Mixing
Branching process with $Z_0 sim operatornamePoisson(lambda)$
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The problem is the following:
Suppose we have a branching process so that the initial generation $Z_0sim operatornamePoisson(lambda)$, and the number of offsprings of each individual is also Poisson with parameter $lambda$. Find a function $f$ such that if $H_n(x)$ is the (probability) generating function of the total number of individuals $Z_0 + Z_1 + dots + Z_n$, then
$$H_n(s) = fleft (sH_n-1(s)right )
$$
My attempt so far:
Let's look at $H_1$ first and then try to discover a generalised solution. Let $G_X$ denote the probability generating function of an r.v. $X$. Let, also, $C_1, C_2, dots C_Z_0$ be the offsprings of $Z_0$ s.t.
$$
Z_1 = C_1 + dots + C_Z_0
$$
$$
H_1(s) = G_Z_0+Z_1(s) = G_Z_0 +C_1 + dots + C_Z_0 (s)
$$
Here I make the following claim even though I am not sure I can, so please do let me know if it is valid or not : $Z_0, C_1, dots , C_Z_0$ are independent.
Hence:
$$
H_1(s) = G_Z_0(s)cdot G_C_1 + dots + C_Z_0 (s) = G_Z_0(s)cdot G_Z_0left (G(s)right ),
$$
where $G(s)$ is the probability generating function of the (identical) offsprings. Given that $Z_0$ and the offsprings have the same distribution, the above equation can be written as:
$$
H_1(s) = G(s)cdot G(G(s))
$$
However I get stuck here. I would appreciate some hints as to how to move forward.
The result they give is that $f(s) = G(s)$, however it is not helpful...
probability stochastic-processes generating-functions
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
Andrei Crisan
1997
add a comment |Â
up vote
2
down vote
favorite
The problem is the following:
Suppose we have a branching process so that the initial generation $Z_0sim operatornamePoisson(lambda)$, and the number of offsprings of each individual is also Poisson with parameter $lambda$. Find a function $f$ such that if $H_n(x)$ is the (probability) generating function of the total number of individuals $Z_0 + Z_1 + dots + Z_n$, then
$$H_n(s) = fleft (sH_n-1(s)right )
$$
My attempt so far:
Let's look at $H_1$ first and then try to discover a generalised solution. Let $G_X$ denote the probability generating function of an r.v. $X$. Let, also, $C_1, C_2, dots C_Z_0$ be the offsprings of $Z_0$ s.t.
$$
Z_1 = C_1 + dots + C_Z_0
$$
$$
H_1(s) = G_Z_0+Z_1(s) = G_Z_0 +C_1 + dots + C_Z_0 (s)
$$
Here I make the following claim even though I am not sure I can, so please do let me know if it is valid or not : $Z_0, C_1, dots , C_Z_0$ are independent.
Hence:
$$
H_1(s) = G_Z_0(s)cdot G_C_1 + dots + C_Z_0 (s) = G_Z_0(s)cdot G_Z_0left (G(s)right ),
$$
where $G(s)$ is the probability generating function of the (identical) offsprings. Given that $Z_0$ and the offsprings have the same distribution, the above equation can be written as:
$$
H_1(s) = G(s)cdot G(G(s))
$$
However I get stuck here. I would appreciate some hints as to how to move forward.
The result they give is that $f(s) = G(s)$, however it is not helpful...
probability stochastic-processes generating-functions
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
Andrei Crisan
1997
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The problem is the following:
Suppose we have a branching process so that the initial generation $Z_0sim operatornamePoisson(lambda)$, and the number of offsprings of each individual is also Poisson with parameter $lambda$. Find a function $f$ such that if $H_n(x)$ is the (probability) generating function of the total number of individuals $Z_0 + Z_1 + dots + Z_n$, then
$$H_n(s) = fleft (sH_n-1(s)right )
$$
My attempt so far:
Let's look at $H_1$ first and then try to discover a generalised solution. Let $G_X$ denote the probability generating function of an r.v. $X$. Let, also, $C_1, C_2, dots C_Z_0$ be the offsprings of $Z_0$ s.t.
$$
Z_1 = C_1 + dots + C_Z_0
$$
$$
H_1(s) = G_Z_0+Z_1(s) = G_Z_0 +C_1 + dots + C_Z_0 (s)
$$
Here I make the following claim even though I am not sure I can, so please do let me know if it is valid or not : $Z_0, C_1, dots , C_Z_0$ are independent.
