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# 06_04 - Method of probability density function Theorem Let...

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p. 6-31 Y =( Y 1 , …, Y n )= g ( X ), where g is 1-to-1, so that its inverse exists and is denoted by x = g 1 ( y ) = w ( y ) = ( w 1 ( y ), w 2 ( y ), …, w n ( y )). Assume w have continuous partial derivatives, and let Then f Y ( y ) = f X ( g 1 ( y )) | J |, for y s.t. y = g ( x ) for some x , and f Y ( y )=0, otherwise. Method of probability density function Theorem. Let X =( X 1 , …, X n ) be continuous random variables with the joint pdf f X . Let Proof . ( Q : What is the role of | J |?) Y 1 Y 2 Y = g ( X ) X 1 X 2 p. 6-32 Remark . When the dimensionality of Y (denoted by k ) is less than n , we can choose another n k transformations Z such that ( Y , Z )= g ( X ) satisfy the assumptions in above theorem. By integrating out the last n k arguments in the pdf of ( Y , Z ), the pdf of Y can be obtained. It then follows from an exercise in advanced calculus that Example. X 1 and X 2 are random variables with joint pdf f X ( x 1 , x 2 ). Find the distribution of Y 1 = X 1 /( X 1 + X 2 ). Let Y 2 = X 1 + X 2 , then f Y ( y 1 ,...,y n )= n y 1 ··· y n F Y ( y 1 n ) = f X ( w 1 ( y ) ,...,w n ( y )) ×| J | . Since w 1 y 1 = y 2 , w 1 y 2 = y 1 , w 2 y 1 = y 2 , w 2 y 2 =1 y 1 , x 1 = y 1 y 2 w 1 ( y 1 ,y 2 ) x 2 = y 2 y 1 y 2 w 2 ( y 1 2 ) .

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p. 6-33 Therefore, and, Proof . For x 1 , x 2 0, the joint pdf of X is f X ( x 1 ,x 2 )= λ α 1 Γ ( α 1 ) x α 1 1 1 e λ x 1 × λ α 2 Γ ( α 2 ) x α 2 1 2 e λ x 2 = λ α 1 + α 2 Γ ( α 1 ) Γ ( α 2 ) x α 1 1 1 x α 2 1 2 e λ ( x 1 + x 2 ) .
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06_04 - Method of probability density function Theorem Let...

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