Find three Poisson-distributed random variables, pairwise independent but not mutually independent
$begingroup$
I am asked to give an example of three Poisson-distributed random variables which are pairwise independent, but are not mutually independent.
I thought of the example about the Intersection where cars are coming from one side and they can go to one of three directions left, right or keep straight, with probabilities $p $, $q $ and $1-p-q$, respectively.
However, I find it hard to prove.
probability-theory poisson-distribution independence
edited Dec 7 '16 at 16:25
Did
248k23224461
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
$endgroup$
add a comment |
$begingroup$
I am asked to give an example of three Poisson-distributed random variables which are pairwise independent, but are not mutually independent.
I thought of the example about the Intersection where cars are coming from one side and they can go to one of three directions left, right or keep straight, with probabilities $p $, $q $ and $1-p-q$, respectively.
However, I find it hard to prove.
probability-theory poisson-distribution independence
edited Dec 7 '16 at 16:25
Did
248k23224461
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
$endgroup$
add a comment |
$begingroup$
I am asked to give an example of three Poisson-distributed random variables which are pairwise independent, but are not mutually independent.
I thought of the example about the Intersection where cars are coming from one side and they can go to one of three directions left, right or keep straight, with probabilities $p $, $q $ and $1-p-q$, respectively.
However, I find it hard to prove.
probability-theory poisson-distribution independence
edited Dec 7 '16 at 16:25
Did
248k23224461
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
$endgroup$
I am asked to give an example of three Poisson-distributed random variables which are pairwise independent, but are not mutually independent.
I thought of the example about the Intersection where cars are coming from one side and they can go to one of three directions left, right or keep straight, with probabilities $p $, $q $ and $1-p-q$, respectively.
However, I find it hard to prove.
probability-theory poisson-distribution independence
probability-theory poisson-distribution independence
edited Dec 7 '16 at 16:25
Did
248k23224461
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
edited Dec 7 '16 at 16:25
Did
248k23224461
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
edited Dec 7 '16 at 16:25
Did
248k23224461
edited Dec 7 '16 at 16:25
Did
248k23224461
edited Dec 7 '16 at 16:25
Did
248k23224461
248k23224461
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
asked Dec 7 '16 at 6:01
Don FanucciDon Fanucci
1,320420
1,320420
add a comment |
add a comment |
1 Answer
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Let $p(x,y,z)$ be the joint PMF for independent Poisson random variables with parameters $lambda_1, lambda_2, lambda_3$. I will modify this slightly to
obtain a joint PMF $q(x,y,z)$.
The requirement for the marginals $X,Y,Z$ to have the correct distributions and be pairwise independent is that
$$P(X=x,Y=y) = sum_{z=0}^infty q(x,y,z) = sum_{z=0}^infty p(x,y,z)$$
and similarly for the sums over $x$ and $y$. Of course we need all $q(x,y,z)$ to be nonnegative. We can manage this by starting with $q = p$ and slightly changing $8$ of the entries: add $epsilon$ to $q(0,0,1), q(0,1,0),q(1,0,0)$ and $q(1,1,1)$ and subtract $epsilon$ from $q(0,0,0), q(0,1,1), q(1,0,1)$ and $q(1,1,0)$, where $epsilon$ is small enough that all of these are still nonnegative.
Of course there is nothing very special about the Poisson distributions here: you could do something similar for any nontrivial discrete distributions.
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
$endgroup$
$begingroup$
Whoa ... if I understand this correctly, this procedure can be readily generalized to create a set of 4 random variables that are not mutually independent but contain independent triplets (and so on with higher orders)? Even though this is shuffling the probability mass around without any context (modeling), it seems like a really important stuff that should be included in the standard curriculum!
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:27
$begingroup$
I have two more questions: (1) the random variables $X,Y,Z,ldots$ etc don't have to be non-negative, right? I don't see a reason why they should. (2) This system procedure is applicable to a large class of continuous distributions as well, isn't it? Just recently there was this post (math.stackexchange.com/questions/2029257) with isolated ideas of different flavors, and maybe this method can be modified to appear over there as well?
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:30
$begingroup$
(1) Right. (2) Yes.
$endgroup$
– Robert Israel
Dec 8 '16 at 6:42
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $p(x,y,z)$ be the joint PMF for independent Poisson random variables with parameters $lambda_1, lambda_2, lambda_3$. I will modify this slightly to
obtain a joint PMF $q(x,y,z)$.
