Almost complete convergence
$begingroup$
In the book nonparametric functionnal data analysis, page 232, i don't understand why:
The almost complete convergence of $Y_n$ to $lne 0$ implies that there
exists some $delta > 0$ (choose for instance $delta = l/2$) such that
$$sum_{ige 0}mathsf{P}(|Y_n| le delta) < infty.$$
Cordially.
probability-theory statistics measure-theory
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
asked Dec 14 '18 at 8:49
estimationestimation
12
$endgroup$
add a comment |
$begingroup$
In the book nonparametric functionnal data analysis, page 232, i don't understand why:
The almost complete convergence of $Y_n$ to $lne 0$ implies that there
exists some $delta > 0$ (choose for instance $delta = l/2$) such that
$$sum_{ige 0}mathsf{P}(|Y_n| le delta) < infty.$$
Cordially.
probability-theory statistics measure-theory
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
asked Dec 14 '18 at 8:49
estimationestimation
12
$endgroup$
1
$begingroup$
What does almost complete convergence mean?
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:52
$begingroup$
Thank you very much
$endgroup$
– estimation
Dec 14 '18 at 13:00
$begingroup$
almost complete convergence of (Yn) to Y MEAN for each δ > 0: $$∑n∈N P(|Yn-Y | >δ)<∞.$$
$endgroup$
– estimation
Dec 14 '18 at 13:06
$begingroup$
the sum over n∈N, of P(|Yn−Y|>δ) is finite.
$endgroup$
– estimation
Dec 14 '18 at 13:22
add a comment |
$begingroup$
In the book nonparametric functionnal data analysis, page 232, i don't understand why:
The almost complete convergence of $Y_n$ to $lne 0$ implies that there
exists some $delta > 0$ (choose for instance $delta = l/2$) such that
$$sum_{ige 0}mathsf{P}(|Y_n| le delta) < infty.$$
Cordially.
probability-theory statistics measure-theory
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
asked Dec 14 '18 at 8:49
estimationestimation
12
$endgroup$
In the book nonparametric functionnal data analysis, page 232, i don't understand why:
The almost complete convergence of $Y_n$ to $lne 0$ implies that there
exists some $delta > 0$ (choose for instance $delta = l/2$) such that
$$sum_{ige 0}mathsf{P}(|Y_n| le delta) < infty.$$
Cordially.
probability-theory statistics measure-theory
probability-theory statistics measure-theory
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
asked Dec 14 '18 at 8:49
estimationestimation
12
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
asked Dec 14 '18 at 8:49
estimationestimation
12
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
edited Dec 14 '18 at 9:33
d.k.o.
9,520628
9,520628
asked Dec 14 '18 at 8:49
estimationestimation
12
asked Dec 14 '18 at 8:49
estimationestimation
12
asked Dec 14 '18 at 8:49
estimationestimation
12
12
1
$begingroup$
What does almost complete convergence mean?
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:52
$begingroup$
Thank you very much
$endgroup$
– estimation
Dec 14 '18 at 13:00
$begingroup$
almost complete convergence of (Yn) to Y MEAN for each δ > 0: $$∑n∈N P(|Yn-Y | >δ)<∞.$$
$endgroup$
– estimation
Dec 14 '18 at 13:06
$begingroup$
the sum over n∈N, of P(|Yn−Y|>δ) is finite.
$endgroup$
– estimation
Dec 14 '18 at 13:22
add a comment |
1
$begingroup$
What does almost complete convergence mean?
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:52
$begingroup$
Thank you very much
$endgroup$
– estimation
Dec 14 '18 at 13:00
$begingroup$
almost complete convergence of (Yn) to Y MEAN for each δ > 0: $$∑n∈N P(|Yn-Y | >δ)<∞.$$
$endgroup$
– estimation
Dec 14 '18 at 13:06
$begingroup$
the sum over n∈N, of P(|Yn−Y|>δ) is finite.
$endgroup$
– estimation
Dec 14 '18 at 13:22
1
1
$begingroup$
What does almost complete convergence mean?
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:52
$begingroup$
What does almost complete convergence mean?
