probability of multivariate

Question

given a joint cdf $$ \dfrac{1}{1+e^{-x}}\dfrac{1}{1+e^{-y}}$$ where $x\in \mathbb{R}$ and $y\in \mathbb{R}$. Find $P(X \geq 0, Y\geq 0)$. I take the integral and find that it diverges but the notes i have ask me to find an answer (Answer with a fraction.) am i not seeing something? Thanks :)!

Y

Answer 1

In your case $X,Y$ are independent:

$$F_{XY}=F_X F_Y$$

Thus

$$P(X>0,Y>0)=(1-F_X(0))(1-F_Y(0))=1/4$$

Answer 2

The expression is the cdf, NOT pdf. You don't integrate. The probability you want is one - probability that either is negative $=(1-\frac{1}{1+e^0})^2=\frac{1}{4}$.