Let $p(x)$ and $q(x)$ be two density functions.
Show that the cross entropy $S(p,q) (=\int_{\mathbb{R}}-p(x) \ln (q(x))dx)$ follows the following inequality: $S(p,q) \geq 1 - \int_{\mathbb{R}}p(x) q(x)dx$
$ \int_{\mathbb{R}}-p(x) \ln (q(x))dx \geq 1 - \int_{\mathbb{R}}p(x) q(x)dx$
So I tried to write 1 as either $\int_{\mathbb{R}}p(x)dx$ or $\int_{\mathbb{R}}q(x)dx$
I tried it first with $1 = \int_{\mathbb{R}}p(x)dx$
Which lead to
$ \int_{\mathbb{R}}-p(x) \ln (q(x))dx \geq \int_ {\mathbb{R}}p(x) dx - \int_{\mathbb{R}}p(x) q(x)dx $
$ \int_{\mathbb{R}}-p(x) \ln (q(x))dx \geq \int_ {\mathbb{R}}p(x)(1- q(x))dx $
$ \int_{\mathbb{R}}p(x) \ln (\frac{1}{q(x)})dx \geq \int_ {\mathbb{R}}p(x)(1- q(x))dx $.
Since $0 \leq q(x) \leq 1$ we know that $ ln(\frac{1}{x}) \geq 1-x $ for $x \in [0,1]$
However I do not know whether this is the right track and this ultimatively leads me to the solution.
RHS is nothing but $\int (1-q(x))p(x)dx$ and the inequality follows from the fact that $\ln (q(x)) \le q(x)-1$ or $-\ln (q(x)) \geq 1-q(x)$.