CS 4/5789: Introduction to Reinforcement Learning

Lecture 2: MDPs and Imitation Learning

Prof. Sarah Dean

MW 2:45-4pm
255 Olin Hall

Announcements

Questions about waitlist/enrollment?
- https://www.cs.cornell.edu/courseinfo/enrollment
Homework released next week
- Problem Set 1 due 1 week later
- Programming Assignment 1 due 2 weeks later

Agenda

1. Recap: Markov Decision Process

2. Imitation Learning

3. Trajectories and Distributions

Markov Decision Process (MDP)

action $a_t$

state $s_t$

$\sim \pi(s_t)$

reward

$r_t\sim r(s_t, a_t)$

$s_{t+1}\sim P(s_t, a_t)$

Agent observes state of environment
Agent takes action
- depending on state according to policy
Environment state updates according to transition function

state
$s_t$

Markovian Assumption: Conditioned on $s_t,a_t$, the reward $r_t$ and next state $s_{t+1}$ are independent of the past.

When state transition are stochastic, we will write either:

$s'\sim P(s,a)$ as a random variable
$P(s'|s,a)$ as a number between $0$ and $1$

Notation: Stochastic Maps

$0$

$1$

Example:

$\mathcal A=\{$stay,switch$\}$
From state $0$, action always works
From state $1$, staying works with probability $p_1$ and switching with probability $p_2$

Notation: Stochastic Maps

$0$

$1$

From state $0$, action always works
From state $1$, staying works with probability $p_1$ and switching with probability $p_2$

$P(0,$stay$)=\mathbf{1}_{0}$
$P(1,$stay$)=\mathsf{Bernoulli}(p_1)$

$P(0\mid 0,$stay$)=$
$P(1\mid 0,$stay$)=$
$P(0\mid 1,$stay$)=$
$P(1\mid 1,$stay$)=$

$P(0\mid 0,$switch$)=$
$P(1\mid 0,$switch$)=$
$P(0\mid 1,$switch$)=$
$P(1\mid 1,$switch$)=$

$1$
$0$
$1-p_1$
$p_1$

$0$
$1$
$p_2$
$1-p_2$

$P(0,$switch$)=\mathbf{1}_{1}$
$P(1,$switch$)=\mathsf{Bernoulli}(1-p_2)$

Infinite Horizon Discounted MDP

$\mathcal{S}, \mathcal{A}$ state and action space
$r$ reward function: stochastic map from (state, action) to scalar reward
$P$ transition function: stochastic map from current state and action to next state
$\gamma$ discount factor between $0$ and $1$

Goal: achieve high cumulative reward:

$$\sum_{t=0}^\infty \gamma^t r_t$$

maximize $\displaystyle \mathbb E\left[\sum_{i=1}^\infty \gamma^t r(s_t, a_t)\right]$

s.t. $s_{t+1}\sim P(s_t, a_t), ~~a_t\sim \pi(s_t)$

$\pi$

$\mathcal M = \{\mathcal{S}, \mathcal{A}, r, P, \gamma\}$

Infinite Horizon Discounted MDP

If $|\mathcal S|$=S and $|\mathcal A|$=A, then how many deterministic policies are there? PollEv.com/sarahdean011

If $A=1$ there is only one policy: a constant policy
If $S=1$ there are $A$ policies: map the state to each action
For general $S$, there are $A$ ways to map state $1$ times $A$ ways to map state $2$ times $A$ ways to map state $3$ .... and so on to $S$

maximize $\displaystyle \mathbb E\left[\sum_{i=1}^\infty \gamma^t r(s_t, a_t)\right]$

s.t. $s_{t+1}\sim P(s_t, a_t), ~~a_t\sim \pi(s_t)$

$\pi$

We will find policies using optimization & learning

Agenda

1. Recap: Markov Decision Process

2. Imitation Learning

3. Trajectories and Distributions

Helicopter Acrobatics (Stanford)

LittleDog Robot (LAIRLab at CMU)

An Autonomous Land Vehicle In A Neural Network [Pomerleau, NIPS ‘88]

