# Beacon Chain Validator Rewards

The core of Ethereum’s Proof of Stake consensus mechanism launched in December 2020, finally delivering the first phase of the transition away from Proof of Work — a vision set out in 2014, a full year before Ethereum’s mainnet launch. In Ethereum’s Proof of Stake, the blockchain is secured by validators who provide a security deposit of 32 ETH and are given the duty of coming to consensus on the state of a new blockchain called the beacon chain. They are incentivised to do this reliably with rewards for carrying out their duties, and penalties for failing to do this properly. There are also severe penalties (slashing) for validators who can be proven to have acted maliciously.

These validator rewards and penalties are a little more complex than the the block rewards familiar from Proof of Work blockchains. While some excellent resources such as beaconcha.in can help validators to understand how much they are earning, it may not be obvious why. This article will try and give validators a feeling for how much they can earn, and how this may vary.

You might have seen an “estimated APR” for running a validator, such as the graphic on the Proof of Stake launchpad and wonder whether those numbers are realistic — where do they come from anyway? In fact, all the APR estimations currently available rest on a series of assumptions (often unstated), and in some cases may be using an outdated version of the beacon chain spec. So to start off, let’s look at what validators would earn if they — and all other validators — participated perfectly. I’ll be using Ben Edgington’s invaluable annotated spec to do this. Note: we’ll be using the mainnet spec used for the December 2020 launch of the beacon chain. Some refinements to validator rewards and penalties are planned for the upcoming Altair beacon chain fork.

# Available Rewards

Ethereum validators receive rewards for participating every epoch (6.4 minutes). The rewards that they receive are a multiple of something called the base_reward. One base_reward is paid to each validator for accurately voting on (or, in the jargon, attesting to) current values of each of 3 aspects of the beacon chain. I’ll refer to these three rewards collectively as the accuracy rewards. They get a fourth reward, the inclusion reward, if their vote (attestation) is included promptly into a beacon chain block. So the maximum reward a validator can receive each epoch is 4 * base_reward. To work out how much validators can earn, then, we need to know how base_reward is determined.

The level of base_reward is determined by the number of active validators on the network, and tuned to incentivise a validator set of the right size. We want to incentivise a good number of validators to join the validator set, without paying out more issuance than necessary. If there aren’t many validators, the protocol needs to offer a high return, to encourage more validators to join. However if there is already a large number of validators, the protocol can afford to pay less, and save on issuance. The function which does this for the beacon chain is an inverse square root — that is, the level of the reward is divided by the square root of currently validating Ether (the reasoning for choosing an inverse square root relationship is explained in Vitalik Buterin’s design rationale document).

From the annotated spec we have:

base_reward = effective_balance * BASE_REWARD_FACTOR // integer_squareroot(total_balance) // BASE_REWARDS_PER_EPOCH

Which calculates the base_reward for each validator, measured in Gwei (=10-9 ETH), where the terms have the following meanings:

Term Meaning
effective_balance An integer number of ETH according to each validator’s balance (measured in Gwei). It has a maximum value of 32 × 109.
BASE_REWARD_FACTOR A constant value of 64 in the eth2 spec, it is the factor used to tune the overall issuance of ETH — some of the rationale for this value was given in the pull request in which it was set.
total_balance The sum of effective_balance for all active validators.
BASE_REWARDS_PER_EPOCH This is a constant value of 4, corresponding to the 4 components of validator reward, each worth one base_reward, which can be earned by each validator, each epoch.

For our ideal reward case, this all simplifies down, since all validators have an effective balance of 32 ETH. So the ideal base reward for n validators is:

base_reward = 512e9 // integer_squareroot(n * 32e9)

(NB again, in the beacon chain spec, this is measured in Gwei. We can just divide by 109 to convert to ETH)

# The Ideal Case

From the information above, we can calculate the maximum rewards available if all validators participated perfectly (i.e. validators receive 4*base_reward every epoch), and we can turn this into an annualised ideal reward, by multiplying by the number epochs in a year. This is simply the number of seconds in a year (31556952) divided by the number of seconds in an epoch (384) to give approximately 82180 epochs per year. The ideal annual per-validator reward is plotted below, as a function of total staked Ether.

