Probability and independence of 2 gaussians

Question

Let X and Z two independent normal random variables centered reduced. I want to calculate $ P(X+Z<0,Z>0) $, so i have done : $$ P(X+Z<0,Z>0)=P(|Z|<|X|,Z>0,X<0) $$ And I am blocked here.

But the correction says only that it is equal to $ 1/8 $ because the r.vs are independants and centered (and no more details).

However my question is : Could we split like that $$ P(|Z|<|X|,Z>0,X<0)=P(|Z|<|X|)P(Z>0)P(X<0) $$ ? And if yes, why ?

Answer 1

Maybe a simulation will help you visualize the relationships among variables. I simulated 100,000 realizations of $X \sim Norm(0,1)$ and independently the same number of realizations of $Z \sim Norm(0,1)$ in R statistical software.

Then I plotted the points with $X + Z < 0$ in orange. The points of interest to you are the orange ones above the x-axis. (Of course, you can draw a similar sketch without any simulation, if you understand the symmetry of the bivariate uncorrelated standard normal distribution.)

m = 10^5;  x = rnorm(m);  z = rnorm(m)
plot(x, z, pch=".")
  cond = (x + z < 0)
  points(x[cond], z[cond], pch=".", col="orange")
  abline(h = 0, col="green", lwd=2)
  abline(v = 0, col="green", lwd=2)
mean(z > 0);  mean(x + z < 0)
## 0.49889  # aprx P(Z > 0) = 1/2
## 0.49951  # aprx P(X + Z < 0) = 1/2
mean(x + z < 0 & z > 0); 1/8 
## 0.12254  # aprx P(X + Z = 0, Z > 0) = 1/8
## 0.125

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{\sigma > 0}$, the answer is given by the following expression: \begin{align} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \bracks{{1 \over \root{2\pi}\sigma}\, \exp\pars{-\,{x^{2} \over 2\sigma^{2}}}} \bracks{{1 \over \root{2\pi}\sigma}\, \exp\pars{-\,{z^{2} \over 2\sigma^{2}}}}\bracks{x + z < 0}\bracks{z > 0} \dd x\,\dd z \\[5mm] = &\ {1 \over 2\pi\sigma^{2}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp\pars{-\,{{x^{2} + z^{2} \over 2\sigma^{2}}}} \bracks{0 < z < -x}\dd x\,\dd z \\[5mm] = &\ {1 \over \pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \expo{-x^{2}\ -\ z^{2}}\,\,\bracks{0 < \root{2}\sigma z < -\root{2}\sigma x} \dd x\,\dd z \\[5mm] = &\ {1 \over \pi}\int_{0}^{2\pi}\int_{0}^{\infty} \expo{-r^{2}}\,\,\bracks{0 < r\sin\pars{\theta} < -r\cos\pars{\theta}}r \,\dd r\,\dd\theta \\[5mm] = &\ {1 \over \pi}\int_{0}^{2\pi}\bracks{0 < \sin\pars{\theta} < -\cos\pars{\theta}}\ \underbrace{\int_{0}^{\infty}\expo{-r^{2}}r\,\dd r}_{\ds{1 \over 2}}\ \dd\theta = {1 \over 2\pi}\int_{0}^{\pi} \bracks{\sin\pars{\theta} < -\cos\pars{\theta}}\,\dd\theta \\[5mm] = &\ {1 \over 2\pi}\int_{3\pi/4}^{\pi}\,\dd\theta =\ \bbox[#ffe,5px,border:1px dotted navy]{\ds{1 \over 8}} \end{align}