Timeline for Radius of convergence of cumulant generating function
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2021 at 16:38 | comment | added | Max Alekseyev | I'm not sure what you're trying to say. If you like Alexandre's answer, just go for it. | |
| Jun 24, 2021 at 15:53 | comment | added | Boby | @MaxAlekseyev I don't see why the radius of convergence is relevant here. $M(t) =E[e^{tX}]>0$ for all $t$ since $e^{u}>0$ for all $u$. | |
| Jun 24, 2021 at 14:28 | comment | added | Max Alekseyev | For all real $t$ within the radius of convergence of $M$. | |
| Jun 24, 2021 at 13:58 | comment | added | Boby | @MaxAlekseyev But $M(t)>0$ for all real $t$, right? Can you explain this a bit more. | |
| Jun 24, 2021 at 13:26 | comment | added | Max Alekseyev | Well, the above argument is valid with addition that we want $M(t)>0$ (to be in the $\log$ domain) for convergence of $K(t)$. If this inequality always holds, then the two radii are the same; otherwise the smallest by absolute vale zero of $M(t)$ comes into play. | |
| Jun 24, 2021 at 12:44 | comment | added | Boby | @MaxAlekseyev See the answer below. | |
| Jun 24, 2021 at 2:40 | review | Close votes | |||
| Jul 16, 2021 at 3:01 | |||||
| Jun 24, 2021 at 2:33 | answer | added | Alexandre Eremenko | timeline score: 4 | |
| Jun 24, 2021 at 2:21 | comment | added | Max Alekseyev | If $M_x(t)$ converges for some $t$, then $K_X(t)$ is simply computed by the given formula $K_X(t) = \log M_X(t)$. And vice versa, $M_X(t)$ can be computed as $\exp K_X(t)$ if $K_X(t)$ converges. So, the slides are correct. | |
| Jun 24, 2021 at 1:45 | history | edited | Boby | CC BY-SA 4.0 |
added 1 character in body
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| Jun 24, 2021 at 1:37 | history | asked | Boby | CC BY-SA 4.0 |