input 1 [click to view code]
# define annualised base reward (measured in ETH) for n validators
# assuming all validators have an effective balance of 32 ETH
import math

EPOCHS_PER_YEAR = 82180

def annualised_base_reward(n):
return EPOCHS_PER_YEAR * 512 / math.sqrt(n * 32e9)

input 2 [click to view code]
# plot ideal ETH staking return

import matplotlib.pyplot as plt

n_validators = [n for n in range(524288//32,int(10e6)//32,3200)]
ideal_reward = [4 * annualised_base_reward(n) for n in n_validators]

fig = plt.figure(figsize=(12, 8))

ax1.plot(n_validators, ideal_reward)
ax2.plot([n * 32e-6 for n in n_validators], [100 * r / 32 for r in ideal_reward])

ax1.set_xlabel('Number of validators')
ax1.set_ylabel('Ideal annual per-validator reward (ETH)')

ax2.set_title('Ideal annual validator returns')
ax2.xaxis.tick_top()
ax2.yaxis.tick_right()
ax2.xaxis.set_label_position('top')
ax2.yaxis.set_label_position('right')
ax2.set_xlabel('Total ETH staked (millions)')
ax2.set_ylabel('Annual yield on 32 ETH deposit (%)');


A few values from this graph are tabulated below for reference

input 3 [click to view code]
# tabulate a few values for validator return

import pandas as pd

n_validators = [524288 // 32, 50000, 100000, 150000, 200000, 250000, 300000, 10000000 // 32]
staked = [32 * n for n in n_validators]
ideal_reward = [4 * annualised_base_reward(n) for n in n_validators]
annual_yield = [100 * r / 32 for r in ideal_reward]
data = {
'n_validators': n_validators,
'total_staked (ETH)': staked,
'annual_reward (ETH)': ideal_reward,
'annual_yield (%)': annual_yield
}

df = pd.DataFrame(data)

pd.options.display.float_format = '{:,.2f}'.format
df.set_index('n_validators')

n_validators total_staked (ETH) annual_reward (ETH) annual_yield (%)
16384 524288 7.35 22.97
50000 1600000 4.21 13.15
100000 3200000 2.98 9.30
150000 4800000 2.43 7.59
200000 6400000 2.10 6.57
250000 8000000 1.88 5.88
300000 9600000 1.72 5.37
312500 10000000 1.68 5.26

# Block Rewards

The figure and table above give us a highly idealised version of what validators would earn if all validators participated perfectly, and all rewards were distributed evenly. However, there is an important component missing from this picture, which is the beacon chain block reward.

The block reward doesn’t change the total amount of Ether paid to validators, but it does mean that a portion of of the available reward is allocated to the block producer. For every slot (a slot is 12 seconds — there are 32 slots in an epoch), one validator, chosen at random, is responsible for producing a block. The block is made up of beacon chain attestations submitted by the other validators, and the block producer is rewarded with a proportion of all the inclusion rewards from attestations in the block. This means that the block producer has an incentive to include all the valid attestations they can gather for the block, maximising their profits and those of all the attesting validators they include.

In the beacon chain spec, the proportion of the inclusion rewards given to the block producer is determined by a constant called PROPOSER_REWARD_QUOTIENT, which has a value of 8. That is, ⅛ of the inclusion reward (or equivalently ¹⁄₃₂ of the full per-epoch reward) goes to the block producer, and ⅞ goes to the validators whose attestations are included in the block.

Since every validator has an equal chance of being selected to produce a block, in the long run the rewards should even out, matching the graph above. But over any finite timescale, there will be variability in rewards, since some validators will be lucky and be given the opportunity to propose a greater than average number of blocks, and some unlucky, proposing fewer.

To work out how significant is the element of luck in block proposal frequency we can apply some basic statistics. Every validator has an equal chance of being selected to propose each slot, and there are 31556952 / 12 = 2629746 slots per year. If there are 100,000 validators then the chance of being selected to propose each slot is 10-5. The number of block proposal opportunities each validator gets will be governed by the binomial distribution.

