The density function of an exponentially distributed random variable $X$ is given by $$ f_X(x) = \begin{cases} \lambda e^{-\lambda x} & x \ge 0, \\ 0 & x < 0. \end{cases} $$ The density function of $Y=X^3$ could be evaluated in the following way $$ f_Y(y) =f_X(y^{1/3})\Bigl|\frac d{\mathrm dy} y^{1/3}\Bigr| = \begin{cases} 0 & y<0,\\ \frac\lambda3 e^{-\lambda y^{1/3}}y^{-2/3}& y>0. \end{cases} $$ Strictly speaking, the density function $f_Y$ is still not defined since the value at $0$ is not defined (we have that $\lim_{y\downarrow 0}f_Y(y)=\infty$).

What is the value of the density function $f_Y$ at $0$? How can we asign it?

Probably, we can assign any real number since it is just a single point and the Lebesgue measure of this set is zero. Am I right?

Any help is much appreciated!


Correct. The presence of a density function implies that the distribution is absolutely continuous with respect to Lebesgue measure, and in that case the density (i.e., the Radon-Nikodym derivative of the distribution) is only defined up to sets of Lebesgue measure zero.


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