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Showing $rho = -1 + int_0^infty frace^-y1+delta y dy = sum_k=1^infty (-1)^k delta^k k!$
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Trying to understand the expansion in the title, i.e.
$$
rho = -1 + int_0^infty frace^-y1+delta y dy = sum_k=1^infty (-1)^k delta^k k!.
$$
The paper claims to "expand in increasing powers of $delta y$"; we have $0 leq delta leq 1$.
Any hints are greatly appreciated.
Background information (can skip):
$rho$ is the coefficient of correlation between two random variables $X$ and $Y$, where $(X, Y)$ follows a specific bivariate distribution, with dependence given through $delta$. Above equivalence serves the purpose to proof that the limit for $deltarightarrow 0$ of the correlation ratio
$$eta^2 = fracdelta3 - frac16-fracrho6delta$$
is $0$, since from $rho = sum_k=1^infty (-1)^k delta^k k!$ we have for $delta=0$ that $fracrhodelta = -1$.
real-analysis sequences-and-series taylor-expansion
asked Jul 20 at 22:30
InterestedStudent
306
add a comment |Â
up vote
0
down vote
favorite
Trying to understand the expansion in the title, i.e.
$$
rho = -1 + int_0^infty frace^-y1+delta y dy = sum_k=1^infty (-1)^k delta^k k!.
$$
The paper claims to "expand in increasing powers of $delta y$"; we have $0 leq delta leq 1$.
Any hints are greatly appreciated.
Background information (can skip):
$rho$ is the coefficient of correlation between two random variables $X$ and $Y$, where $(X, Y)$ follows a specific bivariate distribution, with dependence given through $delta$. Above equivalence serves the purpose to proof that the limit for $deltarightarrow 0$ of the correlation ratio
$$eta^2 = fracdelta3 - frac16-fracrho6delta$$
is $0$, since from $rho = sum_k=1^infty (-1)^k delta^k k!$ we have for $delta=0$ that $fracrhodelta = -1$.
real-analysis sequences-and-series taylor-expansion
asked Jul 20 at 22:30
InterestedStudent
306
2
The integral converges; the series diverges. They cannot be equal.
– Mark Viola
Jul 20 at 22:54
1
The suggested expansion is not true as convergent series as pointed out by several users. Rather, it can be understood as a much weaker statement that: For each fixed $N$, we have $$-1 + int_0^infty frace^-y1+delta y , dy = sum_k=1^N (-1)^k delta^k k! + mathcalO(delta^N+1) quad textas delta to 0^+.$$
– Sangchul Lee
Jul 21 at 8:24
Did you get this result by using integration by parts along the lines of Chappers' answer?
– InterestedStudent
Jul 22 at 10:18
1
That is one possible option, but in this case we can exploit the following formula for a quick proof: $$ frac11+delta y=sum_k=0^N(-delta y)^k + frac(-delta y)^N+11+delta y. $$
– Sangchul Lee
Jul 23 at 13:19
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Trying to understand the expansion in the title, i.e.
$$
rho = -1 + int_0^infty frace^-y1+delta y dy = sum_k=1^infty (-1)^k delta^k k!.
$$
The paper claims to "expand in increasing powers of $delta y$"; we have $0 leq delta leq 1$.
Any hints are greatly appreciated.
Background information (can skip):
$rho$ is the coefficient of correlation between two random variables $X$ and $Y$, where $(X, Y)$ follows a specific bivariate distribution, with dependence given through $delta$. Above equivalence serves the purpose to proof that the limit for $deltarightarrow 0$ of the correlation ratio
$$eta^2 = fracdelta3 - frac16-fracrho6delta$$
is $0$, since from $rho = sum_k=1^infty (-1)^k delta^k k!$ we have for $delta=0$ that $fracrhodelta = -1$.
real-analysis sequences-and-series taylor-expansion
asked Jul 20 at 22:30
InterestedStudent
306
Trying to understand the expansion in the title, i.e.
$$
rho = -1 + int_0^infty frace^-y1+delta y dy = sum_k=1^infty (-1)^k delta^k k!.
$$
The paper claims to "expand in increasing powers of $delta y$"; we have $0 leq delta leq 1$.
Any hints are greatly appreciated.
