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












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