Mixing
What is wrong with this solution of the extinction problem?
$begingroup$
I am considering a standard extinction problem, that is:
A bacterial colony consists of individual bacteria. One of the following
happens with each bacterium each second:
- The bacterium dies.
- The bacterium remains without a change.
- The bacterium splits in two bacteria.
- The bacterium splits in three bacteria.
Each of the above happens with equal probability $frac{1}{4}$.
Estimate probability that the colony that initially consist of one
bacterium will never die.
There is a well documented approach for solving this, e.g. https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch14.pdf,
http://math.uchicago.edu/~may/REU2015/REUPapers/Csernica.pdf.
This approach is working with probability that the colony has died by timestep $n$.
However there is a slightly variant of this approach that has to be wrong for a reason I will explain below.
That is, let $p=P(text{colony of 1 bacteria dies at any future timestep})$ (as opposed to "dies by timstep n"). Then we can write
begin{equation}
p = frac{1}{4} + frac{1}{4} p + frac{1}{4} p^2+ frac{1}{4} p^3
end{equation}
as colony of one bacteria can either instantly die, then remain unchanged, then can split into two bacteria and then they would both have to die independently, etc.
Then we could solve that for $p$, but we would, as per usual solution, get multiple positive roots $p_1 = sqrt{2} - 1$, $p_2 = 1$. The above doesn't explain which root to choose.
So clearly we actually we cannot write the above equation.
Why?
probability-theory markov-process
edited Jan 26 at 9:26
Christian Blatter
175k8115327
asked Jan 25 at 21:27
zen-devzen-dev
424
$endgroup$
add a comment |
$begingroup$
I am considering a standard extinction problem, that is:
A bacterial colony consists of individual bacteria. One of the following
happens with each bacterium each second:
- The bacterium dies.
- The bacterium remains without a change.
- The bacterium splits in two bacteria.
- The bacterium splits in three bacteria.
Each of the above happens with equal probability $frac{1}{4}$.
Estimate probability that the colony that initially consist of one
bacterium will never die.
There is a well documented approach for solving this, e.g. https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch14.pdf,
http://math.uchicago.edu/~may/REU2015/REUPapers/Csernica.pdf.
This approach is working with probability that the colony has died by timestep $n$.
However there is a slightly variant of this approach that has to be wrong for a reason I will explain below.
That is, let $p=P(text{colony of 1 bacteria dies at any future timestep})$ (as opposed to "dies by timstep n"). Then we can write
begin{equation}
p = frac{1}{4} + frac{1}{4} p + frac{1}{4} p^2+ frac{1}{4} p^3
end{equation}
as colony of one bacteria can either instantly die, then remain unchanged, then can split into two bacteria and then they would both have to die independently, etc.
Then we could solve that for $p$, but we would, as per usual solution, get multiple positive roots $p_1 = sqrt{2} - 1$, $p_2 = 1$. The above doesn't explain which root to choose.
So clearly we actually we cannot write the above equation.
Why?
probability-theory markov-process
edited Jan 26 at 9:26
Christian Blatter
175k8115327
asked Jan 25 at 21:27
zen-devzen-dev
424
$endgroup$
add a comment |
$begingroup$
I am considering a standard extinction problem, that is:
A bacterial colony consists of individual bacteria. One of the following
happens with each bacterium each second:
- The bacterium dies.
- The bacterium remains without a change.
- The bacterium splits in two bacteria.
- The bacterium splits in three bacteria.
Each of the above happens with equal probability $frac{1}{4}$.
Estimate probability that the colony that initially consist of one
bacterium will never die.
There is a well documented approach for solving this, e.g. https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch14.pdf,
http://math.uchicago.edu/~may/REU2015/REUPapers/Csernica.pdf.
This approach is working with probability that the colony has died by timestep $n$.
However there is a slightly variant of this approach that has to be wrong for a reason I will explain below.
That is, let $p=P(text{colony of 1 bacteria dies at any future timestep})$ (as opposed to "dies by timstep n"). Then we can write
begin{equation}
p = frac{1}{4} + frac{1}{4} p + frac{1}{4} p^2+ frac{1}{4} p^3
end{equation}
as colony of one bacteria can either instantly die, then remain unchanged, then can split into two bacteria and then they would both have to die independently, etc.
Then we could solve that for $p$, but we would, as per usual solution, get multiple positive roots $p_1 = sqrt{2} - 1$, $p_2 = 1$. The above doesn't explain which root to choose.
So clearly we actually we cannot write the above equation.
