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Understanding an Identity in Arthreya's “Branching Processes”

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I'm currently reading from Chapter 1: The Galton-Watson Process in Arthreya's "Branching Processes," and I'm not sure that I'm understanding this identity correctly. We're given our generating function for the linear fractional Galton-Watson Process as $f$. We denote $mathbb{E}Z_1 = f'(1) = m$.

We have observed that for any two points $u,v$ that
$$
frac{f(s)-f(u)}{f(s)-f(v)}=frac{s-u}{s-v}frac{1-pv}{1-pu}.
$$

We've shown that $f(s)=s$ has roots $s_0$ and $1$. The text claims the following:

Taking $u=s_0, v=1$, if $mneq 1$ we get
$$
frac{1-p}{1-ps_0}=lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)left(frac{f(s)-1}{s-1}right)^{-1} = frac{1}{m}.
$$

I would like to confirm that my understanding of this is correct. From the first identity, we can rearrange to get the terms that are in the second. Then, we can observe that if $mneq1$ then $s_0 neq 1$, then the left factor has a defined limit as $sto1$, and the right factor is the definition of the derivative at 1, only inverted. So we can break our limit into
$$
lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1}.
$$
Then since $f(s)$ is continuous and $f(1)=1$, we have
$$
lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} =
left(frac{1-s_0}{1-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} = lim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} = frac{1}{m}.
$$

Is this a correct assessment of the argument?

probability-theory generating-functions

share|cite|improve this question

asked Mar 20 at 18:10

perpetuallyconfusedperpetuallyconfused

126

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0

$begingroup$

I'm currently reading from Chapter 1: The Galton-Watson Process in Arthreya's "Branching Processes," and I'm not sure that I'm understanding this identity correctly. We're given our generating function for the linear fractional Galton-Watson Process as $f$. We denote $mathbb{E}Z_1 = f'(1) = m$.

We have observed that for any two points $u,v$ that
$$
frac{f(s)-f(u)}{f(s)-f(v)}=frac{s-u}{s-v}frac{1-pv}{1-pu}.
$$

We've shown that $f(s)=s$ has roots $s_0$ and $1$. The text claims the following:

Taking $u=s_0, v=1$, if $mneq 1$ we get
$$
frac{1-p}{1-ps_0}=lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)left(frac{f(s)-1}{s-1}right)^{-1} = frac{1}{m}.
$$

I would like to confirm that my understanding of this is correct. From the first identity, we can rearrange to get the terms that are in the second. Then, we can observe that if $mneq1$ then $s_0 neq 1$, then the left factor has a defined limit as $sto1$, and the right factor is the definition of the derivative at 1, only inverted. So we can break our limit into
$$
lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1}.
$$
Then since $f(s)$ is continuous and $f(1)=1$, we have
$$
lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} =
left(frac{1-s_0}{1-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} = lim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} = frac{1}{m}.
$$

Is this a correct assessment of the argument?

probability-theory generating-functions

share|cite|improve this question

asked Mar 20 at 18:10

perpetuallyconfusedperpetuallyconfused

126

$endgroup$

add a comment | 

$begingroup$

I'm currently reading from Chapter 1: The Galton-Watson Process in Arthreya's "Branching Processes," and I'm not sure that I'm understanding this identity correctly. We're given our generating function for the linear fractional Galton-Watson Process as $f$. We denote $mathbb{E}Z_1 = f'(1) = m$.

We have observed that for any two points $u,v$ that
$$
frac{f(s)-f(u)}{f(s)-f(v)}=frac{s-u}{s-v}frac{1-pv}{1-pu}.
$$

We've shown that $f(s)=s$ has roots $s_0$ and $1$. The text claims the following:

Taking $u=s_0, v=1$, if $mneq 1$ we get
$$
frac{1-p}{1-ps_0}=lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)left(frac{f(s)-1}{s-1}right)^{-1} = frac{1}{m}.
$$

I would like to confirm that my understanding of this is correct. From the first identity, we can rearrange to get the terms that are in the second. Then, we can observe that if $mneq1$ then $s_0 neq 1$, then the left factor has a defined limit as $sto1$, and the right factor is the definition of the derivative at 1, only inverted. So we can break our limit into
$$
lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1}.
$$
Then since $f(s)$ is continuous and $f(1)=1$, we have
$$
lim_{sto 1}left(frac{f(s)-s_0}{s-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} =
left(frac{1-s_0}{1-s_0}right)cdotlim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} = lim_{sto 1}left(frac{f(s)-1}{s-1}right)^{-1} = frac{1}{m}.
$$

Is this a correct assessment of the argument?

probability-theory generating-functions

share|cite|improve this question

asked Mar 20 at 18:10

perpetuallyconfusedperpetuallyconfused

126

$endgroup$

share|cite|improve this question

asked Mar 20 at 18:10

perpetuallyconfusedperpetuallyconfused

126

share|cite|improve this question

asked Mar 20 at 18:10

perpetuallyconfusedperpetuallyconfused

126

asked Mar 20 at 18:10

perpetuallyconfusedperpetuallyconfused

126

asked Mar 20 at 18:10

perpetuallyconfusedperpetuallyconfused

126