# TsviBT comments on Concise Open Problem in Logical Uncertainty

• Could you spell out the step

every iteration where mean(ᵧ4[ᵩ9ᵩBᵨEᵩF:ᵩDᵩ1ᵩ2ᵩC])≥2/​5 will cause bound—bad1() to grow exponentially (by a factor of 11/​10=1+(1/​2)(−1+2/​5ᵩ9ᵿ7))

a little more? I don’t follow. (I think I follow the overall structure of the proof, and if I believed this step I would believe the proof.)

We have that eps is about (2/​3)(1-exp([bad1() - bound]/​(next-this))), or at least half that, but I don’t see how to get a lower bound on the decrease of bad1() (as a fraction of bound-bad1() ).

• You are correct that you use the fact that 1+eps is at approximately e^(eps).

The concrete way this is used in this proof is replacing the ln(1+3eps) you subtract from bad1 when the environment is a 1 with 3eps=(bound—bad1) /​ (next—this), and replacing the ln(1-3eps/​2) you subtract from bad1 when the environment is a 0 with −3eps/​2=-(bound—bad1) /​ (next—this)/​2

I believe the inequality is the wrong direction to just use e^(eps) as a bound for 1+eps, but when next-this gets big, the approximation gets close enough.

• In case anyone shared my confusion:

The while loop where we ensure that eps is small enough so that

bound > bad1() + (next—this) * log((1 - p1) /​ (1 - p1 - eps))

is technically necessary to ensure that bad1() doesn’t surpass bound, but it is immaterial in the limit. Solving

bound = bad1() + (next—this) * log((1 - p1) /​ (1 - p1 - eps))

gives

eps >= (1/​3) (1 - e^{ -[bound—bad1()] /​ [next—this]] })

which, using the log(1+x) = x approximation, is about