Unformatted text preview: Z ∞∞ B 1+ ( ω σ ) 2 dω = σπB, Z ∞∞ D 1+ ( ωz τ ) 2 dω = τπD and Z ∞∞ " Aω 1+ ( ω σ ) 2Cω 1+ ( ωz τ ) 2 # dω = Z ∞∞ " Aω 1+ ( ω σ ) 2C ( ωz ) 1+ ( ωz τ ) 2 # dωCz Z ∞∞ 1 1+ ( ωz τ ) 2 dω = A σ 2 2 log ´ 1+ ± ω σ ² 2 µCτ 2 2 log " 1+ ¶ ωz τ · 2 #³ ³ ³ ³ ³ ∞∞τπCz. The integral is ﬁnite and equal to zero if A = M 2 σ 2 , C = M 2 τ 2 for some constant M . Hence f Z ( z ) = 1 π 2 στ ´ σπBτπD2 πMz τ µ = 1 π ( σ + τ ) 1 1+ ( z/ ( σ + τ )) 2 , if B = τ σ + τ , D = σ σ + τ ) , M =στ 2 2 z ( σ + τ ) 1 1+ ( z σ + τ ) 2 ....
View
Full Document
 Spring '12
 Dr.Hackney
 Statistics, Binomial, Probability, Blindness, Color blindness, color blind person

Click to edit the document details