The probability mass function plotted below gives us a visual sense of how many block proposal opportunities validators can expect to get:

input 4 [click to view code]
# plot pdf

from scipy.stats import binom

x = [el for el in range(51)]
y = binom.pmf(x, 31556952/12, 1e-5)

fig, ax = plt.subplots(figsize=(12, 8))
ax.bar(x, y)
ax.set_xlim(xmin=0)
ax.set_ylim(ymin=0)
ax.set_title('Probability mass function (100,000 validators) — number of block proposal opportunities per year')
ax.set_xlabel('Number of block proposal opportunities in a year')
ax.set_ylabel('Probability')

lmu = binom.ppf([0.01, 0.5, 0.99],31556952/12, 1e-5)
avg = 31556952 / (12 * 100000)
print(f"With 100,000 validators, the mean number of blocks proposed per validator per year is {avg:.2f}\n")
print(f"The unluckiest 1% of validators will have the opportunity to produce at most {int(lmu[0])} blocks in a year")
print(f"The median (average) validator will have the opportunity to produce {int(lmu[1])} blocks in a year")
print(f"The luckiest 1% of validators will have the opportunity to produce at least {int(lmu[2])} blocks in a year")

output:
With 100,000 validators, the mean number of blocks proposed per validator per year is 26.30

The unluckiest 1% of validators will have the opportunity to produce at most 15 blocks in a year
The median (average) validator will have the opportunity to produce 26 blocks in a year
The luckiest 1% of validators will have the opportunity to produce at least 39 blocks in a year


So we can see that there is significant variability in the number of block proposal opportunities each validator will receive, based on luck alone. The luckiest 1% of validators will propose well over twice as many blocks, in the course of a year, as the unluckiest 1%. Over a long enough time, this effect evens out. However, as the number of validators increases and the probability of proposing a block goes down, the unevenness increases.

To show the impact of this effect on validator rewards, I’ve recalculated the ideal validator returns graph, but now I’m going to plot lines to show the luckiest and unluckiest 1% of validators. Another way of looking at this is that 98% of validators will receive an annual reward that sits between the two lines.

input 5 [click to view code]
# plot ideal ETH staking return with interpercentile range

n_validators = [n for n in range(50000,int(10e6)//32,1000)]
full_reward = [4 * annualised_base_reward(n) for n in n_validators]
attestation_reward = [0.75 * f for f in full_reward]
inclusion_reward = [0.25 * f for f in full_reward]
p = [1/n for n in n_validators]

# calculate lower and upper quartiles for block proposal opportunities
l_bpo = [int(binom.ppf(0.01,31556952/12, 1/n)) for n in n_validators]
mean_bpo = [float(binom.mean(31556952/12, 1/n)) for n in n_validators]
u_bpo = [int(binom.ppf(0.99,31556952/12, 1/n)) for n in n_validators]

# calculate lower and upper quartiles for ideal reward, based on block proposal opportunties
l_reward, u_reward = [], []
for i in range(len(full_reward)):
r_att = attestation_reward[i]
r_inc = inclusion_reward[i]
l_reward.append(r_att + r_inc * ((7/8) + (1/8) * l_bpo[i] / mean_bpo[i]))
u_reward.append(r_att + r_inc * ((7/8) + (1/8) * u_bpo[i] / mean_bpo[i]))

fig, ax = plt.subplots(figsize=(12, 8))

ax.plot(n_validators, u_reward, label='Luckiest 1% of validators')
ax.plot(n_validators, l_reward, label='Unluckiest 1% of validators')

ax.set_xlabel('Number of validators')
ax.set_ylabel('Ideal annual per-validator reward (ETH)')
ax.set_title('Ideal annual validator rewards')
leg = ax.legend()

ratio0 = 100 * (u_reward[0] / full_reward[0] - 1)
ratio1 = 100 * (1 - l_reward[0] / full_reward[0])
print(f"With 50,000 validators:\nthe luckiest 1% of validators receive {ratio0:.1f}% greater reward than average"
f"\nthe unluckiest 1% of validators receive {ratio1:.1f}% smaller reward than average")

