# Bounded Probability Density Function

## Main Question or Discussion Point

Let the random variable X have the probability density function f(x). Suppose f(x) is
continuous over its domain and Var[X] is bounded away from zero: 0 < c < Var[X].

Claim: f(x) is bounded over its domain.

Is this claim true?

I don't think a counterexample like X ~ ChiSq_1 applies because, while f(0) is not bounded, the domain is x > 0. The question of continuity at 0 doesn't arise since 0 is not part of the domain, and therefore the lack of boundedness at 0 isn't relevant. But for any given c > 0, I can find an M such that f(c) < M. This example does show, however, why one can't claim "uniformly bounded."

Similarly, a degenerate r.v. (all probability mass at the point X = c, say) is continuous (because it is discrete), but the variance is not bounded away from zero.

So the question remains: is the claim true? How might one prove it?

Thanks.

## Answers and Replies

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quadraphonics
Similarly, a degenerate r.v. (all probability mass at the point X = c, say) is continuous (because it is discrete), but the variance is not bounded away from zero.
That example isn't relevant, because a degenerate r.v. does not have a p.d.f.

statdad
Homework Helper
Let the random variable X have the probability density function f(x). Suppose f(x) is
continuous over its domain and Var[X] is bounded away from zero: 0 < c < Var[X].

Claim: f(x) is bounded over its domain.

Is this claim true?
No, it is not, and the $$\chi^2_1$$ distribution you mention shows it. You do not need to worry about the pdf at $$x = 0$$, since the domain is only $$(0, \infty)$$. The variance of $$\chi^2_k$$ is $$\sigma^2 = 2k$$, so the variance of $$\chi^2_1$$ is bounded away from zero. However, the density is not bounded on its domain.
Statdad