That is the function we use to transform is actually

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That is, the function we use to transform is actually the c.d.f of X . Then g 0 ( x ) = F 0 X ( x ) = f X ( x ) and f Y ( y ) = f X ( x ) | g 0 ( x ) | = f X ( x ) f X ( x ) = 1 0 < y < 1 That is, Y is uniformly distributed. In the image processing literature, this is called histogram equalization . 2 Example 7 Let X ∼ U (0 , 1) , and let Y have a specified c.d.f F Y ( y ) , which we take to be continuous. Let g ( x ) = F - 1 Y ( x ) , 0 < x < 1 , so x = F Y ( y ) . Then f Y ( y ) = f X ( g - 1 ( y )) dx dy Since f X is 1 for all values of y , and since dx dy = | f Y ( y ) | = f Y ( y ) ,

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ECE 6010: Lecture 4 – Change of Variables 3 putting all the pieces together we find f Y ( y ) = f Y ( y ) . That is, Y has the desired distribution! The point of this is that if we can generate U (0 , 1) , we can (in principle!) transform it to produce any other continuous distribution. 2 Multiple inverses It may happen that g is not a uniquely invertible function. That is, for a given y there may be more than one value of x such that y = g ( x ) . For example, y = g ( x ) = x 2 : then x = y and x = - y are both inverses. We will prove the concept for two solutions, Let y = g ( x 1 ) = g ( x 2 ) , assuming to be specific that the slope is positive at x 1 and negative at x 2 . P ( y < Y < y + dy ) = P ( x 1 < X < x 1 + dx 1 ) + P ( x 2 + dx 2 < X < x 2 ) That is f Y ( y ) dy = f X ( x 1 ) dx 1 + f X ( x 1 ) | dx 2 | From this, f Y ( y ) = f X ( x 1 ) dx 1 dy + f X ( x 2 ) dx 2 dy This is sometimes written f Y ( y ) = f X ( x 1 ) | g 0 ( x 1 ) | + f X ( x 1 ) | g 0 ( x 1 ) | In general, with n solutions x 1 , x 2 , . . . , x n we have f Y ( y ) = f X ( x 1 ) | g 0 ( x 1 ) | + f X ( x 1 ) | g 0 ( x 1 ) | + · · · + f X ( x n ) | g 0 ( x n ) | Example 8 Suppose X ∼ U ( - π, π ) and Y = a sin( X + θ ) . Generally the sin function has an infinite number of inverses, but there are only inverses in the stated range of X . We have g 0 ( x ) = a cos( x + θ ) = p a 2 - y 2 . The density of X is f X ( x ) = 1 2 π , for x [ - π, π ] . Then f Y ( y ) = 1 2 π 1 p a 2 - y 2 + 1 2 π 1 p a 2 - y 2 = 1 π p a 2 - y 2 , | y | < a. 2 g ( X ) constant in an interval If the function g ( X ) is constant over any interval, then there is no inverse, nor even multiple inverses. However, we can still compute the distribution. Let g ( x ) = y 1 for x 0 < x x 1 (i.e., constant). Then P ( Y = y 1 ) = P ( x 0 < X x 1 ) = F X ( x 1 ) - F X ( x 0 ) .
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