Why?
probability-theory markov-process
edited Jan 26 at 9:26
Christian Blatter
175k8115327
asked Jan 25 at 21:27
zen-devzen-dev
424
$endgroup$
I am considering a standard extinction problem, that is:
A bacterial colony consists of individual bacteria. One of the following
happens with each bacterium each second:
- The bacterium dies.
- The bacterium remains without a change.
- The bacterium splits in two bacteria.
- The bacterium splits in three bacteria.
Each of the above happens with equal probability $frac{1}{4}$.
Estimate probability that the colony that initially consist of one
bacterium will never die.
There is a well documented approach for solving this, e.g. https://www.math.ucdavis.edu/~gravner/MAT135B/materials/ch14.pdf,
http://math.uchicago.edu/~may/REU2015/REUPapers/Csernica.pdf.
This approach is working with probability that the colony has died by timestep $n$.
However there is a slightly variant of this approach that has to be wrong for a reason I will explain below.
That is, let $p=P(text{colony of 1 bacteria dies at any future timestep})$ (as opposed to "dies by timstep n"). Then we can write
begin{equation}
p = frac{1}{4} + frac{1}{4} p + frac{1}{4} p^2+ frac{1}{4} p^3
end{equation}
as colony of one bacteria can either instantly die, then remain unchanged, then can split into two bacteria and then they would both have to die independently, etc.
Then we could solve that for $p$, but we would, as per usual solution, get multiple positive roots $p_1 = sqrt{2} - 1$, $p_2 = 1$. The above doesn't explain which root to choose.
So clearly we actually we cannot write the above equation.
Why?
probability-theory markov-process
probability-theory markov-process
edited Jan 26 at 9:26
Christian Blatter
175k8115327
asked Jan 25 at 21:27
zen-devzen-dev
424
edited Jan 26 at 9:26
Christian Blatter
175k8115327
asked Jan 25 at 21:27
zen-devzen-dev
424
edited Jan 26 at 9:26
Christian Blatter
175k8115327
edited Jan 26 at 9:26
Christian Blatter
175k8115327
edited Jan 26 at 9:26
Christian Blatter
175k8115327
175k8115327
asked Jan 25 at 21:27
zen-devzen-dev
424
asked Jan 25 at 21:27
zen-devzen-dev
424
asked Jan 25 at 21:27
zen-devzen-dev
424
424
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The classical theory of branching processes says that the extinction probability is the smallest nonnegative fixed point of the generating function of the offspring distribution, which in your case is $f(x)=frac{1}{4}+frac{1}{4}x+frac{1}{4}x^2+frac{1}{4}x^3$. So, exactly as you have stated, the extinction probability satisfies $p=frac{1}{4}+frac{1}{4}p+frac{1}{4}p^2+frac{1}{4}p^3$. This equation will always have the solution $p=1$, but as you have found, there is also a smaller fixed point in this case: $p=sqrt{2}-1$. Therefore $p=sqrt{2}-1$ is the extinction probability in this case. However, I don't know of an elementary argument that tells you that the smaller root is correct without going through the generating function proof.
answered Jan 25 at 21:36
kccukccu
10.6k11229
$endgroup$
$begingroup$
This unfortunately answers different question. It is well known how to solve extinction problem in general, the question is what is wrong with the solution outlined above.
$endgroup$
– zen-dev
Jan 26 at 0:39
$begingroup$
Nothing is wrong with it, but without more work it doesn't tell you which value of $p$ is correct.
$endgroup$
– kccu
Jan 26 at 2:41
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
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active
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$begingroup$
The classical theory of branching processes says that the extinction probability is the smallest nonnegative fixed point of the generating function of the offspring distribution, which in your case is $f(x)=frac{1}{4}+frac{1}{4}x+frac{1}{4}x^2+frac{1}{4}x^3$. So, exactly as you have stated, the extinction probability satisfies $p=frac{1}{4}+frac{1}{4}p+frac{1}{4}p^2+frac{1}{4}p^3$. This equation will always have the solution $p=1$, but as you have found, there is also a smaller fixed point in this case: $p=sqrt{2}-1$. Therefore $p=sqrt{2}-1$ is the extinction probability in this case. However, I don't know of an elementary argument that tells you that the smaller root is correct without going through the generating function proof.
answered Jan 25 at 21:36
kccukccu
10.6k11229
$endgroup$
$begingroup$
This unfortunately answers different question. It is well known how to solve extinction problem in general, the question is what is wrong with the solution outlined above.
$endgroup$
– zen-dev
Jan 26 at 0:39
$begingroup$
Nothing is wrong with it, but without more work it doesn't tell you which value of $p$ is correct.
