Let $X$ be a normal standard random variable and $a>0$ compute the characteristic function of $ Y:= X 1_{|X|\leq a}-X 1_{|X|>a}$

Question

Let $X$ be a normal standard random variable and $a>0$ compute the characteristic function of $$ Y:= X 1_{|X|\leq a}-X 1_{|X|>a}$$

For definition $$\varphi_Y (t)=E[e^{itY}]= \int_{-\infty}^{\infty} e^{it(X 1_{|x|\leq a}-X 1_{|x|>a})}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx$$ $$\Rightarrow \varphi_Y (t)=\int_{-a}^{a} e^{itx}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx+\int_{-\infty}^{-a} e^{-itx}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx+\int_{a}^{\infty}e^{-itx}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx$$

I try to compare these expresion with $$\varphi_X(t)=\int_{-a}^{a} e^{itx}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx+\int_{-\infty}^{-a} e^{itx}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx+\int_{a}^{\infty}e^{itx}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx$$ but i not sure to these is the right way any hint or help i will be very grateful.

Answer 1

You are on the right track. Just notice that if we make the substitution $x\mapsto -x$ in the second and third integrals, we have

\begin{align} \varphi_Y(t)&=\frac{1}{\sqrt{2\pi}}\int_{-a}^{a} e^{itx-\frac{x^2}{2}}dx+ \frac{1}{\sqrt{2\pi}}\int_{a}^{\infty} e^{itx-\frac{x^2}{2}}dx+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{-a}e^{itx-\frac{x^2}{2}}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{itx-\frac{x^2}{2}}dx\\ &=e^{-\frac{t^2}{2}}. \end{align}

So, in fact, we have that $Y$ has the same distribution as $X$, since we know that this is the characteristic function of a standard normal random variable.

Can you see why we do not have that $\varphi_Y(t)=\varphi_{X1_{|X|\leq a}}(t)\varphi_{X1_{|X|>a}}(-t)$?