0) = 1/7$. Therefore, in our case, $V(0) = \frac{1/3 + 1/7}{2}$. | |
| Dec 2, 2015 at 1:19 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |
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| Dec 2, 2015 at 1:13 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 3.0 |