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note 8 - MA2216/ST2131 Probability Notes 8 Multidimensional...

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MA2216/ST2131 Probability Notes 8 Multidimensional Changes of Variables and Bivariate Normal Distribution § 1. Changes of Variables Let X 1 and X 2 be jointly distributed random variables with joint prob- ability density function f X 1 ,X 2 . It is sometimes necessary to obtain the joint distribution of the random variables Y 1 and Y 2 , which arise as func- tions of X 1 and X 2 . Specifically, suppose that Y 1 = g 1 ( X 1 , X 2 ) and Y 2 = g 2 ( X 1 , X 2 ) . For example, X 1 and X 2 are independent exponentially distributed ran- dom variables, and we are interested to know the joint probability density function of Y 1 = X 1 + X 2 , and Y 2 = X 1 X 1 + X 2 . In this case, g 1 ( x 1 , x 2 ) = x 1 + x 2 and g 2 ( x 1 , x 2 ) = x 1 / ( x 1 + x 2 ) . (The details can be found in item 4 below.) Note. The above is a typical example of change-of-variables for a bi- variate case. In item 6, we will also touch upon briefly the formula for change of variables for multi-variate cases. 1
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1. Assumptions: (i) Let X 1 and X 2 be jointly continuous random variables with known joint p.d.f. f X 1 ,X 2 ( x 1 , x 2 ). (ii) Let Y 1 and Y 2 be given functions of X 1 and X 2 in the form: Y 1 = g 1 ( X 1 , X 2 ) and Y 2 = g 2 ( X 1 , X 2 ) . We can uniquely solve for x 1 and x 2 in terms of y 1 and y 2 with solutions given by, say, x 1 = h 1 ( y 1 , y 2 ) and x 2 = h 2 ( y 1 , y 2 ) . In other words, the inverse transformation exists. (iii) The functions g 1 and g 2 have continuous partial derivatives at all points ( x 1 , x 2 ) and are such that the following 2 × 2 determinant J ( x 1 , x 2 ) = fl fl fl fl fl fl fl ∂g 1 ∂x 1 ∂g 1 ∂x 2 ∂g 2 ∂x 1 ∂g 2 ∂x 2 fl fl fl fl fl fl fl = ∂g 1 ∂x 1 ∂g 2 ∂x 2 - ∂g 1 ∂x 2 ∂g 2 ∂x 1 6 = 0 (1 . 1) at all points ( x 1 , x 2 ). (Note that J ( x 1 , x 2 ) is called the Jacobian .) Question: We would like to find the joint p.d.f. of Y 1 and Y 2 in terms of the joint p.d.f. f X 1 ,X 2 together with the transformations g 1 and g 2 . Conclusion: Under the above conditions, it can be shown that the random variables Y 1 and Y 2 are jointly continuous with joint p.d.f. given by f Y 1 ,Y 2 ( y 1 , y 2 ) = 1 | J ( x 1 , x 2 ) | f X 1 ,X 2 ( x 1 , x 2 ) (1 . 2) where x 1 = h 1 ( y 1 , y 2 ) and x 2 = h 2 ( y 1 , y 2 ). 2
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2. Example. Let X 1 and X 2 be jointly continuous random variables. Find the joint p.d.f. of Y 1 = X 1 + X 2 and Y 2 = X 1 - X 2 . Sol. Generally speaking, one should first specify the transformation, which is in this example given below: y 1 = g 1 ( x 1 , x 2 ) = x 1 + x 2 , y 2 = g 2 ( x 1 , x 2 ) = x 1 - x 2 . Second , determine the inverse transformation, which is expressible as follows: x 1 = h 1 ( y 1 , y 2 ) = y 1 + y 2 2 , x 2 = h 2 ( y 1 , y 2 ) = y 1 - y 2 2 . Third , one should determine the values that Y i takes for i = 1 , 2, and if necessary, sketch the transformed region. The fourth step should be to determine the Jacobian. Note that y 1 = g 1 ( x 1 , x 2 ) = x 1 + x 2 , y 2 = g 2 ( x 1 , x 2 ) = x 1 - x 2 .
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