Finding $\mathbb{P}(|U|Let $X\perp Y$ with $X,Y\sim N(0,1)$. Let $U=\frac{(X+Y)}{\sqrt{2}}$ and $V=\frac{(X-Y)}{\sqrt{2}}$.

*

*Find the law of $(U,V)$. What is the value of $\mathbb{E}(X)$, $Var(U)$ and $Cov(U,V)$?

$\rightarrow \left\{\begin{matrix}
\frac{x+y}{\sqrt{2}}=u\\ \frac{x-y}{\sqrt{2}}=v
\end{matrix}\right. \Rightarrow \left\{\begin{matrix}
y=\frac{\sqrt{2}u-\sqrt{2}v}{2}\\ x=\frac{\sqrt{2}u+\sqrt{2}v}{2}
\end{matrix}\right.$ with $|J|=1\Rightarrow f_{UV}(u,v)=f_X(\frac{(\sqrt{2}u+\sqrt{2}v)}{2})f_Y(\frac{(\sqrt{2}u-\sqrt{2}v)}{2})|J|=\frac{1}{2\pi}e^{-\frac{(u^2+v^2)}{2}}\Rightarrow U\sim N(0,1)$ and $V\sim N(0,1)$; $\mathbb{E}(U)=0,Var(U)=1,Cov(U,V)=0\Rightarrow U\perp V$


*Calculate $\mathbb{P}(|U|<z\sqrt{2},|V|<z\sqrt{2})$.

$\rightarrow \mathbb{P}(|U|<z\sqrt{2},|V|<z\sqrt{2})=\mathbb{P}(-z\sqrt{2}<U,V<z\sqrt{2})=$
$\frac{1}{2\pi}\int_{-z\sqrt{2}}^{z\sqrt{2}}e^{-\frac{v^2}{2}}[\int_{-z\sqrt{2}}^{z\sqrt{2}}e^{-\frac{u^2}{2}}du]dv$
but now I'm stuck. How can I solve this integral?


*$\mathbb{P}(X+Y<z|X>0,Y>0)$.


For point 3) I have no idea.  Any suggests? Thanks in advance for any help!
 A: hint:observe that $cov(U,V)=0$ therefore $U$ and $V$ are independent random variable. So use that fact to solve point (2).
and for point(3), use result of total probability,
$$P(X+Y<z)= P(X+Y<z,A,B)+P(X+Y<z,A,B^c)+P(X+Y<z,A^c,B)+P(X+Y<z,A^c,B^c)$$
where $A=\{X>0\}$ and $B=\{Y>0\}$.
In the right-hand side all four probabilities are identical so you have to compute only $P(X+Y>z)$ and $P(X>0)$ and $P(Y>0)$.
A: In general, if $W\sim\mathcal N(\mu,\sigma^2)$ then $W$ has moment generating function
$$
\varphi_W(\theta) := \mathbb E[e^{t W}] = \int_{-\infty}^\infty \frac1{\sqrt{2\pi\sigma^2}} e^{-\frac12\left(\frac{w-\mu}{\sigma}\right)^2}e^{wt} \ \mathsf dw = e^{\mu\theta +\frac12 \sigma^2t^2}.
$$
Also, if $U\sim\mathcal N(\mu_U,\sigma_U^2)$ and $V\sim\mathcal N(\mu_V, \sigma_V^2)$ are independent, then $$U+V\sim\mathcal N(\mu_U+\mu_V,\sigma_U^2+\sigma_V^2)$$ and $$U-V\sim\mathcal N(\mu_U-\mu_V,\sigma_U^2+\sigma_V^2).$$ Recall too that for any random variable $Z$ with finite variance, we have for all $c\in\mathbb R$ $$\mathrm{Var}(cW) = c^2\mathrm{Var}(W). $$ It follows then that $U:=\frac{X+Y}{\sqrt 2}$ and $V:=\frac{X-Y}{\sqrt 2}$ have $\mathcal N(0,1)$ distribution, and so $\varphi_U(s) = e^{\frac12 s^2}$ and $\varphi_V(t)= e^{\frac12 t^2}$. Computing the joint moment generating function of $(U,V)$, we have
\begin{align}
\varphi_{(U,V)}(s,t) &= \mathbb E\left[e^{s\left(\frac{X+Y}{\sqrt 2}\right)+t\left(\frac{X-Y}{\sqrt 2}\right)}\right]\\
&= \mathbb E\left[e^{(s+t)\frac X{\sqrt 2}}e^{(s-t)\frac Y{\sqrt 2}} \right]\\
&= \mathbb E\left[e^{(s+t)\frac X{\sqrt 2}} \right]E\left[e^{(s-t)\frac Y{\sqrt 2}} \right]\\
&= \varphi_{\frac X{\sqrt 2}}(s+t)\varphi_{\frac Y{\sqrt 2}}(s-t)\\
&= e^{\frac14(s+t)^2}e^{\frac14(s-t)^2}\\
&= e^{\frac12 s^2}e^{\frac12 t^2}\\
&= \varphi_U(s)\varphi_V(t),
\end{align}
which shows that $U$ and $V$ are independent. Thus $(U,V)$ follows a multivariate normal distribution with density
$$
f_{(U,V)}(u,v) = f_U(u)f_V(v) = \frac1{2\pi}e^{-\frac12(u^2+v^2)}.
$$
The above computations make it clear that
$$
\mathbb E[U] = 0,\quad \mathrm{Var}(U) = 1,\quad \mathrm{Cov}(U,V) = 0.
$$
For $2.$ we make use of the fact that two random variables $X$ and $Y$ are independent if and only if for any pair of measurable real-valued functions $f$ and $g$, the random variables $f(X)$ and $g(Y)$ are independent. Here we take $f(\cdot)=g(\cdot)=|\cdot|$, and so $|U|$ and $|V|$ are independent. From this we have for $z>0$:
$$
\mathbb P(|U|<z\sqrt 2,|V|<z\sqrt 2) =\mathbb P(|U|<z\sqrt 2)\cdot \mathbb P(|V|<z\sqrt 2) = \mathbb P(|U|<z\sqrt 2)^2.
$$ (The probability is clearly zero if $z\leqslant 0$.) Now, recall that if $W\sim\mathcal N(\mu,\sigma^2)$, then the absolute value of $W$ has a folded normal distribution with density
$$
f_W(w) = \frac1{\sqrt{2\pi\sigma^2}}\left(e^{-\frac12\left(\frac{w-\mu}{\sigma}\right)^2}+ e^{-\frac12\left(\frac{w+\mu}{\sigma}\right)^2}\right)\cdot\mathsf 1_{(0\infty)}(w).
$$
Since $U$ and $V$ have standard normal distributions, it follows that the density of $|U|$ is given by
$$
f_{|U|}(u) = \sqrt{\frac{2}{\pi }} e^{-\frac{u^2}{2}}\cdot\mathsf 1_{(0,\infty)}(u).
$$
We compute the probability in question by integrating the density:
\begin{align}
\mathbb P(|U|<z\sqrt 2) &= \int_0^{z\sqrt 2} f_{|U|}(u)\ \mathsf du\\
&= \int_0^{z\sqrt 2}\sqrt{\frac 2\pi}e^{-\frac{u^2}{2}}\ \mathsf du\\
&= \mathrm{erf}(z),
\end{align}
where $\mathrm{erf}(\cdot)$ denotes the error function. Therefore we have
$$
\mathbb P(|U|<z\sqrt 2,|V|<z\sqrt 2) = \mathrm{erf}(z)^2.
$$
For $3.$ I will defer to this post: Conditional Distribution of The Sum of Two Standard Normal Random Variables
