Branching process and posterior distribution.












1












$begingroup$


I have the following branching process:



enter image description here



(I added the $Z$'s myself for the number of individuals in each gen, hope it's correct.)




Assume the offspring distribution is Poisson with expectation
$lambda$ and also assume that we use an improper prior
$pi(lambda)propto_{lambda}1/lambda$ for $lambda$. Using the
information in the figure, compute the posterior distribution for
$lambda.$




Using Baye's theorem with an observed count $k$, I get



$$pi(lambda|k)propto_{lambda}pi(k|lambda)pi(lambda)propto_{lambda}e^{-lambda}cdotlambda^{k}cdotfrac{1}{lambda}=e^{-lambda}lambda^{k-1}.$$



So we see that this posterior is $pi(lambda|k)simGamma(lambda;k,1).$



However, how do I know what $k$ is based on the figure?






statistics stochastic-processes markov-process







share|cite|improve this question









$endgroup$












  • $begingroup$
    Assuming the next generations of the tree are not represented in the figure (not that they are empty), the figure tells you that the numbers of offspring for this tree are, starting from the root, $$1, 2, 3, 2, 0, 0, 1, 4, 2$$ hence the posterior knowing this tree is $$pi(lambdamid k)proptopi(lambda)prod_kpi(kmidlambda)$$ that is, $$pi(lambdamid k)proptolambda^{-1}(e^{-lambda}lambda)(e^{-lambda}lambda^2)(e^{-lambda}lambda^3)(e^{-lambda}lambda^2)(e^{-lambda}lambda^0)(e^{-lambda}lambda^0)(e^{-lambda}lambda^1)(e^{-lambda}lambda^4)(e^{-lambda}lambda^2)$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20










  • $begingroup$
    ...or, more simply, $$pi(lambdamid k)propto e^{-9lambda}lambda^{14}$$ Assuming the next generations are empty, one arrives at $$pi(lambdamid k)propto e^{-16lambda}lambda^{14}$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20












  • $begingroup$
    @Did Thank you for the answer. Could you elaborate on what you mean by empty and non empty generation? I also don't see how you obtained the number of offsprings? How did you find $1,2,3,2,0,0,1,4,2$?
    $endgroup$
    – Parseval
    Dec 3 '18 at 15:50












  • $begingroup$
    I looked at each node and counted its descendants.
    $endgroup$
    – Did
    Dec 3 '18 at 16:43










  • $begingroup$
    @ Did Okay I understnad that now. So can one then say that $$pi(lambda|k)simGamma(lambda;15,9)?$$
    $endgroup$
    – Parseval
    Dec 3 '18 at 18:03


















1












$begingroup$


I have the following branching process:



enter image description here



(I added the $Z$'s myself for the number of individuals in each gen, hope it's correct.)




Assume the offspring distribution is Poisson with expectation
$lambda$ and also assume that we use an improper prior
$pi(lambda)propto_{lambda}1/lambda$ for $lambda$. Using the
information in the figure, compute the posterior distribution for
$lambda.$




Using Baye's theorem with an observed count $k$, I get



$$pi(lambda|k)propto_{lambda}pi(k|lambda)pi(lambda)propto_{lambda}e^{-lambda}cdotlambda^{k}cdotfrac{1}{lambda}=e^{-lambda}lambda^{k-1}.$$



So we see that this posterior is $pi(lambda|k)simGamma(lambda;k,1).$



However, how do I know what $k$ is based on the figure?






statistics stochastic-processes markov-process







share|cite|improve this question









$endgroup$












  • $begingroup$
    Assuming the next generations of the tree are not represented in the figure (not that they are empty), the figure tells you that the numbers of offspring for this tree are, starting from the root, $$1, 2, 3, 2, 0, 0, 1, 4, 2$$ hence the posterior knowing this tree is $$pi(lambdamid k)proptopi(lambda)prod_kpi(kmidlambda)$$ that is, $$pi(lambdamid k)proptolambda^{-1}(e^{-lambda}lambda)(e^{-lambda}lambda^2)(e^{-lambda}lambda^3)(e^{-lambda}lambda^2)(e^{-lambda}lambda^0)(e^{-lambda}lambda^0)(e^{-lambda}lambda^1)(e^{-lambda}lambda^4)(e^{-lambda}lambda^2)$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20










  • $begingroup$
    ...or, more simply, $$pi(lambdamid k)propto e^{-9lambda}lambda^{14}$$ Assuming the next generations are empty, one arrives at $$pi(lambdamid k)propto e^{-16lambda}lambda^{14}$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20












  • $begingroup$
    @Did Thank you for the answer. Could you elaborate on what you mean by empty and non empty generation? I also don't see how you obtained the number of offsprings? How did you find $1,2,3,2,0,0,1,4,2$?
    $endgroup$
    – Parseval
    Dec 3 '18 at 15:50












  • $begingroup$
    I looked at each node and counted its descendants.
    $endgroup$
    – Did
    Dec 3 '18 at 16:43










  • $begingroup$
    @ Did Okay I understnad that now. So can one then say that $$pi(lambda|k)simGamma(lambda;15,9)?$$
    $endgroup$
    – Parseval
    Dec 3 '18 at 18:03
















1












1








1





$begingroup$


I have the following branching process:



enter image description here



(I added the $Z$'s myself for the number of individuals in each gen, hope it's correct.)




