Random variable equals to its opposite

Question

i'm working on an exercice on probability but i'm not really sure if i understood the problem so here is what I did in order to solve the problem :

let a vector $X=(X_1,X_2)$ and let an introduced bivariate version of logistic law introduced by Gumbel for joint distribution function : $F_{\textbf{X}}(x_1,x_2)=\frac{1}{1+exp(-x_{1})+exp(-x_{2})}$ for $x \in \mathbb{R^{2}} $

What is aked : Demonstrate that the probability of $X_1$ being opposite sign of $X_2$ is equal to $\frac{1}{3}$

What I did :

• 1st attempt : I calculated $F_X(x_1,x_2)=P(X_1\leq x_1,X_2\leq -x_1)=\frac{1}{1+exp(-x_1)+exp(x_1)}$

i ended up to get : $F_X(x_1,x_2)=\frac{exp(x_1)}{exp(2x_1)+exp(x_1)+1}$ which is not what i'm looking for so i tried another approach

• 2nd attempt : I'd like to calculate $P(X_1 + X_2 =0)$ as $X_1$ is opposite sign of $X_2$, but I'm a bit confused because for me, logistic law is supposed to be a continuous random variable and not a discrete one (as logistic law admits a density) so it meant for me to calculate the convolution of these 2 random variable in this way : $P(X_1+X_2\leq 0)$ but there again summing up to zero doesn't make sense in the formula of convolution product ...

I'm not looking for any solution but if possible to have a clue how to proceed next ?

Thank you in advance for your reading and your help

Answer 1

Thank you for your help, now here is my answer which solves the question :

$P(X_1>0>X_2)+P(X_2>0>X_1)=P(X_1>0,X_2<0)+P(X_2>0,X_1<0)$

with, $P(X_1>0,X_2<0)= P(X_2<0)-P(X_1<0,X_2<0)$ and, $P(X_2>0,X_1<0)=P(X_1<0)+P(X_2<0,X_1<0)$

with, $F_X(x)=\frac{1}{1+exp(-x)}$ for univariate logistic law and the Gumbel formula written in my first post for the bivariate one

so it equals : $P(X_1<0)+P(X_2<0)-2P(X_1<0,X_2<0)=\frac{1}{2}+\frac{1}{2}-2*\frac{1}{3}=\frac{1}{3}$

I hope this one is good, thank you again for your help!

Answer 2

$ P(X_1<0)=1/2$ $P( X_2>0|X_1<0)=\frac{ P(X_1<0,X_2>0)}{ P(X_1<0)}=\frac{P(X_1<0)-P(X_1<0,X_2<0)}{P(X_1<0)}$