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Thread: Dependence and Independence Relations (Probability and Stats)

  1. #1
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    Dependence and Independence Relations (Probability and Stats)

    A fair, six-sided dice is rolled two times. Let X1 and X2 denote the number of points showing on the first and second rolls, respectively. Let U=X1-X2, and V=X1+X2.
    Show that U and V are not independent.

    I want to show that the probability of the intersection of two events, is not equal to the product of two individual probabilities.
    So, I am using this formula: P(U and V)= P(U)*P(V). I want to show that the left hand side of the formula does not hold.
    How do I go from there? I changed the U and V with the values but I am stuck from here. Can anyone please explain or direct me?

    Thanks,
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    Re: Dependence and Independence Relations (Probability and Stats)

    One example should suffice to disprove independence.

    $P[U=12 \wedge V=1]=0$ as $(6,6)$ is the only roll that produces $U=12$ and this produces $V=0$

    $P[U=12]=\dfrac {1}{36}$

    $P[V=1]=\dfrac{5}{11}$

    you can check this yourself, it's rolls $(6,5),(5,4),(4,3),(3,2),(2,1)$

    and clearly

    $0=P[U=12 \wedge V=1] \neq P[U=12]P[V=1]=\dfrac{5}{396}$
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    Re: Dependence and Independence Relations (Probability and Stats)

    Quote Originally Posted by SDAS View Post
    A fair, six-sided dice is rolled two times. Let X1 and X2 denote the number of points showing on the first and second rolls, respectively. Let U=X1-X2, and V=X1+X2.
    Show that U and V are not independent.
    \begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{(1  ,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2,  4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3,4  )}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4)  }&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}  &{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&  {(6,5)}&{(6,6)}\end{array}.

    $\begin{array}{*{20}{c}} U&|&{ - 5}&{ - 4}&{ - 3}&{ - 2}&{ - 1}&0&1&2&3&4&5 \\ \hline
    \# &|&1&2&3&4&5&6&5&4&3&2&1 \end{array}$

    $\begin{array}{*{20}{c}} V&|&2&3&4&5&6&7&8&9&{10}&{11}&{12} \\ \hline
    \# &|&1&2&3&4&5&6&5&4&3&2&1 \end{array}$

    Now that I have done the 'dirty' work, you proceed. Or tell us what you think.
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    Re: Dependence and Independence Relations (Probability and Stats)

    Quote Originally Posted by romsek View Post
    One example should suffice to disprove independence.
    $P[V=1]=\dfrac{5}{11}$
    @romsek $\Large V\ne 1$ It must be $\large 2\le V\le 12$.
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    Re: Dependence and Independence Relations (Probability and Stats)

    I think the final formula to prove the dependence relationship should be:

    P (U=12 intersect V=2)=0?
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    Re: Dependence and Independence Relations (Probability and Stats)

    Quote Originally Posted by Plato View Post
    @romsek $\Large V\ne 1$ It must be $\large 2\le V\le 12$.
    oh.. I got the two backwards.

    Just swap U for V in my answer then.
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    Re: Dependence and Independence Relations (Probability and Stats)

    Quote Originally Posted by SDAS View Post
    I think the final formula to prove the dependence relationship should be:
    P (U=12 intersect V=2)=0?
    First of all, it is impossible for $U=12$ in fact $-5\le U\le 5$. $U=X_1-X_2$ it is the value on the first die minus the value on the second die.
    $\mathscr{P}(U=-4)=\dfrac{2}{36}$ if one of these pairs appear: $\{(2,6),(1,5)\}$
    $\mathscr{P}(V=6)=\dfrac{5}{36}$ if one of these pairs appear: $\{(1,5),(5,1),(4,2),(2,4),(3,3)\}$
    $\mathscr{P}(U=-4\wedge V=6)=\dfrac{1}{36}$ if this pair appears: $\{(1,5)\}$
    So what is your conclusion?
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    Re: Dependence and Independence Relations (Probability and Stats)

    Conclusion: P(U intersects V)#P(U)*V
    Plato, so -5<=U<=5, and 2<=V<=12

    I am a bit confused by romsek's answers.
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    Re: Dependence and Independence Relations (Probability and Stats)

    Quote Originally Posted by romsek View Post
    One example should suffice to disprove independence.

    $P[V=12 \wedge U=1]=0$ as $(6,6)$ is the only roll that produces $V=12$ and this produces $U=0$

    $P[V=12]=\dfrac {1}{36}$

    $P[U=1]=\dfrac{5}{36}$

    you can check this yourself, it's rolls $(6,5),(5,4),(4,3),(3,2),(2,1)$

    and clearly

    $0=P[V=12 \wedge U=1] \neq P[V=12]P[U=1]=\dfrac {1}{36}\dfrac{5}{36} = \dfrac{5}{1296}$
    there, I've corrected the fact that I swapped U and V, and another small error.
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    Re: Dependence and Independence Relations (Probability and Stats)

    Quote Originally Posted by SDAS View Post
    Conclusion: P(U intersects V)#P(U)*V
    Plato, so -5<=U<=5, and 2<=V<=12
    I am a bit confused by romsek's answers.
    You'r confused by Romsek's posts? I am totally confused.
    Being charitable, I conclude that surely he misread or read too quickly the OP.

    I posted the outcome set.
    \begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{(1  ,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2,  4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3,4  )}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4)  }&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4)}  &{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)}&  {(6,5)}&{(6,6)}\end{array}

    There are thirty-six pairs. Each of those pairs contributes to both $U~\&~V$

    Picking a point at random, say $(4,5)$ then $U=-1~\&~V=9$ for that point.
    But of course, $V=9$ on the set $\{(3,6),(6,3),(4,5),(5,4)\}$ so by simple counting we know $\mathscr{P}(V=9)=\dfrac{4}{36}$

    And $U=-1$ on the set $\{(5,6),(2,3),(4,5),(3,4),(1,2)\}$ so by simple counting we know $\mathscr{P}(U=-1)=\dfrac{5}{36}$

    Now we see that $V=9\wedge U=-1$ on the set $\{(4,5)\}$ so by simple counting we know $\mathscr{P}(V=9\wedge U=-1)=\dfrac{1}{36}$
    Last edited by Plato; Sep 11th 2017 at 04:10 PM.
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  11. #11
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    Re: Dependence and Independence Relations (Probability and Stats)

    Thank you so much! I sincerely appreciate. I used the two tables you include in your first posts by picking random data and applying your examples and I have a good understanding of showing dependence and independence.
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