Mixing
Clustering computation in pair approximation model
$begingroup$
Let's consider a square lattice of cells. Each cell can be either occupied by a species (1 or 2) or be empty (0).
Each cell can be either in state 1, 2 or 0.
In the pair approximation model, I would like to compute the clustering of the
species, i.e. the clustering of the occupied cells ($+$). It is define as:
- $C_{++} = frac{q_{+|+}}{rho_{+}} = frac{rho_{++}}{rho_{+}^2}$
Where
$q_{+|+}$ is the conditional probability to find an occupied cell in the surrounding cells knowing that the focal cell is occupied
$rho_+$ is the density of occupied cells in the landscape, defines as: $rho_+ = rho_1 + rho_2$$rho_{++}$ is the density of occupied cell pairs in the landscape, defines as: $rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
$q_{i|j} = frac{rho_{ij}}{rho_{i}}$
One approach successfully describes the clustering but the second does not and I
do not find why.
A first approach (Works)
$q_{+|+} = frac{rho_{++}}{rho_{+}}$
$rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
Knowing that $rho_{12} = rho_{21}$:
- $q_{+|+} = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}}$
Hence:
begin{align}
C_{++} & = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}} times frac{1}{rho_{+}} \
& = frac{rho_{11} + 2rho_{12} + rho_{22}}{(rho_{1} + rho_{2})^2}
end{align}
This approach works, I have checked it by running simulation of cellular
automata.
A second approach (Do not works)
Let's define $q_{+|+}$.
It is the probability for one cell of state $1$ to be surrounded by occupied
cells + the probability for one cell of state $2$ to be surrounded by occupied
cells. It means that:
$$
q_{+|+} = q_{+|1} + q_{+|2}
$$
and q_{+|1} and q_{+|2} can be defined as:
- $q_{+|1} = q_{1|1} + q_{2|1}$
- $q_{+|2} = q_{1|2} + q_{2|2}$
So:
begin{align}
q_{+|+} & = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2} \
& = frac{rho_{11}}{rho_{1}} frac{rho_{12}}{rho_{1}} + frac{rho_{21}}{rho_{2}} + frac{rho_{22}}{rho_{2}} \
& = frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12})}{rho_{1}rho_{2}} + frac{rho_{1}(rho_{21} + rho_{22})}{rho_{1}rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12}) + rho_{1}(rho_{21} + rho_{22}) }{rho_{1}rho_{2}} \
end{align}
Obviously this formulation of $q_{+|+}$ is different for the first one. In
simulation, this formulation gives $C_{++}$ twice higher than with a first
approach.
I do not know what is the mistake in this approach but I guess that the following
assertion is false:
$$q_{+|+} = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2}$$
I would be very helpful for me to know why.
ordinary-differential-equations conditional-probability clustering cellular-automata
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
$endgroup$
add a comment |
$begingroup$
Let's consider a square lattice of cells. Each cell can be either occupied by a species (1 or 2) or be empty (0).
Each cell can be either in state 1, 2 or 0.
In the pair approximation model, I would like to compute the clustering of the
species, i.e. the clustering of the occupied cells ($+$). It is define as:
- $C_{++} = frac{q_{+|+}}{rho_{+}} = frac{rho_{++}}{rho_{+}^2}$
Where
$q_{+|+}$ is the conditional probability to find an occupied cell in the surrounding cells knowing that the focal cell is occupied
$rho_+$ is the density of occupied cells in the landscape, defines as: $rho_+ = rho_1 + rho_2$$rho_{++}$ is the density of occupied cell pairs in the landscape, defines as: $rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
$q_{i|j} = frac{rho_{ij}}{rho_{i}}$
One approach successfully describes the clustering but the second does not and I
do not find why.
A first approach (Works)
$q_{+|+} = frac{rho_{++}}{rho_{+}}$
$rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
Knowing that $rho_{12} = rho_{21}$:
- $q_{+|+} = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}}$
Hence:
begin{align}
C_{++} & = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}} times frac{1}{rho_{+}} \
& = frac{rho_{11} + 2rho_{12} + rho_{22}}{(rho_{1} + rho_{2})^2}
end{align}
This approach works, I have checked it by running simulation of cellular
automata.
