Show that, $Z$ is $\mathcal N(0,1)$

Question

If $Y\sim\mathcal N(0,1)$ and let $a>0$. Let $$Z=\ \begin{cases} Y&\text{if } |Y|\le a\\ -Y &\text{if }|Y|> a\\ \end{cases}\ $$ Show that $Z\sim\mathcal N(0,1)$

I tried to prove it using the characteristic function

$$\phi_Z(u)=\phi_{(Y\mathbf1_{|Y|\le a}-Y\mathbf1_{|Y|>a })}(u)\overset{?}=\phi_{Y\mathbf1_{|Y|\le a}(u)}\phi_{-Y\mathbf1_{|Y|> a}}(u)$$

$$=$\phi_{Y\mathbf1_{|Y|\le a}(u)}\phi_{Y\mathbf1_{|Y|> a}}(-u)=\phi_{Y\mathbf1_{|Y|\le a}}(u)\overline{\phi_{Y\mathbf1_{|Y|> a}}(u)}$$

and since $Y$ is symmetric,which means that it is real-valued, we obtain;

$$\overline{\phi_{Y\mathbf1_{|Y|> a}}}(u)=\phi_{Y\mathbf1_{|Y|> a}}(u)$$

Hence;

$$\phi_{Y\mathbf1_{|Y|\le a}}(u)\phi_{Y\mathbf1_{|Y|> a}}(u)=\ \begin{cases} E(e^{iuY})E(e^{iu0})=E(e^{iuY})&\text{if } |Y|\le a\\ E(e^{iu0})E(e^{iuY})=E(e^{iuY})&\text{if }|Y|> a\\ \end{cases}\ $$

$\textbf{My Question is:}$

Is the first step allowed, are $Y\mathbf1_{|Y|\le a}$ and $-Y\mathbf1_{|Y|>a }$ independent ?

Answer 1

Let $h$ be an arbitrary bounded Borel function then $$\eqalign{ \mathbb{E}(h(Z))&=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+ \underbrace{\frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(-y)e^{-y^2/2}dy}_{y\leftarrow-y}\cr &=\frac{1}{\sqrt{2\pi}}\int_{|y|\leq a}h(y)e^{-y^2/2}dy+ \frac{1}{\sqrt{2\pi}}\int_{|y|>a}h(y)e^{-y^2/2}dy\cr &=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}h(y)e^{-y^2/2}dy\cr &=\mathbb{E}(h(Y)) } $$ It follows that $Z$ and $Y$ have the same distribution, since $h$ is arbitrary, and we are done.

Answer 2

$\textbf{My Question is:}$ (...) are $Y\mathbf1_{|Y|\le a}$ and $-Y\mathbf1_{|Y|>a }$ independent ?

Of course not. Let $U=Y\mathbf1_{|Y|\le a}$ and $V=-Y\mathbf1_{|Y|>a }$, then $[U=0]=[|Y|\gt a]\cup[Y=0]$, $[V=0]=[|Y|\leqslant a]$, and $[U=V=0]=[Y=0]$. What does all this tell you about the possibility that $(U,V)$ is independent?

An extended version of the result at the beginning of the post is the following.

Let $\theta:[0,\infty)\to\{-1,1\}$ denote any measurable function and $Y=SX$ where $(X,S)$ is independent with $X\geqslant0$ almost surely and $P(S=1)=P(S=-1)=\frac12$.

Then $\theta(|Y|)\cdot Y$ is distributed like $Y$.

Which in turn follows from the factoid below.

Assume that $P(S=1)=P(S=-1)=\frac12$ and that $T$ is independent of $S$ with $|T|=1$ almost surely.

Then $ST$ is distributed like $S$.