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50 4. SEARCH PRUNING IN VIDEO SURVEILLANCE SYSTEMSWe proceed to analyze these issues and first recall that for a given authentication group v,the categorization algorithm identifies set S of all unpruned subjects, defined as S = {v ∈ v :Ĉ(v) = 1}. We are here interested in the size of the search after pruning, specifically in theparameterτ := |S| , 0 ≤ τ ≤ ρ, (4.10)n/ρwhich represents 2 a relative deviation of |S| from a baseline n/ρ. It can be seen that the typical,i.e., common, value ofτ is (see also Section 4.5.1)|S|ρ∑τ 0 := E vn/ρ = ρ p f ɛ f . (4.11)We are now interested in the entire tail behavior (not just the typical part of it), i.e., we are interestedin understanding the probability of having an authentication groupvthat results in atypicallyunhelpful pruning (τ > τ 0 ), or atypically helpful pruning (τ < τ 0 ).Towards this letα 0,f (v) := |C f|n/ρ , (4.12)let a 0 (v) = {α 0,f (v)} ρ f=1describe the instantaneous normalized distribution (histogram) of{|C f |} ρ f=1for the specific, randomly chosen and fixed authentication group v, and letf=1p := {p f } ρ f=1 = {E |C f |vn }ρ f=1 , (4.13)denote the normalized statistical population distribution of {|C f |} ρ f=1 .Furthermore, for a given v, letα 1,f (v) := |C f ∩S|, 0 ≤ α 1,f ≤ ρ, (4.14)n/ρlet α 1 (v) := {a 1,f (v)} ρ f=1 , and α(v) := {α 0(v),α 1 (v)}, and let 3V(τ) := { 0 ≤ α 1,f ≤ min(τ,α 0,f ),ρ∑α 1,f = τ } , (4.15)denote the set of valid α for a given τ, i.e., describe the set of all possible authentication groupsand categorization errors that can result in|S| = τ n ρ .Given the information that α 1 has on α 0 , given that τ is implied by α 1 , and given that thealgorithms here categorize a subject independently of other subjects, it can be seen that for anyα ∈ V(τ), it is the case thatf=1P(α,τ) = P(α 0 ,α 1 ) = P(α 0 )P(α 1 |α 0 ) (4.16)ρ∏ ρ∏= P(α 0,f ) P(α 1,f |α 0,f ). (4.17)f=12. Note the small change in notation compared to Section 4.2. This change is meant to make the derivations moreconcise.3. For simplicity of notation we will henceforth use α 0,α 1,α,α 0,f ,α 1,f and let the association tov be impliedf=1