print3 - 1 Lecture 1 Transformation of Random Variables...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 Fnd the density or distribution function of Y .W e illustrate the technique for the example in ±igure 1.1. - 1 2 e -x 1/2 -1 f (x) x-axis X Y y X -Sqrt[y] Sqrt[y] Y = X 2 ±igure 1.1 The distribution function method Fnds F Y directly, and then f Y by di²erentiation. We have F Y ( y )=0for y< 0. If y 0, then P { Y y } = P {− y x y } . Case 1 .0 y 1 (±igure 1.2). Then F Y ( y )= 1 2 y + Z y 0 1 2 e x dx = 1 2 y + 1 2 (1 e y ) . x-axis f (x) X ±igure 1.2 Case 2 . y> 1 (±igure 1.3). Then F Y ( y 1 2 + Z y 0 1 2 e x dx = 1 2 + 1 2 (1 e y ) . The density of Y is 0 for 0 and
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 1 2 y+ 1 2 (1-e - y ) 1 2 + 1 2 - y ) ' Figure 1.4 The density function method ±nds 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 | y .) We have f Y ( y ) = 0 for y< 0, and: Case 1 .0 1 (see Figure 1.2). f Y ( y (1 / 2) e y 2 y + 1 / 2 2 y = 1 4 y (1 + e y ) .
Background image of page 2
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 } = ZZ x + y z f XY ( x,y ) dxdy with f ( f X ( x ) f Y ( y ) by independence. Now F Z ( z )=0for z< 0 and F Z ( z )=1 for z> 2 (because 0 Z 2). Case 1 .I f0 z 1, then F Z ( z ) is the shaded area in Figure 1.6, which is z 2 / 2. Case 2 f1 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 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 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 23

print3 - 1 Lecture 1 Transformation of Random Variables...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online