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

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 might be allotted within the channels) or equivalently one can select to sum over all doable observable outcomes. The p_i(O) is the chance 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 may be sufficient to reply your query.
You get totally different solutions as a result of you’ve gotten constructed your observables in another way.

Your observable is the sum of two flows x that goes via S-A-R with 1 sat and y that goes via 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 consequence is unbiased of the end result of y. That is the case in case you are sending two single path funds.

In the event you as a substitute take into account 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 instances) 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 will not be 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 is able to 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 much like case D however the math is flawed.
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 chance.

Let’s examine all doable 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 instances which correspond to having sufficient liquidity for two sat)
  which is identical 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 circumstances.

• Case A is correct for those who ship a two move atomic cost,
• Case B is correct for those who ship two single path funds,
• Case C is flawed,
• Case D is correct for those who ship three single path funds.

I’m assured that for those who run the experiments you will verify.