Assume the offspring distribution is Poisson with expectation
$lambda$ and also assume that we use an improper prior
$pi(lambda)propto_{lambda}1/lambda$ for $lambda$. Using the
information in the figure, compute the posterior distribution for
$lambda.$




Using Baye's theorem with an observed count $k$, I get



$$pi(lambda|k)propto_{lambda}pi(k|lambda)pi(lambda)propto_{lambda}e^{-lambda}cdotlambda^{k}cdotfrac{1}{lambda}=e^{-lambda}lambda^{k-1}.$$



So we see that this posterior is $pi(lambda|k)simGamma(lambda;k,1).$



However, how do I know what $k$ is based on the figure?






statistics stochastic-processes markov-process







share|cite|improve this question









$endgroup$




I have the following branching process:



enter image description here



(I added the $Z$'s myself for the number of individuals in each gen, hope it's correct.)




Assume the offspring distribution is Poisson with expectation
$lambda$ and also assume that we use an improper prior
$pi(lambda)propto_{lambda}1/lambda$ for $lambda$. Using the
information in the figure, compute the posterior distribution for
$lambda.$




Using Baye's theorem with an observed count $k$, I get



$$pi(lambda|k)propto_{lambda}pi(k|lambda)pi(lambda)propto_{lambda}e^{-lambda}cdotlambda^{k}cdotfrac{1}{lambda}=e^{-lambda}lambda^{k-1}.$$



So we see that this posterior is $pi(lambda|k)simGamma(lambda;k,1).$



However, how do I know what $k$ is based on the figure?






statistics stochastic-processes markov-process




statistics stochastic-processes markov-process






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 13:16









ParsevalParseval

2,7901718




2,7901718












  • $begingroup$
    Assuming the next generations of the tree are not represented in the figure (not that they are empty), the figure tells you that the numbers of offspring for this tree are, starting from the root, $$1, 2, 3, 2, 0, 0, 1, 4, 2$$ hence the posterior knowing this tree is $$pi(lambdamid k)proptopi(lambda)prod_kpi(kmidlambda)$$ that is, $$pi(lambdamid k)proptolambda^{-1}(e^{-lambda}lambda)(e^{-lambda}lambda^2)(e^{-lambda}lambda^3)(e^{-lambda}lambda^2)(e^{-lambda}lambda^0)(e^{-lambda}lambda^0)(e^{-lambda}lambda^1)(e^{-lambda}lambda^4)(e^{-lambda}lambda^2)$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20










  • $begingroup$
    ...or, more simply, $$pi(lambdamid k)propto e^{-9lambda}lambda^{14}$$ Assuming the next generations are empty, one arrives at $$pi(lambdamid k)propto e^{-16lambda}lambda^{14}$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20












  • $begingroup$
    @Did Thank you for the answer. Could you elaborate on what you mean by empty and non empty generation? I also don't see how you obtained the number of offsprings? How did you find $1,2,3,2,0,0,1,4,2$?
    $endgroup$
    – Parseval
    Dec 3 '18 at 15:50












  • $begingroup$
    I looked at each node and counted its descendants.
    $endgroup$
    – Did
    Dec 3 '18 at 16:43










  • $begingroup$
    @ Did Okay I understnad that now. So can one then say that $$pi(lambda|k)simGamma(lambda;15,9)?$$
    $endgroup$
    – Parseval
    Dec 3 '18 at 18:03




















  • $begingroup$
    Assuming the next generations of the tree are not represented in the figure (not that they are empty), the figure tells you that the numbers of offspring for this tree are, starting from the root, $$1, 2, 3, 2, 0, 0, 1, 4, 2$$ hence the posterior knowing this tree is $$pi(lambdamid k)proptopi(lambda)prod_kpi(kmidlambda)$$ that is, $$pi(lambdamid k)proptolambda^{-1}(e^{-lambda}lambda)(e^{-lambda}lambda^2)(e^{-lambda}lambda^3)(e^{-lambda}lambda^2)(e^{-lambda}lambda^0)(e^{-lambda}lambda^0)(e^{-lambda}lambda^1)(e^{-lambda}lambda^4)(e^{-lambda}lambda^2)$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20










