Timeline for Positivity of linear combination of gaussian variables
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 28, 2023 at 21:42 | comment | added | happyle | @DieterKadelka sorry it took a while. My goal is to derive the probability $X>0$, which is strictly positive, not nonnegative. But to calculate the integral $\int_{0,\infty}pdf dx$ would give probability $P(X\geq 0)$. | |
| May 26, 2023 at 3:19 | comment | added | Michael Hardy | Possibly you are misled by a "false friend": the word "eventually" in English does not mean the same thing that "eventuell" means in French and German. | |
| May 24, 2023 at 10:51 | comment | added | happyle | sorry, just one second ago I found my question does not make sense. I will revise again | |
| May 24, 2023 at 10:51 | comment | added | Dieter Kadelka | yes you are right. But I still don't know why are you working with estimates and not with the exact expression. $\Phi$ is monotone with the consequence thart you can easily convert any bound for the probability into bounds for the coefficients $a,b,c,\ldots$. | |
| May 24, 2023 at 10:27 | comment | added | happyle | $\mathbb{P}(f<0)$ is equal to $\mathbb{P}(f>2b)$, since $f$ is $\mathcal{N}(b,p^2c^2)$ and its symmetric to $x=b$ am I right? | |
| May 24, 2023 at 10:21 | comment | added | Dieter Kadelka | I still don't understand your question. Maybe you are working with an erroneous equation: $\mathbb{P}(f > 0) = 1 - \mathbb{P}(f > 2b)$? | |
| May 24, 2023 at 10:20 | comment | added | happyle | (I found involving $p$ is misleading, I have delete $p$ in my original question.) | |
| May 24, 2023 at 8:28 | history | answered | Dieter Kadelka | CC BY-SA 4.0 |