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Joint cumulative distribution function

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Compute the joint cumulative distribution function $F_XY(x,y)$?
    2. Relevant equations
    The marginal distribution function $F_X(x)$
    \begin{equation}
    F_X(x)=P(X\leq x)=
    \begin{cases}
    0,x<0\\
    0.6,0\leq x<1\\
    1,x\geq 1
    \end{cases}
    \end{equation}
    and $F_Y(y)$
    \begin{equation}
    F_Y=
    \begin{cases}
    0,y<0\\
    0.3,0\leq y<1\\
    0.7,1\leq y <2\\
    1,y\geq 2
    \end{cases}
    \end{equation}
    3. The attempt at a solution
    For independent (I know the are not) random variables X and Y
    \begin{equation}
    F_XY(x,y)=F_X(x)F_Y(y)=\\
    [0.6u(x)+0.4u(x-1)][0.3u(y)+0.4u(y-1)+0.3u(y-2)]=\\
    0.6*0.3u(x)u(y)+0.6*0.4u(x)u(y-1)+0.6*0.3u(x)u(y-2)+0.4*0.3u(x-1)u(y)+0.4*0.4u(x-1)u(y-1)0.4*0.3u(x-1)u(y-2)=\\
    0.18u(x)u(y)+0.24u(x)u(y-1)+0.18u(x)u(y-2)+0.12u(x-1)u(y)+0.16u(x-1)u(y-1)+0.12u(x-1)u(y-2)
    \end{equation}
     
  2. jcsd
  3. Dec 9, 2015 #2

    ElijahRockers

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    Gold Member

    Why are these variables not independent?

    If they aren't, then Fxy is not equal to FxFy
     
  4. Dec 10, 2015 #3
    My teacher told they are not independent even though I wish they were :frown:
     
  5. Dec 10, 2015 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    If all you are told are the two marginals, then it is impossible to give the joint distribution. Are you not told anything else at all about the two random variables?
     
  6. Dec 10, 2015 #5

    WWGD

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    Gold Member

  7. Dec 10, 2015 #6
    Well, the whole question reads like in the attached picture but I already did the first part!
     

    Attached Files:

  8. Dec 10, 2015 #7

    Ray Vickson

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    Homework Helper

    Just apply the DEFINITION of the joint cdf ##F_{XY}(x,y)##. You will be able to present the results in a ##2 \times 3## table of ##F(x,y)## values, corresponding to ##x = 0,1## and ##y = 0,1,2##.
     
  9. Dec 10, 2015 #8
    I guess you mean
    \begin{equation}
    F_{XY}(x,y)=
    \begin{cases}
    (0.2+0.3+0.1)(0.2+0.1),x=0;y=0\\
    (0.2+0.1+0.1)(0.3+0.1),x=1;y=1\\
    0.2+0.1,y=2
    \end{cases}
    \end{equation}
     
  10. Dec 10, 2015 #9

    Ray Vickson

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    Science Advisor
    Homework Helper

    No, I do not mean that. For one thing, it is completely wrong.

    Let me repeat my previous question: what is the DEFINITION of ##F_{XY}(x,y)##?

    Expanded question: for a given pair ##(x,y)##, how would you compute that?
     
  11. Dec 14, 2015 #10
    I think i finally get it. For a given pair i would have that
    $$
    F_{XY}(x,y)=
    \begin{cases}
    0,x<0,y<0\\
    0.2+0.1,0\leq x<1,0\leq y<1\\
    0.2+0.1+0.3,0\leq x<1,1\leq y<2\\
    0.2+0.1+0.3+0.1,1\leq x<2,1\leq y<2\\
    0.2+0.1+0.3+0.1+0.2+0.1,1\leq x,2\leq y
    \end{cases}
    $$
    or
    $$
    F_{XY}(x,y)=
    \begin{cases}
    0,x<0,y<0\\
    0.3,0\leq x<1,0\leq y<1\\
    0.6,0\leq x<1,1\leq y<2\\
    0.7,1\leq x<2,1\leq y<2\\
    1,1\leq x,2\leq y
    \end{cases}
    $$
    Thanks.
     
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