Convergence of third moments when second moments are uniformly bounded.
Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.
integration probability-theory measure-theory weak-convergence
asked Nov 27 '18 at 21:30
M. Ost
483
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Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.
integration probability-theory measure-theory weak-convergence
asked Nov 27 '18 at 21:30
M. Ost
483
add a comment |
Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.
integration probability-theory measure-theory weak-convergence
asked Nov 27 '18 at 21:30
M. Ost
483
Suppose $lambda_n, ngeq 1 $ are $sigma $-finite measures on $mathbb{R}$ such that $$ sup_{ngeq 1} int_{[-1,1]}x^2 lambda_n(dx) < infty $$ and $$forall delta > 0, exists n_0 geq 1, forall n geq n_0: lambda_n(mathbb{R} setminus [-delta,delta]) = 0. $$ I am trying to prove that $$lim_{nto infty} int_{mathbb{R}}vert xvert ^3 lambda_n(dx) = 0.$$ I came across this result in "Probability in Banach Spaces - Stable and Infinitely Divisible Distributions" by Werner Linde. It is stated in Proposition 5.6.1.
integration probability-theory measure-theory weak-convergence
integration probability-theory measure-theory weak-convergence
asked Nov 27 '18 at 21:30
M. Ost
483
asked Nov 27 '18 at 21:30
M. Ost
483
asked Nov 27 '18 at 21:30
M. Ost
483
asked Nov 27 '18 at 21:30
M. Ost
483
asked Nov 27 '18 at 21:30
M. Ost
483
483
add a comment |
add a comment |
1 Answer
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Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
$$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
$$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
Thus, for $ngeq max{ N, n_0(N) } $
$$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
leq 0 + frac{c}{n} < varepsilon. $$
Hence, we get
$$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
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1 Answer
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Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
$$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
$$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
Thus, for $ngeq max{ N, n_0(N) } $
$$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
leq 0 + frac{c}{n} < varepsilon. $$
Hence, we get
$$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
add a comment |
Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
$$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
$$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
Thus, for $ngeq max{ N, n_0(N) } $
$$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
leq 0 + frac{c}{n} < varepsilon. $$
Hence, we get
$$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
add a comment |
Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
$$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
$$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
Thus, for $ngeq max{ N, n_0(N) } $
$$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
leq 0 + frac{c}{n} < varepsilon. $$
Hence, we get
$$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
Let $c:=sup_{ngeq 1} int_{[-1,1]} x^2 lambda_n(dx) $. Then we have
$$ int_{[-1/m, 1/m]} vert x vert^3 lambda_n(dx) leq int_{[-1/m, 1/m]} frac{1}{m}vert x vert^2 lambda_n(dx) leq frac{c}{m} $$
On the other hand there exists $n_0(m)in mathbb{N}$ such that for all $ngeq n_0$ holds
$$ lambda_n(mathbb{R}setminus [-frac{1}{m}, frac{1}{m}]) =0. $$
Let $varepsilon>0$ and pick $Nin mathbb{N}$ such that $c/N<varepsilon$.
Thus, for $ngeq max{ N, n_0(N) } $
$$ int_mathbb{R} vert x vert^3 lambda_n(dx) = int_{mathbb{R}setminus [-1/n, 1/n]} vert x vert^3 lambda_n(dx) + int_{[-1/n, 1/n]} vert x vert^3 lambda_n(dx)
leq 0 + frac{c}{n} < varepsilon. $$
Hence, we get
$$ lim_{nrightarrow infty} int_mathbb{R} vert x vert^3 lambda_n(dx) = 0. $$
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
edited Nov 27 '18 at 22:27
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
answered Nov 27 '18 at 21:56
Severin Schraven
5,8281934
5,8281934
add a comment |
add a comment |
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