Discrete Random Variable: Independent or Dependent?
Question: You are given a fair red die and a fair blue die. You roll each die once, independently of each other. Let $(i, j)$ be the outcome, where $i$ is the result of the red die and $j$ is the result of the blue die. Define:
$X = i+j$ and $Y = i-j$
Are these independent or not?
Attempt:
Just by looking at it, it seems that they are dependent on one another but I’m not sure how to prove it.
If I take, $i=6$ and $j=4$ then $X=10$, and $Y =2$.
I think I need to use this formula: $Pr$ ($X=x$ $land$ $Y=y$) $=$ $Pr(X=x) $ . $ Pr(Y=y)$
Wouldn’t the probability be $frac{1}{6}$ for each one? Not sure how to use the formula to prove dependence.
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 30 at 6:09
Toby
1577
add a comment |
Question: You are given a fair red die and a fair blue die. You roll each die once, independently of each other. Let $(i, j)$ be the outcome, where $i$ is the result of the red die and $j$ is the result of the blue die. Define:
$X = i+j$ and $Y = i-j$
Are these independent or not?
Attempt:
Just by looking at it, it seems that they are dependent on one another but I’m not sure how to prove it.
If I take, $i=6$ and $j=4$ then $X=10$, and $Y =2$.
I think I need to use this formula: $Pr$ ($X=x$ $land$ $Y=y$) $=$ $Pr(X=x) $ . $ Pr(Y=y)$
Wouldn’t the probability be $frac{1}{6}$ for each one? Not sure how to use the formula to prove dependence.
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 30 at 6:09
Toby
1577
add a comment |
Question: You are given a fair red die and a fair blue die. You roll each die once, independently of each other. Let $(i, j)$ be the outcome, where $i$ is the result of the red die and $j$ is the result of the blue die. Define:
$X = i+j$ and $Y = i-j$
Are these independent or not?
Attempt:
Just by looking at it, it seems that they are dependent on one another but I’m not sure how to prove it.
If I take, $i=6$ and $j=4$ then $X=10$, and $Y =2$.
I think I need to use this formula: $Pr$ ($X=x$ $land$ $Y=y$) $=$ $Pr(X=x) $ . $ Pr(Y=y)$
Wouldn’t the probability be $frac{1}{6}$ for each one? Not sure how to use the formula to prove dependence.
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 30 at 6:09
Toby
1577
Question: You are given a fair red die and a fair blue die. You roll each die once, independently of each other. Let $(i, j)$ be the outcome, where $i$ is the result of the red die and $j$ is the result of the blue die. Define:
$X = i+j$ and $Y = i-j$
Are these independent or not?
Attempt:
Just by looking at it, it seems that they are dependent on one another but I’m not sure how to prove it.
If I take, $i=6$ and $j=4$ then $X=10$, and $Y =2$.
I think I need to use this formula: $Pr$ ($X=x$ $land$ $Y=y$) $=$ $Pr(X=x) $ . $ Pr(Y=y)$
Wouldn’t the probability be $frac{1}{6}$ for each one? Not sure how to use the formula to prove dependence.
probability-theory discrete-mathematics probability-distributions random-variables
probability-theory discrete-mathematics probability-distributions random-variables
asked Nov 30 at 6:09
Toby
1577
asked Nov 30 at 6:09
Toby
1577
asked Nov 30 at 6:09
Toby
1577
asked Nov 30 at 6:09
Toby
1577
asked Nov 30 at 6:09
Toby
1577
1577
add a comment |
add a comment |
1 Answer
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If you want to show that two random variables are not independent, you would show that there are $x,y$ such that $Pr(X = x land Y = y) neq Pr(X = x) cdot Pr(Y = y)$. Here, there's a lot of choices you can take. For example, an easy one would be $x = 12, y = -5$. Note that both $X = 12$ and $Y = -5$ can happen with probability $frac{1}{36}$: the first requires rolling two sixes, and the second requires rolling a 1 then a 6. But it is impossible for both two happen together. Hence,
$$Pr(X = x land Y = y) = 0 neq frac{1}{36} times frac{1}{36} = Pr(X = x) cdot Pr(Y = y)$$
So the two are not independent. Of course, there are many choices of $x$ and $y$ where this relationship fails to holds; any such pair suffices.
answered Nov 30 at 6:22
platty
3,360320
I am a little confused about how they are impossible to roll together. Like, if the question stated that X=|i-j| and Y=max(i,j) with same conditions. If I take (i,j) to be (6,2), then I would have X=4, Y=2. To get X=4, I would need to roll to get two 2's and for Y I would need to roll two 1's. In this case, is it also impossible to get both of these to happen together? And this would also be dependent?
