Mixing
A Cauchy-Schwarz-type inequality for $intprod_n|f_n|$
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If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $
If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?
real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
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add a comment |
$begingroup$
If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $
If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?
real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
$endgroup$
2
$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
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– b00n heT
Jan 2 at 8:33
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Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
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– enthdegree
Jan 2 at 8:36
1
$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37
add a comment |
$begingroup$
If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $
If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?
real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
$endgroup$
If $X_1,X_2$ have finite second moments then Cauchy-Schwarz gives $langle |X_1||X_2|rangle^2 leq langle |X_1|^2rangle langle |X_2|^2rangle $
If $(X_n)_{n=1}^N$ have their $N$th moments is it so that $langleprod_n|X_n|rangle^N leq prod_nlangle |X_n|^Nrangle $?
real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality
real-analysis probability-theory reference-request cauchy-schwarz-inequality holder-inequality
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
edited Jan 2 at 8:36
enthdegree
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
asked Jan 2 at 8:13
enthdegreeenthdegree
2,61821335
2,61821335
2
$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
$endgroup$
– b00n heT
Jan 2 at 8:33
$begingroup$
Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
$endgroup$
– enthdegree
Jan 2 at 8:36
1
$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37
add a comment |
2
$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
$endgroup$
– b00n heT
Jan 2 at 8:33
$begingroup$
Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
$endgroup$
– enthdegree
Jan 2 at 8:36
1
$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37
2
2
$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
$endgroup$
– b00n heT
Jan 2 at 8:33
$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
$endgroup$
– b00n heT
Jan 2 at 8:33
$begingroup$
Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
$endgroup$
– enthdegree
Jan 2 at 8:36
$begingroup$
Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
$endgroup$
– enthdegree
Jan 2 at 8:36
1
1
$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37
$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37
add a comment |
1 Answer
1
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This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
begin{align*}
leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
&leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
end{align*}
where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.
Taking all this to the power $N+1$ you get your desired result.
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
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1 Answer
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1 Answer
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$begingroup$
This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
begin{align*}
leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
&leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
end{align*}
where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.
Taking all this to the power $N+1$ you get your desired result.
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
$endgroup$
add a comment |
$begingroup$
This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
begin{align*}
leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
&leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
end{align*}
where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.
Taking all this to the power $N+1$ you get your desired result.
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
$endgroup$
add a comment |
$begingroup$
This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
begin{align*}
leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
&leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
end{align*}
where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.
Taking all this to the power $N+1$ you get your desired result.
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
$endgroup$
This can be proven using Holder's inequality. Let's do it by induction, obviously this is true for $N=1,2$ so I focus on the induction step. Suppose we are given $(X_n)_{n=1}^{N+1}$ and have finite $N+1$'th moments, then
begin{align*}
leftlangle prod_{n=1}^{N+1} |X_n|rightrangle&leq leftlangle prod_{n=1}^N |X_n|^{frac{N+1}{N}} rightrangle^{frac{N}{N+1}}cdotlangle |X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
&leq prod_{n=1}^{N+1} langle |X_n|^{N+1} rangle^{frac{1}{N+1}} cdot langle|X_{N+1}|^{N+1} rangle^{frac{1}{N+1}}\
end{align*}
where the first inequality is Holder's inequality with $p=frac{N+1}{N}$ and $q=N+1$, the second one is induction hypothesis.
Taking all this to the power $N+1$ you get your desired result.
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
answered Jan 2 at 8:40
P. QuintonP. Quinton
1,603213
1,603213
add a comment |
add a comment |
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$begingroup$
I think this is probably what you are looking for: en.wikipedia.org/wiki/…
$endgroup$
– b00n heT
Jan 2 at 8:33
$begingroup$
Thanks, it does appear much closer to Holder's inequality. I will think about it in the morning.
$endgroup$
– enthdegree
Jan 2 at 8:36
1
$begingroup$
Cauchy-Schwarz is indeed a special case of Hölder's inequality. Good luck :)
$endgroup$
– b00n heT
Jan 2 at 8:37