A second approach (Do not works)
Let's define $q_{+|+}$.
It is the probability for one cell of state $1$ to be surrounded by occupied
cells + the probability for one cell of state $2$ to be surrounded by occupied
cells. It means that:
$$
q_{+|+} = q_{+|1} + q_{+|2}
$$
and q_{+|1} and q_{+|2} can be defined as:
- $q_{+|1} = q_{1|1} + q_{2|1}$
- $q_{+|2} = q_{1|2} + q_{2|2}$
So:
begin{align}
q_{+|+} & = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2} \
& = frac{rho_{11}}{rho_{1}} frac{rho_{12}}{rho_{1}} + frac{rho_{21}}{rho_{2}} + frac{rho_{22}}{rho_{2}} \
& = frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12})}{rho_{1}rho_{2}} + frac{rho_{1}(rho_{21} + rho_{22})}{rho_{1}rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12}) + rho_{1}(rho_{21} + rho_{22}) }{rho_{1}rho_{2}} \
end{align}
Obviously this formulation of $q_{+|+}$ is different for the first one. In
simulation, this formulation gives $C_{++}$ twice higher than with a first
approach.
I do not know what is the mistake in this approach but I guess that the following
assertion is false:
$$q_{+|+} = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2}$$
I would be very helpful for me to know why.
ordinary-differential-equations conditional-probability clustering cellular-automata
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
$endgroup$
add a comment |
$begingroup$
Let's consider a square lattice of cells. Each cell can be either occupied by a species (1 or 2) or be empty (0).
Each cell can be either in state 1, 2 or 0.
In the pair approximation model, I would like to compute the clustering of the
species, i.e. the clustering of the occupied cells ($+$). It is define as:
- $C_{++} = frac{q_{+|+}}{rho_{+}} = frac{rho_{++}}{rho_{+}^2}$
Where
$q_{+|+}$ is the conditional probability to find an occupied cell in the surrounding cells knowing that the focal cell is occupied
$rho_+$ is the density of occupied cells in the landscape, defines as: $rho_+ = rho_1 + rho_2$$rho_{++}$ is the density of occupied cell pairs in the landscape, defines as: $rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
$q_{i|j} = frac{rho_{ij}}{rho_{i}}$
One approach successfully describes the clustering but the second does not and I
do not find why.
A first approach (Works)
$q_{+|+} = frac{rho_{++}}{rho_{+}}$
$rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
Knowing that $rho_{12} = rho_{21}$:
- $q_{+|+} = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}}$
Hence:
begin{align}
C_{++} & = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}} times frac{1}{rho_{+}} \
& = frac{rho_{11} + 2rho_{12} + rho_{22}}{(rho_{1} + rho_{2})^2}
end{align}
This approach works, I have checked it by running simulation of cellular
automata.
A second approach (Do not works)
Let's define $q_{+|+}$.
It is the probability for one cell of state $1$ to be surrounded by occupied
cells + the probability for one cell of state $2$ to be surrounded by occupied
cells. It means that:
$$
q_{+|+} = q_{+|1} + q_{+|2}
$$
and q_{+|1} and q_{+|2} can be defined as:
- $q_{+|1} = q_{1|1} + q_{2|1}$
- $q_{+|2} = q_{1|2} + q_{2|2}$
So:
begin{align}
q_{+|+} & = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2} \
& = frac{rho_{11}}{rho_{1}} frac{rho_{12}}{rho_{1}} + frac{rho_{21}}{rho_{2}} + frac{rho_{22}}{rho_{2}} \
& = frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12})}{rho_{1}rho_{2}} + frac{rho_{1}(rho_{21} + rho_{22})}{rho_{1}rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12}) + rho_{1}(rho_{21} + rho_{22}) }{rho_{1}rho_{2}} \
end{align}
Obviously this formulation of $q_{+|+}$ is different for the first one. In
simulation, this formulation gives $C_{++}$ twice higher than with a first
approach.
I do not know what is the mistake in this approach but I guess that the following
assertion is false:
$$q_{+|+} = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2}$$
I would be very helpful for me to know why.
ordinary-differential-equations conditional-probability clustering cellular-automata
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
$endgroup$
Let's consider a square lattice of cells. Each cell can be either occupied by a species (1 or 2) or be empty (0).
