# zz-3 - ECE 6010 Lecture 4 Change of Variables Reading from...

This preview shows pages 1–3. Sign up to view the full content.

ECE 6010 Lecture 4 – Change of Variables Changing variables: One dimension A simply invertible function Let Y = g ( X ) , where X is a continuous r.v. and g is a one-to-one, onto, measurable function. Then F Y ( y ) = P ( Y y ) = P ( g ( X ) y ) = P ( X g - 1 ( y )) = F X ( g - 1 ( y )) So we can determine the distribution of Y . Let us now take a different point of view that will allow us to generalize to higher dimensions and develop and understand a commonly used formula. Consider an interval along the X axis, P ( x X x + dx ) f X ( x ) dx Suppose the function g ( x ) has a positive derivative. The interval along the Y axis, when Y = g ( X ) is dy dy dx dx at the point x . The probability that X falls in its interval is the probability that Y falls in its interval: P ( x X x + dx ) = P ( y Y y + dy ) where y = g ( x ) , or equivalently, x = g - 1 ( y ) . Then f X ( x ) dx f Y ( y ) dy That is f Y ( y ) = f X ( x ) dx dy ± ± ± ± x = g - 1 ( y ) = f X ( g - 1 ( y )) dx dy . If we take the other case that g ( x ) has a negative derivative, we have to take f X ( x ) dx = f Y ( y )( - dy ) . Combining these together we obtain f Y ( y ) = f X ( g - 1 ( y )) ± ± ± ± dx dy ± ± ± ± = f X ( g - 1 ( y )) | dy/dx | = f X ( g - 1 ( y ) | g 0 ( g - 1 ( x )) . Example 1 Let Y = aX + b . Then f Y ( y ) = 1 | a | f X (( y - b ) /a ) . 2 Example 2 Suppose f X ( x ) = a/π x 2 + a 2 (Cauchy). Let Y = 1 /X . Then f Y ( y ) = 1 /aπ y 2 + 1 /a 2 (Cauchy) 2

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

View Full Document
ECE6010:Lecture4–ChangeofVariables 2 Example 3 Suppose X ∼ U ( a, b ) , with 0 < a < b . Then f X ( x ) = 1 b - a for x [ a, b ] . Let Y = 1 /X . Then f Y ( y ) = 1 ( b - a ) y 2 for 1 b < y < 1 a . 2 Example 4 Let Y = e X . Then f Y ( Y ) = 1 y f X (ln y ) y > 0 . If X ∼ N ( μ, σ 2 ) then f Y ( y ) = 1 2 π e - (ln y - μ ) 2 / 2 σ 2 This density is log-normal. 2 Example 5 Suppose y = g ( x ) = tan x , or x = tan - 1 y . This has an infinite number of solutions. However, if we take x ( - π/ 2 , π/ 2) , then there is a unique inverse. We have g 0 ( x ) = 1 cos 2 x = 1 + y 2 . Now let
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

zz-3 - ECE 6010 Lecture 4 Change of Variables Reading from...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online