Mixing
Distribution for population growth
Clash Royale CLAN TAG#URR8PPP
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For the exponential population growth with a constant growth rate of $R$,
$$
n(t+1) = R*n(t)
$$
where $n$ is the number of individuals in the population. Therefore, we can sample from a Poisson distribution since the sum of $n$ numbers drawn from a Poisson distribution with mean $R$ is known to follow a Poisson distribution with mean $Rn$.
If $n=K$ remains constant and $R$ is a function of $t$, what distribution would,
$$
R(t+1) = R(t)*K
$$
would follow? Note that $R$ need not be an integer and so Poisson may not apply.
probability-distributions stochastic-processes
asked Aug 2 at 19:13
user2167741
1019
add a comment |Â
up vote
0
down vote
favorite
For the exponential population growth with a constant growth rate of $R$,
$$
n(t+1) = R*n(t)
$$
where $n$ is the number of individuals in the population. Therefore, we can sample from a Poisson distribution since the sum of $n$ numbers drawn from a Poisson distribution with mean $R$ is known to follow a Poisson distribution with mean $Rn$.
If $n=K$ remains constant and $R$ is a function of $t$, what distribution would,
$$
R(t+1) = R(t)*K
$$
would follow? Note that $R$ need not be an integer and so Poisson may not apply.
probability-distributions stochastic-processes
asked Aug 2 at 19:13
user2167741
1019
Please clarify! Do you mean $R(t)=R(0)K^t$? (Assuming $t$ an integer).
– herb steinberg
Aug 2 at 21:30
In fact, $R(t)= alpha log fracMG(t)$ for some function $G(t)$. $alpha$ and $M$ are constants.
– user2167741
Aug 3 at 1:48
1
You have $fracR(t+1)R(t)=K$, a constant. How do you get anything other than what I had ($R(t)=R(0)K^t$) in my previous comment?
– herb steinberg
Aug 3 at 2:59
Yes, you are right. I made some calculations and the function $G$ is such that $R(t)=R(0)K^t$.
– user2167741
Aug 3 at 11:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
For the exponential population growth with a constant growth rate of $R$,
$$
n(t+1) = R*n(t)
$$
where $n$ is the number of individuals in the population. Therefore, we can sample from a Poisson distribution since the sum of $n$ numbers drawn from a Poisson distribution with mean $R$ is known to follow a Poisson distribution with mean $Rn$.
If $n=K$ remains constant and $R$ is a function of $t$, what distribution would,
$$
R(t+1) = R(t)*K
$$
would follow? Note that $R$ need not be an integer and so Poisson may not apply.
probability-distributions stochastic-processes
asked Aug 2 at 19:13
user2167741
1019
For the exponential population growth with a constant growth rate of $R$,
$$
n(t+1) = R*n(t)
$$
where $n$ is the number of individuals in the population. Therefore, we can sample from a Poisson distribution since the sum of $n$ numbers drawn from a Poisson distribution with mean $R$ is known to follow a Poisson distribution with mean $Rn$.
If $n=K$ remains constant and $R$ is a function of $t$, what distribution would,
$$
R(t+1) = R(t)*K
$$
would follow? Note that $R$ need not be an integer and so Poisson may not apply.
probability-distributions stochastic-processes
asked Aug 2 at 19:13
user2167741
1019
asked Aug 2 at 19:13
user2167741
1019
asked Aug 2 at 19:13
user2167741
1019
asked Aug 2 at 19:13
user2167741
1019
1019
Please clarify! Do you mean $R(t)=R(0)K^t$? (Assuming $t$ an integer).
– herb steinberg
Aug 2 at 21:30
In fact, $R(t)= alpha log fracMG(t)$ for some function $G(t)$. $alpha$ and $M$ are constants.
– user2167741
Aug 3 at 1:48
1
You have $fracR(t+1)R(t)=K$, a constant. How do you get anything other than what I had ($R(t)=R(0)K^t$) in my previous comment?
– herb steinberg
Aug 3 at 2:59
Yes, you are right. I made some calculations and the function $G$ is such that $R(t)=R(0)K^t$.
– user2167741
Aug 3 at 11:07
add a comment |Â
Please clarify! Do you mean $R(t)=R(0)K^t$? (Assuming $t$ an integer).
– herb steinberg
Aug 2 at 21:30
In fact, $R(t)= alpha log fracMG(t)$ for some function $G(t)$. $alpha$ and $M$ are constants.
– user2167741
Aug 3 at 1:48
1
You have $fracR(t+1)R(t)=K$, a constant. How do you get anything other than what I had ($R(t)=R(0)K^t$) in my previous comment?
– herb steinberg
Aug 3 at 2:59
Yes, you are right. I made some calculations and the function $G$ is such that $R(t)=R(0)K^t$.
– user2167741
Aug 3 at 11:07
Please clarify! Do you mean $R(t)=R(0)K^t$? (Assuming $t$ an integer).
– herb steinberg
Aug 2 at 21:30
Please clarify! Do you mean $R(t)=R(0)K^t$? (Assuming $t$ an integer).
– herb steinberg
Aug 2 at 21:30
In fact, $R(t)= alpha log fracMG(t)$ for some function $G(t)$. $alpha$ and $M$ are constants.
– user2167741
Aug 3 at 1:48
In fact, $R(t)= alpha log fracMG(t)$ for some function $G(t)$. $alpha$ and $M$ are constants.
– user2167741
Aug 3 at 1:48
1
1
You have $fracR(t+1)R(t)=K$, a constant. How do you get anything other than what I had ($R(t)=R(0)K^t$) in my previous comment?
– herb steinberg
Aug 3 at 2:59
You have $fracR(t+1)R(t)=K$, a constant. How do you get anything other than what I had ($R(t)=R(0)K^t$) in my previous comment?
– herb steinberg
Aug 3 at 2:59
Yes, you are right. I made some calculations and the function $G$ is such that $R(t)=R(0)K^t$.
– user2167741
Aug 3 at 11:07
Yes, you are right. I made some calculations and the function $G$ is such that $R(t)=R(0)K^t$.
– user2167741
Aug 3 at 11:07
add a comment |Â
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Please clarify! Do you mean $R(t)=R(0)K^t$? (Assuming $t$ an integer).
– herb steinberg
Aug 2 at 21:30
In fact, $R(t)= alpha log fracMG(t)$ for some function $G(t)$. $alpha$ and $M$ are constants.
– user2167741
Aug 3 at 1:48
1
You have $fracR(t+1)R(t)=K$, a constant. How do you get anything other than what I had ($R(t)=R(0)K^t$) in my previous comment?
– herb steinberg
Aug 3 at 2:59
Yes, you are right. I made some calculations and the function $G$ is such that $R(t)=R(0)K^t$.
– user2167741
Aug 3 at 11:07