Hence:
$$
H_1(s) = G_Z_0(s)cdot G_C_1 + dots + C_Z_0 (s) = G_Z_0(s)cdot G_Z_0left (G(s)right ),
$$
where $G(s)$ is the probability generating function of the (identical) offsprings. Given that $Z_0$ and the offsprings have the same distribution, the above equation can be written as:
$$
H_1(s) = G(s)cdot G(G(s))
$$
However I get stuck here. I would appreciate some hints as to how to move forward.
The result they give is that $f(s) = G(s)$, however it is not helpful...
probability stochastic-processes generating-functions
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
Andrei Crisan
1997
The problem is the following:
Suppose we have a branching process so that the initial generation $Z_0sim operatornamePoisson(lambda)$, and the number of offsprings of each individual is also Poisson with parameter $lambda$. Find a function $f$ such that if $H_n(x)$ is the (probability) generating function of the total number of individuals $Z_0 + Z_1 + dots + Z_n$, then
$$H_n(s) = fleft (sH_n-1(s)right )
$$
My attempt so far:
Let's look at $H_1$ first and then try to discover a generalised solution. Let $G_X$ denote the probability generating function of an r.v. $X$. Let, also, $C_1, C_2, dots C_Z_0$ be the offsprings of $Z_0$ s.t.
$$
Z_1 = C_1 + dots + C_Z_0
$$
$$
H_1(s) = G_Z_0+Z_1(s) = G_Z_0 +C_1 + dots + C_Z_0 (s)
$$
Here I make the following claim even though I am not sure I can, so please do let me know if it is valid or not : $Z_0, C_1, dots , C_Z_0$ are independent.
Hence:
$$
H_1(s) = G_Z_0(s)cdot G_C_1 + dots + C_Z_0 (s) = G_Z_0(s)cdot G_Z_0left (G(s)right ),
$$
where $G(s)$ is the probability generating function of the (identical) offsprings. Given that $Z_0$ and the offsprings have the same distribution, the above equation can be written as:
$$
H_1(s) = G(s)cdot G(G(s))
$$
However I get stuck here. I would appreciate some hints as to how to move forward.
The result they give is that $f(s) = G(s)$, however it is not helpful...
probability stochastic-processes generating-functions
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
Andrei Crisan
1997
edited 2 days ago
Michael Hardy
204k23185460
edited 2 days ago
Michael Hardy
204k23185460
edited 2 days ago
Michael Hardy
204k23185460
204k23185460
asked 2 days ago
Andrei Crisan
1997
asked 2 days ago
Andrei Crisan
1997
asked 2 days ago
Andrei Crisan
1997
1997
add a comment |Â
add a comment |Â
1 Answer
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I mean, your claim is strange when put in that form since $Z_0$ determines how many of the $C_i$s you have. With that said, you can consider $C_i$ to be $X_i 1_i leq Z_0$, where $X_i$ is independent from $Z_0$. Now, I will do the base case:
$H_1(s) = mathbbE s^Z_0 + Z_1 = mathbbE[ mathbbE[ s^Z_0 + Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^sum_i=1^infty X_i 1_i leq Z_0 mid Z_0]] = mathbbE[ s^Z_0 G(s)^Z_0] = G(sG(s))$.