Background information (can skip):
$rho$ is the coefficient of correlation between two random variables $X$ and $Y$, where $(X, Y)$ follows a specific bivariate distribution, with dependence given through $delta$. Above equivalence serves the purpose to proof that the limit for $deltarightarrow 0$ of the correlation ratio
$$eta^2 = fracdelta3 - frac16-fracrho6delta$$
is $0$, since from $rho = sum_k=1^infty (-1)^k delta^k k!$ we have for $delta=0$ that $fracrhodelta = -1$.
real-analysis sequences-and-series taylor-expansion
asked Jul 20 at 22:30
InterestedStudent
306
edited Jul 22 at 9:34
asked Jul 20 at 22:30
InterestedStudent
306
asked Jul 20 at 22:30
InterestedStudent
306
asked Jul 20 at 22:30
InterestedStudent
306
306
2
The integral converges; the series diverges. They cannot be equal.
– Mark Viola
Jul 20 at 22:54
1
The suggested expansion is not true as convergent series as pointed out by several users. Rather, it can be understood as a much weaker statement that: For each fixed $N$, we have $$-1 + int_0^infty frace^-y1+delta y , dy = sum_k=1^N (-1)^k delta^k k! + mathcalO(delta^N+1) quad textas delta to 0^+.$$
– Sangchul Lee
Jul 21 at 8:24
Did you get this result by using integration by parts along the lines of Chappers' answer?
– InterestedStudent
Jul 22 at 10:18
1
That is one possible option, but in this case we can exploit the following formula for a quick proof: $$ frac11+delta y=sum_k=0^N(-delta y)^k + frac(-delta y)^N+11+delta y. $$
– Sangchul Lee
Jul 23 at 13:19
add a comment |Â
2
The integral converges; the series diverges. They cannot be equal.
– Mark Viola
Jul 20 at 22:54
1
The suggested expansion is not true as convergent series as pointed out by several users. Rather, it can be understood as a much weaker statement that: For each fixed $N$, we have $$-1 + int_0^infty frace^-y1+delta y , dy = sum_k=1^N (-1)^k delta^k k! + mathcalO(delta^N+1) quad textas delta to 0^+.$$
– Sangchul Lee
Jul 21 at 8:24
Did you get this result by using integration by parts along the lines of Chappers' answer?
– InterestedStudent
Jul 22 at 10:18
1
That is one possible option, but in this case we can exploit the following formula for a quick proof: $$ frac11+delta y=sum_k=0^N(-delta y)^k + frac(-delta y)^N+11+delta y. $$
– Sangchul Lee
Jul 23 at 13:19
2
2
The integral converges; the series diverges. They cannot be equal.
– Mark Viola
Jul 20 at 22:54
The integral converges; the series diverges. They cannot be equal.
– Mark Viola
Jul 20 at 22:54
1
1
The suggested expansion is not true as convergent series as pointed out by several users. Rather, it can be understood as a much weaker statement that: For each fixed $N$, we have $$-1 + int_0^infty frace^-y1+delta y , dy = sum_k=1^N (-1)^k delta^k k! + mathcalO(delta^N+1) quad textas delta to 0^+.$$
– Sangchul Lee
Jul 21 at 8:24
The suggested expansion is not true as convergent series as pointed out by several users. Rather, it can be understood as a much weaker statement that: For each fixed $N$, we have $$-1 + int_0^infty frace^-y1+delta y , dy = sum_k=1^N (-1)^k delta^k k! + mathcalO(delta^N+1) quad textas delta to 0^+.$$
– Sangchul Lee
Jul 21 at 8:24
Did you get this result by using integration by parts along the lines of Chappers' answer?
– InterestedStudent
Jul 22 at 10:18
Did you get this result by using integration by parts along the lines of Chappers' answer?
– InterestedStudent
Jul 22 at 10:18
1
1
That is one possible option, but in this case we can exploit the following formula for a quick proof: $$ frac11+delta y=sum_k=0^N(-delta y)^k + frac(-delta y)^N+11+delta y. $$
– Sangchul Lee
Jul 23 at 13:19
That is one possible option, but in this case we can exploit the following formula for a quick proof: $$ frac11+delta y=sum_k=0^N(-delta y)^k + frac(-delta y)^N+11+delta y. $$
– Sangchul Lee
Jul 23 at 13:19
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
As stated the first equation is simply wrong: it holds only for $delta=0$, otherwise the right-hand side is a divergent series. The correct thing to do to examine the behaviour for small $delta$ in these situations is to integrate by parts: we have
$$ -1+ int_0^infty frace^-y1+delta y , dy = left[ frac-e^-y1+delta y right]_0^infty - delta int_0^infty frace^-y(1+delta y)^2 , dy \
= -1+1 - delta left( left[ frac-e^-y(1+delta y)^2 right]_0^infty - delta int_0^infty frac2e^-y(1+delta y)^3 , dy right) \
= -delta +delta^2 int_0^infty frac2e^-y(1+delta y)^3 , dy $$
This last integral is finite, and for $delta>0$, the denominator is larger than $1$, so it is bounded above by $int_0^infty 2e^-y , dy = 2$, so as $delta downarrow 0$,
$$ rho = -delta + O(delta^2), $$
which is enough to take the limit you need.
