Timeline for Definition of weak conditional convergence of random variables
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 6, 2023 at 5:23 | comment | added | Nate River | @SydAmerikaner Yes the full proof is rather difficult to do. I’ll see if I have time to try to work it out. | |
| Dec 5, 2023 at 16:11 | comment | added | Syd Amerikaner | More details: I would have to check that $\mu_X(\{\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}\text{ for all $\mathbf x$}\}) = 1$ holds. So what I would do is to check $\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}$ for all $\mathbf x$ and then compute the probability of those $\mathbf x$ for which no convergence could be shown. However, this feels like it's very difficult (exhaustive) to do, too. In the last paragraph you gave a nice heuristic: $Z_n$ converge conditional to normal and LLN applies to $X_i$. I think this argument spelled out rigorously is what I am looking for. | |
| Dec 5, 2023 at 16:04 | comment | added | Syd Amerikaner | "$Z_n\rightarrow Z$ weakly conditional on $X_i$ if for $\mu_X$-almost every $\mathbf x$, we have that $\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}$ weakly in the usual sense". I think this is what confuses me: How can we have $\mu^{\mathbf x}_{Z_n} \rightarrow\mu_Z^{\mathbf x}$ weakly in the usual sense if a LLN at work at the same time. | |
| Dec 5, 2023 at 8:43 | history | edited | Nate River | CC BY-SA 4.0 |
added 85 characters in body
|
| Dec 5, 2023 at 8:02 | history | answered | Nate River | CC BY-SA 4.0 |