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

Let’s evaluation 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 through which liquidity may very well 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 might be sufficient to reply your query.
You get totally different solutions as a result of you may have constructed your observables in a different 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 end result of y. That is the case in case you are sending two single path funds.

If you happen to as a substitute contemplate an atomic multi-path fee through 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 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 hooked up to one another, y would possibly succeed after which z may fail.

Ordinary 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 much like case D however the math is mistaken.
You’re accurately computing E(x,O)=2/3 and E(y,O)=4/5, however with
E(z,O) you’re messing up with the conditional chance.

Let’s have a look at 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 true when you ship a two stream atomic fee,
• Case B is true when you ship two single path funds,
• Case C is mistaken,
• Case D is true when you ship three single path funds.

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