– Toby
Dec 1 at 0:49
add a comment |
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If you want to show that two random variables are not independent, you would show that there are $x,y$ such that $Pr(X = x land Y = y) neq Pr(X = x) cdot Pr(Y = y)$. Here, there's a lot of choices you can take. For example, an easy one would be $x = 12, y = -5$. Note that both $X = 12$ and $Y = -5$ can happen with probability $frac{1}{36}$: the first requires rolling two sixes, and the second requires rolling a 1 then a 6. But it is impossible for both two happen together. Hence,
$$Pr(X = x land Y = y) = 0 neq frac{1}{36} times frac{1}{36} = Pr(X = x) cdot Pr(Y = y)$$
So the two are not independent. Of course, there are many choices of $x$ and $y$ where this relationship fails to holds; any such pair suffices.
answered Nov 30 at 6:22
platty
3,360320
I am a little confused about how they are impossible to roll together. Like, if the question stated that X=|i-j| and Y=max(i,j) with same conditions. If I take (i,j) to be (6,2), then I would have X=4, Y=2. To get X=4, I would need to roll to get two 2's and for Y I would need to roll two 1's. In this case, is it also impossible to get both of these to happen together? And this would also be dependent?
– Toby
Dec 1 at 0:49
add a comment |
If you want to show that two random variables are not independent, you would show that there are $x,y$ such that $Pr(X = x land Y = y) neq Pr(X = x) cdot Pr(Y = y)$. Here, there's a lot of choices you can take. For example, an easy one would be $x = 12, y = -5$. Note that both $X = 12$ and $Y = -5$ can happen with probability $frac{1}{36}$: the first requires rolling two sixes, and the second requires rolling a 1 then a 6. But it is impossible for both two happen together. Hence,
$$Pr(X = x land Y = y) = 0 neq frac{1}{36} times frac{1}{36} = Pr(X = x) cdot Pr(Y = y)$$
So the two are not independent. Of course, there are many choices of $x$ and $y$ where this relationship fails to holds; any such pair suffices.
answered Nov 30 at 6:22
platty
3,360320
I am a little confused about how they are impossible to roll together. Like, if the question stated that X=|i-j| and Y=max(i,j) with same conditions. If I take (i,j) to be (6,2), then I would have X=4, Y=2. To get X=4, I would need to roll to get two 2's and for Y I would need to roll two 1's. In this case, is it also impossible to get both of these to happen together? And this would also be dependent?
– Toby
Dec 1 at 0:49
add a comment |
If you want to show that two random variables are not independent, you would show that there are $x,y$ such that $Pr(X = x land Y = y) neq Pr(X = x) cdot Pr(Y = y)$. Here, there's a lot of choices you can take. For example, an easy one would be $x = 12, y = -5$. Note that both $X = 12$ and $Y = -5$ can happen with probability $frac{1}{36}$: the first requires rolling two sixes, and the second requires rolling a 1 then a 6. But it is impossible for both two happen together. Hence,
$$Pr(X = x land Y = y) = 0 neq frac{1}{36} times frac{1}{36} = Pr(X = x) cdot Pr(Y = y)$$
So the two are not independent. Of course, there are many choices of $x$ and $y$ where this relationship fails to holds; any such pair suffices.
answered Nov 30 at 6:22
platty
3,360320
If you want to show that two random variables are not independent, you would show that there are $x,y$ such that $Pr(X = x land Y = y) neq Pr(X = x) cdot Pr(Y = y)$. Here, there's a lot of choices you can take. For example, an easy one would be $x = 12, y = -5$. Note that both $X = 12$ and $Y = -5$ can happen with probability $frac{1}{36}$: the first requires rolling two sixes, and the second requires rolling a 1 then a 6. But it is impossible for both two happen together. Hence,
$$Pr(X = x land Y = y) = 0 neq frac{1}{36} times frac{1}{36} = Pr(X = x) cdot Pr(Y = y)$$
So the two are not independent. Of course, there are many choices of $x$ and $y$ where this relationship fails to holds; any such pair suffices.
answered Nov 30 at 6:22
platty
3,360320
answered Nov 30 at 6:22
platty
3,360320
answered Nov 30 at 6:22
platty
3,360320
answered Nov 30 at 6:22
platty
3,360320
3,360320
I am a little confused about how they are impossible to roll together. Like, if the question stated that X=|i-j| and Y=max(i,j) with same conditions. If I take (i,j) to be (6,2), then I would have X=4, Y=2. To get X=4, I would need to roll to get two 2's and for Y I would need to roll two 1's. In this case, is it also impossible to get both of these to happen together? And this would also be dependent?
– Toby
Dec 1 at 0:49
add a comment |
I am a little confused about how they are impossible to roll together. Like, if the question stated that X=|i-j| and Y=max(i,j) with same conditions. If I take (i,j) to be (6,2), then I would have X=4, Y=2. To get X=4, I would need to roll to get two 2's and for Y I would need to roll two 1's. In this case, is it also impossible to get both of these to happen together? And this would also be dependent?
– Toby
Dec 1 at 0:49
I am a little confused about how they are impossible to roll together. Like, if the question stated that X=|i-j| and Y=max(i,j) with same conditions. If I take (i,j) to be (6,2), then I would have X=4, Y=2. To get X=4, I would need to roll to get two 2's and for Y I would need to roll two 1's. In this case, is it also impossible to get both of these to happen together? And this would also be dependent?
– Toby
Dec 1 at 0:49
I am a little confused about how they are impossible to roll together. Like, if the question stated that X=|i-j| and Y=max(i,j) with same conditions. If I take (i,j) to be (6,2), then I would have X=4, Y=2. To get X=4, I would need to roll to get two 2's and for Y I would need to roll two 1's. In this case, is it also impossible to get both of these to happen together? And this would also be dependent?
– Toby
Dec 1 at 0:49
add a comment |
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