ratio0 = 100 * (u_reward[50] / full_reward[50] - 1)
ratio1 = 100 * (1 - l_reward[50] / full_reward[50])
print(f"\nWith 100,000 validators:\nthe luckiest 1% of validators receive {ratio0:.1f}% greater reward than average"
f"\nthe unluckiest 1% of validators receive {ratio1:.1f}% smaller reward than average")

ratio0 = 100 * (u_reward[150] / full_reward[150] - 1)
ratio1 = 100 * (1 - l_reward[150] / full_reward[150])
print(f"\nWith 200,000 validators:\nthe luckiest 1% of validators receive {ratio0:.1f}% greater reward than average"
f"\nthe unluckiest 1% of validators receive {ratio1:.1f}% smaller reward than average")

output:
With 50,000 validators:
the luckiest 1% of validators receive 1.0% greater reward than average
the unluckiest 1% of validators receive 1.0% smaller reward than average

With 100,000 validators:
the luckiest 1% of validators receive 1.5% greater reward than average
the unluckiest 1% of validators receive 1.3% smaller reward than average

With 200,000 validators:
the luckiest 1% of validators receive 2.1% greater reward than average
the unluckiest 1% of validators receive 1.7% smaller reward than average


As can be seen from the graph and stats calculated above, there is a potential variation in reward in the range of a few percent over the course of a year. Remember, this applies even though in our ideal case, every validator performs all their duties perfectly. The effect gradually increases as the validator set grows, due to the reduced per-slot probability of block proposal for each validator.

While this level of variation is perhaps not worrying from the point of view of the investment risk for validators, it is worth bearing in mind as we delve into the actual performance of validators in the network. Marginal differences in performance can easily be swallowed up — even over the course of a whole year, as modelled here — by the random variation in block proposal opportunities given to validators.

# Modelling Imperfect Participation

Our model so far has assumed that all validators carry out their duties perfectly. This allows us to set an upper bound on the rewards available, subject to the variation in block producer opportunities explained above. To get a bit closer to modelling real-world rewards however, we need to consider the impact of a less-than-perfect validator set. Even if you are running a perfect validator with zero downtime, your rewards will be impacted if the rest of the network falls short of the ideal case. The beacon chain incentives are designed such that rewards for everyone are maximised if overall network performance is optimised. This helpfully disincentivises adversarial behaviour (such as trying to take other validators offline to maximise your own rewards), but it does mean that individual validators’ rewards can be reduced through no fault of their own.

There are lots of reason why validators might fail to produce attestations, or fail to propagate them to the network, or fail to produce blocks, or fail to have their blocks accepted by the network. Modelling all of these factors would be very difficult, so we’ll try something simpler. We’ll assume that either validators are online and operating perfectly, or they are offline and not fulfilling out their duties at all. In our model, the level of network performance can be captured by a single number called the participation rate, which corresponds to the proportion of validators who are online.

## Perfect Validator on an Imperfect Network

Considering a perfect validator operating within an imperfect validator set, there are two mechanisms that will tend to reduce the rewards they receive:

1. For the accuracy rewards, the reward is scaled by the proportion of active validators who voted the same way. So if 99% of the validators are online and vote correctly, the rewards are scaled by 0.99.
2. If our perfect validator’s attestation is included late — for example because the block producer in the first slot after our attestation fails to produce a block because it is offline — then the inclusion reward declines in inverse proportion to the inclusion delay (i.e. it scales by ½, then ⅓, then ¼ etc.).

So it’s easy to work out that if 99% of the network participates correctly in one epoch, the impact on the accuracy rewards for our perfect validator is a 1% reduction from the ideal case. The inclusion reward is a little more tricky however. Again, it comes down to luck — if the block producer who should have included our attestation at the next slot just happens to be part of the 1% of validators who are offline, then our inclusion reward will be cut in half. If we are exceptionally unlucky and this happens twice in a row, then our inclusion delay will be reduced by a factor of three.

