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visits | member for | 3 years, 5 months |
seen | Feb 8 '13 at 15:53 | |
stats | profile views | 36 |
Aug 18 |
awarded | Popular Question |
Sep 29 |
comment |
Linear programming - uniqueness of optimal solution
Yes, I need any vertex of the polytope due to its properties that I want to use: at every vertex there are at least d constraints that hold with equality. |
Sep 28 |
comment |
Linear programming - uniqueness of optimal solution
OK, so it does not suit to me... What I need is a vertex (even any vertex). Thank you in any case! |
Sep 28 |
comment |
Linear programming - uniqueness of optimal solution
Brendan, only now I realized that the last comment was yours. So, thank you too for the help! |
Sep 28 |
accepted | Linear programming - uniqueness of optimal solution |
Sep 28 |
comment |
Linear programming - uniqueness of optimal solution
Thanks a lot Igor, I think this will solve my problem. |
Sep 28 |
awarded | Commentator |
Sep 28 |
awarded | Supporter |
Sep 28 |
comment |
Linear programming - uniqueness of optimal solution
Thank you Survit, I'll look at this paper. But can I be sure that returned solution (with the smallest L-2 norm) will be a vertex of the original polytope? |
Sep 28 |
comment |
Linear programming - uniqueness of optimal solution
Igor, thanks. You mean iterating the ellipsoid with any objective every time? In that case, every iteration will lead to decrease in dimension (at least one constraint will hold with equality), so finally I'll end with dimension 0, i.e., vertex. Correct? |
Sep 28 |
asked | Linear programming - uniqueness of optimal solution |
Nov 12 |
comment |
Open Jackson network with deterministic arrivals.
I mean the following: service time at every queue is distributed exponentially, and the external arrivals to the system are coming at fixed intervals (i.e., instead of Poisson($\lambda$), arrivals occur at fixed time interval of $\tfrac{1}{\lambda}$). The time is continuous. Thank you. |
Nov 11 |
asked | Open Jackson network with deterministic arrivals. |
Nov 7 |
accepted | Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v. |
Nov 7 |
comment |
Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v.
This is what I wrote in the last comment of the previous answer. So, in general, if the variables are not identical, the sum is not strongly concentrated around the $E[X]$ if $E[X]=\Theta(n)$. Thanks a lot! |
Nov 6 |
awarded | Student |
Nov 6 |
revised |
Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v.
added 38 characters in body |
Nov 6 |
comment |
Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v.
Maybe I should write this in the question: Yes, my $E[X]=\Theta(n)$. As I commented in the another answer, I think there is no such bound for this case (linear expectation and exponential bound). |
Nov 6 |
comment |
Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v.
I mean, $X_n$ is distributed Geom($p=\tfrac{1}{n}$) |
Nov 6 |
comment |
Probability Theory, Chernoff Bounds, Sum of Independent (but not identically distributed) r.v.
I think now that there is no such a bound. For example, if $X_1,...,X_{n-1}$ are distributed Geom(p=1), and $X_n$ is distributed $X_n$. Then, $E[X]=(n-1)\cdot 1 + n\approx 2n$. But X now is not much concentrated around $E[X]$. To obtain an exponential high probability, we have to repeat the experience $\Omega(n)$ times. If we want only $\tfrac{1}{n}$ high probability, we need to repeat it $log n$ times. So, the sum of $n$ indepemdent but not identical geometric r.v. is not concentrated around $\alpha\cdot E[x]$, where $\alpha=const$. |