why is it equal to $\int^{\infty}_{1} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx$ directly

Question

$1$.$X_n$ is iid Gaussian process and $U_n$ is iid binary random process with $Pr${$U_n=-1$}=$Pr${$U_n=1$}$=0.5$.$X_n$ and $U_n$ are independent

$2.Y_n=X_n+U_n$,and $\hat U_n=Q(Y_n),$where $ Q(Y_n)=\begin{cases} 1, & \text{$r \ge 0$} \\ -1, & \text{$r \lt 0$} \end{cases}$, and $\epsilon_n=\begin{cases} 0, & \text{$\hat U_n=U_n$} \\ 1, & \text{$\hat U_n \neq U_n$} \end{cases}$

\begin{align} E[\epsilon_n]&=1 \times P[\hat U_n\neq U_n] \\ &= P[\hat U_n=-1,U_n=1]+P[\hat U_n=1,U_n=-1]\\ &= P[U_n=1]P[X_n +U_n \lt 0|U_n=1]+P[U_n=-1]P[X_n +U_n \ge 0|U_n=-1]\\ &= \int^{\infty}_{1} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx \end{align}

I don't understand why $P[U_n=1]P[X_n +U_n \lt 0|U_n=1]+P[U_n=-1]P[X_n +U_n \ge 0|U_n=-1]$ can be equal to $\int^{\infty}_{1} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx$ directly,can anyone explain it to me ?

Answer 1

Let $X_n$ be Gaussian Process, mean $0$, variance $\sigma^2$. We have $P(X_n<-a)=P(X_n>a) \ \ \forall \ a\in \Bbb R$ by the symmetric property of the random variable $X_n$.

$$P[U_n=1]P[X_n +U_n \lt 0|U_n=1]+P[U_n=-1]P[X_n +U_n \ge 0|U_n=-1]$$ $$=0.5 \cdot P(X_n+1<0)+0.5*P(X_n-1\ge0)$$ $$=0.5 \cdot P(X_n<-1)+0.5 \cdot P(X_n\ge 1)$$ $$=0.5 \cdot P(X_n \ge 1)+0.5 \cdot P(X_n\ge 1) \quad \text {By symmetry}$$ $$=2 \cdot 0.5 \cdot P(X_n \ge 1)$$ $$=\int^{\infty}_{1} \frac{1}{\sqrt{2\pi \sigma^2}}e^{-\frac{x^2}{2\sigma^2}}dx \qquad \text {As required by definition of the CDF.}$$