Calculating the expected (i.e. average) inclusion reward, taking account of every possible delay is given by the sum of a geometric series (this was previously explained in the context of beacon chain modelling by Herman Junge). If B is the base_reward and P is the participation rate, we calculate the expected reward as:

\begin{align} E(P) &= \frac{7}{8}B\left(P + \frac{P(1-P)}{2} + \frac{P(1-P)^2}{3} + \ldots\right) \\\ &= \frac{7}{8}BP\sum_{i=1}^{\infty}\frac{(1-P)^{i-1}}{i} \\\ &= \frac{7}{8}BP\frac{\ln(P)}{P-1} \end{align}

Note that the latest that an attestation can be included into the chain is a delay of 32 slots. Properly therefore, we should only sum up to 32 rather than infinity as in the above form. However if the participation rate $P$ is high (e.g. in the 99% range as in the current network), the error that results from summing to infinity rather than 32 is vanishingly small.

So now let’s plot the expected rewards for a perfect validator operating within a range of different participation rate scenarios, bearing in mind that in the 4 months since the genesis of the beacon chain, the participation rate has almost never dipped below 96%.

input 6 [click to view code]
# plot reward for perfect validator in several participation level contexts

participation_rate = [1,0.99,0.98,0.97,0.96]

n_validators = [n for n in range(50000,int(10e6)//32,1000)]
base_reward = [annualised_base_reward(n) for n in n_validators]

fig, ax = plt.subplots(figsize=(12, 8))

r_100000 = []
for P in participation_rate:
accuracy_rewards = [P * 3 * r for r in base_reward]
if P < 1:
inclusion_reward = [(7/8) * r * P * math.log(P) / (P-1) for r in base_reward]
else:
inclusion_reward = [(7/8) * r for r in base_reward]

block_reward = [(1/8) * r * P for r in base_reward]
total_reward = [accuracy_rewards[i] + inclusion_reward[i] + block_reward[i]
for i in range(len(block_reward))]
ax.plot(n_validators, total_reward, label=f'P = {P:.2f}')
r_100000.append(total_reward[50])

ax.set_xlabel('Number of validators')
ax.set_ylabel('Ideal annual per-validator reward (ETH)')
ax.set_title('Ideal annual validator rewards')
leg = ax.legend()

print(f'at P = {participation_rate[1]:.2f}, rewards fall by {100 * (1 - r_100000[1] / r_100000[0]):.2f}%')
print(f'at P = {participation_rate[2]:.2f}, rewards fall by {100 * (1 - r_100000[2] / r_100000[0]):.2f}%')
print(f'at P = {participation_rate[3]:.2f}, rewards fall by {100 * (1 - r_100000[3] / r_100000[0]):.2f}%')
print(f'at P = {participation_rate[4]:.2f}, rewards fall by {100 * (1 - r_100000[4] / r_100000[0]):.2f}%')

output:
at P = 0.99, rewards fall by 0.89%
at P = 0.98, rewards fall by 1.78%
at P = 0.97, rewards fall by 2.68%
at P = 0.96, rewards fall by 3.57%


As shown in the graph and figures above, rewards for our perfect validator fall slightly less than the drop in participation in the network in general.

## Imperfect Validator on a Perfect Network

If our own validator is in fact offline some of the time then it will miss out on some rewards, and will also receive penalties for missed attestations. The accuracy rewards each have counterpart penalties — providing incorrect attestations, or failing to provide an attestation at all, is penalised by a single base_reward for each part of the attestation that was incorrect. This means that missed attestations result in a penalty of 3 * base_reward. To see what impact this has, imagine that you are running a validator which has some level of downtime, but that every other validator on the network behaves perfectly. We’ll ignore the fact that the network participation rate is actually just below 100% due to our lone imperfect validator and instead model the network as being ‘perfect’ whilst our validator occasionally goes offline. What would the impact on our profitability be, taking into account the penalties we will accrue for going offline?

