The Bellman update using sampling weighted by Q values (equation 3) is not guaranteed to be a contraction in expectation. Here's the detailed reasoning:

Key Background: Contraction Mappings in RL

The standard Q-learning update (equation 1) leverages the max operator, which is non-expansive. This property ensures that the Bellman operator ( T ) (defined as ( T(Q)(s,a) = r(s,a) + \gamma ...

Short answer: No. Replacing the max by “sampling with probability proportional to Q” does not give a non-expansive map, hence the corresponding Bellman operator is not a contraction in general (except in very special parameter ranges such as very small discount).

Why your inequality does not apply

• Your inequality |Eρ[f] − Eρ[g]| ≤ Eρ[|f − g|] holds only when the sampling distribution ρ is the s...

No – replacing the max–operator by a “draw-an-action-with-probability proportional to its current Q value’’ destroys the non-expansiveness that makes the usual Bellman optimality operator a γ–contraction.
The idea looks harmless – we only average the Q values instead of taking their maximum – but the weighting term itself depends on Q and can amplify differences between two Q–tables.

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First, let us restate precisely what you are comparing:

1. The optimality Bellman operator
   (T⋆Q)(s,a) = R(s,a) + γ ∑{s′}P(s′|s,a) max{a′}Q(s′,a′)
   is a γ‐contraction in the sup‐norm because
   ∥T⋆Q₁ − T⋆Q₂∥∞
   = γ ∥max_a′Q₁(·,a′) − max_a′Q₂(·,a′)∥∞
   ≤ γ ∥Q₁ − Q₂∥∞.

2. The policy‐evaluation Bellman operator for a fixed policy π
   (T^πQ)(s...

Notation
For one state–transition (s\rightarrow s^{\prime}) write the vector of the action-values in the next state as

q = \bigl(Q(s',1),\ldots ,Q(s',n)\bigr)\in\mathbb R^{n}_{+}; .

With ordinary Q–learning we use

max(q)=\max_{i}q_i,

while in the rule you propose we first turn the vector itself into a probability distribution

π_q(i)=\dfrac{q_i}{\sum_{j}q_j}\qquad(i=1,\dots ,n)

and...

<think> We are given the standard Q-learning Bellman update (equation 1) and know that the max operator is non-expansive (equation 2). The question is about a variant (equation 3) where instead of taking the max over the next state actions, we sample an action from the next state proportional to the Q-values. Specifically, we define:

Excellent question. This gets to the heart of the convergence proofs for value-based reinforcement learning.

The short answer is no, the Bellman operator using a weighted sample is not a contraction in the max-norm (), and your proof attempt has a subtle but critical flaw.

Let's break down why.

1. Formalizing the Operator

First, let's formalize the operator you're proposing. ...

Understanding the Bellman Equation with Q-value Weighted Sampling

Your question investigates whether a modified Bellman update that uses sampling weighted by Q-values (instead of the max operation) is a contraction mapping, which would guarantee convergence.

The Issue with Your Proof Approach

Your inequality approach correctly shows:

This is an interesting question about whether a modified Bellman operator that uses sampling instead of max remains a contraction. Let me analyze this carefully.

The Issue with Your Current Approach

Your proof attempt has a fundamental problem. You're trying to show:

However, this doesn't ...