Mean number of replications in a birth-death process conditioned on extinction












0












$begingroup$


Background: We begin with a population of one individual. Each individual has a probability $p$ to survive until it replicates into two independent and identical individuals.



Then the probability $pi$ that the population ultimately goes extinct can be found by first step analysis over the first event:
$$
pi=ppi^2+1-p quadRightarrowquad pi=begin{cases}1, & p<frac12 \ p^{-1}-1, & pgeqfrac12.end{cases}
$$



Likewise, if $p<frac12$ such that extinction is assured, then the average number of replications $R$ that occur prior to extinction can be found by the same approach:
$$
R=p(2R+1) quadRightarrowquad R=begin{cases}frac{1}{p^{-1}-2}, & p<frac12 \ infty, & pgeqfrac12,end{cases}
$$



where the average number of replications $R$ is infinite if the extinction probability $pi$ is nonzero. Instead, I would like to condition on extinction so that $R$ is always finite except at precisely $p=frac12$.



Question: For $p>frac12$, conditional on a population that ultimately goes extinct, what is the average number of replications $R$ that occur prior to extinction?



Numerics: In case it might be useful, I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values:



Table of Values






probability-theory stochastic-processes markov-chains







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    When $pgeqslantfrac12$, the process of parameter $p$ conditioned by its extinction has the distribution of the process of parameter $1-p$. Thus, for every $p>frac12$, $$E_p(Rmidtext{extinction})=E_{1-p}(R)=frac{1-p}{1-2(1-p)}=frac{1-p}{2p-1}$$
    $endgroup$
    – Did
    Dec 18 '18 at 7:51










  • $begingroup$
    "I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values" You really obtained these values of R by numerical simulation? Sorry but you have to show us how you did to obtain such a supernatural accuracy...
    $endgroup$
    – Did
    Dec 18 '18 at 7:52










  • $begingroup$
    Hi Did, thank you for your answer! Your answer agrees with the numerical results, so it seems like it is correct. Would you mind explaining how you knew that the result would be symmetric about $p=frac12$?
    $endgroup$
    – Alex
    Dec 18 '18 at 17:06










  • $begingroup$
    As for the accuracy of the numerical simulations, I used the following Matlab script, which runs in a matter of seconds: imgur.com/xjtCPjc
    $endgroup$
    – Alex
    Dec 18 '18 at 17:08












  • $begingroup$
    The result is not symmetric with respect to $p$, rather $P_p( midtext{extinction})=P_{1-p}$ for every $pgeqslantfrac12$.
    $endgroup$
    – Did
    Dec 18 '18 at 19:58
















0












$begingroup$


Background: We begin with a population of one individual. Each individual has a probability $p$ to survive until it replicates into two independent and identical individuals.



Then the probability $pi$ that the population ultimately goes extinct can be found by first step analysis over the first event:
$$
pi=ppi^2+1-p quadRightarrowquad pi=begin{cases}1, & p<frac12 \ p^{-1}-1, & pgeqfrac12.end{cases}
$$



Likewise, if $p<frac12$ such that extinction is assured, then the average number of replications $R$ that occur prior to extinction can be found by the same approach:
$$
R=p(2R+1) quadRightarrowquad R=begin{cases}frac{1}{p^{-1}-2}, & p<frac12 \ infty, & pgeqfrac12,end{cases}
$$



where the average number of replications $R$ is infinite if the extinction probability $pi$ is nonzero. Instead, I would like to condition on extinction so that $R$ is always finite except at precisely $p=frac12$.



Question: For $p>frac12$, conditional on a population that ultimately goes extinct, what is the average number of replications $R$ that occur prior to extinction?



Numerics: In case it might be useful, I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values:



Table of Values






probability-theory stochastic-processes markov-chains







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    When $pgeqslantfrac12$, the process of parameter $p$ conditioned by its extinction has the distribution of the process of parameter $1-p$. Thus, for every $p>frac12$, $$E_p(Rmidtext{extinction})=E_{1-p}(R)=frac{1-p}{1-2(1-p)}=frac{1-p}{2p-1}$$
    $endgroup$
    – Did
    Dec 18 '18 at 7:51










  • $begingroup$
    "I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values" You really obtained these values of R by numerical simulation? Sorry but you have to show us how you did to obtain such a supernatural accuracy...
    $endgroup$
    – Did
    Dec 18 '18 at 7:52










