Supplementary Materials for ``Optimal Decisions from Probabilistic Models: the Intersection-over-Union Case''

Supplementary materials to the main paper. 1. Proofs Proof of Proposition 4 Proof. We have E[Sk] = ∑ i∈V Ezi [1{zi=k∧yi=k}] = ∑ i∈V pi(k)1{yi=k}. Likewise we have E[Tk] = ∑ i∈V Ezi [1{zi=k∨yi=k}] = ∑ i∈V Ezi [1{zi=k} + 1{yi=k} − 1{zi=k∧yi=k}] = ∑ i∈V (pi(k) + 1{yi=k} − pi(k) 1{yi=k}) = ∑ i∈V (1{yi=k} + pi(k) 1{yi 6=k}). Proof of Proposition 5 Proof. In (5) the first central moment 〈Tk〉 is always zero, hence the only term that could be non-zero is 〈Sk, Tk〉 = E[(Sk − ESk)(Tk − ETk)] = E[SkTk]− ESkETk − ETkESk + ESkETk = E[SkTk]− ESkETk. We will show that these two terms are equal and hence their difference is zero. To this end we first expand the product as E[SkTk] = E [ ( ∑ i∈V 1{zi=k∧yi=k})( ∑ i∈V 1{zi=k∨yi=k}) ] = E ∑ i∈V ∑ j∈V 1{zi=k∧yi=k}1{zj=k∨yj=k}  .(1) In order to expand the expectation operator in (1) we would like to use the conditional independence assumption pij(zi, zj) = pi(zi) pj(zj) for i 6= j. For this, we split the sum into the cases i = j and i 6= j as follows. = ∑ i∈V ∑ zi∈Y pi(zi) 1{zi=k∧yi=k}1{zi=k∨yi=k} + ∑ i∈V ∑ j∈V\{i} ∑ zi∈Y ∑ zj∈Y pi(zi) pj(zj) 1{zi=k∧yi=k}1{zj=k∨yj=k} The first sum can be simplified by observing that 1{zi=k∧yi=k} 1{zj=k∨yj=k} = 1{zi=k∧yi=k} and by observing that all terms in the sum are zero except for the case when zi = k. For the second sum we reorder terms, obtaining = ∑ i∈V pi(k) 1{yi=k} + ∑ i∈V ∑ zi∈Y pi(zi) 1{zi=k∧yi=k}  ∑ j∈V\{i} ∑ zj∈Y pj(zj) 1{zj=k∨yj=k}  . We perform the same simplification as before on the second