Imitation Learning

Expert Demonstrations

Supervised ML Algorithm

Policy $\pi$

ex - SVM, Gaussian Process, Kernel Ridge Regression, Deep Networks

maps states to actions

Behavioral Cloning

Dataset from expert policy $\pi_\star$: $$ \{(s_i, a_i)\}_{i=1}^N \sim \mathcal D_\star $$

maximize $\displaystyle \mathbb E\left[\sum_{i=1}^\infty \gamma^t r(s_t, a_t)\right]$

s.t. $s_{t+1}\sim P(s_t, a_t), ~~a_t\sim \pi(s_t)$

$\pi$

rather than optimize,

imitate!

minimize $\sum_{i=1}^N \ell(\pi(s_i), a_i)$

$\pi$

Behavioral Cloning

Policy class $\Pi$: usually parametrized by some $w\in\mathbb R^d$, e.g. weights of deep network, SVM, etc
Loss function $\ell(\cdot,\cdot)$: quantify accuracy
Optimization Algorithm: gradient descent, interior point methods, sklearn, torch

minimize $\sum_{i=1}^N \ell(\pi(s_i), a_i)$

$\pi\in\Pi$

Supervised learning with empirical risk minimization (ERM)

Behavioral Cloning

In this class, we assume that supervised learning works!

minimize $\sum_{i=1}^N \ell(\pi(s_i), a_i)$

$\pi\in\Pi$

Supervised learning with empirical risk minimization (ERM)

i.e. we successfully optimize and generalize, so that the population loss is small: $\displaystyle \mathbb E_{s,a\sim\mathcal D_\star}[\ell(\pi(s), a)]\leq \epsilon$

For many loss functions, this means that
$\displaystyle \mathbb E_{s\sim\mathcal D_\star}[\mathbb 1\{\pi(s)\neq \pi_\star(s)\}]\leq \epsilon$

Ex: Learning to Drive

Policy $\pi$

Input: Camera Image

Output: Steering Angle

Ex: Learning to Drive

Supervised Learning

Policy

Dataset of expert trajectory

$(x, y)$

...

$\pi$( ) =

Ex: Learning? to Drive

expert trajectory

learned policy

No training data of "recovery" behavior!

Ex: Learning? to Drive

What about assumption $\displaystyle \mathbb E_{s\sim\mathcal D_\star}[\mathbb 1\{\pi(s)\neq \pi_\star(s)\}]\leq \epsilon$?

An Autonomous Land Vehicle In A Neural Network [Pomerleau, NIPS ‘88]

“If the network is not presented with sufficient variability in its training exemplars to cover the conditions it is likely to encounter...[it] will perform poorly”

Agenda

1. Recap: Markov Decision Process

2. Imitation Learning

3. Trajectories and Distributions

Trajectory

$s_0$

$a_0$

$s_1$

$a_1$

$s_2$

$a_2$

...

When we deploy or roll out a policy, we will observe a sequence of states and actions
- define the initial state distribution $\mu_0\in\Delta(\mathcal S)$
This sequence is called a trajectory: $$ \tau = (s_0, a_0, s_1, a_1, ... ) $$

observe $s_0\sim \mu_0$
play $a_0\sim \pi(s_0)$
observe $s_1\sim P(s_0, a_0)$
play $a_1\sim \pi(s_1)$
...

Trajectory Example

$0$

$1$

stay: $1$

switch: $1$

stay: $p_1$

switch: $1-p_2$

stay: $1-p_1$

switch: $p_2$

Probability of a Trajectory

Probability of trajectory $\tau =(s_0, a_0, s_1, ... s_t, a_t)$ under policy $\pi$ starting from initial distribution $\mu_0$:
$\mathbb{P}_{\mu_0}^\pi (\tau)=$
- $=\mu_0(s_0)\pi(a_0 \mid s_0) {P}(s_1 \mid s_{0}, a_{0})\pi(a_1 \mid s_1) {P}(s_2 \mid s_{1}, a_{1})...$
- $=\mu_0(s_0)\pi(a_0 \mid s_0)\displaystyle\prod_{i=1}^t {P}(s_i \mid s_{i-1}, a_{i-1}) \pi(a_i \mid s_i)$

$s_0$

$a_0$

$s_1$

$a_1$

$s_2$

$a_2$

...