Each cell can be either in state 1, 2 or 0.
In the pair approximation model, I would like to compute the clustering of the
species, i.e. the clustering of the occupied cells ($+$). It is define as:
- $C_{++} = frac{q_{+|+}}{rho_{+}} = frac{rho_{++}}{rho_{+}^2}$
Where
$q_{+|+}$ is the conditional probability to find an occupied cell in the surrounding cells knowing that the focal cell is occupied
$rho_+$ is the density of occupied cells in the landscape, defines as: $rho_+ = rho_1 + rho_2$$rho_{++}$ is the density of occupied cell pairs in the landscape, defines as: $rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
$q_{i|j} = frac{rho_{ij}}{rho_{i}}$
One approach successfully describes the clustering but the second does not and I
do not find why.
A first approach (Works)
$q_{+|+} = frac{rho_{++}}{rho_{+}}$
$rho_{++} = rho_{11} + rho_{12} + rho_{21} + rho_{22}$
Knowing that $rho_{12} = rho_{21}$:
- $q_{+|+} = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}}$
Hence:
begin{align}
C_{++} & = frac{rho_{11} + 2rho_{12} + rho_{22}}{rho_{1} + rho_{2}} times frac{1}{rho_{+}} \
& = frac{rho_{11} + 2rho_{12} + rho_{22}}{(rho_{1} + rho_{2})^2}
end{align}
This approach works, I have checked it by running simulation of cellular
automata.
A second approach (Do not works)
Let's define $q_{+|+}$.
It is the probability for one cell of state $1$ to be surrounded by occupied
cells + the probability for one cell of state $2$ to be surrounded by occupied
cells. It means that:
$$
q_{+|+} = q_{+|1} + q_{+|2}
$$
and q_{+|1} and q_{+|2} can be defined as:
- $q_{+|1} = q_{1|1} + q_{2|1}$
- $q_{+|2} = q_{1|2} + q_{2|2}$
So:
begin{align}
q_{+|+} & = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2} \
& = frac{rho_{11}}{rho_{1}} frac{rho_{12}}{rho_{1}} + frac{rho_{21}}{rho_{2}} + frac{rho_{22}}{rho_{2}} \
& = frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12})}{rho_{1}rho_{2}} + frac{rho_{1}(rho_{21} + rho_{22})}{rho_{1}rho_{2}} \
& = frac{rho_{2}(rho_{11} + rho_{12}) + rho_{1}(rho_{21} + rho_{22}) }{rho_{1}rho_{2}} \
end{align}
Obviously this formulation of $q_{+|+}$ is different for the first one. In
simulation, this formulation gives $C_{++}$ twice higher than with a first
approach.
I do not know what is the mistake in this approach but I guess that the following
assertion is false:
$$q_{+|+} = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2}$$
I would be very helpful for me to know why.
ordinary-differential-equations conditional-probability clustering cellular-automata
ordinary-differential-equations conditional-probability clustering cellular-automata
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
edited Oct 3 '18 at 12:20
Alain Danet
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
asked Oct 2 '18 at 13:03
Alain DanetAlain Danet
85
85
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Indeed, your assertion is wrong:
$$q_{+|+}
= frac{rho_{++}}{rho_{+}}
= frac{rho_{11} + rho_{12} + rho_{21} + rho_{22}}{rho_{1} + rho_{2}}
ne frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}}
= frac{rho_{+1}}{rho_{1}} + frac{rho_{+2}}{rho_{2}}
= q_{+|1} + q_{+|2}.
$$
In particular, if $rho_{1} = rho_{2}$, then $rho_{1} + rho_{2} = 2rho_{1} = 2rho_{2}$, and so the left hand side will work out to exactly $frac12$ times the right hand side.
(The same will also happen if $q_{+|1} = q_{+|2}$; in particular, that means that the assertion will always be off by exactly a factor of $frac12$ for well mixed systems, where $q_{a|b} = rho_a$ for all states $a$ and $b$. On the other extreme, if the states 1 and 2 are highly clustered such that $rho_{12} = rho_{21} approx 0$, you can basically have $q_{+|1}$ and $q_{+|2}$ take any arbitrary values independently of each other and have $q_{+|+}$ be anywhere in between them, depending on the ratio of $rho_1$ and $rho_2$.)
