Mixing
How to prove that $r_n=r_{n-1} - frac12 b r_{n-1}^2 $ is asymptotic to $frac{1}{n}$?
So my question is:
How do we prove that $r_n=r_{n-1} -frac12 b r_{n-1}^2 $ is asymptotic to $frac{1}{n}$ when we have an offspring distribution $p_i:=P(xi=i)$ and b is the variance
$$b=sum_{igeq 2}i(i-1) p_i< infty$$ and $r_n:=1-q^n(0)$ for $$q^n(x):=q(q^{n-1}(x)), quad q(x):=p_0+p_1x+p_2x^2+ldots,$$
what gives us $$r_1=1-p_0.$$
Also we have
$$sum_{igeq 1}i p_i=1.$$
Best regards!
Edit: In fact it is asymptotic to $frac{2}{bn}$.
asymptotics recursion
asked Nov 20 '18 at 17:26
cptflint
208
|
show 9 more comments
So my question is:
How do we prove that $r_n=r_{n-1} -frac12 b r_{n-1}^2 $ is asymptotic to $frac{1}{n}$ when we have an offspring distribution $p_i:=P(xi=i)$ and b is the variance
$$b=sum_{igeq 2}i(i-1) p_i< infty$$ and $r_n:=1-q^n(0)$ for $$q^n(x):=q(q^{n-1}(x)), quad q(x):=p_0+p_1x+p_2x^2+ldots,$$
what gives us $$r_1=1-p_0.$$
Also we have
$$sum_{igeq 1}i p_i=1.$$
Best regards!
Edit: In fact it is asymptotic to $frac{2}{bn}$.
asymptotics recursion
asked Nov 20 '18 at 17:26
cptflint
208
1
If $b$ is sufficiently large (say $b = 5$ for $r_1 = 1$) then this will definitely just explode to negative infinity.
– Mees de Vries
Nov 20 '18 at 17:31
Yeah, right... Mhm. I am reading an old paper from Kolmogorov (On the solution of a problem in biology) and what he says is this: $frac{r_n}{r_{n-1}}=1-frac12 br_{n-1} + O(r_{n-2}^2)$ gives us $r_n sim frac{2}{nb}$. In this special case $b$ is the variance of a critical offspring distribution. $r_n $ is defined by $r_n=1-q^{(n)}(0)$ and $q^{(n+1)}(x):=q(q^n(x))$ for $q^1(x):=p_0+p_1x+p_2x^2+...$.
– cptflint
Nov 20 '18 at 17:40
This looks true under the form "asymptotic to $K dfrac{1}{n}$" (K=constant) for the very narrow range $b in (0,1)$ (and $r_1 in (0,1)$). Do like (good) journalists : check your sources !
– Jean Marie
Nov 20 '18 at 17:45
Ok so $r_1in (0,1)$ instead of $[0,1]$, thx! But: $bin(0,1)$ seems like a problem. Kolmogorov only chooses $b>0$. So I am missing something or that's an old fault...
– cptflint
Nov 20 '18 at 17:46
For $r_1$ it is unimportant, but you cannot say $b in (0,infty)$ : if I take b=3, 4, 5, 6..., the sequence explodes
– Jean Marie
Nov 20 '18 at 17:50
|
show 9 more comments
So my question is:
How do we prove that $r_n=r_{n-1} -frac12 b r_{n-1}^2 $ is asymptotic to $frac{1}{n}$ when we have an offspring distribution $p_i:=P(xi=i)$ and b is the variance
$$b=sum_{igeq 2}i(i-1) p_i< infty$$ and $r_n:=1-q^n(0)$ for $$q^n(x):=q(q^{n-1}(x)), quad q(x):=p_0+p_1x+p_2x^2+ldots,$$
what gives us $$r_1=1-p_0.$$
Also we have
$$sum_{igeq 1}i p_i=1.$$
Best regards!
