Timeline for Interplay between CLT and convergence in Total Variation
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 9, 2022 at 14:32 | comment | added | Cain | Not in general. But at least when the X_i are Rademacher, there is no convergence in Total variation. In this case, F_n is always supported on a lattice, which has Gaussian measure 0. | |
| Dec 7, 2022 at 5:59 | comment | added | Amit Portnoy | Ha you managed to find a solution here? In may case the X are Rademacher variable and I'm also trying to characterize A_n's properties | |
| Jan 8, 2016 at 19:12 | comment | added | Cain | The thing is, it's not obvious how to 'smear' a discrete random variable into one which a Poincare inequality. Satisfying such inequalities, means the density cannot be too 'bumpy', in the sense that it cannot be 0 at any interval (otherwise just choose $f$ to be a bump function on the complement of the interval). | |
| Jan 7, 2016 at 12:17 | comment | added | Douglas Zare | Ok. I'm not sure what that implies for an absolutely continuous distribution. The Berry-Esseen bound is sharp (within a constant) when a random variable is well-approximated by lattice random variables. If you can smear a lattice random variable slightly (by a decreasing amount as you move out on the tails) so that it satisfies a Poincare inequality, then a candidate event would be that the sum is far from the lattice, say that the fractional part is between $1/4$ and $3/4$. | |
| Jan 7, 2016 at 7:15 | comment | added | Cain | Well, $X$ is fixed. The sequence is $a_iX$ in essence. Sp, they are not identical, but up to re-scaling all elements of the sequence are like $X$. A random variable satisfies Poincare inequality if there is a positive constant $c$ such that for every (differentiable) $f$, $c\mathrm{Var}(f(X)) \leq E[f'(x)]^2$. Basically, what is important here, that $X$ cannot be supported on 'atoms'. | |
| Jan 7, 2016 at 5:17 | comment | added | Douglas Zare | By $X$, you mean a sequence of distributions since you are talking about the non-identically distributed case, right? What does it mean for $X$ to satisfy a Poincare-type inequality? | |
| Jan 6, 2016 at 23:09 | history | asked | Cain | CC BY-SA 3.0 |