# Comparing noisy truncated RV with noisy regular RV

For some reason, I'm having difficulties proving something that is intuitively simple. Assuming I have two a random variable, $$x$$ and $$x^{truncated}$$, where $$x^{truncated}$$ is the truncated version of $$x$$, i.e., it is distributed as, $$X^{truncated}\sim X | X \in[-b,b]$$ for some $$b>0$$, and $$x$$ is distributed symmetrically around zero. I want to prove that, $$Pr(X^{truncated}+\Lambda>a)a)$$ for some $$a>0$$ where $$\Lambda\sim N(0,\sigma^2)$$. I denote $$Q(a) = Pr(x>a)$$. Initialy I said that the $$x^{truncated}$$ has the following property, \begin{align} Pr(x^{truncated}>a) = \frac{Q(a)-Q(b)}{1-2Q(b)} \end{align} And then the proof is rather simple, but this is not true for $$a>b$$ or $$a<-b$$ and anything I try with this restriction is not working. The (wrong) proof assuming $$Pr(x^{truncated}a) = \frac{Q(a)-Q(b)}{1-2Q(b)}$$ for all $$a$$ is as follows,

\begin{align} &Pr\left(x^{tr}+\Lambda>a\right) - Pr\left(x+\Lambda>a\right)\\ &=\int_{-\infty}^{\infty} f_{\Lambda}(\lambda)\left[Pr\left(x^{tr}>\left(a-\lambda\right)\right) - Pr\left(x>(a-\lambda)\right)\right]d\lambda \label{eq:tr_vs_normal_err}\\ &= \int_{-\infty}^{\infty} f_{\Lambda}(\lambda)\left[\frac{Q\left(a-\lambda\right) - Q\left(b\right)}{1-2Q\left(b\right)} - Q\left(a-\lambda\right)\right]d\lambda \label{eq:def_C}\\ &=\int_{-\infty}^{\infty}f_{\Lambda}(\lambda)\frac{2Q\left(b\right)Q\left(a-\lambda\right)-Q\left(b\right)}{1-2Q\left(b\right)}d\lambda\\ &=\frac{Q\left(b\right)}{1-2Q\left(b\right)}\int_{-\infty}^{\infty} f_{\Lambda}\left(\lambda\right)\left(2Q\left(a-\lambda\right)-1\right)d\lambda = \frac{Q\left(b\right)}{1-2Q\left(b\right)}\int_{-\infty}^{\infty} f_{\Lambda}\left(\lambda\right)g\left(\lambda\right)d\lambda \end{align} we defined $$g(\lambda) =2Q\left(a-\lambda\right)-1$$. We note that $$g(\lambda)$$ is anti symmetric around $$a$$, i.e., $$g(a+x)=-g(a-x)$$, thus we can rewrite the integral using $$u=\lambda - a$$, \begin{align} \int_{-\infty}^{\infty} f_{\Lambda}\left(\lambda\right)g\left(\lambda\right)d\lambda &= \int_{-\infty}^{\infty} f_{\Lambda}\left(u+a\right)g\left(u+a\right)du\\ &=\int_0^{\infty}f_{\Lambda}\left(u+a\right)g\left(u+a\right)du + \int_{-\infty}^{0}f_{\Lambda}\left(u+a\right)g\left(u+a\right)du\\ &=\int_0^{\infty}f_{\Lambda}\left(u+a\right)g\left(u+a\right)du - \int_{-\infty}^{0}f_{\Lambda}\left(u+a\right)g\left(a-u\right)du\\ \end{align} In the right integral we can use $$x = -u$$. such that, \begin{align} \int_{-\infty}^{\infty} f_{\Lambda}\left(\lambda\right)g\left(\lambda\right)d\lambda &=\int_0^{\infty}f_{\Lambda}\left(u+a\right)g\left(u+a\right)du + \int_{\infty}^{0}f_{\Lambda}\left(-x+a\right)g\left(a+x\right)dx\\ &=\int_0^{\infty}f_{\Lambda}\left(u+a\right)g\left(u+a\right)du - \int_{0}^{\infty}f_{\Lambda}\left(-x+a\right)g\left(a+x\right)dx\\ &= \int_{0}^{\infty}g\left(a+x\right)\left(f_{\Lambda}\left(x+a\right) - f_{\Lambda}\left(-x+a\right)\right)dx \end{align} Finally, since $$\Lambda$$ is a Gaussian distribution with zero mean and $$a>0$$, we have, \begin{align} f_{\Lambda}(a+x) - f_{\Lambda}(a-x) < 0 \,\forall x>0 \end{align} Moreover, for $$x>0$$ we have $$g\left(x+a\right) =2Q\left(-x\right)-1>0$$, and thus the proof follows.

Does anybody have an idea that can help?

Thanks

$$\newcommand{\R}{\mathbb{R}}$$ Let $$Y:=X^{truncated}$$ and $$Z:=\Lambda$$. Then, by rescaling, without loss of generality $$Z\sim N(0,1)$$.
We have to show that $$P(Y+Z>a)\le P(X+Z>a)$$ for all $$a>0$$. Here we need to assume that $$Z$$ is independent of $$X$$ and $$Y$$ (I gather this assumption is missing in your question, and without this assumption the desired conclusion will not hold in general). In view of the symmetry of the distribution of $$X$$, \begin{align*} 2P(X+Z>a)&=2\int_\R P(X\in dx)P(Z>a-x) \\ & =\int_\R P(X\in dx)P(Z>a-x)+\int_\R P(-X\in dx)P(Z>a+x) \\ &= \int_\R P(X\in dx)g(x)=Eg(X), \end{align*} where $$\begin{equation*} g(x):=g_a(x):=P(Z>a-x)+P(Z>a+x). \end{equation*}$$ Similarly, $$2P(Y+Z>a)=Eg(Y)$$. Therefore and because the function $$g$$ is even, it suffices to show that $$Eg(|Y|)\le Eg(|X|)$$.
Since $$\begin{equation*} Eg(|Y|)=\frac{Eg(|X|)1_{|X|\le b}}{E1_{|X|\le b}}, \end{equation*}$$ the problem reduces to the inequality $$\begin{equation*} Eg(|X|)h(|X|)\le Eg(|X|)\,Eh(|X|), \tag{1} \end{equation*}$$ where $$h(x):=1_{|x|\le b}$$.
Note that the function $$g$$ is increasing on $$[0,\infty)$$, because $$\begin{equation*} g'(x)=\frac1{\sqrt{2\pi}}\,(e^{-(x-a)^2/2}-e^{-(x+a)^2/2})\ge0 \end{equation*}$$ if $$a,x\ge0$$. Also, it is obvious that the function $$h$$ is decreasing on $$[0,\infty)$$.
• Nice. At the end you also need that g is decreasing on $[-\infty,0]$ and h is increasing there right? Because x can also be negative – MRm Sep 16 '19 at 6:17
• @MRm : I am glad you liked the answer. We don't actually need to know how $g$ and $h$ behave on $(-\infty,0)$, because in (1) the value of the arguments of $g$ and $h$ is the value of $|X|$, which is $\ge0$. So, the answer is quite complete. – Iosif Pinelis Sep 16 '19 at 13:09