Mixing
proofing the strong law of large numbers?
$begingroup$
I beg your pardon. But I am not going to write down the statement of the law proofs and so on, because I think every one knows what I am talking about and because I am a hobby mathemation I do not think I would write this text correct anyway .
When i looked up the proof for the weak law it ended like this:
$Pr( left| overline{X}_n-mu right| geq varepsilon) leq frac{sigma^2}{nvarepsilon^2}rightarrow 0$, where the random variable $overline{X}_n$ has a variance of $frac{sigma^2}{n}$.
An equivalent definition of the weak law of large numbers, using chebychev inequality for the proof is:
A random variable $X_n$ with $mu=0$ and $var(X_n)=sigma^2$
converges in probability to zero if
$var(X_n)rightarrow0$.
Now I have the following idea:
$var(X^n)=E[X^{2n}]-E[X^n]^2 geq 0 iff frac{var(X^n)}{E[X^{2n}]}=1-frac{E[X^n]^2}{E[X^{2n}]} geq 0 implieslimsup frac{E[X^n]^2}{E[X^{2n}]} leq 1 $.
Now $var(X^2)implies frac{E[X^n]^2}{E[X^{2n}]}=frac{var(X)^2}{E[X^{4}]} iff var(X)^2 = O(E[X^{4}]) $
Now we use the inequality and the variance used in the proof of the weak law.
$Pr(|X_n|>epsilon)=Pr(|X_n|^2>epsilon^2 leq frac{E[X^{4}]}{epsilon^4} $.
Now summing up the probabilities
$O(sum_{}^{infty} frac{E[X^{4}]}{epsilon^4}) =sum_{}^{infty}frac{var(x)^2}{epsilon^4}=sum_{}^{infty}frac{sigma^4}{n^2epsilon^4}<infty$ which implies almost sure convergance.
I hope you can understand what I was writing and would like to know if my idea is somehow correct. Thanks
probability-theory
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
$endgroup$
add a comment |
$begingroup$
I beg your pardon. But I am not going to write down the statement of the law proofs and so on, because I think every one knows what I am talking about and because I am a hobby mathemation I do not think I would write this text correct anyway .
When i looked up the proof for the weak law it ended like this:
$Pr( left| overline{X}_n-mu right| geq varepsilon) leq frac{sigma^2}{nvarepsilon^2}rightarrow 0$, where the random variable $overline{X}_n$ has a variance of $frac{sigma^2}{n}$.
An equivalent definition of the weak law of large numbers, using chebychev inequality for the proof is:
A random variable $X_n$ with $mu=0$ and $var(X_n)=sigma^2$
converges in probability to zero if
$var(X_n)rightarrow0$.
Now I have the following idea:
$var(X^n)=E[X^{2n}]-E[X^n]^2 geq 0 iff frac{var(X^n)}{E[X^{2n}]}=1-frac{E[X^n]^2}{E[X^{2n}]} geq 0 implieslimsup frac{E[X^n]^2}{E[X^{2n}]} leq 1 $.
Now $var(X^2)implies frac{E[X^n]^2}{E[X^{2n}]}=frac{var(X)^2}{E[X^{4}]} iff var(X)^2 = O(E[X^{4}]) $
Now we use the inequality and the variance used in the proof of the weak law.
$Pr(|X_n|>epsilon)=Pr(|X_n|^2>epsilon^2 leq frac{E[X^{4}]}{epsilon^4} $.
Now summing up the probabilities
$O(sum_{}^{infty} frac{E[X^{4}]}{epsilon^4}) =sum_{}^{infty}frac{var(x)^2}{epsilon^4}=sum_{}^{infty}frac{sigma^4}{n^2epsilon^4}<infty$ which implies almost sure convergance.
I hope you can understand what I was writing and would like to know if my idea is somehow correct. Thanks
probability-theory
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
$endgroup$
1
$begingroup$
There is a very elementary proof of the strong law of large numbers under the assumption of finite fourth moments (as you seem to have assumed). However, your argument isn't intelligible to me... too many $implies$'s and $iff$'s and very few words, and no clear statement of the theorem and the assumptions. The fact that only $sigma$ appears in the final sum shows that there is at least a minor error here. $E(X^4)$ is a fourth moment quantity, not a second moment quantity (perhaps you confused $var(x^2)$ for $var(x)^2$?)
