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Iosif Pinelis
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$\newcommand{\R}{\mathbb{R}}$ 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*}\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.

$\newcommand{\R}{\mathbb{R}}$ 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, which completes the proof.

$\newcommand{\R}{\mathbb{R}}$ 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.

Source Link
Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229

$\newcommand{\R}{\mathbb{R}}$ 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, which completes the proof.