Distributional RL
Pavel Temirchev
Distributional - not distributed
Reminder: Q-function
Q^\pi(s, a) = \mathbb{E}\big[ \sum_t \gamma^t r(s_t, a_t) | s_0 =s, a_0 = a \big]
By definition:
Motivation
V^*(s) = +1
V^*(s) = -1
Motivation
V^*(s) = 0.5* (1 - 1) = 0
V^*(s) = 0
We expect DQN to learn two features:
- is rocket? (then -1)
- is finish? (then +1)
But it can learn arbitrary staff, since we are learning average value! (which is zero)
Rewards are not equal
R_1(a)
R_1(a)
R_2(a)
-8
2
12
\mathbb{E} R_1(a) = 2
\mathbb{E} R_2(a) = 2
Intuitively, which one is better?
Utility functions
Von Neumann-Morgenstern theorem:
A rational, in some sense, agent maximises mathematical expectation of a utility function.
In our case, the utility function is the return
Z(s, a) = \sum_t \gamma^t R(s_t, a_t)
Q(s, a) = \mathbb{E}Z(s, a)
from where the randomness comes under the expectation?
Maximization of the Q-function is enough to obtain an optimal policy.
And Q-function is defined in the terms of expected rewards.
Q(s, a) = \mathbb{E} R(s, a) + \gamma \mathbb{E}Q(s', a')
Expectations are enough - we don't need to know the distribution of rewards to succeed
Let's try to make learning easier
Deep Q-network should return a distribution over the outputs!
s
a
DQN
Q(s, a)
was
what we want
p(Q = z_1|s, a)
p(Q = z_2|s, a)
p(Q = z_n|s, a)
\dots
This will work in a simple discrete setting:
Q(s, a) \in \{z_1, z_2, \dots, z_n \}
We will use a projection step to extend this idea to a broad class of RL problems
Q(s, a) = \sum_i z_i p(Q = z_i|s, a)
Reminder
Q^\pi(s, a) = \mathbb{E} R(s, a) + \gamma \mathbb{E}_{p, \pi} Q^\pi(s', a')
Q^*(s, a) = \mathbb{E} R(s, a) + \gamma \mathbb{E}_{p} \max_{a'} Q^*(s', a')
[\mathcal{T}^\pi Q](s, a) = \mathbb{E} R(s, a) + \gamma \mathbb{E}_{p, \pi} Q(s', a')
[\mathcal{T} Q](s, a) = \mathbb{E} R(s, a) + \gamma \mathbb{E}_{p} \max_{a'} Q(s', a')
Bellman equation
Bellman optimality equation
In the form of operators:
contractions in
||\cdot||_\infty
Distributional perspective on RL
\mathcal{T}^\pi Z(s, a) := R(s, a) + \gamma P^\pi Z(s, a)
Distributional Bellman operator
S' \sim p(\cdot | s, a)
A' \sim \pi(\cdot | S')
where
\mathcal{T}^\pi Z(s, a) := R(s, a) + \gamma Z(S', A')
Is the distributional operator a contraction?
Wasserstein metric
d_p(U, V) = \Big( \inf_{\mu \in \Gamma(U, V)} \mathbb{E}_{(u, v) \sim \mu} ||u-v||^p_p \Big)^{1/p}
or
d_p(U, V) = \Big(\int^1_0 |F^{-1}_U(\omega)-F^{-1}_V(\omega)|^p d\omega\Big)^{1/p}
Is the distributional operator a contraction?
Wasserstein metric's properties
||A||_p = (\mathbb{E}||A||_p^p)^{1/p}
What about the optimality distributional Bellman operator?
Lemma
Operator is not a contraction.
\mathcal{T}
Lemma
|| \mathbb{E} \mathcal{T} Z_1 - \mathbb{E} \mathcal{T} Z_2||_\infty \le \gamma ||\mathbb{E}Z_1 - \mathbb{E}Z_2||_\infty
\mathcal{T} Z(s, a) := R(s, a) + \gamma P^{\pi\in \mathcal{G}_Z}Z(s, a)
\mathcal{G}_Z := \{ \pi: \mathbb{E}_\pi \mathbb{E}Z(s,a) = max_{a'} \mathbb{E}Z(s,a') \}
|| \mathbb{E} \mathcal{T} Z_1 - \mathbb{E} \mathcal{T} Z_2||_\infty = || \mathcal{T}_E \mathbb{E} Z_1 - \mathcal{T}_E \mathbb{E} Z_2||_\infty \le \gamma ||\mathbb{E}Z_1 - \mathbb{E}Z_2||_\infty
Optimality distributional Bellman operator
is not a contraction!
