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4ns

@fourenes.bsky.social

Stats @ Duke. Stochastic process enthusiast

26 Followers  |  238 Following  |  7 Posts  |  Joined: 09.11.2024  |  1.4136

Latest posts by fourenes.bsky.social on Bluesky

Wenge Kong karakoe - All Alone On Christmas Eve [Ping Pong the Animation]
YouTube video by Pluview Wenge Kong karakoe - All Alone On Christmas Eve [Ping Pong the Animation]

Annual www.youtube.com/watch?v=8i93...

25.12.2024 01:43 β€” πŸ‘ 0    πŸ” 0    πŸ’¬ 0    πŸ“Œ 0

Although of course both examples are perfectly fine. I just think the first is a better illustration of the desired result.

19.11.2024 16:12 β€” πŸ‘ 0    πŸ” 0    πŸ’¬ 0    πŸ“Œ 0

I think it's much more intuitive, because we can more easily confirm that X_n = 1 must always happen again, no matter where we are. In the 1/n example, we have to accept that there will be infinitely many additional 1's, even when the probability reaches arbitrarily small values

19.11.2024 16:12 β€” πŸ‘ 0    πŸ” 0    πŸ’¬ 1    πŸ“Œ 0

So this fulfills convergence in probability β€œperfectly” in my opinion, since X_n = 1 clearly becomes much less likely, but it’s equally clear that it will ALWAYS happen again. Compared to the usual example of P(X_n = 1) = 1/n, which can also easily be verified to converge in p but not a.s.,...

19.11.2024 16:12 β€” πŸ‘ 0    πŸ” 0    πŸ’¬ 1    πŸ“Œ 0

The idea behind convergence in probability is that it becomes increasingly unlikely that X_n > epsilon, while almost sure convergence states that X_n > epsilon is impossible past a certain point.

19.11.2024 16:12 β€” πŸ‘ 0    πŸ” 0    πŸ’¬ 1    πŸ“Œ 0
Post image

Want to share my favorite example ever. Perfectly shows that convergence in p doesn't imply convergence a.s. The idea is that, though clearly we are converging to 0 in probability, we can always find omega and m >= n (for all n) such that X_m(omega) = 1 (*I think LHS of last line needs ->0 in {})

19.11.2024 15:57 β€” πŸ‘ 0    πŸ” 0    πŸ’¬ 1    πŸ“Œ 0

Yo

09.11.2024 01:29 β€” πŸ‘ 0    πŸ” 0    πŸ’¬ 0    πŸ“Œ 0

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