answered Jul 20 at 22:57
Chappers
55k74190
add a comment |Â
up vote
0
down vote
What could be useful (I hope) is to consider (simple substitution $y=t-frac1delta $)
$$I=int frace^-y1+delta y dy=frace^frac1delta delta int frace^-tt dt=frace^frac1delta delta textEileft(-tright)=frace^frac1delta delta textEileft(-y-frac1delta right)$$ where appears the exponential integral function.
$$J=int_0^infty frace^-y1+delta y dy=frace^frac1delta delta Gamma left(0,frac1delta right)$$ where appears the incomplete gamma function.
Now, using the asymptotics of $Gamma left(0,zright)$ when $z to infty$
$$J=sum_k=0^infty (-1)^k, k!, delta ^k$$ and then the result.
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
Should your first line not be $$frace^frac1deltadelta int_frac1delta^infty frace^-tt dt = - frace^frac1deltadelta textEileft(-frac1deltaright) = frace^frac1deltadelta Gammaleft(0, frac1deltaright)?$$ Just confused me a bit with that $t$ and $y$ in the argument of Ei. Can you elaborate on how I can use the asymptotics of $Gamma(0, z)$ to end up with the result?
– InterestedStudent
Jul 22 at 10:13
Or better: [...] to end up with the correct result pointed out by Chappers and Sangchul Lee.
– InterestedStudent
Jul 22 at 10:24
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As stated the first equation is simply wrong: it holds only for $delta=0$, otherwise the right-hand side is a divergent series. The correct thing to do to examine the behaviour for small $delta$ in these situations is to integrate by parts: we have
$$ -1+ int_0^infty frace^-y1+delta y , dy = left[ frac-e^-y1+delta y right]_0^infty - delta int_0^infty frace^-y(1+delta y)^2 , dy \
= -1+1 - delta left( left[ frac-e^-y(1+delta y)^2 right]_0^infty - delta int_0^infty frac2e^-y(1+delta y)^3 , dy right) \
= -delta +delta^2 int_0^infty frac2e^-y(1+delta y)^3 , dy $$
This last integral is finite, and for $delta>0$, the denominator is larger than $1$, so it is bounded above by $int_0^infty 2e^-y , dy = 2$, so as $delta downarrow 0$,
$$ rho = -delta + O(delta^2), $$
which is enough to take the limit you need.
answered Jul 20 at 22:57
Chappers
55k74190
add a comment |Â
up vote
1
down vote
accepted
As stated the first equation is simply wrong: it holds only for $delta=0$, otherwise the right-hand side is a divergent series. The correct thing to do to examine the behaviour for small $delta$ in these situations is to integrate by parts: we have
$$ -1+ int_0^infty frace^-y1+delta y , dy = left[ frac-e^-y1+delta y right]_0^infty - delta int_0^infty frace^-y(1+delta y)^2 , dy \
= -1+1 - delta left( left[ frac-e^-y(1+delta y)^2 right]_0^infty - delta int_0^infty frac2e^-y(1+delta y)^3 , dy right) \
= -delta +delta^2 int_0^infty frac2e^-y(1+delta y)^3 , dy $$
This last integral is finite, and for $delta>0$, the denominator is larger than $1$, so it is bounded above by $int_0^infty 2e^-y , dy = 2$, so as $delta downarrow 0$,
$$ rho = -delta + O(delta^2), $$
which is enough to take the limit you need.
answered Jul 20 at 22:57
Chappers
55k74190
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As stated the first equation is simply wrong: it holds only for $delta=0$, otherwise the right-hand side is a divergent series. The correct thing to do to examine the behaviour for small $delta$ in these situations is to integrate by parts: we have
$$ -1+ int_0^infty frace^-y1+delta y , dy = left[ frac-e^-y1+delta y right]_0^infty - delta int_0^infty frace^-y(1+delta y)^2 , dy \
= -1+1 - delta left( left[ frac-e^-y(1+delta y)^2 right]_0^infty - delta int_0^infty frac2e^-y(1+delta y)^3 , dy right) \
= -delta +delta^2 int_0^infty frac2e^-y(1+delta y)^3 , dy $$
This last integral is finite, and for $delta>0$, the denominator is larger than $1$, so it is bounded above by $int_0^infty 2e^-y , dy = 2$, so as $delta downarrow 0$,
$$ rho = -delta + O(delta^2), $$
which is enough to take the limit you need.
