Probability of finding nodes in a moving circle
Clash Royale CLAN TAG#URR8PPP
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This seems to be a simple problem to me, but am not getting right results.
As attached in the figure (not drawn to scale). A circle centered at $X$ (X in Blue) with initial position at time $t=t_0$ moves to next location (X in red) at $t=t_1$ (all within a square of sides $a$).
Some items (called 'nodes' henceforth) are distributed uniformly with density $lambda$ in the square (not shown in fig.). $X$ has to discover each item within its communication range. Hence every circle has $N = lambdapi r^2$ nodes, where $r$ is the radius around $X$. ($X$ is called master).
If $P_iN = NP$ (not to be confused by NP-hard etc.) are the nodes that $X$ discover within in communication range at the time $t=i$. Where $P$ is the probability that $X$ discovers a single node.
Then After time $T = t_0+t_1$ what is the probability of finding all discovered nodes.
My solution is:
$$P_T = P_0N + P_1N - P'_01N'$$
where $P'_01$ is the probability of nodes that might repeat in the shaded area.
I find this shaded area as $A = A_circle - A_Cresent = pi r^2 - fracpi r w2$.
I just want to know if my strategy is right. If someone is interested, I can share the formulas for $P$ as well.
So $N' = lambda times (pi r^2 - fracpi r w2)$
probability geometry geometric-probability
asked Aug 24 at 1:40
Kashan
383110
add a comment |Â
up vote
0
down vote
favorite
This seems to be a simple problem to me, but am not getting right results.
As attached in the figure (not drawn to scale). A circle centered at $X$ (X in Blue) with initial position at time $t=t_0$ moves to next location (X in red) at $t=t_1$ (all within a square of sides $a$).
Some items (called 'nodes' henceforth) are distributed uniformly with density $lambda$ in the square (not shown in fig.). $X$ has to discover each item within its communication range. Hence every circle has $N = lambdapi r^2$ nodes, where $r$ is the radius around $X$. ($X$ is called master).
If $P_iN = NP$ (not to be confused by NP-hard etc.) are the nodes that $X$ discover within in communication range at the time $t=i$. Where $P$ is the probability that $X$ discovers a single node.
Then After time $T = t_0+t_1$ what is the probability of finding all discovered nodes.
My solution is:
$$P_T = P_0N + P_1N - P'_01N'$$
where $P'_01$ is the probability of nodes that might repeat in the shaded area.
I find this shaded area as $A = A_circle - A_Cresent = pi r^2 - fracpi r w2$.
I just want to know if my strategy is right. If someone is interested, I can share the formulas for $P$ as well.
So $N' = lambda times (pi r^2 - fracpi r w2)$
probability geometry geometric-probability
asked Aug 24 at 1:40
Kashan
383110
You have to explain more carefully. $P_iN$ cannot be a probability, as it may be greater than 1. I guess $P_iN$ is the expected number of nodes that can be discovered. Also saying that each disc has $N=lambda pi r^2$ is a huge approximation (specially for low $lambda$). Lastly, I guess $P_T$ is once again an expected value. It would be really helpfull if you pose the problem and then show how you try to solve it. As it is written I find it very confusing.
– David Jaramillo
Aug 24 at 1:51
Ah right.... I will make this correction. It is not probability. But the number of discovered devices. That was typo on my side in the question. But the rest remain the same.
– Kashan
Aug 24 at 2:13
By the way. Since $lambda$ is the density of uniformly distributed nodes (as in PPP). Why do you say $N = lambda pi r^2$ is an approximation? as $r$ is constant, so does $lambda$. Am I missing something?
– Kashan
Aug 24 at 2:17
well that depends what you mean with uniformly distributed. For instance if that means they are kind of on a grid or something like that. then the circles may touch a couple more or a couple less depending on the position. It you mean that the probability that a given node is some patch of a given area is proportional to the area, then there is a chance that there are some overpopulated patch and some underpopulated. Due to pure chance. It can be seen that it is an approximation because $lambdapi r^2$ might not be an integer. I still find a little confusing but...
– David Jaramillo
Aug 24 at 2:26
I think I understand it now. What you computed is the expected value of nodes found. I think you are idea is correct (not sure how you compute it the area of the crescent). I also find quite confusing what you use time when, as I understand it, is just two moments when you make the measurements.
– David Jaramillo
Aug 24 at 2:29
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This seems to be a simple problem to me, but am not getting right results.
As attached in the figure (not drawn to scale). A circle centered at $X$ (X in Blue) with initial position at time $t=t_0$ moves to next location (X in red) at $t=t_1$ (all within a square of sides $a$).
Some items (called 'nodes' henceforth) are distributed uniformly with density $lambda$ in the square (not shown in fig.). $X$ has to discover each item within its communication range. Hence every circle has $N = lambdapi r^2$ nodes, where $r$ is the radius around $X$. ($X$ is called master).
If $P_iN = NP$ (not to be confused by NP-hard etc.) are the nodes that $X$ discover within in communication range at the time $t=i$. Where $P$ is the probability that $X$ discovers a single node.