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:18
1
$begingroup$
Squinting a little harder, assuming you used $sigma^4$ as a notation for the fourth moment, and reading between the lines and typos a bit, it resembles a correct argument.
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:50
add a comment |
$begingroup$
I beg your pardon. But I am not going to write down the statement of the law proofs and so on, because I think every one knows what I am talking about and because I am a hobby mathemation I do not think I would write this text correct anyway .
When i looked up the proof for the weak law it ended like this:
$Pr( left| overline{X}_n-mu right| geq varepsilon) leq frac{sigma^2}{nvarepsilon^2}rightarrow 0$, where the random variable $overline{X}_n$ has a variance of $frac{sigma^2}{n}$.
An equivalent definition of the weak law of large numbers, using chebychev inequality for the proof is:
A random variable $X_n$ with $mu=0$ and $var(X_n)=sigma^2$
converges in probability to zero if
$var(X_n)rightarrow0$.
Now I have the following idea:
$var(X^n)=E[X^{2n}]-E[X^n]^2 geq 0 iff frac{var(X^n)}{E[X^{2n}]}=1-frac{E[X^n]^2}{E[X^{2n}]} geq 0 implieslimsup frac{E[X^n]^2}{E[X^{2n}]} leq 1 $.
Now $var(X^2)implies frac{E[X^n]^2}{E[X^{2n}]}=frac{var(X)^2}{E[X^{4}]} iff var(X)^2 = O(E[X^{4}]) $
Now we use the inequality and the variance used in the proof of the weak law.
$Pr(|X_n|>epsilon)=Pr(|X_n|^2>epsilon^2 leq frac{E[X^{4}]}{epsilon^4} $.
Now summing up the probabilities
$O(sum_{}^{infty} frac{E[X^{4}]}{epsilon^4}) =sum_{}^{infty}frac{var(x)^2}{epsilon^4}=sum_{}^{infty}frac{sigma^4}{n^2epsilon^4}<infty$ which implies almost sure convergance.
I hope you can understand what I was writing and would like to know if my idea is somehow correct. Thanks
probability-theory
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
$endgroup$
I beg your pardon. But I am not going to write down the statement of the law proofs and so on, because I think every one knows what I am talking about and because I am a hobby mathemation I do not think I would write this text correct anyway .
When i looked up the proof for the weak law it ended like this:
$Pr( left| overline{X}_n-mu right| geq varepsilon) leq frac{sigma^2}{nvarepsilon^2}rightarrow 0$, where the random variable $overline{X}_n$ has a variance of $frac{sigma^2}{n}$.
An equivalent definition of the weak law of large numbers, using chebychev inequality for the proof is:
A random variable $X_n$ with $mu=0$ and $var(X_n)=sigma^2$
converges in probability to zero if
$var(X_n)rightarrow0$.
Now I have the following idea:
$var(X^n)=E[X^{2n}]-E[X^n]^2 geq 0 iff frac{var(X^n)}{E[X^{2n}]}=1-frac{E[X^n]^2}{E[X^{2n}]} geq 0 implieslimsup frac{E[X^n]^2}{E[X^{2n}]} leq 1 $.
Now $var(X^2)implies frac{E[X^n]^2}{E[X^{2n}]}=frac{var(X)^2}{E[X^{4}]} iff var(X)^2 = O(E[X^{4}]) $
Now we use the inequality and the variance used in the proof of the weak law.
$Pr(|X_n|>epsilon)=Pr(|X_n|^2>epsilon^2 leq frac{E[X^{4}]}{epsilon^4} $.
Now summing up the probabilities
$O(sum_{}^{infty} frac{E[X^{4}]}{epsilon^4}) =sum_{}^{infty}frac{var(x)^2}{epsilon^4}=sum_{}^{infty}frac{sigma^4}{n^2epsilon^4}<infty$ which implies almost sure convergance.
I hope you can understand what I was writing and would like to know if my idea is somehow correct. Thanks
probability-theory
probability-theory
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
asked Jan 10 at 1:06
Markus KrumplMarkus Krumpl
61
61
1
$begingroup$
There is a very elementary proof of the strong law of large numbers under the assumption of finite fourth moments (as you seem to have assumed). However, your argument isn't intelligible to me... too many $implies$'s and $iff$'s and very few words, and no clear statement of the theorem and the assumptions. The fact that only $sigma$ appears in the final sum shows that there is at least a minor error here. $E(X^4)$ is a fourth moment quantity, not a second moment quantity (perhaps you confused $var(x^2)$ for $var(x)^2$?)