answered Jul 20 at 22:57
Chappers
55k74190
As stated the first equation is simply wrong: it holds only for $delta=0$, otherwise the right-hand side is a divergent series. The correct thing to do to examine the behaviour for small $delta$ in these situations is to integrate by parts: we have
$$ -1+ int_0^infty frace^-y1+delta y , dy = left[ frac-e^-y1+delta y right]_0^infty - delta int_0^infty frace^-y(1+delta y)^2 , dy \
= -1+1 - delta left( left[ frac-e^-y(1+delta y)^2 right]_0^infty - delta int_0^infty frac2e^-y(1+delta y)^3 , dy right) \
= -delta +delta^2 int_0^infty frac2e^-y(1+delta y)^3 , dy $$
This last integral is finite, and for $delta>0$, the denominator is larger than $1$, so it is bounded above by $int_0^infty 2e^-y , dy = 2$, so as $delta downarrow 0$,
$$ rho = -delta + O(delta^2), $$
which is enough to take the limit you need.
answered Jul 20 at 22:57
Chappers
55k74190
answered Jul 20 at 22:57
Chappers
55k74190
answered Jul 20 at 22:57
Chappers
55k74190
answered Jul 20 at 22:57
Chappers
55k74190
55k74190
add a comment |Â
add a comment |Â
up vote
0
down vote
What could be useful (I hope) is to consider (simple substitution $y=t-frac1delta $)
$$I=int frace^-y1+delta y dy=frace^frac1delta delta int frace^-tt dt=frace^frac1delta delta textEileft(-tright)=frace^frac1delta delta textEileft(-y-frac1delta right)$$ where appears the exponential integral function.
$$J=int_0^infty frace^-y1+delta y dy=frace^frac1delta delta Gamma left(0,frac1delta right)$$ where appears the incomplete gamma function.
Now, using the asymptotics of $Gamma left(0,zright)$ when $z to infty$
$$J=sum_k=0^infty (-1)^k, k!, delta ^k$$ and then the result.
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
Should your first line not be $$frace^frac1deltadelta int_frac1delta^infty frace^-tt dt = - frace^frac1deltadelta textEileft(-frac1deltaright) = frace^frac1deltadelta Gammaleft(0, frac1deltaright)?$$ Just confused me a bit with that $t$ and $y$ in the argument of Ei. Can you elaborate on how I can use the asymptotics of $Gamma(0, z)$ to end up with the result?
– InterestedStudent
Jul 22 at 10:13
Or better: [...] to end up with the correct result pointed out by Chappers and Sangchul Lee.
– InterestedStudent
Jul 22 at 10:24
add a comment |Â
up vote
0
down vote
What could be useful (I hope) is to consider (simple substitution $y=t-frac1delta $)
$$I=int frace^-y1+delta y dy=frace^frac1delta delta int frace^-tt dt=frace^frac1delta delta textEileft(-tright)=frace^frac1delta delta textEileft(-y-frac1delta right)$$ where appears the exponential integral function.
$$J=int_0^infty frace^-y1+delta y dy=frace^frac1delta delta Gamma left(0,frac1delta right)$$ where appears the incomplete gamma function.
Now, using the asymptotics of $Gamma left(0,zright)$ when $z to infty$
$$J=sum_k=0^infty (-1)^k, k!, delta ^k$$ and then the result.
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
Should your first line not be $$frace^frac1deltadelta int_frac1delta^infty frace^-tt dt = - frace^frac1deltadelta textEileft(-frac1deltaright) = frace^frac1deltadelta Gammaleft(0, frac1deltaright)?$$ Just confused me a bit with that $t$ and $y$ in the argument of Ei. Can you elaborate on how I can use the asymptotics of $Gamma(0, z)$ to end up with the result?
– InterestedStudent
Jul 22 at 10:13
Or better: [...] to end up with the correct result pointed out by Chappers and Sangchul Lee.
– InterestedStudent
Jul 22 at 10:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What could be useful (I hope) is to consider (simple substitution $y=t-frac1delta $)
$$I=int frace^-y1+delta y dy=frace^frac1delta delta int frace^-tt dt=frace^frac1delta delta textEileft(-tright)=frace^frac1delta delta textEileft(-y-frac1delta right)$$ where appears the exponential integral function.
$$J=int_0^infty frace^-y1+delta y dy=frace^frac1delta delta Gamma left(0,frac1delta right)$$ where appears the incomplete gamma function.