The requirement for the marginals $X,Y,Z$ to have the correct distributions and be pairwise independent is that
$$P(X=x,Y=y) = sum_{z=0}^infty q(x,y,z) = sum_{z=0}^infty p(x,y,z)$$
and similarly for the sums over $x$ and $y$. Of course we need all $q(x,y,z)$ to be nonnegative. We can manage this by starting with $q = p$ and slightly changing $8$ of the entries: add $epsilon$ to $q(0,0,1), q(0,1,0),q(1,0,0)$ and $q(1,1,1)$ and subtract $epsilon$ from $q(0,0,0), q(0,1,1), q(1,0,1)$ and $q(1,1,0)$, where $epsilon$ is small enough that all of these are still nonnegative.
Of course there is nothing very special about the Poisson distributions here: you could do something similar for any nontrivial discrete distributions.
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
$endgroup$
$begingroup$
Whoa ... if I understand this correctly, this procedure can be readily generalized to create a set of 4 random variables that are not mutually independent but contain independent triplets (and so on with higher orders)? Even though this is shuffling the probability mass around without any context (modeling), it seems like a really important stuff that should be included in the standard curriculum!
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:27
$begingroup$
I have two more questions: (1) the random variables $X,Y,Z,ldots$ etc don't have to be non-negative, right? I don't see a reason why they should. (2) This system procedure is applicable to a large class of continuous distributions as well, isn't it? Just recently there was this post (math.stackexchange.com/questions/2029257) with isolated ideas of different flavors, and maybe this method can be modified to appear over there as well?
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:30
$begingroup$
(1) Right. (2) Yes.
$endgroup$
– Robert Israel
Dec 8 '16 at 6:42
add a comment |
$begingroup$
Let $p(x,y,z)$ be the joint PMF for independent Poisson random variables with parameters $lambda_1, lambda_2, lambda_3$. I will modify this slightly to
obtain a joint PMF $q(x,y,z)$.
The requirement for the marginals $X,Y,Z$ to have the correct distributions and be pairwise independent is that
$$P(X=x,Y=y) = sum_{z=0}^infty q(x,y,z) = sum_{z=0}^infty p(x,y,z)$$
and similarly for the sums over $x$ and $y$. Of course we need all $q(x,y,z)$ to be nonnegative. We can manage this by starting with $q = p$ and slightly changing $8$ of the entries: add $epsilon$ to $q(0,0,1), q(0,1,0),q(1,0,0)$ and $q(1,1,1)$ and subtract $epsilon$ from $q(0,0,0), q(0,1,1), q(1,0,1)$ and $q(1,1,0)$, where $epsilon$ is small enough that all of these are still nonnegative.
Of course there is nothing very special about the Poisson distributions here: you could do something similar for any nontrivial discrete distributions.
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
$endgroup$
$begingroup$
Whoa ... if I understand this correctly, this procedure can be readily generalized to create a set of 4 random variables that are not mutually independent but contain independent triplets (and so on with higher orders)? Even though this is shuffling the probability mass around without any context (modeling), it seems like a really important stuff that should be included in the standard curriculum!
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:27
$begingroup$
I have two more questions: (1) the random variables $X,Y,Z,ldots$ etc don't have to be non-negative, right? I don't see a reason why they should. (2) This system procedure is applicable to a large class of continuous distributions as well, isn't it? Just recently there was this post (math.stackexchange.com/questions/2029257) with isolated ideas of different flavors, and maybe this method can be modified to appear over there as well?
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:30
$begingroup$
(1) Right. (2) Yes.
$endgroup$
– Robert Israel
Dec 8 '16 at 6:42
add a comment |
$begingroup$
Let $p(x,y,z)$ be the joint PMF for independent Poisson random variables with parameters $lambda_1, lambda_2, lambda_3$. I will modify this slightly to
obtain a joint PMF $q(x,y,z)$.
The requirement for the marginals $X,Y,Z$ to have the correct distributions and be pairwise independent is that
$$P(X=x,Y=y) = sum_{z=0}^infty q(x,y,z) = sum_{z=0}^infty p(x,y,z)$$
and similarly for the sums over $x$ and $y$. Of course we need all $q(x,y,z)$ to be nonnegative. We can manage this by starting with $q = p$ and slightly changing $8$ of the entries: add $epsilon$ to $q(0,0,1), q(0,1,0),q(1,0,0)$ and $q(1,1,1)$ and subtract $epsilon$ from $q(0,0,0), q(0,1,1), q(1,0,1)$ and $q(1,1,0)$, where $epsilon$ is small enough that all of these are still nonnegative.
Of course there is nothing very special about the Poisson distributions here: you could do something similar for any nontrivial discrete distributions.
