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I am studying random variables.

Question is this: if rv X & a function g is known, what is the pdf of random variable Y = g(x)?

in the textbook answer is explained as follows.

P[y ≤ Y ≤ y + dy] = P[x ≤ X ≤ x + dx]

F_y(y + dy) - F_y(y) / dy dx = F_x(x + dx) - F_x(x) / dx dy

why is left side of dx & right side of dy exists in above equation?

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closed as off-topic by Ricardo Andrade, j.c., Andrey Rekalo, Daniel Moskovich, Olivier Benoist Nov 28 '13 at 8:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, j.c., Andrey Rekalo, Daniel Moskovich, Olivier Benoist
If this question can be reworded to fit the rules in the help center, please edit the question.

Please capitalise your title sensibly, and preferably actually ask a question, rather than just giving the subject area. – Scott Morrison Oct 22 '09 at 2:48
up vote 4 down vote accepted

What you're looking at is known as "the transformation theorem" and is just an integral change of variables written in probability notation.

Suppose g is an increasing function and Y = g(X). Then

F_Y(y) = P( g(X) < y ) = P( X < g^{-1}(y) ) = F_X( g^{-1}(y) )

To obtain the PDF, differentiate both sides of the equation above:

f_Y(y) = f_X( g^{-1}(y) ) D_y ( g^{-1}(y) )

where D_y means derivative with respect to y. Now if g were a decreasing function we'd have

F_Y(y) = P( g(X) < y ) = P( X > g^{-1}(y) ) = 1 - F_X( g^{-1}(y) )


f_Y(y) = f_X( g^{-1}(y) ) | D_y ( g^{-1}(y) ) |.

In the last line we would have -D_y. Since g is a decreasing function, it's derivative is negative and so the absolute values take care of the negative sign.

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