Edit: In fact it is asymptotic to $frac{2}{bn}$.
asymptotics recursion
asked Nov 20 '18 at 17:26
cptflint
208
So my question is:
How do we prove that $r_n=r_{n-1} -frac12 b r_{n-1}^2 $ is asymptotic to $frac{1}{n}$ when we have an offspring distribution $p_i:=P(xi=i)$ and b is the variance
$$b=sum_{igeq 2}i(i-1) p_i< infty$$ and $r_n:=1-q^n(0)$ for $$q^n(x):=q(q^{n-1}(x)), quad q(x):=p_0+p_1x+p_2x^2+ldots,$$
what gives us $$r_1=1-p_0.$$
Also we have
$$sum_{igeq 1}i p_i=1.$$
Best regards!
Edit: In fact it is asymptotic to $frac{2}{bn}$.
asymptotics recursion
asymptotics recursion
asked Nov 20 '18 at 17:26
cptflint
208
asked Nov 20 '18 at 17:26
cptflint
208
edited Nov 21 '18 at 22:19
asked Nov 20 '18 at 17:26
cptflint
208
asked Nov 20 '18 at 17:26
cptflint
208
asked Nov 20 '18 at 17:26
cptflint
208
208
1
If $b$ is sufficiently large (say $b = 5$ for $r_1 = 1$) then this will definitely just explode to negative infinity.
– Mees de Vries
Nov 20 '18 at 17:31
Yeah, right... Mhm. I am reading an old paper from Kolmogorov (On the solution of a problem in biology) and what he says is this: $frac{r_n}{r_{n-1}}=1-frac12 br_{n-1} + O(r_{n-2}^2)$ gives us $r_n sim frac{2}{nb}$. In this special case $b$ is the variance of a critical offspring distribution. $r_n $ is defined by $r_n=1-q^{(n)}(0)$ and $q^{(n+1)}(x):=q(q^n(x))$ for $q^1(x):=p_0+p_1x+p_2x^2+...$.
– cptflint
Nov 20 '18 at 17:40
This looks true under the form "asymptotic to $K dfrac{1}{n}$" (K=constant) for the very narrow range $b in (0,1)$ (and $r_1 in (0,1)$). Do like (good) journalists : check your sources !
– Jean Marie
Nov 20 '18 at 17:45
Ok so $r_1in (0,1)$ instead of $[0,1]$, thx! But: $bin(0,1)$ seems like a problem. Kolmogorov only chooses $b>0$. So I am missing something or that's an old fault...
– cptflint
Nov 20 '18 at 17:46
For $r_1$ it is unimportant, but you cannot say $b in (0,infty)$ : if I take b=3, 4, 5, 6..., the sequence explodes
– Jean Marie
Nov 20 '18 at 17:50
|
show 9 more comments
1
If $b$ is sufficiently large (say $b = 5$ for $r_1 = 1$) then this will definitely just explode to negative infinity.
– Mees de Vries
Nov 20 '18 at 17:31
Yeah, right... Mhm. I am reading an old paper from Kolmogorov (On the solution of a problem in biology) and what he says is this: $frac{r_n}{r_{n-1}}=1-frac12 br_{n-1} + O(r_{n-2}^2)$ gives us $r_n sim frac{2}{nb}$. In this special case $b$ is the variance of a critical offspring distribution. $r_n $ is defined by $r_n=1-q^{(n)}(0)$ and $q^{(n+1)}(x):=q(q^n(x))$ for $q^1(x):=p_0+p_1x+p_2x^2+...$.
– cptflint
Nov 20 '18 at 17:40
This looks true under the form "asymptotic to $K dfrac{1}{n}$" (K=constant) for the very narrow range $b in (0,1)$ (and $r_1 in (0,1)$). Do like (good) journalists : check your sources !
– Jean Marie
Nov 20 '18 at 17:45
Ok so $r_1in (0,1)$ instead of $[0,1]$, thx! But: $bin(0,1)$ seems like a problem. Kolmogorov only chooses $b>0$. So I am missing something or that's an old fault...
– cptflint
Nov 20 '18 at 17:46
For $r_1$ it is unimportant, but you cannot say $b in (0,infty)$ : if I take b=3, 4, 5, 6..., the sequence explodes
– Jean Marie
Nov 20 '18 at 17:50
1
1
If $b$ is sufficiently large (say $b = 5$ for $r_1 = 1$) then this will definitely just explode to negative infinity.
– Mees de Vries
Nov 20 '18 at 17:31
If $b$ is sufficiently large (say $b = 5$ for $r_1 = 1$) then this will definitely just explode to negative infinity.
