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, $Pr(X^{truncated}+\Lambda>a)<Pr(X+\Lambda>a)$ 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
 A: $\newcommand{\R}{\mathbb{R}}$
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 $[0,\infty)$. 
Thus, (1) follows by the integral Chebyshev inequality (see e.g. inequality (1.1)), which completes the proof. 
