lightning community – How one can compute the anticipated variety of sats to reach in a probabilistic cost movement?

Let’s evaluate the definition of anticipated worth.
The anticipated worth of the random variable X given the state of the system O,
denoted as E(X,O) is computed as:

E(X,O) = sum_i p_i(O) X_i

The sum is over all microstates (all methods wherein liquidity could possibly be allotted within the channels) or equivalently one can select to sum over all attainable observable outcomes. The p_i(O) is the likelihood of verifying i given the state O, and X_i is the worth that X takes if i is verified.
Utilizing this definition, one instantly sees that E(.,O) is a linear operator:

E(X+a*Y,O) = E(X,O) + a*E(Y,O)

That will be sufficient to reply your query.
You get totally different solutions as a result of you could have constructed your observables in another way.

Your observable is the sum of two flows x that goes by way of S-A-R with 1 sat and y that goes by way of S-B-R with 2 sat.

E(x+y,O) = E(x,O) + E(y,O)

Now, x both fails (prob. 1/3) getting us 0 sat or it succeeds giving us 1 sat (prob. 2/3).

E(x,O) = 0*1/3 + 1*2/3 = 2/3

Equally with y

E(y,O) = 0*2/5 + 2*3/5 = 6/5

Including as much as

E(x+y,O) = 2/3 + 6/5 = 28/15

However watch out, that right here we’re assuming that x final result is impartial of the result of y. That is the case if you’re sending two single path funds.

For those who as an alternative contemplate an atomic multi-path cost wherein both each x and y succeed or none will, then the 2 outcomes for x are once more 1 sat and 0 sat, however with chances 2/3*3/5=2/5 (each x and y succeed)
and three/5 (all different circumstances) respectively:

E(x,O)= 1*2/5 + 0*3/5 = 2/5

equally for y

E(y,O)= 2*2/5 + 0*3/5 = 4/5

Including as much as

E(x+y,O) = 2/5 + 4/5 = 6/5 = 18/15

You’re constructing your observable because the sum of three single path flows (non-atomic):
x representing 1 sat over S-A-R, y representing 1 sat over S-B-R
and z representing 1 sat over S-B-R AFTER y. That is totally different from case B as a result of y and z usually are not connected to one another, y may succeed after which z might fail.

Normal computations

E(x,O) = 0*1/3 + 1*2/3 = 2/3

for y

E(y,O) = 0*1/5 + 1*4/5 = 4/5

Then comes z, which can succeed provided that there’s sufficient liquidity for two sats on channel B-R, then

E(z,O) = 0*2/5 + 1*3/5= 3/5

Including up:

E(x+y+z,O) = 2/3+4/5+3/5 = 31/15

Is just like case D however the math is unsuitable.
You’re appropriately computing E(x,O)=2/3 and E(y,O)=4/5, however with
E(z,O) you might be messing up with the conditional likelihood.

Let’s examine all attainable outcomes:

• y fails, then additionally z fails, prob. 1/5, (having precisely 0 sat liquidity)
• y succeeds, however z fails, prob. 1/5, (having precisely 1 sat of liquidity)
• y succeeds, z succeeds, prob. 3/5, (all different circumstances which correspond to having sufficient liquidity for two sat)
  which is similar because the multiplication of y succeeding and the conditional prob. of z succeeding after y does (3/5 = 4/5 * 3/4).

E(z,O) = 0*1/5 + 0*1/5 + 1*3/5 = 3/5

You will need to state that z is tried after y or we get into race situations.

• Case A is true when you ship a two movement atomic cost,
• Case B is true when you ship two single path funds,
• Case C is unsuitable,
• Case D is true when you ship three single path funds.

I’m assured that when you run the experiments you may affirm.