prob.part36.73_74 - 2.2. CONTINUOUS DENSITY FUNCTIONS 2 65...

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2.2. CONTINUOUS DENSITY FUNCTIONS 65 -1 -0.5 0.5 1 1.5 2 0.2 0.4 0.6 0.8 1 -1 -0.5 0 0.5 1 1.5 2 0.25 0.5 0.75 1 1.25 1.5 1.75 2 F (z) Z f (z) Z Figure 2.17: Distribution and density for Z = X 2 + Y 2 . E be the event { Z z } . Then the distribution function F Z of Z (see Figure 2.16) is given by F Z ( z ) = P ( Z z ) = Area of E Area of target . Thus, we easily compute that F Z ( z ) = 0 , if z 0 , z 2 , if 0 z 1 , 1 , if z > 1 . The density f Z ( z ) is given again by the derivative of F Z ( z ): f Z ( z ) = 0 , if z 0 , 2 z, if 0 z 1 , 0 , if z > 1 . The reader is referred to Figure 2.17 for the graphs of these functions. We can verify this result by simulation, as follows: We choose values for X and Y at random from [0 , 1] with uniform distribution, calculate Z = X 2 + Y 2 , check whether 0 Z 1, and present the results in a bar graph (see Figure 2.18). ± Example 2.16
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This note was uploaded on 09/15/2009 for the course SCF scf taught by Professor Scf during the Spring '09 term at Indian Institute Of Management, Ahmedabad.

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prob.part36.73_74 - 2.2. CONTINUOUS DENSITY FUNCTIONS 2 65...

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