Prove a random variable has normal distribution

Question

Let $X$ be a standard normal random variable and $a>0$ a constant. Define:

$Y = \begin{cases} \phantom{-}X & \text{if $\,|X| < a$}; \\ -X & \text{otherwise}. \end{cases}$

Show that $Y$ is a standard normal distribution and the vector $(X,Y)$ is not two-dimensional normal distributed.

I tried approaching this using the expectation saying for $g$ a measurable function \begin{align*} E[g(Y)] &= \int_{\mathbb{R}} g(y)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx \\ &= \int_{-\infty}^{-a} g(x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx + \int_{-a}^a g(-x)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx + \int_a^{\infty} g(x) \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} dx \end{align*}

as $x^2 = (-x)^2$ I could reason that $Y$ must have normal distribution, but this just doesn't seem clean enough.

As to the second question I thought I could express $(X,Y)$ as a function of $X$ and then: $E[g(X,Y)] = E[g(h(X))] = E[f(X)]$ where $f = g \circ h$ a measurable function at which point I was confused by the dimensions.

Answer 1

The distribution of a real-valued random variable $Y$ is determined by its cdf $F := \mathbb{P}(Y \leq y)$ (because sets of the form $\{(-\infty, y]\}$ generate the Borel $\sigma$-algebra on $\mathbb{R}$). Let $N$ denote a standard normal random variable, $Y = -X$ and $y \in \mathbb{R}$; there are three cases: (i) $y \leq -a$, (ii) $-a < y \leq a$, and (iii) $a < y$. Consider case (ii):

\begin{align*} P(Y \leq y) &= P(Y \leq -a) + P(-a < Y \leq y) \\ &= P(-X \leq -a) + P(-a < X \leq y) \\ &= P(X \geq a) + P(-a < N \leq y) \\ &= P(N \geq a) + P(-a < N \leq y) \\ &= P(N \leq -a) + P(-a < N \leq y) \\ &= P(N \leq y). \end{align*} The first equality follows by disjointness and finite additivity; the second because $\{\omega : Y \leq -a\} = \{ \omega : -X \leq -a\}$ and $\{-a < Y \leq y\} = \{-a < X \leq y\}$ for $-a < y \leq a$. The second-to-last line follows by symmetry of $N$.

The case (i) is easy and (iii) is symmetric.

For the second question, have you drawn the support of the measure on $\mathbb{R}^2$?

Answer 2

$\textbf{i)First part}$

$\Pr(Z\in A)=\Pr(Y\in A, |Y|\le a)+\Pr(-Y\in A, |Y|> a)$

Since $\displaystyle\Pr(-Y\in A, |Y|> a)=\Pr(-(-Y)\in A, |-Y|> a)=\Pr(Y\in A, |Y|> a)$

Hence

$\Pr(Z\in A)=\Pr(Y\in A, |Y|\le a)+\Pr(Y\in A, |Y|> a)=\Pr(Y\in A)$

$\textbf{ii)Second part}$

Recall that $Cov(X,Y)=E(XY)-E(X)E(Y)$

and we say that two random variables $X,Y$ are uncorrelated if $Cov(X,Y)=0$

this is equivalent to $E(XY)=E(X)E(Y)$, this is of course true if $X$ and $Y$ are independent, but the converse is wrong, however it is true for the Multivariate Normal (in this case Bivariate Normal)

$X,Y$ are standard normal $\Rightarrow E(X)=E(Y)=0$, set $f(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$

and $\displaystyle E(XY)=\int_{-a}^{a}x^2f(x)-\int_{-\infty}^{-a}x^2f(x)-\int_{a}^{\infty}x^2f(x)=1-4\int_{a}^{\infty}x^2f(x)$

Now you can choose $a>0$ such that the expression above is zero (http://inci.ca/ujb6c1_2vp)

which means that $X$ and $Y$ are independent, but this is false since;

$\Pr(X\in(-a,a);Y\in(-a,a))=\Pr(X\in(-a,a))\neq\Pr(X\in(-a,a))\Pr(Y\in(-a,a))$