Mean number of replications in a birth-death process conditioned on extinction
$begingroup$
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:
probability-theory stochastic-processes markov-chains
asked Dec 18 '18 at 4:44
AlexAlex
1018
$endgroup$
|
show 3 more comments
$begingroup$
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:
probability-theory stochastic-processes markov-chains
asked Dec 18 '18 at 4:44
AlexAlex
1018
$endgroup$
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
|
show 3 more comments
$begingroup$
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:
probability-theory stochastic-processes markov-chains
asked Dec 18 '18 at 4:44
AlexAlex
1018
$endgroup$
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:
probability-theory stochastic-processes markov-chains
probability-theory stochastic-processes markov-chains
asked Dec 18 '18 at 4:44
AlexAlex
1018
asked Dec 18 '18 at 4:44
AlexAlex
1018
edited Dec 19 '18 at 20:43
Alex
asked Dec 18 '18 at 4:44
AlexAlex
1018
asked Dec 18 '18 at 4:44
AlexAlex
1018
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
|
show 3 more comments
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
|
show 3 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044781%2fmean-number-of-replications-in-a-birth-death-process-conditioned-on-extinction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3044781%2fmean-number-of-replications-in-a-birth-death-process-conditioned-on-extinction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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