# Thread: Dependence and Independence Relations (Probability and Stats)

1. ## 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,

2. ## 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

$0=P[U=12 \wedge V=1] \neq P[U=12]P[V=1]=\dfrac{5}{396}$

3. ## Re: Dependence and Independence Relations (Probability and Stats)

Originally Posted by SDAS
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.

4. ## Re: Dependence and Independence Relations (Probability and Stats)

Originally Posted by romsek
One example should suffice to disprove independence.
$P[V=1]=\dfrac{5}{11}$
@romsek $\Large V\ne 1$ It must be $\large 2\le V\le 12$.

5. ## 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?

6. ## Re: Dependence and Independence Relations (Probability and Stats)

Originally Posted by Plato
@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.

7. ## Re: Dependence and Independence Relations (Probability and Stats)

Originally Posted by SDAS
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: $\{(1,5)\}$

8. ## 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.

9. ## Re: Dependence and Independence Relations (Probability and Stats)

Originally Posted by romsek
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.

10. ## Re: Dependence and Independence Relations (Probability and Stats)

Originally Posted by SDAS
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 $\mathscr{P}(V=9\wedge U=-1)=\dfrac{1}{36}$

11. ## 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.