Take the 2-minute tour ×  Let $f, \hat{f}, g,$ and $\hat{g}$ be
continuous probability densities. Define probability densities
$p \propto fg$ and $\hat{p} \propto \hat{f}\hat{g}$. Is it true
that \begin{align*} ||p - \hat{p}||_{1} \le || f - \hat{f}||
_{1} + ||g - \hat{g}||_1 \end{align*}  share|improve this
question Why would this inequality imply something of interest,
if true? In short, what are your motivations for asking this? – 
Did Oct 29 '11 at 7:13  1 Answer 1  up vote 1 down vote accepted
Here is a real counterexample, as verified by mathematica: so we
take the same discrete two point space as below, and let
$f=(u_1,u_2), \hat{f} = (v_1,v_2), g = (a_1,a_2),
\hat{g}=(b_1,b_2)$, with the relation that $x_1 + x_2 = 1$ where
$x \in \{u,v,a,b\}$. So your claim becomes in this special case
$$ |\frac{u_1 a_1}{u_1 p_1 + u_2 a_2} - \frac{v_1 b_1}{v_1 b_1 +
v_2 b_2}| + | \frac{u_2 a_2}{u_1 a_1 + u_2 a_2} - \frac{v_2
b_2}{v_1 b_1 + v_2 b_2}|$$ $$\le |u_1 - v_1| + |u_2 - v_2| +
|a_1 - b_1| + |a_2 - b_2|$$  No