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I am reading this tutorial http://math.arizona.edu/~jwatkins/f-transform.pdf which explains transformations of random variables.

It says that the PDF of a transformed random variable $Y = g(X)$ can be found easily when $g(\cdot)$ is an increasing or decreasing function and one to one.

However, is the important part here that $g^{-1}(\cdot)$ is increasing/decreasing, which just happens to be the case if the above two conditions are also met?

For example, consider $Y = \sin(X)$. Now $\sin(\cdot)$ is not one-to-one but $\arcsin$ is. Is this sufficient to use the CDF method for deriving the PDF?

Or would one split the sine into increasing and decreasing parts?

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  • Regarding the sin stuff:

    The arcsin is only one-to-one in a certain specific definition of that function (e.g compare https://en.wikipedia.org/wiki/Multivalued_function ). But in that specific case $arcsin \neq sin^{-1}$.

    If $y=g(x)$ is monotonically increasing/decreasing then $x=g^{-1}(x)$ has to be monotonically increasing/decreasing as well. Since the slopes of the two dy/dx and dx/dy are related and have the same sign (however, you can have a jump where dy/dx is not well defined, e.g consider tan(x) which has asymptotes every half period of the circle, but this may not be problematic if the function is one-to-one, e.g. consider 1/x).

    So you can never have the case that $g(x)$ is not monotonically increasing/decreasing while $g^{-1}(x)$ is. Except when you have these piece-wise definitions/evaluations of the functions like is done with arcsin.

  • Why the pdf for transformed variables is found easily for one-to-one functions? Or, what is the problem with functions that are not one-to-one?:

    The "problem" that you deal with is that Y=sin(X) maps multiple X to a single Y. So, the probability for a certain Y is not anymore defined in a simple way, using a single formula with $f_Y= f_X(g^{-1}(y)) \partial_y g^{-1}(y)$, but instead requires a sum of this evaluation in all the points where Y and X are related.

    here is an example case of a transform from X to Y where two X values map to one Y value: https://stats.stackexchange.com/a/319362/164061

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  • $\begingroup$ So if I limit the domain of $\arcsin$ to $[-1, 1]$ I could still use this method? $\endgroup$
    – mmh
    Commented Feb 19, 2018 at 9:21
  • $\begingroup$ It might be. But certainly not in general. It depends on X. In this way (the domain of arcsin) you'd say that the domain of X=arcsin(Y) is [-1,1], which may be or not be correct (depending on the true domain of X). You will have to consider the domain of your variable X at hand and not some artificial conventional sub-domain of the arcsin function. $\endgroup$ Commented Feb 19, 2018 at 9:36

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