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