# Dividing miner reward over a k-length jumping window

Nowadays, a miner can gain more than another one even when they mine the same amount of blocks due to variance in tx fees. It's expected that, especially once the coinbase reward is gone, this will result in miners competing for "juicy" txs and hence damaging the stability of the system. See https://dl.acm.org/citation.cfm?id=2978408 for future reference.

I've been thinking if it's possible to overcome this simply by changing the rewarding mechanisms of Bitcoin. So let's say we keep a k-length jumping window, take sum of fees in that window and divide it evenly to miners. So if we have k=4 and 1st block has 20 in fees, 2nd has 20 in fees, 3rd has 10 in fees and 4th has 30 in fees each miner will get (20+20+10+30)/4 = 20 in fees.

What would be shortcomings/downsides of such a mechanism ?

• When using an aggregative method as you have suggested, how would you ensure that the total amount of bitcoin paid in fees, is the same as the total amount of bitcoin collected from fees? The moving averages that are collected wouldn't sum to exactly the sum of the bitcoin paid. Jan 8, 2018 at 17:50
• `The moving averages that are collected wouldn't sum to exactly the sum of the bitcoin paid.` Why is that? I don't see. Fee of each block is simply Input sum - Output sum. Is there anything more to this than that ? Jan 8, 2018 at 18:44
• For each block yes. But not over an aggregation of blocks. Jan 8, 2018 at 18:54
• But if we add each block individually, what do we miss ? Surely, we can go back k blocks in the chain and aggregate each block, no ? Jan 8, 2018 at 19:05
• Hmm, I've misread your question... Yes, what you have described is possible. You may have found a solution to Undercutting attacks. Jan 8, 2018 at 19:21