Timeline for Bound moments wrt. known initial and final moments
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2021 at 18:24 | vote | accept | Philipp Wacker | ||
| Apr 20, 2021 at 12:00 | comment | added | Iosif Pinelis | (ii) The inequality $E|(1-\sqrt t)X+\sqrt t\,(X+Y)|^p\le(1-\sqrt t)^p E|X|^p+(\sqrt t)^p E|X+Y|^p$ follows from the inequality $(a+b)^p\le a^p+b^p$ for $p\in(0,1]$ and nonnegative $a,b$, and standard properties of the expectation. | |
| Apr 20, 2021 at 11:56 | comment | added | Iosif Pinelis | @MercuryBench : (i) The path here is quite irrelevant. The expectation of a given function of two random variables (r.v.'s) (here of $X$ and $W_t$) depends only on the joint distribution of the r.v.'s. In this case, $X$ is independent of each $W_t$, so that the joint distribution of $X$ and $W_t$ is determined by the individual distributions of $X$ and of $W_t$. Also, $W_t$ has the same distribution as $\sqrt t\,W_1$. So, the joint distribution of $X$ and $W_t$ is the same as that of $X$ and $\sqrt t\,W_1$. So, $E|X+W_t|^p=E|X+\sqrt t\,W_1|^p=E|X+\sqrt t\,Y|^p$. | |
| Apr 20, 2021 at 6:16 | comment | added | Philipp Wacker | And, also, why does $E|(1-\sqrt t)X+\sqrt t\,(X+Y)|^p \le(1-\sqrt t)^p E|X|^p+(\sqrt t)^p E|X+Y|^p$ hold? For $p<1$ we do not have the Minkowski inequality, right? | |
| Apr 20, 2021 at 6:07 | comment | added | Philipp Wacker | And I can just swap random variables out with other ones having the same distribution? I am worried that I can't do that here because the path of $W_t$ and $\sqrt t W_1$ are very different. | |
| Apr 20, 2021 at 5:54 | comment | added | Anthony Quas | $W_t$ has the same distribution as $\sqrt t W_1$. | |
| Apr 20, 2021 at 5:32 | comment | added | Philipp Wacker | But $W_t \neq \sqrt t W_1$, so I don't understand why the first term is valid. | |
| Apr 19, 2021 at 23:29 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |