No. The *integral* of the density is between 0 and 1, but the density
itself is not constrained to be in (0, 1). This is a common misnomer.
For an obvious counterexample, consider the following densities:
* Uniform(0, 1)
* Uniform(0, 0.5)
* Uniform(0, 0.01)
If a random variable X has a Uniform(a, b) distribution, its density is
f(x) = 1/(b - a) for a <= x <= b
and zero otherwise. It is easy from here to show that the latter two
uniforms above have density > 1 over their respective regions of
support, but they are legitimate probability densities, so the total
area under their respective density functions is 1.
Dennis
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