The source of your confusion seems to be a basic misunderstanding of conditional probabilities. In particular, while it's true that $mathrm{Pr}[A text{ or } B mid C] = mathrm{Pr}[A mid C] + mathrm{Pr}[B mid C]$ whenever $A$ and $B$ are mutually exclusive events, this additivity only holds when the probabilities are conditioned on the same event $C$. In deriving your incorrect formula, you've basically asserted that $mathrm{Pr}[C mid A text{ or } B] = mathrm{Pr}[C mid A] + mathrm{Pr}[C mid B]$, which does not hold except in degenerate cases.
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
$endgroup$
add a comment |
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$begingroup$
Indeed, your assertion is wrong:
$$q_{+|+}
= frac{rho_{++}}{rho_{+}}
= frac{rho_{11} + rho_{12} + rho_{21} + rho_{22}}{rho_{1} + rho_{2}}
ne frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}}
= frac{rho_{+1}}{rho_{1}} + frac{rho_{+2}}{rho_{2}}
= q_{+|1} + q_{+|2}.
$$
In particular, if $rho_{1} = rho_{2}$, then $rho_{1} + rho_{2} = 2rho_{1} = 2rho_{2}$, and so the left hand side will work out to exactly $frac12$ times the right hand side.
(The same will also happen if $q_{+|1} = q_{+|2}$; in particular, that means that the assertion will always be off by exactly a factor of $frac12$ for well mixed systems, where $q_{a|b} = rho_a$ for all states $a$ and $b$. On the other extreme, if the states 1 and 2 are highly clustered such that $rho_{12} = rho_{21} approx 0$, you can basically have $q_{+|1}$ and $q_{+|2}$ take any arbitrary values independently of each other and have $q_{+|+}$ be anywhere in between them, depending on the ratio of $rho_1$ and $rho_2$.)
The source of your confusion seems to be a basic misunderstanding of conditional probabilities. In particular, while it's true that $mathrm{Pr}[A text{ or } B mid C] = mathrm{Pr}[A mid C] + mathrm{Pr}[B mid C]$ whenever $A$ and $B$ are mutually exclusive events, this additivity only holds when the probabilities are conditioned on the same event $C$. In deriving your incorrect formula, you've basically asserted that $mathrm{Pr}[C mid A text{ or } B] = mathrm{Pr}[C mid A] + mathrm{Pr}[C mid B]$, which does not hold except in degenerate cases.
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
$endgroup$
add a comment |
$begingroup$
Indeed, your assertion is wrong:
$$q_{+|+}
= frac{rho_{++}}{rho_{+}}
= frac{rho_{11} + rho_{12} + rho_{21} + rho_{22}}{rho_{1} + rho_{2}}
ne frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}}
= frac{rho_{+1}}{rho_{1}} + frac{rho_{+2}}{rho_{2}}
= q_{+|1} + q_{+|2}.
$$
In particular, if $rho_{1} = rho_{2}$, then $rho_{1} + rho_{2} = 2rho_{1} = 2rho_{2}$, and so the left hand side will work out to exactly $frac12$ times the right hand side.
(The same will also happen if $q_{+|1} = q_{+|2}$; in particular, that means that the assertion will always be off by exactly a factor of $frac12$ for well mixed systems, where $q_{a|b} = rho_a$ for all states $a$ and $b$. On the other extreme, if the states 1 and 2 are highly clustered such that $rho_{12} = rho_{21} approx 0$, you can basically have $q_{+|1}$ and $q_{+|2}$ take any arbitrary values independently of each other and have $q_{+|+}$ be anywhere in between them, depending on the ratio of $rho_1$ and $rho_2$.)
The source of your confusion seems to be a basic misunderstanding of conditional probabilities. In particular, while it's true that $mathrm{Pr}[A text{ or } B mid C] = mathrm{Pr}[A mid C] + mathrm{Pr}[B mid C]$ whenever $A$ and $B$ are mutually exclusive events, this additivity only holds when the probabilities are conditioned on the same event $C$. In deriving your incorrect formula, you've basically asserted that $mathrm{Pr}[C mid A text{ or } B] = mathrm{Pr}[C mid A] + mathrm{Pr}[C mid B]$, which does not hold except in degenerate cases.