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:52
$begingroup$
Thank you very much
$endgroup$
– estimation
Dec 14 '18 at 13:00
$begingroup$
Thank you very much
$endgroup$
– estimation
Dec 14 '18 at 13:00
$begingroup$
almost complete convergence of (Yn) to Y MEAN for each δ > 0: $$∑n∈N P(|Yn-Y | >δ)<∞.$$
$endgroup$
– estimation
Dec 14 '18 at 13:06
$begingroup$
almost complete convergence of (Yn) to Y MEAN for each δ > 0: $$∑n∈N P(|Yn-Y | >δ)<∞.$$
$endgroup$
– estimation
Dec 14 '18 at 13:06
$begingroup$
the sum over n∈N, of P(|Yn−Y|>δ) is finite.
$endgroup$
– estimation
Dec 14 '18 at 13:22
$begingroup$
the sum over n∈N, of P(|Yn−Y|>δ) is finite.
$endgroup$
– estimation
Dec 14 '18 at 13:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $l>0$. Since
begin{align}
mathsf{P}(|Y_n|le l/2)&=mathsf{P}(-l/2le Y_nle l/2)\
&le mathsf{P}(Y_nle l-l/2)le mathsf{P}(|Y_n-l|ge l/2),
end{align}
$$
sum_{nge 1}mathsf{P}(|Y_n|le l/2)le sum_{nge 1}mathsf{P}(|Y_n-l|ge l/2)<infty.
$$
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose that $l>0$. Since
begin{align}
mathsf{P}(|Y_n|le l/2)&=mathsf{P}(-l/2le Y_nle l/2)\
&le mathsf{P}(Y_nle l-l/2)le mathsf{P}(|Y_n-l|ge l/2),
end{align}
$$
sum_{nge 1}mathsf{P}(|Y_n|le l/2)le sum_{nge 1}mathsf{P}(|Y_n-l|ge l/2)<infty.
$$
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
$endgroup$
add a comment |
$begingroup$
Suppose that $l>0$. Since
begin{align}
mathsf{P}(|Y_n|le l/2)&=mathsf{P}(-l/2le Y_nle l/2)\
&le mathsf{P}(Y_nle l-l/2)le mathsf{P}(|Y_n-l|ge l/2),
end{align}
$$
sum_{nge 1}mathsf{P}(|Y_n|le l/2)le sum_{nge 1}mathsf{P}(|Y_n-l|ge l/2)<infty.
$$
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
$endgroup$
add a comment |
$begingroup$
Suppose that $l>0$. Since
begin{align}
mathsf{P}(|Y_n|le l/2)&=mathsf{P}(-l/2le Y_nle l/2)\
&le mathsf{P}(Y_nle l-l/2)le mathsf{P}(|Y_n-l|ge l/2),
end{align}
$$
sum_{nge 1}mathsf{P}(|Y_n|le l/2)le sum_{nge 1}mathsf{P}(|Y_n-l|ge l/2)<infty.
$$
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
$endgroup$
Suppose that $l>0$. Since
begin{align}
mathsf{P}(|Y_n|le l/2)&=mathsf{P}(-l/2le Y_nle l/2)\
&le mathsf{P}(Y_nle l-l/2)le mathsf{P}(|Y_n-l|ge l/2),
end{align}
$$
sum_{nge 1}mathsf{P}(|Y_n|le l/2)le sum_{nge 1}mathsf{P}(|Y_n-l|ge l/2)<infty.
$$
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
answered Dec 14 '18 at 9:31
d.k.o.d.k.o.
9,520628
9,520628
add a comment |
add a comment |
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1
$begingroup$
What does almost complete convergence mean?
$endgroup$
– Kavi Rama Murthy
Dec 14 '18 at 8:52
$begingroup$
Thank you very much
$endgroup$
– estimation
Dec 14 '18 at 13:00
$begingroup$
almost complete convergence of (Yn) to Y MEAN for each δ > 0: $$∑n∈N P(|Yn-Y | >δ)<∞.$$
$endgroup$
– estimation
Dec 14 '18 at 13:06
$begingroup$
the sum over n∈N, of P(|Yn−Y|>δ) is finite.
$endgroup$
– estimation
Dec 14 '18 at 13:22