1. ## [SOLVED] Jointly distribution

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

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

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

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

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

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

We know that $\{X>a\} \cup \{X \leq a\}=\Omega$
So
\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 :
$\mathbb{P}(Y\leq b)=\mathbb{P}([Y\leq b,X>a] \text{ or } [Y \leq b,X\leq a])$
But the two events $\{Y \leq b\} \cap \{X \leq a\}$ and $\{Y \leq b\} \cap \{X>a\}$ are disjoint (because {X>a} and {X<a} are disjoint)

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

Can you finish it ?

3. Originally Posted by Moo
Hello,

Let $\Omega$ be the probability space.

We know that $\{X>a\} \cup \{X \leq a\}=\Omega$
So
\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 :
$\mathbb{P}(Y\leq b)=\mathbb{P}([Y\leq b,X>a] \text{ or } [Y \leq b,X\leq a])$
But the two events $\{Y \leq b\} \cap \{X \leq a\}$ and $\{Y \leq b\} \cap \{X>a\}$ are disjoint (because {X>a} and {X<a} are disjoint)

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

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

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