– Mees de Vries
Nov 20 '18 at 17:31
Yeah, right... Mhm. I am reading an old paper from Kolmogorov (On the solution of a problem in biology) and what he says is this: $frac{r_n}{r_{n-1}}=1-frac12 br_{n-1} + O(r_{n-2}^2)$ gives us $r_n sim frac{2}{nb}$. In this special case $b$ is the variance of a critical offspring distribution. $r_n $ is defined by $r_n=1-q^{(n)}(0)$ and $q^{(n+1)}(x):=q(q^n(x))$ for $q^1(x):=p_0+p_1x+p_2x^2+...$.
– cptflint
Nov 20 '18 at 17:40
Yeah, right... Mhm. I am reading an old paper from Kolmogorov (On the solution of a problem in biology) and what he says is this: $frac{r_n}{r_{n-1}}=1-frac12 br_{n-1} + O(r_{n-2}^2)$ gives us $r_n sim frac{2}{nb}$. In this special case $b$ is the variance of a critical offspring distribution. $r_n $ is defined by $r_n=1-q^{(n)}(0)$ and $q^{(n+1)}(x):=q(q^n(x))$ for $q^1(x):=p_0+p_1x+p_2x^2+...$.
– cptflint
Nov 20 '18 at 17:40
This looks true under the form "asymptotic to $K dfrac{1}{n}$" (K=constant) for the very narrow range $b in (0,1)$ (and $r_1 in (0,1)$). Do like (good) journalists : check your sources !
– Jean Marie
Nov 20 '18 at 17:45
This looks true under the form "asymptotic to $K dfrac{1}{n}$" (K=constant) for the very narrow range $b in (0,1)$ (and $r_1 in (0,1)$). Do like (good) journalists : check your sources !
– Jean Marie
Nov 20 '18 at 17:45
Ok so $r_1in (0,1)$ instead of $[0,1]$, thx! But: $bin(0,1)$ seems like a problem. Kolmogorov only chooses $b>0$. So I am missing something or that's an old fault...
– cptflint
Nov 20 '18 at 17:46
Ok so $r_1in (0,1)$ instead of $[0,1]$, thx! But: $bin(0,1)$ seems like a problem. Kolmogorov only chooses $b>0$. So I am missing something or that's an old fault...
– cptflint
Nov 20 '18 at 17:46
For $r_1$ it is unimportant, but you cannot say $b in (0,infty)$ : if I take b=3, 4, 5, 6..., the sequence explodes
– Jean Marie
Nov 20 '18 at 17:50
For $r_1$ it is unimportant, but you cannot say $b in (0,infty)$ : if I take b=3, 4, 5, 6..., the sequence explodes
– Jean Marie
Nov 20 '18 at 17:50
|
show 9 more comments
1 Answer
1
active
oldest
votes
Write $q(x) = mathbf{E}[x^{xi}]$. If $mathbf{E}[xi] = 1$ and $b = mathbf{Var}(xi) in (0, infty)$, then
$r_n = 1 - q^{circ n}(0)$ solves $r_n = 1 - q(1 - r_{n-1})$, hence decreases to $0$ as $ntoinfty$.
By the Taylor's theorem, $1-q(1-h) = h - left(frac{b}{2}+o(1)right)h^2$.
So, if we write $a_n = frac{1}{r_n}$, then
$$a_{n}
= a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1}
= a_{n-1} + frac{b}{2} + o(1). $$
Using this, we can prove that $a_n = left( frac{b}{2} + o(1) right) n$ as $ntoinfty$.
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
How do you get $a_{n} = a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1} = a_{n-1} + frac{b}{2} + o(1). $? Should it be $a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)$? But I get: $a_n=frac{1}{r_n}= frac{1}{r_{n-1} - frac12 br_{n-1}^2 } = a_{n-1} cdot frac{1}{1 - frac12 br_{n-1} }$
– cptflint
Nov 20 '18 at 19:19
@cptflint, I utilized $1/(1-x)=1+x+x^2+cdots$.
– Sangchul Lee
Nov 20 '18 at 19:23
Okay! Thank you!
– cptflint
Nov 20 '18 at 19:30
Mhm, I think we missed something. In every step we add $a_{n-1}o(1)$ so how can we proove the asymptotic behavior?