$endgroup$
– kccu
Jan 26 at 2:41
add a comment |
$begingroup$
The classical theory of branching processes says that the extinction probability is the smallest nonnegative fixed point of the generating function of the offspring distribution, which in your case is $f(x)=frac{1}{4}+frac{1}{4}x+frac{1}{4}x^2+frac{1}{4}x^3$. So, exactly as you have stated, the extinction probability satisfies $p=frac{1}{4}+frac{1}{4}p+frac{1}{4}p^2+frac{1}{4}p^3$. This equation will always have the solution $p=1$, but as you have found, there is also a smaller fixed point in this case: $p=sqrt{2}-1$. Therefore $p=sqrt{2}-1$ is the extinction probability in this case. However, I don't know of an elementary argument that tells you that the smaller root is correct without going through the generating function proof.
answered Jan 25 at 21:36
kccukccu
10.6k11229
$endgroup$
$begingroup$
This unfortunately answers different question. It is well known how to solve extinction problem in general, the question is what is wrong with the solution outlined above.
$endgroup$
– zen-dev
Jan 26 at 0:39
$begingroup$
Nothing is wrong with it, but without more work it doesn't tell you which value of $p$ is correct.
$endgroup$
– kccu
Jan 26 at 2:41
add a comment |
$begingroup$
The classical theory of branching processes says that the extinction probability is the smallest nonnegative fixed point of the generating function of the offspring distribution, which in your case is $f(x)=frac{1}{4}+frac{1}{4}x+frac{1}{4}x^2+frac{1}{4}x^3$. So, exactly as you have stated, the extinction probability satisfies $p=frac{1}{4}+frac{1}{4}p+frac{1}{4}p^2+frac{1}{4}p^3$. This equation will always have the solution $p=1$, but as you have found, there is also a smaller fixed point in this case: $p=sqrt{2}-1$. Therefore $p=sqrt{2}-1$ is the extinction probability in this case. However, I don't know of an elementary argument that tells you that the smaller root is correct without going through the generating function proof.
answered Jan 25 at 21:36
kccukccu
10.6k11229
$endgroup$
The classical theory of branching processes says that the extinction probability is the smallest nonnegative fixed point of the generating function of the offspring distribution, which in your case is $f(x)=frac{1}{4}+frac{1}{4}x+frac{1}{4}x^2+frac{1}{4}x^3$. So, exactly as you have stated, the extinction probability satisfies $p=frac{1}{4}+frac{1}{4}p+frac{1}{4}p^2+frac{1}{4}p^3$. This equation will always have the solution $p=1$, but as you have found, there is also a smaller fixed point in this case: $p=sqrt{2}-1$. Therefore $p=sqrt{2}-1$ is the extinction probability in this case. However, I don't know of an elementary argument that tells you that the smaller root is correct without going through the generating function proof.
answered Jan 25 at 21:36
kccukccu
10.6k11229
answered Jan 25 at 21:36
kccukccu
10.6k11229
answered Jan 25 at 21:36
kccukccu
10.6k11229
answered Jan 25 at 21:36
kccukccu
10.6k11229
10.6k11229
$begingroup$
This unfortunately answers different question. It is well known how to solve extinction problem in general, the question is what is wrong with the solution outlined above.
$endgroup$
– zen-dev
Jan 26 at 0:39
$begingroup$
Nothing is wrong with it, but without more work it doesn't tell you which value of $p$ is correct.
$endgroup$
– kccu
Jan 26 at 2:41
add a comment |
$begingroup$
This unfortunately answers different question. It is well known how to solve extinction problem in general, the question is what is wrong with the solution outlined above.
$endgroup$
– zen-dev
Jan 26 at 0:39
$begingroup$
Nothing is wrong with it, but without more work it doesn't tell you which value of $p$ is correct.
$endgroup$
– kccu
Jan 26 at 2:41
$begingroup$
This unfortunately answers different question. It is well known how to solve extinction problem in general, the question is what is wrong with the solution outlined above.
$endgroup$
– zen-dev
Jan 26 at 0:39
$begingroup$
This unfortunately answers different question. It is well known how to solve extinction problem in general, the question is what is wrong with the solution outlined above.
$endgroup$
– zen-dev
Jan 26 at 0:39
$begingroup$
Nothing is wrong with it, but without more work it doesn't tell you which value of $p$ is correct.
$endgroup$
– kccu
Jan 26 at 2:41
$begingroup$
Nothing is wrong with it, but without more work it doesn't tell you which value of $p$ is correct.
$endgroup$
– kccu
Jan 26 at 2:41
add a comment |
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