First recall Baye's rule: For events $A$ and $B$,

$$ \mathbb P\{A \cap B\} = \mathbb P\{A\}\mathbb P\{B\mid A\} = \mathbb P\{B\} \mathbb P\{B\mid A\}$$

Why?

Then we have that

$$\mathbb P_{\mu_0}^\pi(s_0, a_0) = \mathbb P\{s_0\} \mathbb P\{a_0\mid s_0\} = \mu_0(s_0)\pi(a_0\mid s_0)$$

then

$$\mathbb P_{\mu_0}^\pi(s_0, a_0, s_1) = \mathbb P_{\mu_0}^\pi(s_0, a_0)\underbrace{\mathbb P\{s_1\mid a_0, s_0\}}_{P(s_1\mid s_0, a_0)}$$

and so on

Probability of a Trajectory

For deterministic policy (rest of lecture), actions are determined by states precisely $a_t = \pi(s_t)$
Probability of state trajectory $\tau =(s_0, s_1, ... s_t)$:
- $\mathbb{P}_{\mu_0}^\pi (\tau)=\mu_0(s_0)\displaystyle\prod_{i=1}^t {P}(s_i \mid s_{i-1}, \pi(s_{i-1})) $

$s_0$

$a_0$

$s_1$

$a_1$

$s_2$

$a_2$

...

Probability of a State

Example: $\pi(s)=$stay and $\mu_0$ is each state with probability $1/2$.

What is $\mathbb{P}\{s_t=1\mid \mu_0,\pi\}$?
$\mathbb{P}\{s_0=1\} = 1/2$
$\mathbb{P}\{s_1=1\} = \mathbb{P}^\pi_{\mu_0} (0, 1) + \mathbb{P}^\pi_{\mu_0} (1, 1)=p_1/2$
$\mathbb{P}\{s_1=2\} = \mathbb{P}^\pi_{\mu_0} (1, 1,1)=p_1^2/2$

Probability of state $s$ at time $t$ $$ \mathbb{P}\{s_t=s\mid \mu_0,\pi\} = \displaystyle\sum_{\substack{s_{0:t-1}\in\mathcal S^t}} \mathbb{P}^\pi_{\mu_0} (s_{0:t-1}, s) $$

$1$

$1-p_1$

$p_1$

$0$

$1$

Why? First recall that

$$ \mathbb P\{A \cup B\} = \mathbb P\{A\}+\mathbb P\{B\} - \mathbb P\{A\cap B\}$$

If $A$ and $B$ are disjoint, the final term is $0$ by definition.

If all $A_i$ are disjoint events, then the probability any one of them happens is

$$ \mathbb P\{\cup_{i} A_i\} = \sum_{i} \mathbb P\{A_i\}$$

State Distribution

Keep track of probability for all states $s\in\mathcal S$ with $d_t$
$d_t$ is a distribution over $\mathcal S$
Can be represented as an $S=|\mathcal S|$ dimensional vector $$ d_t[s] = \mathbb{P}\{s_t=s\mid \mu_0,\pi\} $$

$$d_0 = \begin{bmatrix} 1/2\\ 1/2\end{bmatrix} \quad d_1 = \begin{bmatrix} 1-p_1/2\\ p_1/2\end{bmatrix}\quad d_2 = \begin{bmatrix} 1-p_1^2/2\\ p_1^2/2\end{bmatrix}$$

Example: $\pi(s)=$stay and $\mu_0$ is each state with probability $1/2$.

