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Applying ϕ (notice that ϕ ( t ) = arctan t ) to the last inequality and taking into account the increasing character of ϕ, we have α − β ≤ arctan ( tan α − tan β ), or equivalently, arctan x − arctan y = α − β ≤ arctan ( x − y ), and this proves (8).
Without loss of generality, assuming that ν 1 > ν 2, by the increasing character of Ψ in ( 0, ∞ ), we know that μ 0 = μ 1 − μ 2 > 0. By simple computation, it can be shown that E U k 0 = k 0 ( n − k 0 ) n ( μ 1 − μ 2 ) = n τ 0 ( 1 − τ 0 ) μ 0, (14).
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