_index.org

Belief iLQR

Last edited: August 8, 2025

Motivation

  • Imperfect sensors in robot control: partial observations
  • Manipulators face tradeoff between sensing + acting

curse of dimensionality and curse of history.

Belief-Space Planning

Perhaps we should plan over all possible distributions of state space, making a belief-state MDP.

But: this is a nonlinear, stochastic dynamic. In fact: there maybe stochastic events that affects dynamics.

Big problem:

Belief iLQR

“determinize and replan”: simplify the dynamics at each step, plan, take action, and replan

belief-state MDP

Last edited: August 8, 2025

Our belief can be represented as vectors as the probability of us being in each state. If we have that, we can just use our belief vector as our state vector. Now use MDP any solving you’d like, keeping in mind that the reward is just the expected reward:

\begin{equation} \mathbb{E}[R(b,a)] = \sum_{s} R(s,a) b(s) \end{equation}

we can estimate our transition between belief-states like so:

\begin{align} T(b’|b,a) &= P(b’|b,a) \\ &= \sum_{o}^{} P(b’|b,a,o) P(o|b,a) \\ &= \sum_{o}^{} P(b’ = Update(b,a,o)) \sum_{s’}^{}O(o|a,s’) \sum_{s}^{}T(s’|s,a)b(s) \end{align}

bending

Last edited: August 8, 2025

Bending is what happens when you apply a transverse load to an object and it goes wooosh.

That’s cool. Now how does it work? see Euler-Bernoulli Theory

Bernoulli distribution

Last edited: August 8, 2025

Consider a case where there’s only a single binary outcome:

  • “success”, with probability \(p\)
  • “failure”, with probability \(1-p\)

constituents

\begin{equation} X \sim Bern(p) \end{equation}

requirements

the probability mass function:

\begin{equation} P(X=k) = \begin{cases} p,\ if\ k=1\\ 1-p,\ if\ k=0\\ \end{cases} \end{equation}

This is sadly not Differentiable, which is sad for Maximum Likelihood Parameter Learning. Therefore, we write:

\begin{equation} P(X=k) = p^{k} (1-p)^{1-k} \end{equation}

Which emulates the behavior of your function at \(0\) and \(1\) and we kinda don’t care any other place.

Bessel's Equation

Last edited: August 8, 2025

\begin{equation} x^{2}y’’ + xy’ + (x^{2}-n^{2})y = 0 \end{equation}

this function is very useful, they have no well defined elementary result.