lightning community – Methods to compute the anticipated variety of sats to reach in a probabilistic fee circulate?

Let’s assessment 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 by which 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 completely different solutions as a result of you will have constructed your observables otherwise.

Your observable is the sum of two flows x that goes by means of S-A-R with 1 sat and y that goes by means 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 unbiased of the end result of y. That is the case if you’re sending two single path funds.

If you happen to as a substitute think about an atomic multi-path fee by which 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 possibilities 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 might be 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 completely different from case B as a result of y and z should 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 may be 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 incorrect.
You might be accurately 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 instances 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 circumstances.

• Case A is true in case you ship a two circulate atomic fee,
• Case B is true in case you ship two single path funds,
• Case C is incorrect,
• Case D is true in case you ship three single path funds.

I’m assured that in case you run the experiments you may verify.