Consider this question:
If X is a random variable with FX (cumulative distribution function), then X is absolutely continuous if:
a) FX(x) is differentiable for all x in the real numbers.
b) fX(x) (Probability density function) is differentiable for all x in the real numbers.
c) If the integral of fX(x) over all the real numbers is equal to 1.
d) None of the above.
I know that the answer is d, but I don't know why or what even makes X absolutely continuous.
Answer
In probability language, FX is absolutely continuous if there exists an associated probability density function fX. This means there is the relationship FX(x)=∫x−∞fX(y)dy. In this case we usually just say that X is continuous, not absolutely continuous (though we may say something even more precise, like "the distribution of X is absolutely continuous with respect to Lebesgue measure", if there is some possibility of confusion).
This does not require that FX be differentiable at every point; for a familiar example, you can look at FX(x)={0x<01−e−xx≥0. It certainly doesn't require fX to be differentiable at every point, indeed fX can even be nowhere continuous in principle.
Property (c) I don't entirely understand what is meant, seeing as absolute continuity is equivalent to fX merely existing.
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