Then After time $T = t_0+t_1$ what is the probability of finding all discovered nodes.
My solution is:
$$P_T = P_0N + P_1N - P'_01N'$$
where $P'_01$ is the probability of nodes that might repeat in the shaded area.
I find this shaded area as $A = A_circle - A_Cresent = pi r^2 - fracpi r w2$.
I just want to know if my strategy is right. If someone is interested, I can share the formulas for $P$ as well.
So $N' = lambda times (pi r^2 - fracpi r w2)$
probability geometry geometric-probability
asked Aug 24 at 1:40
Kashan
383110
This seems to be a simple problem to me, but am not getting right results.
As attached in the figure (not drawn to scale). A circle centered at $X$ (X in Blue) with initial position at time $t=t_0$ moves to next location (X in red) at $t=t_1$ (all within a square of sides $a$).
Some items (called 'nodes' henceforth) are distributed uniformly with density $lambda$ in the square (not shown in fig.). $X$ has to discover each item within its communication range. Hence every circle has $N = lambdapi r^2$ nodes, where $r$ is the radius around $X$. ($X$ is called master).
If $P_iN = NP$ (not to be confused by NP-hard etc.) are the nodes that $X$ discover within in communication range at the time $t=i$. Where $P$ is the probability that $X$ discovers a single node.
Then After time $T = t_0+t_1$ what is the probability of finding all discovered nodes.
My solution is:
$$P_T = P_0N + P_1N - P'_01N'$$
where $P'_01$ is the probability of nodes that might repeat in the shaded area.
I find this shaded area as $A = A_circle - A_Cresent = pi r^2 - fracpi r w2$.
I just want to know if my strategy is right. If someone is interested, I can share the formulas for $P$ as well.
So $N' = lambda times (pi r^2 - fracpi r w2)$
probability geometry geometric-probability
asked Aug 24 at 1:40
Kashan
383110
edited Aug 24 at 2:14
asked Aug 24 at 1:40
Kashan
383110
asked Aug 24 at 1:40
Kashan
383110
asked Aug 24 at 1:40
Kashan
383110
383110
You have to explain more carefully. $P_iN$ cannot be a probability, as it may be greater than 1. I guess $P_iN$ is the expected number of nodes that can be discovered. Also saying that each disc has $N=lambda pi r^2$ is a huge approximation (specially for low $lambda$). Lastly, I guess $P_T$ is once again an expected value. It would be really helpfull if you pose the problem and then show how you try to solve it. As it is written I find it very confusing.
– David Jaramillo
Aug 24 at 1:51
Ah right.... I will make this correction. It is not probability. But the number of discovered devices. That was typo on my side in the question. But the rest remain the same.
– Kashan
Aug 24 at 2:13
By the way. Since $lambda$ is the density of uniformly distributed nodes (as in PPP). Why do you say $N = lambda pi r^2$ is an approximation? as $r$ is constant, so does $lambda$. Am I missing something?
– Kashan
Aug 24 at 2:17
well that depends what you mean with uniformly distributed. For instance if that means they are kind of on a grid or something like that. then the circles may touch a couple more or a couple less depending on the position. It you mean that the probability that a given node is some patch of a given area is proportional to the area, then there is a chance that there are some overpopulated patch and some underpopulated. Due to pure chance. It can be seen that it is an approximation because $lambdapi r^2$ might not be an integer. I still find a little confusing but...
– David Jaramillo
Aug 24 at 2:26
I think I understand it now. What you computed is the expected value of nodes found. I think you are idea is correct (not sure how you compute it the area of the crescent). I also find quite confusing what you use time when, as I understand it, is just two moments when you make the measurements.
– David Jaramillo
Aug 24 at 2:29
add a comment |Â
You have to explain more carefully. $P_iN$ cannot be a probability, as it may be greater than 1. I guess $P_iN$ is the expected number of nodes that can be discovered. Also saying that each disc has $N=lambda pi r^2$ is a huge approximation (specially for low $lambda$). Lastly, I guess $P_T$ is once again an expected value. It would be really helpfull if you pose the problem and then show how you try to solve it. As it is written I find it very confusing.
– David Jaramillo
Aug 24 at 1:51
Ah right.... I will make this correction. It is not probability. But the number of discovered devices. That was typo on my side in the question. But the rest remain the same.
– Kashan
Aug 24 at 2:13
By the way. Since $lambda$ is the density of uniformly distributed nodes (as in PPP). Why do you say $N = lambda pi r^2$ is an approximation? as $r$ is constant, so does $lambda$. Am I missing something?
– Kashan
Aug 24 at 2:17
well that depends what you mean with uniformly distributed. For instance if that means they are kind of on a grid or something like that. then the circles may touch a couple more or a couple less depending on the position. It you mean that the probability that a given node is some patch of a given area is proportional to the area, then there is a chance that there are some overpopulated patch and some underpopulated. Due to pure chance. It can be seen that it is an approximation because $lambdapi r^2$ might not be an integer. I still find a little confusing but...