– cptflint
Nov 21 '18 at 10:48
@cptflint, This is a typical application of Stolz–Cesàro theorem. Since $a_n - a_{n-1} to frac{b}{2}$ as $ntoinfty$, by Stolz–Cesàro theorem, $$ lim_{ntoinfty} frac{a_n}{n} = lim_{ntoinfty} frac{a_n - a_{n-1}}{n - (n-1)} = frac{b}{2}. $$
– Sangchul Lee
Nov 22 '18 at 1:57
|
show 5 more comments
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1 Answer
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Write $q(x) = mathbf{E}[x^{xi}]$. If $mathbf{E}[xi] = 1$ and $b = mathbf{Var}(xi) in (0, infty)$, then
$r_n = 1 - q^{circ n}(0)$ solves $r_n = 1 - q(1 - r_{n-1})$, hence decreases to $0$ as $ntoinfty$.
By the Taylor's theorem, $1-q(1-h) = h - left(frac{b}{2}+o(1)right)h^2$.
So, if we write $a_n = frac{1}{r_n}$, then
$$a_{n}
= a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1}
= a_{n-1} + frac{b}{2} + o(1). $$
Using this, we can prove that $a_n = left( frac{b}{2} + o(1) right) n$ as $ntoinfty$.
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
How do you get $a_{n} = a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1} = a_{n-1} + frac{b}{2} + o(1). $? Should it be $a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)$? But I get: $a_n=frac{1}{r_n}= frac{1}{r_{n-1} - frac12 br_{n-1}^2 } = a_{n-1} cdot frac{1}{1 - frac12 br_{n-1} }$
– cptflint
Nov 20 '18 at 19:19
@cptflint, I utilized $1/(1-x)=1+x+x^2+cdots$.
– Sangchul Lee
Nov 20 '18 at 19:23
Okay! Thank you!
– cptflint
Nov 20 '18 at 19:30
Mhm, I think we missed something. In every step we add $a_{n-1}o(1)$ so how can we proove the asymptotic behavior?
– cptflint
Nov 21 '18 at 10:48
@cptflint, This is a typical application of Stolz–Cesàro theorem. Since $a_n - a_{n-1} to frac{b}{2}$ as $ntoinfty$, by Stolz–Cesàro theorem, $$ lim_{ntoinfty} frac{a_n}{n} = lim_{ntoinfty} frac{a_n - a_{n-1}}{n - (n-1)} = frac{b}{2}. $$
– Sangchul Lee
Nov 22 '18 at 1:57
|
show 5 more comments
Write $q(x) = mathbf{E}[x^{xi}]$. If $mathbf{E}[xi] = 1$ and $b = mathbf{Var}(xi) in (0, infty)$, then
$r_n = 1 - q^{circ n}(0)$ solves $r_n = 1 - q(1 - r_{n-1})$, hence decreases to $0$ as $ntoinfty$.
By the Taylor's theorem, $1-q(1-h) = h - left(frac{b}{2}+o(1)right)h^2$.
So, if we write $a_n = frac{1}{r_n}$, then
$$a_{n}
= a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1}
= a_{n-1} + frac{b}{2} + o(1). $$
Using this, we can prove that $a_n = left( frac{b}{2} + o(1) right) n$ as $ntoinfty$.
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
How do you get $a_{n} = a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1} = a_{n-1} + frac{b}{2} + o(1). $? Should it be $a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)$? But I get: $a_n=frac{1}{r_n}= frac{1}{r_{n-1} - frac12 br_{n-1}^2 } = a_{n-1} cdot frac{1}{1 - frac12 br_{n-1} }$
– cptflint
Nov 20 '18 at 19:19
@cptflint, I utilized $1/(1-x)=1+x+x^2+cdots$.
– Sangchul Lee
Nov 20 '18 at 19:23
Okay! Thank you!
– cptflint
Nov 20 '18 at 19:30
Mhm, I think we missed something. In every step we add $a_{n-1}o(1)$ so how can we proove the asymptotic behavior?
