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distn fns - 1 Lecture 1 Transformation of Random Variables...

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1 Lecture 1. Transformation of Random Variables Suppose we are given a random variable X with density f X ( x ). We apply a function g to produce a random variable Y = g ( X ). We can think of X as the input to a black box, and Y the output. We wish to find the density or distribution function of Y . We illustrate the technique for the example in Figure 1.1. - 1 2 e -x 1/2 -1 f (x) x-axis X Y y X -Sqrt[y] Sqrt[y] Y = X 2 Figure 1.1 The distribution function method finds F Y directly, and then f Y by differentiation. We have F Y ( y ) = 0 for y < 0. If y 0, then P { Y y } = P {− y x y } . Case 1 . 0 y 1 (Figure 1.2). Then F Y ( y ) = 1 2 y + y 0 1 2 e x dx = 1 2 y + 1 2 (1 e y ) . 1/2 -1 x-axis -Sqrt[y] Sqrt[y] f (x) X Figure 1.2 Case 2 . y > 1 (Figure 1.3). Then F Y ( y ) = 1 2 + y 0 1 2 e x dx = 1 2 + 1 2 (1 e y ) . The density of Y is 0 for y < 0 and
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2 1/2 -1 x-axis -Sqrt[y] Sqrt[y] f (x) X 1 ' Figure 1.3 f Y ( y ) = 1 4 y (1 + e y ) , 0 < y < 1; f Y ( y ) = 1 4 y e y , y > 1 . See Figure 1.4 for a sketch of f Y and F Y . (You can take f Y ( y ) to be anything you like at y = 1 because { Y = 1 } has probability zero.) y f (y) Y 1 ' y F (y) Y 1 ' Figure 1.4 The density function method finds f Y directly, and then F Y by integration; see Figure 1.5. We have f Y ( y ) | dy | = f X ( y ) dx + f X ( y ) dx ; we write | dy | because proba- bilities are never negative. Thus f Y ( y ) = f X ( y ) | dy/dx | x = y + f X ( y ) | dy/dx | x = y with y = x 2 , dy/dx = 2 x , so f Y ( y ) = f X ( y ) 2 y + f X ( y ) 2 y . (Note that | − 2 y | = 2 y .) We have f Y ( y ) = 0 for y < 0, and: Case 1 . 0 < y < 1 (see Figure 1.2). f Y ( y ) = (1 / 2) e y 2 y + 1 / 2 2 y = 1 4 y (1 + e y ) .
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3 Case 2 . y > 1 (see Figure 1.3). f Y ( y ) = (1 / 2) e y 2 y + 0 = 1 4 y e y as before. Y y X y - y Figure 1.5 The distribution function method generalizes to situations where we have a single out- put but more than one input. For example, let X and Y be independent, each uniformly distributed on [0 , 1]. The distribution function of Z = X + Y is F Z ( z ) = P { X + Y z } = x + y z f XY ( x,y ) dxdy with f XY ( x,y ) = f X ( x ) f Y ( y ) by independence. Now F Z ( z ) = 0 for z < 0 and F Z ( z ) = 1 for z > 2 (because 0 Z 2). Case 1 . If 0 z 1, then F Z ( z ) is the shaded area in Figure 1.6, which is z 2 / 2. Case 2 . If 1 z 2, then F Z ( z ) is the shaded area in FIgure 1.7, which is 1 [(2 z ) 2 / 2]. Thus (see Figure 1.8) f Z ( z ) = z, 0 z 1 2 z, 1 z 2 0 elsewhere . Problems 1. Let X,Y,Z be independent, identically distributed (from now on, abbreviated iid) random variables, each with density f ( x ) = 6 x 5 for 0 x 1, and 0 elsewhere. Find the distribution and density functions of the maximum of X,Y and Z . 2. Let X and Y be independent, each with density e x ,x 0. Find the distribution (from now on, an abbreviation for “Find the distribution or density function”) of Z = Y/X .
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