1(Normalization) Let $f$ be a density function and $g$ be a nonnegative function. Then show that:
$$\int|f-\frac{g}{\int g}|\leq \int |f-g|$$
- Let $R(A):=\max(\int_{A}f_2,\int_{A^C}f_1)$, where $f_1,f_2$ are two density functions and $A$ is a set belonging to the sigma algebra. Show that the set $A=${$f_1>f_2$} minimizes $R(A)$.
My idea: for (1), I tried to use the set $Z=${$f>\frac{g}{\int g}$} and simplify the LHS to $2\int_{Z}f-\frac{g}{\int g}$.
For (2), I tried to use Scheffe's identity as $\sup_{A}|\int_{A}f_1-\int_{A}f_2|=\int_{f_1>f_2}f_1-f_2$.
Both of these relations are actually false.
For the first part, take $f(x) = 2x$ and $g(x) = 1.9(1-x)$ over the domain $[0,1]$. The LHS works out to $1$, and the RHS to $<0.976.$
For the second part, take $f_1 = \frac32 \mathbf{1}\{0 \le x\le 2/3\}, f_2 = \mathbf{1}\{0 \le x \le 1\}$. Then $R(\{f_1 > f_2\}) = R([0,2/3]) = 1$. But $$R([0,1/2]) = \max\left(\frac{3}{4} ,\frac{1}{2} \right) = \frac34 < 1.$$