Can PDF be larger than CDF for continuous RV?

Question

Let $f$ be the PDF of a continuous RV $X$, and define $F$ to be its CDF: $F(x)=\int _{-\infty}^xf(t)dt$. Can $F(x)$ be strictly less than $f(x)$? This clearly cannot happen in the discrete case ($\mathbb{P}(X\le x) \ge \mathbb{P}(X=x)$). However, in the continuous case, viewing $f(x)$ as the probability of $X$ falling within the infinitesimal interval $(x,x+dx)$ is it possible that the $f(x) > F(x)$?

If $CDF \ge PDF$ in the continuous case, I don't know how to verify it from the definition.

If only at certain places you may have $f(x)>F(x)$ but as $F(x)$ tends to one as $x$ goes to $+\infty$ you obviously can't have it for all $x$ — H. H. Rugh, Aug 21 '16 at 22:19
related http://math.stackexchange.com/questions/105455/how-can-a-probability-density-be-greater-than-one-and-integrate-to-one — Henry, Aug 21 '16 at 23:06

score 1 · Accepted Answer · answered Aug 21 '16 at 22:22

Yes. The probability density can easily have greater magnitude than the cumulative probability mass. They are measures of different dimensions.

For one thing, a cumulative distribution function cannot be greater than $1$ at any point, while a probability density function has no such restriction.

As an example: a uniform continuous distribution with the support of $(0;\tfrac 12)$ has a probability density of $2$ everywhere within its support. (Outside the support, however...)

Can PDF be larger than CDF for continuous RV?

1 Answers1