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
$endgroup$
add a comment |
$begingroup$
Indeed, your assertion is wrong:
$$q_{+|+}
= frac{rho_{++}}{rho_{+}}
= frac{rho_{11} + rho_{12} + rho_{21} + rho_{22}}{rho_{1} + rho_{2}}
ne frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}}
= frac{rho_{+1}}{rho_{1}} + frac{rho_{+2}}{rho_{2}}
= q_{+|1} + q_{+|2}.
$$
In particular, if $rho_{1} = rho_{2}$, then $rho_{1} + rho_{2} = 2rho_{1} = 2rho_{2}$, and so the left hand side will work out to exactly $frac12$ times the right hand side.
(The same will also happen if $q_{+|1} = q_{+|2}$; in particular, that means that the assertion will always be off by exactly a factor of $frac12$ for well mixed systems, where $q_{a|b} = rho_a$ for all states $a$ and $b$. On the other extreme, if the states 1 and 2 are highly clustered such that $rho_{12} = rho_{21} approx 0$, you can basically have $q_{+|1}$ and $q_{+|2}$ take any arbitrary values independently of each other and have $q_{+|+}$ be anywhere in between them, depending on the ratio of $rho_1$ and $rho_2$.)
The source of your confusion seems to be a basic misunderstanding of conditional probabilities. In particular, while it's true that $mathrm{Pr}[A text{ or } B mid C] = mathrm{Pr}[A mid C] + mathrm{Pr}[B mid C]$ whenever $A$ and $B$ are mutually exclusive events, this additivity only holds when the probabilities are conditioned on the same event $C$. In deriving your incorrect formula, you've basically asserted that $mathrm{Pr}[C mid A text{ or } B] = mathrm{Pr}[C mid A] + mathrm{Pr}[C mid B]$, which does not hold except in degenerate cases.
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
$endgroup$
Indeed, your assertion is wrong:
$$q_{+|+}
= frac{rho_{++}}{rho_{+}}
= frac{rho_{11} + rho_{12} + rho_{21} + rho_{22}}{rho_{1} + rho_{2}}
ne frac{rho_{11} + rho_{12}}{rho_{1}} + frac{rho_{21} + rho_{22}}{rho_{2}}
= frac{rho_{+1}}{rho_{1}} + frac{rho_{+2}}{rho_{2}}
= q_{+|1} + q_{+|2}.
$$
In particular, if $rho_{1} = rho_{2}$, then $rho_{1} + rho_{2} = 2rho_{1} = 2rho_{2}$, and so the left hand side will work out to exactly $frac12$ times the right hand side.
(The same will also happen if $q_{+|1} = q_{+|2}$; in particular, that means that the assertion will always be off by exactly a factor of $frac12$ for well mixed systems, where $q_{a|b} = rho_a$ for all states $a$ and $b$. On the other extreme, if the states 1 and 2 are highly clustered such that $rho_{12} = rho_{21} approx 0$, you can basically have $q_{+|1}$ and $q_{+|2}$ take any arbitrary values independently of each other and have $q_{+|+}$ be anywhere in between them, depending on the ratio of $rho_1$ and $rho_2$.)
The source of your confusion seems to be a basic misunderstanding of conditional probabilities. In particular, while it's true that $mathrm{Pr}[A text{ or } B mid C] = mathrm{Pr}[A mid C] + mathrm{Pr}[B mid C]$ whenever $A$ and $B$ are mutually exclusive events, this additivity only holds when the probabilities are conditioned on the same event $C$. In deriving your incorrect formula, you've basically asserted that $mathrm{Pr}[C mid A text{ or } B] = mathrm{Pr}[C mid A] + mathrm{Pr}[C mid B]$, which does not hold except in degenerate cases.
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
edited Feb 1 at 4:17
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
answered Feb 1 at 3:54
Ilmari KaronenIlmari Karonen
20.2k25186
20.2k25186
add a comment |
add a comment |
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