Oswin So

49 posts

Oswin So

Oswin So

@oswinso

Graduate Researcher with Chuchu Fan at MIT @mit_REALM. Bringing Guarantees to Safe Reinforcement Learning 🇭🇰

Cambridge, Massachusetts เข้าร่วม Nisan 2013
112 กำลังติดตาม165 ผู้ติดตาม
Oswin So รีทวีตแล้ว
Guan-Horng Liu
Guan-Horng Liu@guanhorng_liu·
Adjoint Matching works great for fine-tuning diffusion models with reward gradients. How about #AM for #diffusionLLMs with #nondifferentiable #rewards? Does "discrete adjoint" even exist ... and how? 🤔 📢 Introduce #DiscreteAdjointMatching (#DAM)—a unifying AM for discrete generative models, accepted to #ICLR2026 🇧🇷 Work done with my amazing intern @oswinso and @RickyTQChen, Brian, Chuchu 🙌 📰 arxiv.org/abs/2602.07132
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Oswin So
Oswin So@oswinso·
At #NeurIPS from Dec 2 to Dec 7 in San Diego! Looking forward to catching up and meeting new friends. Excited to chat about safety for robotics, constraint satisfaction in RL, and (stochastic) optimal control. Feel free to DM me to grab coffee or have a chat!
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Isabel Liu
Isabel Liu@YijieIsabelLiu·
Robots can plan, but rarely improvise. How do we move beyond pick-and-place to multi-object, improvisational manipulation without giving up completeness guarantees? We introduce Shortcut Learning for Abstract Planning (SLAP), a new method that uses reinforcement learning (RL) to discover shortcuts in the planning graphs induced by task and motion planning (TAMP) skill libraries. It is a plug-and-play module that can be trained on top of existing planners to speed up execution through learned shortcuts. (1/5)
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Oswin So รีทวีตแล้ว
Huihan Liu
Huihan Liu@huihan_liu·
Meet Casper👻, a friendly robot sidekick who shadows your day, decodes your intents on the fly, and lends a hand while you stay in control! Instead of passively receiving commands, what if a robot actively sense what you need in the background, and step in when confident? (1/n)
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Oswin So
Oswin So@oswinso·
@Almost_Sure Yeah, I was thinking about it from the perspective of using Feller’s test of explosion to show that it either hits zero in finite time or explodes to infinity, and that former happens with positive probability but is not 1 so not almost surely.
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Almost Sure
Almost Sure@Almost_Sure·
@oswinso Almost surely. At first, was thinking that is the same thing as +ve probability over an infinite time horizon, but it’s not true. If the process tends to infinity then it only has finite time to hit 0, so could potentially do it with positive probability, but not almost surely
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Almost Sure@Almost_Sure·
Standard Brownian motion started at a positive value will eventually hit 0. Let’s stop this by adding a positive drift: dXₜ = dBₜ + μ(Xₜ)dt We try μ equal to A: 1/√X B: 1/X C: 1/X² These all go to ∞ as X→0. But which grow sufficiently fast to stop X from reaching 0
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Oswin So
Oswin So@oswinso·
@Almost_Sure Here is an attempt at a proof: Let t be such that Wₜ > 0. Then, we must have that Lₜ > L₀=0 ⟹ Wₜ > Xₜ. Suppose there exists some u. Since Wₜ > 0 occurs at arbitrarily small times, there will always exist some t < u where Wₜ > Xₜ, resulting in a contradiction.
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Almost Sure
Almost Sure@Almost_Sure·
@oswinso Yes, X goes negative at arbitrarily small times if it is started from 0. Also, if X starts from +ve value then, when it first hits 0 we must have X=W=0 and then we have W > X at times arbitrarily soon after this. So the answer to amended question is no
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Oswin So
Oswin So@oswinso·
Suppose now that Xₜ is started from ε: Xₜ ≔ ε + ∫₀ᵗ 1{Wₛ >= 0} dWₛ Since Xₜ = ε + max(W_t,0) - ½ L(t) and L(t) strictly increases only when Wₜ=0, does there exist a time u such that Xᵤ ≥ Wᵤ AND Xᵤ < 0? twitter.com/oswinso/status…
Oswin So@oswinso

More observations and questions on the following stochastic integral: Xₜ ≔ ∫₀ᵗ 1{Wₛ >= 0} dWₛ Numerically simulating this does confirm that E[Xₜ]=0 and Xₜ does go negative. What I did not expect, however, is the distribution of Xₜ to look the way it does.

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Oswin So
Oswin So@oswinso·
@Almost_Sure I think I'm getting confused by the fact that Wₜ goes both negative (and positive) at arbitrarily small times, which (should?) mean that Xₜ also goes negative at arbitrarily small times (and hence inf { t | Xₜ < 0 } = 0 a.s. ?)
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Oswin So
Oswin So@oswinso·
@Almost_Sure Ah. I realized I have another bad typo. The condition should be whether there exists a time u such that Xₛ ≥ Wₛ ∀ s∈[0,u] AND Xᵤ < 0? i.e., while X goes negative, does it stay at or above W the entire time?
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Oswin So รีทวีตแล้ว
Almost Sure
Almost Sure@Almost_Sure·
#almostsure blog post: On the integral ∫I(W ≥ 0) dW This looks at the mentioned integral, which displays properties particular to stochastic integration and which may seem counter-intuitive. almostsuremath.com/2023/11/07/int…
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Oswin So
Oswin So@oswinso·
@Almost_Sure Good point. The second case can be reduced to the first case. How would you show that it holds for ε=0?
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Almost Sure
Almost Sure@Almost_Sure·
@oswinso Not sure what ε has to do with this, since if it holds for ε=0 (which it should) then it holds for small ε > 0
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Oswin So
Oswin So@oswinso·
More observations and questions on the following stochastic integral: Xₜ ≔ ∫₀ᵗ 1{Wₛ >= 0} dWₛ Numerically simulating this does confirm that E[Xₜ]=0 and Xₜ does go negative. What I did not expect, however, is the distribution of Xₜ to look the way it does.
Oswin So tweet media
Oswin So@oswinso

Small question about Ito integrals: Consider Xₜ ≔ ∫₀ᵗ 1{Wₛ >= 0} dWₛ where Wₜ is a Brownian Motion and 1 is the indicator. Xₜ is a martingale, so E[Xₜ] = 0. I would think that Xₜ is non-negative, but that doesn't seem to be true?

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Oswin So
Oswin So@oswinso·
Using intuition from the discrete case, "Xᵤ downcrosses 0 when Wᵤ also downcrosses 0", and so u exists. However, I have no idea whether this holds in the continuous limit... Numerical simulations show that u exists, but I feel like this is due to numerical error?
Oswin So tweet media
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Bia
Bia@BeatrizGietner·
@oswinso /unrelated How are you guys typing equations on Twitter? 👀
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Oswin So
Oswin So@oswinso·
@Almost_Sure Here's a bonus plot of W, X and L on a single path!
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Oswin So@oswinso·
@Almost_Sure Seems about right? I think L(t) =ᵈ | W(t) |, which looks close in the histogram plot
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