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:18
1
$begingroup$
Squinting a little harder, assuming you used $sigma^4$ as a notation for the fourth moment, and reading between the lines and typos a bit, it resembles a correct argument.
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:50
add a comment |
1
$begingroup$
There is a very elementary proof of the strong law of large numbers under the assumption of finite fourth moments (as you seem to have assumed). However, your argument isn't intelligible to me... too many $implies$'s and $iff$'s and very few words, and no clear statement of the theorem and the assumptions. The fact that only $sigma$ appears in the final sum shows that there is at least a minor error here. $E(X^4)$ is a fourth moment quantity, not a second moment quantity (perhaps you confused $var(x^2)$ for $var(x)^2$?)
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:18
1
$begingroup$
Squinting a little harder, assuming you used $sigma^4$ as a notation for the fourth moment, and reading between the lines and typos a bit, it resembles a correct argument.
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:50
1
1
$begingroup$
There is a very elementary proof of the strong law of large numbers under the assumption of finite fourth moments (as you seem to have assumed). However, your argument isn't intelligible to me... too many $implies$'s and $iff$'s and very few words, and no clear statement of the theorem and the assumptions. The fact that only $sigma$ appears in the final sum shows that there is at least a minor error here. $E(X^4)$ is a fourth moment quantity, not a second moment quantity (perhaps you confused $var(x^2)$ for $var(x)^2$?)
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:18
$begingroup$
There is a very elementary proof of the strong law of large numbers under the assumption of finite fourth moments (as you seem to have assumed). However, your argument isn't intelligible to me... too many $implies$'s and $iff$'s and very few words, and no clear statement of the theorem and the assumptions. The fact that only $sigma$ appears in the final sum shows that there is at least a minor error here. $E(X^4)$ is a fourth moment quantity, not a second moment quantity (perhaps you confused $var(x^2)$ for $var(x)^2$?)
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:18
1
1
$begingroup$
Squinting a little harder, assuming you used $sigma^4$ as a notation for the fourth moment, and reading between the lines and typos a bit, it resembles a correct argument.
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:50
$begingroup$
Squinting a little harder, assuming you used $sigma^4$ as a notation for the fourth moment, and reading between the lines and typos a bit, it resembles a correct argument.
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:50
add a comment |
1 Answer
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$begingroup$
As I mentioned in the comments, I find your proof difficult to follow. I will give a quick reiteration of (what I think is) your proof for reference:
Let $X_i$ be i.i.d., with mean zero and finite fourth moment.
By Chebyshev, we have $$ P(|bar X_n|^2>epsilon^2) le frac{operatorname{Var}(bar X_n^2)}{epsilon^4} = frac{operatorname{Var}(S_n^2)}{epsilon^4n^4}le frac{E(S_n^4)}{epsilon^4n^4}.$$
Since the $X_i$ have mean zero, we can multiply things out to get $$ E(S_n^4) = E((X_1+ldots+X_n)^4)=nE(X^4)+3n(n-1)E(X^2)^2,$$ and by Jensen, $E(X^2)^2<E(X^4),$ so we have $$ E(S_n^4)le 3n^2E(X^4).$$
Plugging that into the first inequality, $$ P(|bar X_n|^2>epsilon^2)le frac{3E(X^4)}{epsilon^4n^2}.$$
Since this is summable, Borel Cantelli implies that $bar X_nto 0$ almost surely.
This simple proof relies on the fourth moment assumption. The proof for the optimal case that only assumes $E(|X|)<infty$ is a lot harder.
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
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add a comment |
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$begingroup$
As I mentioned in the comments, I find your proof difficult to follow. I will give a quick reiteration of (what I think is) your proof for reference:
Let $X_i$ be i.i.d., with mean zero and finite fourth moment.
By Chebyshev, we have $$ P(|bar X_n|^2>epsilon^2) le frac{operatorname{Var}(bar X_n^2)}{epsilon^4} = frac{operatorname{Var}(S_n^2)}{epsilon^4n^4}le frac{E(S_n^4)}{epsilon^4n^4}.$$
Since the $X_i$ have mean zero, we can multiply things out to get $$ E(S_n^4) = E((X_1+ldots+X_n)^4)=nE(X^4)+3n(n-1)E(X^2)^2,$$ and by Jensen, $E(X^2)^2<E(X^4),$ so we have $$ E(S_n^4)le 3n^2E(X^4).$$
Plugging that into the first inequality, $$ P(|bar X_n|^2>epsilon^2)le frac{3E(X^4)}{epsilon^4n^2}.$$
Since this is summable, Borel Cantelli implies that $bar X_nto 0$ almost surely.
