Topic: Calculating variance (Read 1114 times)
No, I just chose to believe the majority of people to choose from.
Thanks Meni. Of course you're right, but Multi-PPS would require both major changes in mining software and pool cooperation, would it not? Therefore it would have to find some way around the same winner-take-all problem that affects pools today.
Yes, but I doubt there are any sustainable easier solutions. 
1. The exponential distribution describes the time until a block is found. What you should be looking at is the number of blocks per time period, which follows the Poisson distribution, for which the variance is equal to the mean.
2. Variance / mean is not a very meaningful quantity. You should be looking at Variance / Mean^2.
3.
Let
H = pool's percentage of the network's hashrate
h = the miner's percent of the network
N = the average number of blocks per day (about 144)
B = Bitcoins per block (about 25)
T = number of days in the period we are looking at
Then the miner's average income is hNTB, the variance is h^2NTB/H, which means that the relative variance is 1/(HNT).
4. The network hashrate can be calculated as about 7 million times the difficulty.
5. This is all covered in AoBPMRS.
6. The real answer to the variance counterargument is not "there isn't that much variance in small pools", it's Multi-PPS.
2. Variance / mean is not a very meaningful quantity. You should be looking at Variance / Mean^2.
3.
Let
H = pool's percentage of the network's hashrate
h = the miner's percent of the network
N = the average number of blocks per day (about 144)
B = Bitcoins per block (about 25)
T = number of days in the period we are looking at
Then the miner's average income is hNTB, the variance is h^2NTB/H, which means that the relative variance is 1/(HNT).
4. The network hashrate can be calculated as about 7 million times the difficulty.
5. This is all covered in AoBPMRS.
6. The real answer to the variance counterargument is not "there isn't that much variance in small pools", it's Multi-PPS.
Thanks Meni. Of course you're right, but Multi-PPS would require both major changes in mining software and pool cooperation, would it not? Therefore it would have to find some way around the same winner-take-all problem that affects pools today.
1. The exponential distribution describes the time until a block is found. What you should be looking at is the number of blocks per time period, which follows the Poisson distribution, for which the variance is equal to the mean.
2. Variance / mean is not a very meaningful quantity. You should be looking at Variance / Mean^2.
3.
Let
H = pool's percentage of the network's hashrate
h = the miner's percent of the network
N = the average number of blocks per day (about 144)
B = Bitcoins per block (about 25)
T = number of days in the period we are looking at
Then the miner's average income is hNTB, the variance is h^2NTB/H, which means that the relative variance is 1/(HNT).
4. The network hashrate can be calculated as about 7 million times the difficulty.
5. This is all covered in AoBPMRS.
6. The real answer to the variance counterargument is not "there isn't that much variance in small pools", it's Multi-PPS.
2. Variance / mean is not a very meaningful quantity. You should be looking at Variance / Mean^2.
3.
Let
H = pool's percentage of the network's hashrate
h = the miner's percent of the network
N = the average number of blocks per day (about 144)
B = Bitcoins per block (about 25)
T = number of days in the period we are looking at
Then the miner's average income is hNTB, the variance is h^2NTB/H, which means that the relative variance is 1/(HNT).
4. The network hashrate can be calculated as about 7 million times the difficulty.
5. This is all covered in AoBPMRS.
6. The real answer to the variance counterargument is not "there isn't that much variance in small pools", it's Multi-PPS.
So after reading http://en.wikipedia.org/wiki/Exponential_distribution#Mean.2C_variance.2C_moments_and_median, the variance of any exponential function can be calculated as:
E(x) = 1/lambda.
Var(x) = 1/lambda^2
So, does that mean lambda for any particular expected block time can be expressed as 1/E(x)? And then Var(x) would be 1/(1/E(x))^2.
If this is the case, variance at even a fairly low pool hashrate is a much lower factor than most people seem to think.
For example, a 1PH pool at the current difficulty, has a theoretical variance over a 10-day difficulty adjustment period of only 4.87%, far lower than many other factors that could influence profitability, and for a 5PH pool that figure is lowered to 0.97%. At 5% variance over two Bitcoin difficulty periods, assuming one 'unlucky' period where profits were 5% lower than expected, and one 'lucky' period, where profits were 5% higher than expected, the profit lost would only be 0.5%. At this level, many other factors become more important than a pool's payout variance.
Looking for someone to tell me my math is wrong here.
E(x) = 1/lambda.
Var(x) = 1/lambda^2
So, does that mean lambda for any particular expected block time can be expressed as 1/E(x)? And then Var(x) would be 1/(1/E(x))^2.
If this is the case, variance at even a fairly low pool hashrate is a much lower factor than most people seem to think.
For example, a 1PH pool at the current difficulty, has a theoretical variance over a 10-day difficulty adjustment period of only 4.87%, far lower than many other factors that could influence profitability, and for a 5PH pool that figure is lowered to 0.97%. At 5% variance over two Bitcoin difficulty periods, assuming one 'unlucky' period where profits were 5% lower than expected, and one 'lucky' period, where profits were 5% higher than expected, the profit lost would only be 0.5%. At this level, many other factors become more important than a pool's payout variance.
Looking for someone to tell me my math is wrong here.
So a lot of people have been throwing around the term 'variance' as a defense of why they want to mine at only the largest pool.
Does anyone have a method of calculating the variance, both over a day and over a bitcoin difficulty period (10-12 days) based on difficulty and hashrate? I started looking around on some sites but the math is a bit over my head.
I think if we can show that variance is not really an issue when using a smaller pool (down to a certain threshold, anyway), people will be more likely to move away from giant pools like ghash.
Does anyone have a method of calculating the variance, both over a day and over a bitcoin difficulty period (10-12 days) based on difficulty and hashrate? I started looking around on some sites but the math is a bit over my head.
I think if we can show that variance is not really an issue when using a smaller pool (down to a certain threshold, anyway), people will be more likely to move away from giant pools like ghash.
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