From this, you should be able to generalize (though it seems like for the proof go through, at time $i$, number of people having children is $Z_1 + Z_2 + ... + Z_i-1$, not just $Z_i-1$ which is not a convention I am used to in terms of branching processes; I am more used to the parent "disappearing" after splitting.)
answered yesterday
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1,7901312
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I mean, your claim is strange when put in that form since $Z_0$ determines how many of the $C_i$s you have. With that said, you can consider $C_i$ to be $X_i 1_i leq Z_0$, where $X_i$ is independent from $Z_0$. Now, I will do the base case:
$H_1(s) = mathbbE s^Z_0 + Z_1 = mathbbE[ mathbbE[ s^Z_0 + Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^sum_i=1^infty X_i 1_i leq Z_0 mid Z_0]] = mathbbE[ s^Z_0 G(s)^Z_0] = G(sG(s))$.
From this, you should be able to generalize (though it seems like for the proof go through, at time $i$, number of people having children is $Z_1 + Z_2 + ... + Z_i-1$, not just $Z_i-1$ which is not a convention I am used to in terms of branching processes; I am more used to the parent "disappearing" after splitting.)
answered yesterday
E-A
1,7901312
add a comment |Â
up vote
0
down vote
I mean, your claim is strange when put in that form since $Z_0$ determines how many of the $C_i$s you have. With that said, you can consider $C_i$ to be $X_i 1_i leq Z_0$, where $X_i$ is independent from $Z_0$. Now, I will do the base case:
$H_1(s) = mathbbE s^Z_0 + Z_1 = mathbbE[ mathbbE[ s^Z_0 + Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^sum_i=1^infty X_i 1_i leq Z_0 mid Z_0]] = mathbbE[ s^Z_0 G(s)^Z_0] = G(sG(s))$.
From this, you should be able to generalize (though it seems like for the proof go through, at time $i$, number of people having children is $Z_1 + Z_2 + ... + Z_i-1$, not just $Z_i-1$ which is not a convention I am used to in terms of branching processes; I am more used to the parent "disappearing" after splitting.)
answered yesterday
E-A
1,7901312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I mean, your claim is strange when put in that form since $Z_0$ determines how many of the $C_i$s you have. With that said, you can consider $C_i$ to be $X_i 1_i leq Z_0$, where $X_i$ is independent from $Z_0$. Now, I will do the base case:
$H_1(s) = mathbbE s^Z_0 + Z_1 = mathbbE[ mathbbE[ s^Z_0 + Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^sum_i=1^infty X_i 1_i leq Z_0 mid Z_0]] = mathbbE[ s^Z_0 G(s)^Z_0] = G(sG(s))$.
From this, you should be able to generalize (though it seems like for the proof go through, at time $i$, number of people having children is $Z_1 + Z_2 + ... + Z_i-1$, not just $Z_i-1$ which is not a convention I am used to in terms of branching processes; I am more used to the parent "disappearing" after splitting.)
answered yesterday
E-A
1,7901312
I mean, your claim is strange when put in that form since $Z_0$ determines how many of the $C_i$s you have. With that said, you can consider $C_i$ to be $X_i 1_i leq Z_0$, where $X_i$ is independent from $Z_0$. Now, I will do the base case:
$H_1(s) = mathbbE s^Z_0 + Z_1 = mathbbE[ mathbbE[ s^Z_0 + Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^Z_1 mid Z_0]] = mathbbE[ s^Z_0 mathbbE[ s^sum_i=1^infty X_i 1_i leq Z_0 mid Z_0]] = mathbbE[ s^Z_0 G(s)^Z_0] = G(sG(s))$.
From this, you should be able to generalize (though it seems like for the proof go through, at time $i$, number of people having children is $Z_1 + Z_2 + ... + Z_i-1$, not just $Z_i-1$ which is not a convention I am used to in terms of branching processes; I am more used to the parent "disappearing" after splitting.)
answered yesterday
E-A
1,7901312
answered yesterday
E-A
1,7901312
answered yesterday
E-A
1,7901312
answered yesterday
E-A
1,7901312
1,7901312
add a comment |Â
add a comment |Â
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