Timeline for Distribution of bounded summation of i.i.d random variables
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 20, 2015 at 5:29 | comment | added | Bloodmoon | If you just want to say $P(S_{n-1} \le T, T < S_n \le s) = P(S_{n-1} \le T)*P(T < S_n \le s)$, then it is incorrect. | |
| Jul 20, 2015 at 4:54 | comment | added | Bloodmoon | Ohhh!!! Do you mean $\int_0^T dt\; \big( f_{n-1}(t) \int_T^s dr\; f(r-t) \big)$, rather than $\big( \int_0^T dt\; f_{n-1}(t)\big) \big( \int_T^s dr\; f(r-t) \big)$? | |
| Jul 20, 2015 at 4:52 | comment | added | Bloodmoon | I can understand your preference, but it seems these are two different formulas. It's like $\int f(x)dx g(x)$ and $\int f(x)g(x)dx$, obviously they are not equal. | |
| Jul 20, 2015 at 4:42 | comment | added | Robert Israel | I prefer to write multiple integrals "physics style" with the $d$(variable)'s after the integral sign rather than at the end. It helps to make the connection between variables and endpoints clearer. | |
| Jul 20, 2015 at 2:55 | comment | added | Bloodmoon | sorry, by $x$ I mean $s$. If it is a function of $s$ and $T$, it should be $P(S_K \le s) = \int_T^s f(r) + \sum_{n=2}^\infty \int_0^T f_{n-1}(t) \int_T^s f(r-t) dr\,dt$. $dr$ should be placed at the end of the formula. | |
| Jul 19, 2015 at 18:33 | comment | added | Robert Israel | $x$? What $x$? It's a function of $s$ and $T$ | |
| Jul 19, 2015 at 11:47 | comment | added | Bloodmoon | I am sorry to disturb you again, but I cannot figure out how to compute $\int_0^T dt\; f_{n-1}(t) \int_T^s dr\; f(r-t)$. Because mathematically, as this formula shows, its result is a function of $s$ and $r$. But allowing for the physical meaning, it should be a function of $x$. Could you give me a hand on this? | |
| Jul 13, 2015 at 16:17 | vote | accept | Bloodmoon | ||
| Jul 10, 2015 at 17:06 | comment | added | Bloodmoon | Yes, that's true. My thoughts on this was too naive. Let $N(t)$ denote the counting process, it's easy to see that $N(Z)=K-1$, so $E[N(Z)]=E[K]-1$. Define $m(t)=E[N(t)]$, $m(t)$ is the renewal function, and it can be solved by the renewal equation using Laplace transform, so $m(Z)$ can be computed. In this way, $E[K]$ can be derived by $E[K]=m(Z)+1$. Is this correct? | |
| Jul 10, 2015 at 15:45 | comment | added | Robert Israel | For one thing, there's no reason for $E[K]$ to be an integer. | |
| Jul 10, 2015 at 6:33 | comment | added | Bloodmoon | $T$ will not has a limit in my case. Why $E[K]=\lceil {T/E[X]} \rceil$ is incorrect? This seems straightforward based on the definition of $K$. | |
| Jul 10, 2015 at 5:35 | comment | added | Robert Israel | No, but in the limit as $T \to \infty$ you have the Elementary Renewal Theorem. | |
| Jul 10, 2015 at 3:15 | comment | added | Bloodmoon | Thanks! It's clear that I can compute the expectation of $K$ $E[K]$ according to its CDF. But this still involves convolution. Is there any simpler way to derive $E[K]$? Is $E[K]=\lceil {T/E[X]} \rceil$ correct? | |
| Jul 9, 2015 at 19:20 | comment | added | Robert Israel | 1) memoization. 2) Thats why it's a sum over $n$ rather than an integral. 3) Because $S_n = S_{n-1} + X_n$. | |
| Jul 9, 2015 at 19:16 | history | edited | Robert Israel | CC BY-SA 3.0 |
added 232 characters in body
|
| Jul 9, 2015 at 16:53 | comment | added | Bloodmoon | The CDF of $K$ is straightforward to understand, but since $f_n(t)$ involves convolution, 1) is there any way to reduce the computational complexity for computers? 2) the values of $K$ are actually discrete, should this be considered while deriving the CDF for $S_K$? 3) why $f(r-t)$ rather than $f_n(r)$ is at the end of $P(S_k\leq s)$? | |
| Jul 9, 2015 at 16:14 | history | answered | Robert Israel | CC BY-SA 3.0 |