\mathcal{T} Z(s, a) := R(s, a) + \gamma P^{\pi\in \mathcal{G}_Z}Z(s, a)
\mathcal{G}_Z := \{ \pi: \mathbb{E}_\pi \mathbb{E}Z(s,a) = max_{a'} \mathbb{E}Z(s,a') \}
\pi^*(s_1) = a_1
\pi^*(s_2) = a_2^2
Z^*
Z
\mathcal{T}Z
s_1, a_1
s_2, a_2^1
s_2, a_2^2
\epsilon \pm 1
0
\epsilon \pm 1
s_1
s_2
a_1
R(s_1, a_1) = 0
a_2^1
a_2^2
R(s_2, a_2^1) = 0
R(s_2, a_2^1) = \epsilon \pm 1
-\epsilon \pm 1
0
\epsilon \pm 1
\epsilon \pm 1
0
0
\bar d_1(Z, Z^*) = 2 \epsilon
\bar d_1(\mathcal{T}Z, Z^*) = \frac{1}{2} | 1 - \epsilon| + \frac{1}{2} | 1 + \epsilon|
Ok! Still! Want to use it instead of DQN anyway.
Distribution
What we had in DQN:
(s, a)
\theta
\hat Q (s, a) \in \mathbb{R}
What we need now:
(s, a)
\theta
Z
z_{MIN}
z_{MAX}
z_{0}
z_{1}
p(z_0|s, a)
p(z_1|s, a)
Distributional Bellman update rule
(s, a, r, s')
We have a sample from buffer:
\hat\mathcal{T}z_j = r + \gamma z_j
p(z_0|s', a_0)
p(z_1|s', a_0)
p(z_0|s', a_1)
p(z_1|s', a_1)
p(z_0|s', a_2)
p(z_1|s', a_2)
MEAN
z_0
z_1
a' = \arg\max_a
p(\hat \mathcal{T}z_j|s, a) = p_\theta(z_j|s', a')
Now, minimize the distance from
a guess to the guess
What distance? Wasserstein?
No. It is a bad idea to estimate it from samples:
d_p(\mathbb{E}_i P_i, Q) \le \mathbb{E}_i d_p(P_i, Q)
Kulback-Leibler divergence? Kinda, but it goes to INF if the support is disjoint.
KL \Big ( \hat \mathcal{T}Z_{\tilde \theta}(s, a) \Big|\Big| Z_\theta(s, a) \Big) = KL ( q || p) = \sum_{z: q(z) > 0} q(z) \log \frac{q(z)}{p(z) }
Project the updated distribution
on the domain.
For j := 0 to N-1:
b = \frac{\hat \mathcal{T}z_j - z_{MIN}}{\Delta z}
l = \lfloor b \rfloor ; \;\;\;\;u = \lceil b\rceil
\Phi q(z_l) = \Phi q(z_l) + p_{\tilde\theta}(z_j|s', a')(u - b)
\Phi q(z_u) = \Phi q(z_u) + p_{\tilde\theta}(z_j|s', a')(b - l)
KL \Big ( \Phi \hat \mathcal{T}Z_{\tilde \theta}(s, a) \Big|\Big| Z_\theta(s, a) \Big) = KL ( \Phi q || p_\theta) \propto -\sum_{j=0} ^{N-1} \Phi q(z_j) \log p_\theta(z_j) \rightarrow \min_\theta
\hat \mathcal{T} z_j
z_l
z_u
p_{\tilde\theta}(z_j|s', a')
C51 results
C51 results
C51 results
Cons of C51
-
Hyperparameters \( z_{MIN} , z_{MAX} \) are required in advance
-
We are not minimizing Wasserstein!
-
Strange projection step is required
Let's transpose the parametrization!
p_j = \frac{1}{N}
Let's set a uniform distribution over some returns
\tau_j = \sum_i^j p_i
The minimizer of \( d_1(Z, Z_\theta) \) is such \(\theta : (z_\theta)_j = F^{-1}_Z(\hat \tau_j) \;\; \forall j \)
While the values are from a parametric model.
(s, a)
\theta
z_\theta(s, a)_j
\forall j
Let's transpose the parametrization!
What if and ???
N = 1
\hat \tau = 0.5
and you have only samples from \(Z\)
\theta = \arg\min_\theta \mathbb{E}_{\tilde z \sim Z} \Big [ |\tilde z - z_\theta| \Big]
What if and some ???
N = 1
\hat \tau
\theta = \arg\min_\theta \mathbb{E}_{\tilde z \sim Z} \Big [ \rho_{\hat \tau} (\tilde z - z_\theta) \Big]
What if ???
N > 1
\theta = \arg\min_\theta \mathbb{E}_{\tilde z \sim Z} \sum_j \Big [ \rho_{\hat \tau_j} (\tilde z_j - (z_\theta)_j) \Big]
\rho_\tau(x) = x(\tau - [x<0])
Quantile Regression DQN
Input: s, a, r, s'
Q(s', \hat a) = \frac{1}{N} \sum_j z_\theta(s', \hat a)_j
a' = \arg\max_{\hat a} Q(s', \hat a)
\mathcal{T}z_j = r + \gamma z_\theta(s', a')_j \;\;\; \forall j
\frac{1}{N} \sum_{i=1}^N \sum_{j=1}^N \Big ( \rho_{\hat \tau_i}(\mathcal{T}z_j - z_\theta(s, a)_i \Big)
Output:
QR-DQN results
QR-DQN results
Is the distributional operator a contraction?
