(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 ) .
Research. Envision an aside-of-shipping input x-out with Meters inv = [ We s ? s 0 1 ? s ] , and you can M elizabeth = [ 0 s ? age p ? ] , then function logo are ? age ( x ) = [ z away p ? z e ] , in which p is the product-standard vector defined during 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: In a far more standard instance, z away is going to be modeled since the a haphazard vector that is in addition to the in the-delivery names y = step 1 and you will y = ? 1 and you may environment has actually: z away ? ? y and you can z away ? ? z age . Hence in Eq. 5 i have P ( z away ? y = step 1 ) = P ( z away ? y = https://datingranking.net/asiandate-review/? 1 ) = P ( z aside ) . Then P ( y = step one ? ? away ) = ? ( dos p ? z elizabeth ? + record ? / ( 1 ? ? ) ) , just like in the Eq. 7 . (más…)