  • $begingroup$
    Hi Did, thank you for your answer! Your answer agrees with the numerical results, so it seems like it is correct. Would you mind explaining how you knew that the result would be symmetric about $p=frac12$?
    $endgroup$
    – Alex
    Dec 18 '18 at 17:06










  • $begingroup$
    As for the accuracy of the numerical simulations, I used the following Matlab script, which runs in a matter of seconds: imgur.com/xjtCPjc
    $endgroup$
    – Alex
    Dec 18 '18 at 17:08












  • $begingroup$
    The result is not symmetric with respect to $p$, rather $P_p( midtext{extinction})=P_{1-p}$ for every $pgeqslantfrac12$.
    $endgroup$
    – Did
    Dec 18 '18 at 19:58














0












0








0


1



$begingroup$


Background: We begin with a population of one individual. Each individual has a probability $p$ to survive until it replicates into two independent and identical individuals.



Then the probability $pi$ that the population ultimately goes extinct can be found by first step analysis over the first event:
$$
pi=ppi^2+1-p quadRightarrowquad pi=begin{cases}1, & p<frac12 \ p^{-1}-1, & pgeqfrac12.end{cases}
$$



Likewise, if $p<frac12$ such that extinction is assured, then the average number of replications $R$ that occur prior to extinction can be found by the same approach:
$$
R=p(2R+1) quadRightarrowquad R=begin{cases}frac{1}{p^{-1}-2}, & p<frac12 \ infty, & pgeqfrac12,end{cases}
$$



where the average number of replications $R$ is infinite if the extinction probability $pi$ is nonzero. Instead, I would like to condition on extinction so that $R$ is always finite except at precisely $p=frac12$.



Question: For $p>frac12$, conditional on a population that ultimately goes extinct, what is the average number of replications $R$ that occur prior to extinction?



Numerics: In case it might be useful, I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values:



Table of Values






probability-theory stochastic-processes markov-chains







share|cite|improve this question











$endgroup$




Background: We begin with a population of one individual. Each individual has a probability $p$ to survive until it replicates into two independent and identical individuals.



Then the probability $pi$ that the population ultimately goes extinct can be found by first step analysis over the first event:
$$
pi=ppi^2+1-p quadRightarrowquad pi=begin{cases}1, & p<frac12 \ p^{-1}-1, & pgeqfrac12.end{cases}
$$



Likewise, if $p<frac12$ such that extinction is assured, then the average number of replications $R$ that occur prior to extinction can be found by the same approach:
$$
R=p(2R+1) quadRightarrowquad R=begin{cases}frac{1}{p^{-1}-2}, & p<frac12 \ infty, & pgeqfrac12,end{cases}
$$



where the average number of replications $R$ is infinite if the extinction probability $pi$ is nonzero. Instead, I would like to condition on extinction so that $R$ is always finite except at precisely $p=frac12$.



Question: For $p>frac12$, conditional on a population that ultimately goes extinct, what is the average number of replications $R$ that occur prior to extinction?



Numerics: In case it might be useful, I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values:



Table of Values






probability-theory stochastic-processes markov-chains




probability-theory stochastic-processes markov-chains






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 20:43







Alex

















asked Dec 18 '18 at 4:44









AlexAlex

1018




1018








  • 1




    $begingroup$
    When $pgeqslantfrac12$, the process of parameter $p$ conditioned by its extinction has the distribution of the process of parameter $1-p$. Thus, for every $p>frac12$, $$E_p(Rmidtext{extinction})=E_{1-p}(R)=frac{1-p}{1-2(1-p)}=frac{1-p}{2p-1}$$
    $endgroup$
    – Did
    Dec 18 '18 at 7:51










  • $begingroup$
    "I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values" You really obtained these values of R by numerical simulation? Sorry but you have to show us how you did to obtain such a supernatural accuracy...
    $endgroup$
    – Did
    Dec 18 '18 at 7:52










  • $begingroup$
    Hi Did, thank you for your answer! Your answer agrees with the numerical results, so it seems like it is correct. Would you mind explaining how you knew that the result would be symmetric about $p=frac12$?
    $endgroup$
    – Alex
    Dec 18 '18 at 17:06










  • $begingroup$
    As for the accuracy of the numerical simulations, I used the following Matlab script, which runs in a matter of seconds: imgur.com/xjtCPjc
    $endgroup$
    – Alex
    Dec 18 '18 at 17:08