Proposition:
Wasserstein metric in a maximal form is a distance between value distributions
\bar{d}_p(Z_1, Z_2) = \sup_{s, a} d_p(Z_1(s, a), Z_2(s, a))
Lemma:
Distributional Bellman operator is a contraction in the maximal Wasserstein distance.
\bar{d}_p(\mathcal{T}^\pi Z_1, \mathcal{T}^\pi Z_2) \leq \gamma \bar{d}_p( Z_1, Z_2)
Is the distributional operator a contraction?
Proof:
\bar{d}_p(\mathcal{T}^\pi Z_1, \mathcal{T}^\pi Z_2) \leq \gamma \bar{d}_p( Z_1, Z_2)
\sup_{s, a} d_p(\mathcal{T}^\pi Z_1(s,a), \mathcal{T}^\pi Z_2(s, a)) \leq \gamma \sup_{s,a } d_p( Z_1(s,a), Z_2(s,a))
d_p(\mathcal{T}^\pi Z_1(s,a), \mathcal{T}^\pi Z_2(s, a)) =
= d_p\Big(R(s,a) + \gamma Z_1(S',A'), R(s,a) + \gamma Z_2(S', A')\Big) \leq
\le d_p\Big( \gamma Z_1(S',A'), \gamma Z_2(S', A')\Big) \leq
\le \gamma d_p\Big( Z_1(S',A'), Z_2(S', A')\Big) \leq
\le \gamma \sup_{s', a'} d_p\Big( Z_1(s',a'), Z_2(s', a')\Big)
Is the distributional operator a contraction?
Proof:
d_p\Big( Z_1(X), Z_2(X)\Big) \leq \sup_{x} d_p\Big( Z_1(x), Z_2(x)\Big)
\sup_x d_p\Big( Z_1(x), Z_2(x)\Big) = \sup_x \Big(\inf_{\mu \in \Gamma_x} \mathbb{E}_\mu ||z_1-z_2||^p_p \Big)^{1/p}
Z_1(x) \sim p(z_1|x)
Z_2(x) \sim p(z_2|x)
\Gamma_x(Z_1(x), Z_2(x)) = \Bigg \{ \mu :
\int \mu(z_1, z_2 | x) dz_2 = p(z_1|x)
\int \mu(z_1, z_2| x) dz_1 = p(z_2|x)
\Bigg \}
Is the distributional operator a contraction?
Proof:
d_p\Big( Z_1(X), Z_2(X)\Big) \leq \sup_{x} d_p\Big( Z_1(x), Z_2(x)\Big)
d_p\Big( Z_1(X), Z_2(X)\Big) = \Big(\inf_{\mu \in \Gamma_X} \mathbb{E}_\mu ||z_1-z_2||^p_p \Big)^{1/p}
Z_1(X) \sim \int p(z_1|x) p(x) dx
Z_2(X) \sim \int p(z_2|x) p(x) dx
\Gamma_X(Z_1(X), Z_2(X)) = \Bigg \{ \mu :
\int \mu(z_1, z_2 | x) p(x) dz_2 dx = p(z_1)
\int \mu(z_1, z_2| x) p(x) dz_1 dx = p(z_2)
\Bigg \}
Is the distributional operator a contraction?
Proof:
d_p\Big( Z_1(X), Z_2(X)\Big) \leq \sup_{x} d_p\Big( Z_1(x), Z_2(x)\Big)
d_p\Big( Z_1(X), Z_2(X)\Big) = \Big(\inf_{\mu \in \Gamma} \int \mu(z_1, z_2|x)p(x)||z_1-z_2||^p_pdz_1dz_2dx \Big)^{1/p}
And RHS of the upper inequality will be rewritten as:
\sup_x d_p\Big( Z_1(x), Z_2(x)\Big) = \sup_x \Big(\inf_{\mu \in \Gamma} \int \mu(z_1, z_2|x)||z_1-z_2||^p_pdz_1dz_2 \Big)^{1/p}
LHS:
or, equally:
d_p\Big( Z_1(X), Z_2(X)\Big) = \Big(\inf_{\mu \in \Gamma} \mathbb{E}_x \int \mu(z_1, z_2|x)||z_1-z_2||^p_pdz_1dz_2 \Big)^{1/p}
Is the distributional operator a contraction?
The last stage of the proof:
\Big( \inf_\mu \mathbb{E}_{p(x)} f(\mu, x) \Big)^{1/p} \le \sup_x \Big(\inf_\mu f(\mu, x) \Big) ^{1/p}
Due to the monotonicity of 1/p
\Big( \inf_\mu \mathbb{E}_{p(x)} f(\mu, x) \Big)^{1/p} \le \Big( \sup_x \inf_\mu f(\mu, x) \Big) ^{1/p}
Apply Jensens inequality and voila
\inf_\mu \mathbb{E}_{p(x)} f(\mu, x) \le \sup_x \inf_\mu f(\mu, x)
\inf_\mu \mathbb{E}_{p(x)} f(\mu, x) \le \mathbb{E}_{p(x)} \inf_\mu f(\mu, x) \le \sup_x \inf_\mu f(\mu, x)
Thank you
Distributional RL
By cydoroga
