1. ## [SOLVED] Jointly distribution

Since $\displaystyle F(a,b)=P[X\leq a, Y\leq b]$

$\displaystyle P[X>a, Y>b]= 1-P[(X > a, Y > b)^c]$
$\displaystyle =1-[P(X\leq a) + P(Y \leq b) - P(X \leq a, Y \leq b)]$
$\displaystyle =1-F_X(a)- F_Y(b) + F(a,b)$

What about: $\displaystyle P[X>a, Y \leq b]$ ?

2. Hello,
Originally Posted by noob mathematician
Since $\displaystyle F(a,b)=P[X\leq a, Y\leq b]$

$\displaystyle P[X>a, Y>b]= 1-P[(X > a, Y > b)^c]$
$\displaystyle =1-[P(X\leq a) + P(Y \leq b) - P(X \leq a, Y \leq b)]$
$\displaystyle =1-F_X(a)- F_Y(b) + F(a,b)$

What about: $\displaystyle P[X>a, Y \leq b]$ ?
Let $\displaystyle \Omega$ be the probability space.

We know that $\displaystyle \{X>a\} \cup \{X \leq a\}=\Omega$
So
\displaystyle \begin{aligned} \{Y\leq b\} &=\{Y\leq b\} \cap \left\{\{X>a\} \cup \{X \leq a\}\right\} \\ &=\left\{\{Y \leq b\} \cap \{X>a\}\right\} \cup \left\{\{Y \leq b\} \cap \{X \leq a\}\right\} \end{aligned}

Hence :
$\displaystyle \mathbb{P}(Y\leq b)=\mathbb{P}([Y\leq b,X>a] \text{ or } [Y \leq b,X\leq a])$
But the two events $\displaystyle \{Y \leq b\} \cap \{X \leq a\}$ and $\displaystyle \{Y \leq b\} \cap \{X>a\}$ are disjoint (because {X>a} and {X<a} are disjoint)

Thus $\displaystyle \mathbb{P}(Y\leq b)=\mathbb{P}(Y\leq b,X>a)+\mathbb{P}(Y\leq b,X\leq a)$

Can you finish it ?

3. Originally Posted by Moo
Hello,

Let $\displaystyle \Omega$ be the probability space.

We know that $\displaystyle \{X>a\} \cup \{X \leq a\}=\Omega$
So
\displaystyle \begin{aligned} \{Y\leq b\} &=\{Y\leq b\} \cap \left\{\{X>a\} \cup \{X \leq a\}\right\} \\ &=\left\{\{Y \leq b\} \cap \{X>a\}\right\} \cup \left\{\{Y \leq b\} \cap \{X \leq a\}\right\} \end{aligned}

Hence :
$\displaystyle \mathbb{P}(Y\leq b)=\mathbb{P}([Y\leq b,X>a] \text{ or } [Y \leq b,X\leq a])$
But the two events $\displaystyle \{Y \leq b\} \cap \{X \leq a\}$ and $\displaystyle \{Y \leq b\} \cap \{X>a\}$ are disjoint (because {X>a} and {X<a} are disjoint)

Thus $\displaystyle \mathbb{P}(Y\leq b)=\mathbb{P}(Y\leq b,X>a)+\mathbb{P}(Y\leq b,X\leq a)$

Can you finish it ?
So $\displaystyle \mathbb{P}(Y\leq b,X>a)=\mathbb{P}(Y\leq b)-\mathbb{P}(Y\leq b,X\leq a)$
$\displaystyle =F_Y(b)-F(a,b)$ ??

4. Originally Posted by noob mathematician
So $\displaystyle \mathbb{P}(Y\leq b,X>a)=\mathbb{P}(Y\leq b)-\mathbb{P}(Y\leq b,X\leq a)$
$\displaystyle =F_Y(b)-F(a,b)$ ??
Yup. Nothing more, nothing less