Timeline for A type of stochastic jump process
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Dec 23, 2010 at 16:03 | comment | added | fedja | Well, when I have plenty of fixed parameters ($\mu$ is fixed as well) and write $O(\mu)$, I usually mean that the implied constant is absolute. Also, the estimate $\le (K+1)\mu$ is totally trivial (we get an overshot in first $K+1$ steps for sure and each step can add only $\mu$ since the overshot is not greater than the step itself), so I just couldn't imagine that it was what was asked :) | |
| Dec 23, 2010 at 14:50 | comment | added | Shai Covo | @fedja: Since (in the original question) $K$ is a fixed integer, we do get $O(\mu)$. So, the first sentence in your answer has probably confused some readers. Note also that your answer corresponds to the upper bound indicated in my answer. | |
| Dec 22, 2010 at 2:27 | comment | added | fedja | Sorry, should be $(K/M_{k-1})$ and $K=k^3M_{k-1}$, of course. | |
| Dec 22, 2010 at 2:24 | comment | added | fedja | No luck here either. Let's say we have $M_k$ with probability $p_k$. To have finite $\mu$, all we need is the convergence of $\sum_k p_k M_k$. Now, the same argument gives the expectation of the overshot at least $(K/M_k)p_k (M_k-K)$. Now, if $M_k=k!^4$, we can take $K=k^3 M_k$ and still get $k^3p_kM_k/2$ for the expectation. But if $M_kp_k=k^{-2}$, this tends to infinity. You need some higher moments to put this under control. | |
| Dec 22, 2010 at 2:08 | comment | added | Probabilist | What if $K$ = a\mu for some large positive constant $a$? | |
| Dec 22, 2010 at 1:51 | history | answered | fedja | CC BY-SA 2.5 |