Now, using the asymptotics of $Gamma left(0,zright)$ when $z to infty$
$$J=sum_k=0^infty (-1)^k, k!, delta ^k$$ and then the result.
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
What could be useful (I hope) is to consider (simple substitution $y=t-frac1delta $)
$$I=int frace^-y1+delta y dy=frace^frac1delta delta int frace^-tt dt=frace^frac1delta delta textEileft(-tright)=frace^frac1delta delta textEileft(-y-frac1delta right)$$ where appears the exponential integral function.
$$J=int_0^infty frace^-y1+delta y dy=frace^frac1delta delta Gamma left(0,frac1delta right)$$ where appears the incomplete gamma function.
Now, using the asymptotics of $Gamma left(0,zright)$ when $z to infty$
$$J=sum_k=0^infty (-1)^k, k!, delta ^k$$ and then the result.
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
answered Jul 21 at 3:24
Claude Leibovici
111k1055126
111k1055126
Should your first line not be $$frace^frac1deltadelta int_frac1delta^infty frace^-tt dt = - frace^frac1deltadelta textEileft(-frac1deltaright) = frace^frac1deltadelta Gammaleft(0, frac1deltaright)?$$ Just confused me a bit with that $t$ and $y$ in the argument of Ei. Can you elaborate on how I can use the asymptotics of $Gamma(0, z)$ to end up with the result?
– InterestedStudent
Jul 22 at 10:13
Or better: [...] to end up with the correct result pointed out by Chappers and Sangchul Lee.
– InterestedStudent
Jul 22 at 10:24
add a comment |Â
Should your first line not be $$frace^frac1deltadelta int_frac1delta^infty frace^-tt dt = - frace^frac1deltadelta textEileft(-frac1deltaright) = frace^frac1deltadelta Gammaleft(0, frac1deltaright)?$$ Just confused me a bit with that $t$ and $y$ in the argument of Ei. Can you elaborate on how I can use the asymptotics of $Gamma(0, z)$ to end up with the result?
– InterestedStudent
Jul 22 at 10:13
Or better: [...] to end up with the correct result pointed out by Chappers and Sangchul Lee.
– InterestedStudent
Jul 22 at 10:24
Should your first line not be $$frace^frac1deltadelta int_frac1delta^infty frace^-tt dt = - frace^frac1deltadelta textEileft(-frac1deltaright) = frace^frac1deltadelta Gammaleft(0, frac1deltaright)?$$ Just confused me a bit with that $t$ and $y$ in the argument of Ei. Can you elaborate on how I can use the asymptotics of $Gamma(0, z)$ to end up with the result?
– InterestedStudent
Jul 22 at 10:13
Should your first line not be $$frace^frac1deltadelta int_frac1delta^infty frace^-tt dt = - frace^frac1deltadelta textEileft(-frac1deltaright) = frace^frac1deltadelta Gammaleft(0, frac1deltaright)?$$ Just confused me a bit with that $t$ and $y$ in the argument of Ei. Can you elaborate on how I can use the asymptotics of $Gamma(0, z)$ to end up with the result?
– InterestedStudent
Jul 22 at 10:13
Or better: [...] to end up with the correct result pointed out by Chappers and Sangchul Lee.
– InterestedStudent
Jul 22 at 10:24
Or better: [...] to end up with the correct result pointed out by Chappers and Sangchul Lee.
– InterestedStudent
Jul 22 at 10:24
add a comment |Â
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2
The integral converges; the series diverges. They cannot be equal.
– Mark Viola
Jul 20 at 22:54
1
The suggested expansion is not true as convergent series as pointed out by several users. Rather, it can be understood as a much weaker statement that: For each fixed $N$, we have $$-1 + int_0^infty frace^-y1+delta y , dy = sum_k=1^N (-1)^k delta^k k! + mathcalO(delta^N+1) quad textas delta to 0^+.$$
– Sangchul Lee
Jul 21 at 8:24
Did you get this result by using integration by parts along the lines of Chappers' answer?
– InterestedStudent
Jul 22 at 10:18
1
That is one possible option, but in this case we can exploit the following formula for a quick proof: $$ frac11+delta y=sum_k=0^N(-delta y)^k + frac(-delta y)^N+11+delta y. $$
– Sangchul Lee
Jul 23 at 13:19