  • $begingroup$
    The result is not symmetric with respect to $p$, rather $P_p( midtext{extinction})=P_{1-p}$ for every $pgeqslantfrac12$.
    $endgroup$
    – Did
    Dec 18 '18 at 19:58














  • 1




    $begingroup$
    When $pgeqslantfrac12$, the process of parameter $p$ conditioned by its extinction has the distribution of the process of parameter $1-p$. Thus, for every $p>frac12$, $$E_p(Rmidtext{extinction})=E_{1-p}(R)=frac{1-p}{1-2(1-p)}=frac{1-p}{2p-1}$$
    $endgroup$
    – Did
    Dec 18 '18 at 7:51










  • $begingroup$
    "I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values" You really obtained these values of R by numerical simulation? Sorry but you have to show us how you did to obtain such a supernatural accuracy...
    $endgroup$
    – Did
    Dec 18 '18 at 7:52










  • $begingroup$
    Hi Did, thank you for your answer! Your answer agrees with the numerical results, so it seems like it is correct. Would you mind explaining how you knew that the result would be symmetric about $p=frac12$?
    $endgroup$
    – Alex
    Dec 18 '18 at 17:06










  • $begingroup$
    As for the accuracy of the numerical simulations, I used the following Matlab script, which runs in a matter of seconds: imgur.com/xjtCPjc
    $endgroup$
    – Alex
    Dec 18 '18 at 17:08












  • $begingroup$
    The result is not symmetric with respect to $p$, rather $P_p( midtext{extinction})=P_{1-p}$ for every $pgeqslantfrac12$.
    $endgroup$
    – Did
    Dec 18 '18 at 19:58








1




1




$begingroup$
When $pgeqslantfrac12$, the process of parameter $p$ conditioned by its extinction has the distribution of the process of parameter $1-p$. Thus, for every $p>frac12$, $$E_p(Rmidtext{extinction})=E_{1-p}(R)=frac{1-p}{1-2(1-p)}=frac{1-p}{2p-1}$$
$endgroup$
– Did
Dec 18 '18 at 7:51




$begingroup$
When $pgeqslantfrac12$, the process of parameter $p$ conditioned by its extinction has the distribution of the process of parameter $1-p$. Thus, for every $p>frac12$, $$E_p(Rmidtext{extinction})=E_{1-p}(R)=frac{1-p}{1-2(1-p)}=frac{1-p}{2p-1}$$
$endgroup$
– Did
Dec 18 '18 at 7:51












$begingroup$
"I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values" You really obtained these values of R by numerical simulation? Sorry but you have to show us how you did to obtain such a supernatural accuracy...
$endgroup$
– Did
Dec 18 '18 at 7:52




$begingroup$
"I have simulated this birth-death process numerically and obtained the following table of $p$ and $R$ values" You really obtained these values of R by numerical simulation? Sorry but you have to show us how you did to obtain such a supernatural accuracy...
$endgroup$
– Did
Dec 18 '18 at 7:52












$begingroup$
Hi Did, thank you for your answer! Your answer agrees with the numerical results, so it seems like it is correct. Would you mind explaining how you knew that the result would be symmetric about $p=frac12$?
$endgroup$
– Alex
Dec 18 '18 at 17:06




$begingroup$
Hi Did, thank you for your answer! Your answer agrees with the numerical results, so it seems like it is correct. Would you mind explaining how you knew that the result would be symmetric about $p=frac12$?
$endgroup$
– Alex
Dec 18 '18 at 17:06












$begingroup$
As for the accuracy of the numerical simulations, I used the following Matlab script, which runs in a matter of seconds: imgur.com/xjtCPjc
$endgroup$
– Alex
Dec 18 '18 at 17:08






$begingroup$
As for the accuracy of the numerical simulations, I used the following Matlab script, which runs in a matter of seconds: imgur.com/xjtCPjc
$endgroup$
– Alex
Dec 18 '18 at 17:08














$begingroup$
The result is not symmetric with respect to $p$, rather $P_p( midtext{extinction})=P_{1-p}$ for every $pgeqslantfrac12$.
$endgroup$
– Did
Dec 18 '18 at 19:58




$begingroup$
The result is not symmetric with respect to $p$, rather $P_p( midtext{extinction})=P_{1-p}$ for every $pgeqslantfrac12$.
$endgroup$
– Did
Dec 18 '18 at 19:58










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