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Math Help - [SOLVED] Jointly distribution

  1. #1
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    [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] ?
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  2. #2
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    Hello,
    Quote Originally Posted by noob mathematician View Post
    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}<br />
\{Y\leq b\} &=\{Y\leq b\} \cap \left\{\{X>a\} \cup \{X \leq a\}\right\} \\<br />
&=\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 ?
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    Let \Omega be the probability space.

    We know that \{X>a\} \cup \{X \leq a\}=\Omega
    So
    \begin{aligned}<br />
\{Y\leq b\} &=\{Y\leq b\} \cap \left\{\{X>a\} \cup \{X \leq a\}\right\} \\<br />
&=\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) ??
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  4. #4
    Moo
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    Quote Originally Posted by noob mathematician View Post
    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
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