– cptflint
Nov 21 '18 at 10:48
@cptflint, This is a typical application of Stolz–Cesàro theorem. Since $a_n - a_{n-1} to frac{b}{2}$ as $ntoinfty$, by Stolz–Cesàro theorem, $$ lim_{ntoinfty} frac{a_n}{n} = lim_{ntoinfty} frac{a_n - a_{n-1}}{n - (n-1)} = frac{b}{2}. $$
– Sangchul Lee
Nov 22 '18 at 1:57
|
show 5 more comments
Write $q(x) = mathbf{E}[x^{xi}]$. If $mathbf{E}[xi] = 1$ and $b = mathbf{Var}(xi) in (0, infty)$, then
$r_n = 1 - q^{circ n}(0)$ solves $r_n = 1 - q(1 - r_{n-1})$, hence decreases to $0$ as $ntoinfty$.
By the Taylor's theorem, $1-q(1-h) = h - left(frac{b}{2}+o(1)right)h^2$.
So, if we write $a_n = frac{1}{r_n}$, then
$$a_{n}
= a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1}
= a_{n-1} + frac{b}{2} + o(1). $$
Using this, we can prove that $a_n = left( frac{b}{2} + o(1) right) n$ as $ntoinfty$.
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
Write $q(x) = mathbf{E}[x^{xi}]$. If $mathbf{E}[xi] = 1$ and $b = mathbf{Var}(xi) in (0, infty)$, then
$r_n = 1 - q^{circ n}(0)$ solves $r_n = 1 - q(1 - r_{n-1})$, hence decreases to $0$ as $ntoinfty$.
By the Taylor's theorem, $1-q(1-h) = h - left(frac{b}{2}+o(1)right)h^2$.
So, if we write $a_n = frac{1}{r_n}$, then
$$a_{n}
= a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1}
= a_{n-1} + frac{b}{2} + o(1). $$
Using this, we can prove that $a_n = left( frac{b}{2} + o(1) right) n$ as $ntoinfty$.
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
answered Nov 20 '18 at 19:00
Sangchul Lee
91.4k12164265
91.4k12164265
How do you get $a_{n} = a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1} = a_{n-1} + frac{b}{2} + o(1). $? Should it be $a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)$? But I get: $a_n=frac{1}{r_n}= frac{1}{r_{n-1} - frac12 br_{n-1}^2 } = a_{n-1} cdot frac{1}{1 - frac12 br_{n-1} }$
– cptflint
Nov 20 '18 at 19:19
@cptflint, I utilized $1/(1-x)=1+x+x^2+cdots$.
– Sangchul Lee
Nov 20 '18 at 19:23
Okay! Thank you!
– cptflint
Nov 20 '18 at 19:30
Mhm, I think we missed something. In every step we add $a_{n-1}o(1)$ so how can we proove the asymptotic behavior?
– cptflint
Nov 21 '18 at 10:48
@cptflint, This is a typical application of Stolz–Cesàro theorem. Since $a_n - a_{n-1} to frac{b}{2}$ as $ntoinfty$, by Stolz–Cesàro theorem, $$ lim_{ntoinfty} frac{a_n}{n} = lim_{ntoinfty} frac{a_n - a_{n-1}}{n - (n-1)} = frac{b}{2}. $$
– Sangchul Lee
Nov 22 '18 at 1:57
|
show 5 more comments
How do you get $a_{n} = a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1} = a_{n-1} + frac{b}{2} + o(1). $? Should it be $a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)$? But I get: $a_n=frac{1}{r_n}= frac{1}{r_{n-1} - frac12 br_{n-1}^2 } = a_{n-1} cdot frac{1}{1 - frac12 br_{n-1} }$
– cptflint
Nov 20 '18 at 19:19
@cptflint, I utilized $1/(1-x)=1+x+x^2+cdots$.
– Sangchul Lee
Nov 20 '18 at 19:23
Okay! Thank you!
– cptflint
Nov 20 '18 at 19:30
Mhm, I think we missed something. In every step we add $a_{n-1}o(1)$ so how can we proove the asymptotic behavior?