– David Jaramillo
Aug 24 at 2:26
I think I understand it now. What you computed is the expected value of nodes found. I think you are idea is correct (not sure how you compute it the area of the crescent). I also find quite confusing what you use time when, as I understand it, is just two moments when you make the measurements.
– David Jaramillo
Aug 24 at 2:29
You have to explain more carefully. $P_iN$ cannot be a probability, as it may be greater than 1. I guess $P_iN$ is the expected number of nodes that can be discovered. Also saying that each disc has $N=lambda pi r^2$ is a huge approximation (specially for low $lambda$). Lastly, I guess $P_T$ is once again an expected value. It would be really helpfull if you pose the problem and then show how you try to solve it. As it is written I find it very confusing.
– David Jaramillo
Aug 24 at 1:51
You have to explain more carefully. $P_iN$ cannot be a probability, as it may be greater than 1. I guess $P_iN$ is the expected number of nodes that can be discovered. Also saying that each disc has $N=lambda pi r^2$ is a huge approximation (specially for low $lambda$). Lastly, I guess $P_T$ is once again an expected value. It would be really helpfull if you pose the problem and then show how you try to solve it. As it is written I find it very confusing.
– David Jaramillo
Aug 24 at 1:51
Ah right.... I will make this correction. It is not probability. But the number of discovered devices. That was typo on my side in the question. But the rest remain the same.
– Kashan
Aug 24 at 2:13
Ah right.... I will make this correction. It is not probability. But the number of discovered devices. That was typo on my side in the question. But the rest remain the same.
– Kashan
Aug 24 at 2:13
By the way. Since $lambda$ is the density of uniformly distributed nodes (as in PPP). Why do you say $N = lambda pi r^2$ is an approximation? as $r$ is constant, so does $lambda$. Am I missing something?
– Kashan
Aug 24 at 2:17
By the way. Since $lambda$ is the density of uniformly distributed nodes (as in PPP). Why do you say $N = lambda pi r^2$ is an approximation? as $r$ is constant, so does $lambda$. Am I missing something?
– Kashan
Aug 24 at 2:17
well that depends what you mean with uniformly distributed. For instance if that means they are kind of on a grid or something like that. then the circles may touch a couple more or a couple less depending on the position. It you mean that the probability that a given node is some patch of a given area is proportional to the area, then there is a chance that there are some overpopulated patch and some underpopulated. Due to pure chance. It can be seen that it is an approximation because $lambdapi r^2$ might not be an integer. I still find a little confusing but...
– David Jaramillo
Aug 24 at 2:26
well that depends what you mean with uniformly distributed. For instance if that means they are kind of on a grid or something like that. then the circles may touch a couple more or a couple less depending on the position. It you mean that the probability that a given node is some patch of a given area is proportional to the area, then there is a chance that there are some overpopulated patch and some underpopulated. Due to pure chance. It can be seen that it is an approximation because $lambdapi r^2$ might not be an integer. I still find a little confusing but...
– David Jaramillo
Aug 24 at 2:26
I think I understand it now. What you computed is the expected value of nodes found. I think you are idea is correct (not sure how you compute it the area of the crescent). I also find quite confusing what you use time when, as I understand it, is just two moments when you make the measurements.
– David Jaramillo
Aug 24 at 2:29
I think I understand it now. What you computed is the expected value of nodes found. I think you are idea is correct (not sure how you compute it the area of the crescent). I also find quite confusing what you use time when, as I understand it, is just two moments when you make the measurements.
– David Jaramillo
Aug 24 at 2:29
add a comment |Â
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You have to explain more carefully. $P_iN$ cannot be a probability, as it may be greater than 1. I guess $P_iN$ is the expected number of nodes that can be discovered. Also saying that each disc has $N=lambda pi r^2$ is a huge approximation (specially for low $lambda$). Lastly, I guess $P_T$ is once again an expected value. It would be really helpfull if you pose the problem and then show how you try to solve it. As it is written I find it very confusing.
– David Jaramillo
Aug 24 at 1:51
Ah right.... I will make this correction. It is not probability. But the number of discovered devices. That was typo on my side in the question. But the rest remain the same.
– Kashan
Aug 24 at 2:13
By the way. Since $lambda$ is the density of uniformly distributed nodes (as in PPP). Why do you say $N = lambda pi r^2$ is an approximation? as $r$ is constant, so does $lambda$. Am I missing something?
– Kashan
Aug 24 at 2:17
well that depends what you mean with uniformly distributed. For instance if that means they are kind of on a grid or something like that. then the circles may touch a couple more or a couple less depending on the position. It you mean that the probability that a given node is some patch of a given area is proportional to the area, then there is a chance that there are some overpopulated patch and some underpopulated. Due to pure chance. It can be seen that it is an approximation because $lambdapi r^2$ might not be an integer. I still find a little confusing but...
– David Jaramillo
Aug 24 at 2:26
I think I understand it now. What you computed is the expected value of nodes found. I think you are idea is correct (not sure how you compute it the area of the crescent). I also find quite confusing what you use time when, as I understand it, is just two moments when you make the measurements.
– David Jaramillo
Aug 24 at 2:29