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
$endgroup$
Let $p(x,y,z)$ be the joint PMF for independent Poisson random variables with parameters $lambda_1, lambda_2, lambda_3$. I will modify this slightly to
obtain a joint PMF $q(x,y,z)$.
The requirement for the marginals $X,Y,Z$ to have the correct distributions and be pairwise independent is that
$$P(X=x,Y=y) = sum_{z=0}^infty q(x,y,z) = sum_{z=0}^infty p(x,y,z)$$
and similarly for the sums over $x$ and $y$. Of course we need all $q(x,y,z)$ to be nonnegative. We can manage this by starting with $q = p$ and slightly changing $8$ of the entries: add $epsilon$ to $q(0,0,1), q(0,1,0),q(1,0,0)$ and $q(1,1,1)$ and subtract $epsilon$ from $q(0,0,0), q(0,1,1), q(1,0,1)$ and $q(1,1,0)$, where $epsilon$ is small enough that all of these are still nonnegative.
Of course there is nothing very special about the Poisson distributions here: you could do something similar for any nontrivial discrete distributions.
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
answered Dec 8 '16 at 2:52
Robert IsraelRobert Israel
323k23212466
323k23212466
$begingroup$
Whoa ... if I understand this correctly, this procedure can be readily generalized to create a set of 4 random variables that are not mutually independent but contain independent triplets (and so on with higher orders)? Even though this is shuffling the probability mass around without any context (modeling), it seems like a really important stuff that should be included in the standard curriculum!
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:27
$begingroup$
I have two more questions: (1) the random variables $X,Y,Z,ldots$ etc don't have to be non-negative, right? I don't see a reason why they should. (2) This system procedure is applicable to a large class of continuous distributions as well, isn't it? Just recently there was this post (math.stackexchange.com/questions/2029257) with isolated ideas of different flavors, and maybe this method can be modified to appear over there as well?
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:30
$begingroup$
(1) Right. (2) Yes.
$endgroup$
– Robert Israel
Dec 8 '16 at 6:42
add a comment |
$begingroup$
Whoa ... if I understand this correctly, this procedure can be readily generalized to create a set of 4 random variables that are not mutually independent but contain independent triplets (and so on with higher orders)? Even though this is shuffling the probability mass around without any context (modeling), it seems like a really important stuff that should be included in the standard curriculum!
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:27
$begingroup$
I have two more questions: (1) the random variables $X,Y,Z,ldots$ etc don't have to be non-negative, right? I don't see a reason why they should. (2) This system procedure is applicable to a large class of continuous distributions as well, isn't it? Just recently there was this post (math.stackexchange.com/questions/2029257) with isolated ideas of different flavors, and maybe this method can be modified to appear over there as well?
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:30
$begingroup$
(1) Right. (2) Yes.
$endgroup$
– Robert Israel
Dec 8 '16 at 6:42
$begingroup$
Whoa ... if I understand this correctly, this procedure can be readily generalized to create a set of 4 random variables that are not mutually independent but contain independent triplets (and so on with higher orders)? Even though this is shuffling the probability mass around without any context (modeling), it seems like a really important stuff that should be included in the standard curriculum!
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:27
$begingroup$
Whoa ... if I understand this correctly, this procedure can be readily generalized to create a set of 4 random variables that are not mutually independent but contain independent triplets (and so on with higher orders)? Even though this is shuffling the probability mass around without any context (modeling), it seems like a really important stuff that should be included in the standard curriculum!
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:27
$begingroup$
I have two more questions: (1) the random variables $X,Y,Z,ldots$ etc don't have to be non-negative, right? I don't see a reason why they should. (2) This system procedure is applicable to a large class of continuous distributions as well, isn't it? Just recently there was this post (math.stackexchange.com/questions/2029257) with isolated ideas of different flavors, and maybe this method can be modified to appear over there as well?
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:30
$begingroup$
I have two more questions: (1) the random variables $X,Y,Z,ldots$ etc don't have to be non-negative, right? I don't see a reason why they should. (2) This system procedure is applicable to a large class of continuous distributions as well, isn't it? Just recently there was this post (math.stackexchange.com/questions/2029257) with isolated ideas of different flavors, and maybe this method can be modified to appear over there as well?
$endgroup$
– Lee David Chung Lin
Dec 8 '16 at 5:30
$begingroup$
(1) Right. (2) Yes.
$endgroup$
– Robert Israel
Dec 8 '16 at 6:42
$begingroup$
(1) Right. (2) Yes.
$endgroup$
– Robert Israel
Dec 8 '16 at 6:42
add a comment |
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