Operating as part of an otherwise perfect validator set, our validator earns an expected 4 * base_reward when it is online, and suffers a penalty of 3 * base_reward when it is offline. So for base_reward $B$ and uptime fraction $U$, the net reward $R$ is given by: \begin{align} R &= 4BU - 3B(1-U) \\\ &= B(7U - 3) \end{align} And if the net reward is zero, we have: $$U = \frac{3}{7} \approx 43%$$

Which implies as long as the validator is online at least 43% of the time, it will receive a positive net reward.

input 7 [click to view code]
# plot expected reward for imperfect validator/perfect network at various validator set sizes

n_validators = [50000, 100000, 150000, 200000, 250000, 300000]
uptime = [i / 100 for i in range(101)]

fig, ax = plt.subplots(figsize=(12, 8))
for n in n_validators:
base_reward = annualised_base_reward(n)
net_reward = []
for u in uptime:
rewards = 4 * u * base_reward
penalties = 3 * (1 - u) * base_reward
net_reward.append(rewards - penalties)

ax.plot(range(101), net_reward, label=f'n_validators = {n}')

ax.set_xlabel('Percentage uptime')
ax.set_ylabel('Annual net reward (ETH)')
ax.set_title('Expected annual net rewards against validator downtime\n'
'(for an imperfect validator in a perfect validator set)')
leg = ax.legend()



# Full Model

Putting this all together we have our model for an imperfect validator operating in an imperfect validator set. With a base_reward of $B$, a participation rate of $P$ and uptime for ‘our’ validator of $U$, we have: $$R = \underbrace{\vphantom{\frac{1}{1}}3BPU }_{\text{accuracy}} - \underbrace{\vphantom{\frac{1}{1}}3B(1-U)}_{\text{penalties}} + \underbrace{\frac{7}{8}BPU\frac{\ln(P)}{P-1}}_{\text{inclusion}} + \underbrace{\vphantom{\frac{1}{1}}\frac{1}{8}BPU}_{\text{block reward}}$$

So, for example, if you run a validator with 99% uptime on a network of 100,000 validators with a participation rate of 99%, this model can be used to predict the expected net reward:

input 8 [click to view code]
# calculate annualised expected net reward for given parameters
base_reward = annualised_base_reward(100000)
participation = 0.99
uptime = 0.99
net_reward = 3 * base_reward * participation * uptime \
- 3 * base_reward * (1 - participation)  \
+ (7/8) * base_reward * participation * uptime * math.log(participation) / (participation - 1) \
+ (1/8) * base_reward * participation * uptime

print(f'Net annual reward = {net_reward:.2f} ETH ({100 * net_reward / 32:.2f}% return on 32 ETH stake)')

output:
Net annual reward = 2.90 ETH (9.05% return on 32 ETH stake)


For ease of experimentation, please see this spreadsheet which will allow you to see the effect of different participation rates, uptime levels and validator numbers on the expected net reward.

Again, however, we need to remember that this expected net reward does not take into account the element of luck. As well as the random chance of being assigned as the block producer for a given slot, as our model has grown in complexity, there are additional items in the formula above which are subject to chance. For example, there is the risk of receiving a reduced inclusion reward because the block producer for the next slot happens to be offline, or the risk of missing the opportunity to produce a block, because your validator happens to be offline when allocated the block proposer duty. These additional factors will slightly increase the variability of net rewards around the “expected” case given by the model.

Given the non-linearity in the inclusion reward, it is tricky to combine all the items in the formula above to produce a probability distribution as we did for block proposer opportunities. We could get a good idea what the distribution looks like by running a Monte Carlo simulation (i.e. use a random number generator to simulate lots of validators and then plot their net rewards on a graph), but let’s leave that step until we are comparing our model to the real network…

# Final Notes

Congratulations for making it to the end! Hopefully this article has helped explain why and how beacon chain rewards vary — both when validators behave in an idealised fashion and when they are less than perfectly reliable.

One key takeaway from this modelling seems to be that although the available rewards vary significantly according to the number of active validators on the network, they are not dramatically affected by small amounts of down time. In fact we showed that in an idealised network, validators are profitable as long as they are online more than 43% of the time. We would not expect this minimum uptime figure to be significantly higher in the beacon chain that we observe today (which generally exhibits participation rates around the 99% level). If you’re considering staking your own Ether at home, this should reassure you — even major internet or power outages should not significantly affect your profitability over the course of a year.

# Acknowledgements

This article was written as part of an Ethereum Foundation Staking Community Grant. Many thanks to Lakshman Sankar, Barnabé Monnot and Jim McDonald for suggestions and feedback. Photo by Cookie the Pom on Unsplash.

Updated: