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This question already has an answer here:

Suppose we have a discrete r.v. $X$, take $Y = g(X) $ where $g$ is one-to-one and onto-

If we want to obtain the new pdf for the discrete r.v. we simply notice that $$f_Y(y) = P(Y=y) = f_X(g^{-1}(y)) $$

Now if random variable $X$ is continous and $g$ is always one-to-one and onto why can't we apply this to the continous case also?

that is, why can't we say $$f_Y(y) = f_X(g^{-1}(y)) $$ for a continous random variable?

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marked as duplicate by whuber Jan 11 '15 at 20:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I was unable to comment but the answer is that the values of the probability density function are not probabilities. If you use your logic with the CDF, it is correct. But, when you differentiate the CDF, the chain rule says you have to multiply by the derivative of the argument, so your identity fails.

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