(Failure of OOD detection under invariant classifier) Consider an out-of-distribution input which contains the environmental feature: ? out ( x ) = M inv z out + M e z e , where z out ? ? inv . Given the invariant classifier (cf. Lemma 2), the posterior probability for the OOD input is p ( y = 1 ? ? out ) = ? ( 2 p ? z e ? + log ? / ( 1 ? ? ) ) , where ? is the logistic function. Thus for arbitrary confidence 0 < c : = P ( y = 1 ? ? out ) < 1 , there exists ? out ( x ) with z e such that p ? z e = 1 2 ? log c ( 1 ? ? ) ? ( 1 ? c ) .
Proof. Envision an aside-of-shipments input x out which have Yards inv = [ We s ? s 0 step 1 ? s ] , and you may Meters elizabeth = [ 0 s ? age p ? ] , then your function expression was ? elizabeth ( x ) = [ z out p ? z elizabeth ] , in which p is the tool-standard vector defined inside the Lemma dos .
Then we have P ( y = 1 ? ? out ) = P ( y = 1 ? z out , p ? z e ) = ? ( 2 p ? z e ? + log ? / ( 1 ? ? ) ) , where ? is the logistic function. Thus for arbitrary confidence 0 < c : = P ( y = 1 ? ? out ) < 1 , there exists ? out ( x ) with z e such that p ? z e = 1 2 ? log c ( 1 ? ? ) ? ( 1 ? c ) . ?
Remark: Inside a more general situation, z away will be modeled due to the fact a haphazard vector that’s independent of the when you look at the-shipments names y = step one and you will y = ? step 1 and you can environment keeps: z out ? ? y and you will z out ? ? z elizabeth . For this reason in Eq. 5 you will find P ( z out ? y = 1 ) = P ( z aside ? y = ? step 1 ) = P ( z away ) . Up coming P ( y = step 1 ? ? away ) = ? ( dos p ? z elizabeth ? + record ? / ( 1 ? ? ) ) , same as during the Eq. seven . Ergo the head theorem nevertheless retains under so much more general instance.