Timeline for Estimate on gaussian distribution
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 28, 2014 at 11:42 | vote | accept | splinter123 | ||
| Jan 27, 2014 at 17:42 | answer | added | splinter123 | timeline score: 0 | |
| Jan 27, 2014 at 11:33 | comment | added | splinter123 | @cardinal It is actually not clear to me how you get the bound with your method, either in the singular and in the nonsingular case. If you could post it as an answer it would be more clear I guess. | |
| Jan 26, 2014 at 23:35 | comment | added | cardinal | @Michael: What part of my comment involves numerical methods? The OP only wants a lower bound. Maybe I just need to polish up my original comment and post it. | |
| Jan 26, 2014 at 23:14 | comment | added | Michael Hardy | @cardinal : "Hard" in the sense that you probably have to resort to numerical methods. | |
| Jan 26, 2014 at 18:36 | comment | added | cardinal | @Michael: Just as hard, perhaps...but not hard. | |
| Jan 26, 2014 at 18:16 | comment | added | Michael Hardy | If you do reduce it to a normal distibution whose variance is just an identity matrix, then the constraints will also get transformed so that each will involve more than one of the independent components, so I think you'll still have just as hard a problem. | |
| Jan 26, 2014 at 15:14 | comment | added | cardinal | Perhaps you can take a moment to reread my comment with additional care. I mention nothing about eigenvalues and the argument is readily adaptable to the nonsingular case. (The handling of the lack of inverse should take only a slight amount of care by using the SVD.) | |
| Jan 26, 2014 at 11:45 | answer | added | Carlo Beenakker | timeline score: 5 | |
| Jan 26, 2014 at 8:29 | history | edited | splinter123 | CC BY-SA 3.0 |
added some details about some possible solutions.
|
| Jan 26, 2014 at 8:13 | comment | added | splinter123 | yes of course I can use $x'\Sigma^{-1}x\leq |x|^2/\lambda_1$ (where $\lambda_1$ is the lowest eigenvalue of $\Sigma$) and then use a bound on the standard normal. But again this requires invertibility of $\Sigma$, and I don't want to assume that! | |
| Jan 26, 2014 at 3:42 | comment | added | cardinal | Here is a basic schematic: Find the maximal ellipsoid of the form $\mathcal E_r = \{x: x^T \Sigma^{-1} x \leq r^2\}$ that is inscribed in the hypercube $[-M,M]^d$. The correct $r$ is something like $M/\sigma_1$ where $\sigma_1 = \max_i \sqrt{\Sigma_{ii}}$. This gives a lower bound for your probability in terms of the probability that the norm of a $d$-dimensional standard normal falls within the given radius. This latter quantity can be bounded using a Chernoff bound giving something like $f(M) = 1 - b M^{d/2} \exp(-M/2)$ for constant $b$ (likely, $b = d^{-d/2}\exp(d/2)$ or so). | |
| Jan 25, 2014 at 23:39 | comment | added | splinter123 | not feasible if the rank of the matrix is lower than $d$, which was the main point of the question | |
| Jan 25, 2014 at 22:36 | comment | added | Carlo Beenakker | $f(M)=(2\pi)^{-d/2}||\Sigma||^{-1/2}\int_{-M}^{M}\cdots\int_{-M}^{M}\exp\large(-\frac{1}{2}x\cdot\Sigma^{-1}\cdot x\large)dx_1dx_2\cdots dx_d$ | |
| Jan 25, 2014 at 14:28 | history | asked | splinter123 | CC BY-SA 3.0 |