  • $begingroup$
    ...or, more simply, $$pi(lambdamid k)propto e^{-9lambda}lambda^{14}$$ Assuming the next generations are empty, one arrives at $$pi(lambdamid k)propto e^{-16lambda}lambda^{14}$$
    $endgroup$
    – Did
    Dec 3 '18 at 14:20












  • $begingroup$
    @Did Thank you for the answer. Could you elaborate on what you mean by empty and non empty generation? I also don't see how you obtained the number of offsprings? How did you find $1,2,3,2,0,0,1,4,2$?
    $endgroup$
    – Parseval
    Dec 3 '18 at 15:50












  • $begingroup$
    I looked at each node and counted its descendants.
    $endgroup$
    – Did
    Dec 3 '18 at 16:43










  • $begingroup$
    @ Did Okay I understnad that now. So can one then say that $$pi(lambda|k)simGamma(lambda;15,9)?$$
    $endgroup$
    – Parseval
    Dec 3 '18 at 18:03


















$begingroup$
Assuming the next generations of the tree are not represented in the figure (not that they are empty), the figure tells you that the numbers of offspring for this tree are, starting from the root, $$1, 2, 3, 2, 0, 0, 1, 4, 2$$ hence the posterior knowing this tree is $$pi(lambdamid k)proptopi(lambda)prod_kpi(kmidlambda)$$ that is, $$pi(lambdamid k)proptolambda^{-1}(e^{-lambda}lambda)(e^{-lambda}lambda^2)(e^{-lambda}lambda^3)(e^{-lambda}lambda^2)(e^{-lambda}lambda^0)(e^{-lambda}lambda^0)(e^{-lambda}lambda^1)(e^{-lambda}lambda^4)(e^{-lambda}lambda^2)$$
$endgroup$
– Did
Dec 3 '18 at 14:20




$begingroup$
Assuming the next generations of the tree are not represented in the figure (not that they are empty), the figure tells you that the numbers of offspring for this tree are, starting from the root, $$1, 2, 3, 2, 0, 0, 1, 4, 2$$ hence the posterior knowing this tree is $$pi(lambdamid k)proptopi(lambda)prod_kpi(kmidlambda)$$ that is, $$pi(lambdamid k)proptolambda^{-1}(e^{-lambda}lambda)(e^{-lambda}lambda^2)(e^{-lambda}lambda^3)(e^{-lambda}lambda^2)(e^{-lambda}lambda^0)(e^{-lambda}lambda^0)(e^{-lambda}lambda^1)(e^{-lambda}lambda^4)(e^{-lambda}lambda^2)$$
$endgroup$
– Did
Dec 3 '18 at 14:20












$begingroup$
...or, more simply, $$pi(lambdamid k)propto e^{-9lambda}lambda^{14}$$ Assuming the next generations are empty, one arrives at $$pi(lambdamid k)propto e^{-16lambda}lambda^{14}$$
$endgroup$
– Did
Dec 3 '18 at 14:20






$begingroup$
...or, more simply, $$pi(lambdamid k)propto e^{-9lambda}lambda^{14}$$ Assuming the next generations are empty, one arrives at $$pi(lambdamid k)propto e^{-16lambda}lambda^{14}$$
$endgroup$
– Did
Dec 3 '18 at 14:20














$begingroup$
@Did Thank you for the answer. Could you elaborate on what you mean by empty and non empty generation? I also don't see how you obtained the number of offsprings? How did you find $1,2,3,2,0,0,1,4,2$?
$endgroup$
– Parseval
Dec 3 '18 at 15:50






$begingroup$
@Did Thank you for the answer. Could you elaborate on what you mean by empty and non empty generation? I also don't see how you obtained the number of offsprings? How did you find $1,2,3,2,0,0,1,4,2$?
$endgroup$
– Parseval
Dec 3 '18 at 15:50














$begingroup$
I looked at each node and counted its descendants.
$endgroup$
– Did
Dec 3 '18 at 16:43




$begingroup$
I looked at each node and counted its descendants.
$endgroup$
– Did
Dec 3 '18 at 16:43












$begingroup$
@ Did Okay I understnad that now. So can one then say that $$pi(lambda|k)simGamma(lambda;15,9)?$$
$endgroup$
– Parseval
Dec 3 '18 at 18:03






$begingroup$
@ Did Okay I understnad that now. So can one then say that $$pi(lambda|k)simGamma(lambda;15,9)?$$
$endgroup$
– Parseval
Dec 3 '18 at 18:03












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024052%2fbranching-process-and-posterior-distribution%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
















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024052%2fbranching-process-and-posterior-distribution%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bundesstraße 106

Image
Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
Read more

Le Mesnil-Réaume

Image
Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
Read more

Ida-Boy-Ed-Garten

Image
Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
Read more