Fundamentals of Probability and Statistics for Engineers

152 Fundamentals of Probability and Statistics for EngineersEquation (5.68) now applies and we havef Y1 Y 2…y 1 ; y 2 †ˆf X ‰g 11 …y†ŠjJ 1j‡f X ‰g 12 …y†ŠjJ 2j; …5:74†wheref X …x† ˆf X1…x 1 † f X2…x 2 †ˆ 1 2 exp x 2 1 ‡ x2 2;2…5:75†qg111 qg 111qy 1 qy 2y 1J 1 ˆ J 2 ˆˆqg121 qg1211 ‡ y 2 :2qy 1 qy 2…5:76†On substituting Equations (5.75) and (5.76) into Equation (5.74), we have y 1 2f Y1 Y 2…y 1 ; y 2 †ˆ1 ‡ y 2 22 exp y 2 1 y2 2 ‡ y2 12…1 ‡ y 2 2 †y 1ˆ…1 ‡ y 2 2 † exp y 2 12;; for y 1 0; and 1 < y 2 < 1;ˆ 0; elsewhere: …5:77†We note that the result can again be expressed as the product of f Y 1(y 1 )andf Y 2(y 2 ), with8 y 2 >:0; elsewhere;f Y2 …y 2 †ˆ1…1 ‡ y 2 2 † ; for 1 < y 2 < 1:Thus random variables Y 1 and Y 2 are again independent in this case where Y 1has the so-called Raleigh distribution and Y 2 is Cauchy distributed. We remarkthat the factor (1/ ) is assigned to f Y 2(y 2 ) to make the area under each pdfequal to 1.Ex ample 5. 20. Let us determine the pdf of Y considered in Example 5.11 byusing the formulae developed in this section. The transformation isY ˆ X 1 X 2 :…5:78†TLFeBOOK