Stein Bounds for TV Metric

Recall \[ f_h(x) = e^{x^2/2}\int_{-\infty}^x [h(y)-\mathbb{E}(h(N))]e^{y^2/2} dy \] Note that \[ |h(y)-\mathbb{E}(h(N))|\le 1 \] Note that : \[ \int_{-\infty}^x = \int_{-\infty}^\infty - \int_x^{\infty} = - \int_x^{\infty} \] So, \[ |f_h(x)| \le e^{x^2/2}\int_{-\infty}^x e^{y^2/2} dy \] But also \[ |f_h(x)| \le e^{x^2/2}\int_{x}^\infty e^{y^2/2} dy \] If $x>0$, then \[ |f_h(x)| \le e^{x^2/2}\int_{|x|}^\infty e^{y^2/2} dy \] If $x<0$, then \[ |f_h(x)| \le e^{x^2/2}\int_{-\infty}^{|x|} e^{y^2/2} dy = e^{x^2/2}\int_{|x|}^{\infty} e^{y^2/2} \] We have obtained that \[ |f_h(x)| \le e^{x^2/2}\int_{-\infty}^{|x|} e^{y^2/2} dy = e^{x^2/2}\int_{|x|}^{\infty} e^{-y^2/2} dy \quad \forall x \] The function $x \mapsto e^{x^2/2}\int_{|x|}^{\infty} e^{-y^2/2} dy$ attins its maximum at $x=0$. and \[ \int_0^\infty e^{y^2/2}dy = \sqrt{\pi/2} \] Which proves the result by taking the $\sup$. For the derivative, \[ f_h'(x) = x e^{x^2/2}\int_{-\infty}^x [h(y)-\mathbb{E}(h(N))]e^{-y^2/2}dy + 0 = h(x)-\mathbb{E}(h(N))+xe^{x^2/2}\int ... dy \] Now we take bounds \[ |f'_h(x)| \le 1 + |x|e^{x^2/2}\int_{-\infty}^x e^{-y^2/2} \] \begin{lemma}[Kol Topology > Cvg Cdf Topology] Kolmogorov Metric The topology induced by the Kolmogorov metric is strictly stronger than the topology induced by convergence of CDFs of random variables. \end{lemma}