– cptflint
Nov 21 '18 at 10:48
@cptflint, This is a typical application of Stolz–Cesàro theorem. Since $a_n - a_{n-1} to frac{b}{2}$ as $ntoinfty$, by Stolz–Cesàro theorem, $$ lim_{ntoinfty} frac{a_n}{n} = lim_{ntoinfty} frac{a_n - a_{n-1}}{n - (n-1)} = frac{b}{2}. $$
– Sangchul Lee
Nov 22 '18 at 1:57
How do you get $a_{n} = a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1} = a_{n-1} + frac{b}{2} + o(1). $? Should it be $a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)$? But I get: $a_n=frac{1}{r_n}= frac{1}{r_{n-1} - frac12 br_{n-1}^2 } = a_{n-1} cdot frac{1}{1 - frac12 br_{n-1} }$
– cptflint
Nov 20 '18 at 19:19
How do you get $a_{n} = a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)^{-1} = a_{n-1} + frac{b}{2} + o(1). $? Should it be $a_{n-1} left( 1 - frac{frac{b}{2}+o(1)}{a_{n-1}} right)$? But I get: $a_n=frac{1}{r_n}= frac{1}{r_{n-1} - frac12 br_{n-1}^2 } = a_{n-1} cdot frac{1}{1 - frac12 br_{n-1} }$
– cptflint
Nov 20 '18 at 19:19
@cptflint, I utilized $1/(1-x)=1+x+x^2+cdots$.
– Sangchul Lee
Nov 20 '18 at 19:23
@cptflint, I utilized $1/(1-x)=1+x+x^2+cdots$.
– Sangchul Lee
Nov 20 '18 at 19:23
Okay! Thank you!
– cptflint
Nov 20 '18 at 19:30
Okay! Thank you!
– cptflint
Nov 20 '18 at 19:30
Mhm, I think we missed something. In every step we add $a_{n-1}o(1)$ so how can we proove the asymptotic behavior?
– cptflint
Nov 21 '18 at 10:48
Mhm, I think we missed something. In every step we add $a_{n-1}o(1)$ so how can we proove the asymptotic behavior?
– cptflint
Nov 21 '18 at 10:48
@cptflint, This is a typical application of Stolz–Cesàro theorem. Since $a_n - a_{n-1} to frac{b}{2}$ as $ntoinfty$, by Stolz–Cesàro theorem, $$ lim_{ntoinfty} frac{a_n}{n} = lim_{ntoinfty} frac{a_n - a_{n-1}}{n - (n-1)} = frac{b}{2}. $$
– Sangchul Lee
Nov 22 '18 at 1:57
@cptflint, This is a typical application of Stolz–Cesàro theorem. Since $a_n - a_{n-1} to frac{b}{2}$ as $ntoinfty$, by Stolz–Cesàro theorem, $$ lim_{ntoinfty} frac{a_n}{n} = lim_{ntoinfty} frac{a_n - a_{n-1}}{n - (n-1)} = frac{b}{2}. $$
– Sangchul Lee
Nov 22 '18 at 1:57
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If $b$ is sufficiently large (say $b = 5$ for $r_1 = 1$) then this will definitely just explode to negative infinity.
– Mees de Vries
Nov 20 '18 at 17:31
Yeah, right... Mhm. I am reading an old paper from Kolmogorov (On the solution of a problem in biology) and what he says is this: $frac{r_n}{r_{n-1}}=1-frac12 br_{n-1} + O(r_{n-2}^2)$ gives us $r_n sim frac{2}{nb}$. In this special case $b$ is the variance of a critical offspring distribution. $r_n $ is defined by $r_n=1-q^{(n)}(0)$ and $q^{(n+1)}(x):=q(q^n(x))$ for $q^1(x):=p_0+p_1x+p_2x^2+...$.
– cptflint
Nov 20 '18 at 17:40
This looks true under the form "asymptotic to $K dfrac{1}{n}$" (K=constant) for the very narrow range $b in (0,1)$ (and $r_1 in (0,1)$). Do like (good) journalists : check your sources !
– Jean Marie
Nov 20 '18 at 17:45
Ok so $r_1in (0,1)$ instead of $[0,1]$, thx! But: $bin(0,1)$ seems like a problem. Kolmogorov only chooses $b>0$. So I am missing something or that's an old fault...
– cptflint
Nov 20 '18 at 17:46
For $r_1$ it is unimportant, but you cannot say $b in (0,infty)$ : if I take b=3, 4, 5, 6..., the sequence explodes
– Jean Marie
Nov 20 '18 at 17:50