$1$

$1-p_1$

$p_1$

$0$

$1$

State Distribution Transition

How does state distribution change over time?
- Recall, $s_{t+1}\sim P(s_t,\pi(s_t))$
- i.e. $s_{t+1} = s'$ with probability $P(s'|s_t, \pi(s_t))$
Write as a summation over possible $s_t$:
- $\mathbb{P}\{s_{t+1}=s'\mid \mu_0,\pi\} =\sum_{s\in\mathcal S} P(s'\mid s, \pi(s))\mathbb{P}\{s_{t}=s\mid \mu_0,\pi\}$
In vector notation:
- $d_{t+1}[s'] =\sum_{s\in\mathcal S} P(s'\mid s, \pi(s))d_t[s]$
- $d_{t+1}[s'] =\langle\begin{bmatrix} P(s'\mid 1, \pi(1)) & \dots & P(s'\mid S, \pi(S))\end{bmatrix},d_t\rangle $

Transition Matrix

Given a policy $\pi(\cdot)$ and a transition function $P(\cdot\mid \cdot,\cdot)$

Define the transition matrix $P_\pi \in \mathbb R^{S\times S}$
At row $s$ and column $s'$, entry is $P(s'\mid s,\pi(s))$

$P_\pi=$

$s$

$s'$

$P(s'\mid s,\pi(s))$

State Evolution

Proposition: The state distribution evolves according to $$ d_t = (P_\pi^t)^\top d_0$$

Proof: (by induction)

Base case: when $t=0$, $d_0=d_0$
Induction step:
- Need to prove $d_{k+1} = P_\pi^\top d_k$
- Exercise: review proof below

$s$

$s'$

$P(s'\mid s,\pi(s))$

$P_\pi^\top=$

Proof of claim that $d_{k+1} = P_\pi^\top d_k$

$d_{k+1}[s'] =\sum_{s\in\mathcal S} P(s'\mid s, \pi(s))d_k[s]$
$\textcolor{orange}{d_{k+1}[s']} =\langle$$\begin{bmatrix} P(s'\mid 1, \pi(1)) & \dots & P(s'\mid S, \pi(S))\end{bmatrix}$$,\textcolor{red}{d_k}\rangle $
Each entry of $d_{k+1}$ is the inner product of a column of $P_\pi$ with $d_k$
- in other word, inner product of a row of $P_\pi^\top$ with $d_k$
By the definition of matrix multiplication, $d_{k+1} = P_\pi^\top d_k$

$s$

$s'$

$P(s'\mid s,\pi(s))$

$P_\pi=$

$1$

$1-p_1$

$p_1$

$0$

$1$

$d_0 = \begin{bmatrix} 1/2\\ 1/2\end{bmatrix} $
$d_1 = \begin{bmatrix}1& 1-p_1\\0 & p_1\end{bmatrix} \begin{bmatrix} 1/2\\ 1/2\end{bmatrix} = \begin{bmatrix} 1-p_1/2\\ p_1/2\end{bmatrix}$
$d_2 =\begin{bmatrix}1& 1-p_1\\0 & p_1\end{bmatrix}\begin{bmatrix} 1-p_1/2\\ p_1/2\end{bmatrix} = \begin{bmatrix} 1-p_1^2/2\\ p_1^2/2\end{bmatrix}$

Example: $\pi(s)=$stay and $\mu_0$ is each state with probability $1/2$.

State Evolution Example

$$P_\pi = \begin{bmatrix}1& 0\\ 1-p_1 & p_1\end{bmatrix}$$

Recap

1. Markov Decision Process

2. Imitation Learning

3. Trajectories and Distributions

Sp23 CS 4/5789: Lecture 2

By Sarah Dean

Sp23 CS 4/5789: Lecture 2

Sarah Dean PRO

asst prof in CS at Cornell

sdean.website

CS 4/5789: Introduction to Reinforcement Learning

Lecture 2: MDPs and Imitation Learning

Announcements

Agenda

Markov Decision Process (MDP)

Notation: Stochastic Maps

Notation: Stochastic Maps

Infinite Horizon Discounted MDP

Infinite Horizon Discounted MDP

Agenda

Imitation Learning

Behavioral Cloning

Behavioral Cloning

Behavioral Cloning

Ex: Learning to Drive

Ex: Learning to Drive

Ex: Learning? to Drive

Ex: Learning? to Drive

Agenda

Trajectory

Trajectory Example

Probability of a Trajectory

Probability of a Trajectory

Probability of a State

State Distribution

State Distribution Transition

Transition Matrix

State Evolution

State Evolution Example

Recap

Sp23 CS 4/5789: Lecture 2

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