This simple proof relies on the fourth moment assumption. The proof for the optimal case that only assumes $E(|X|)<infty$ is a lot harder.
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
$endgroup$
add a comment |
$begingroup$
As I mentioned in the comments, I find your proof difficult to follow. I will give a quick reiteration of (what I think is) your proof for reference:
Let $X_i$ be i.i.d., with mean zero and finite fourth moment.
By Chebyshev, we have $$ P(|bar X_n|^2>epsilon^2) le frac{operatorname{Var}(bar X_n^2)}{epsilon^4} = frac{operatorname{Var}(S_n^2)}{epsilon^4n^4}le frac{E(S_n^4)}{epsilon^4n^4}.$$
Since the $X_i$ have mean zero, we can multiply things out to get $$ E(S_n^4) = E((X_1+ldots+X_n)^4)=nE(X^4)+3n(n-1)E(X^2)^2,$$ and by Jensen, $E(X^2)^2<E(X^4),$ so we have $$ E(S_n^4)le 3n^2E(X^4).$$
Plugging that into the first inequality, $$ P(|bar X_n|^2>epsilon^2)le frac{3E(X^4)}{epsilon^4n^2}.$$
Since this is summable, Borel Cantelli implies that $bar X_nto 0$ almost surely.
This simple proof relies on the fourth moment assumption. The proof for the optimal case that only assumes $E(|X|)<infty$ is a lot harder.
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
$endgroup$
add a comment |
$begingroup$
As I mentioned in the comments, I find your proof difficult to follow. I will give a quick reiteration of (what I think is) your proof for reference:
Let $X_i$ be i.i.d., with mean zero and finite fourth moment.
By Chebyshev, we have $$ P(|bar X_n|^2>epsilon^2) le frac{operatorname{Var}(bar X_n^2)}{epsilon^4} = frac{operatorname{Var}(S_n^2)}{epsilon^4n^4}le frac{E(S_n^4)}{epsilon^4n^4}.$$
Since the $X_i$ have mean zero, we can multiply things out to get $$ E(S_n^4) = E((X_1+ldots+X_n)^4)=nE(X^4)+3n(n-1)E(X^2)^2,$$ and by Jensen, $E(X^2)^2<E(X^4),$ so we have $$ E(S_n^4)le 3n^2E(X^4).$$
Plugging that into the first inequality, $$ P(|bar X_n|^2>epsilon^2)le frac{3E(X^4)}{epsilon^4n^2}.$$
Since this is summable, Borel Cantelli implies that $bar X_nto 0$ almost surely.
This simple proof relies on the fourth moment assumption. The proof for the optimal case that only assumes $E(|X|)<infty$ is a lot harder.
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
$endgroup$
As I mentioned in the comments, I find your proof difficult to follow. I will give a quick reiteration of (what I think is) your proof for reference:
Let $X_i$ be i.i.d., with mean zero and finite fourth moment.
By Chebyshev, we have $$ P(|bar X_n|^2>epsilon^2) le frac{operatorname{Var}(bar X_n^2)}{epsilon^4} = frac{operatorname{Var}(S_n^2)}{epsilon^4n^4}le frac{E(S_n^4)}{epsilon^4n^4}.$$
Since the $X_i$ have mean zero, we can multiply things out to get $$ E(S_n^4) = E((X_1+ldots+X_n)^4)=nE(X^4)+3n(n-1)E(X^2)^2,$$ and by Jensen, $E(X^2)^2<E(X^4),$ so we have $$ E(S_n^4)le 3n^2E(X^4).$$
Plugging that into the first inequality, $$ P(|bar X_n|^2>epsilon^2)le frac{3E(X^4)}{epsilon^4n^2}.$$
Since this is summable, Borel Cantelli implies that $bar X_nto 0$ almost surely.
This simple proof relies on the fourth moment assumption. The proof for the optimal case that only assumes $E(|X|)<infty$ is a lot harder.
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
answered Jan 10 at 5:05
spaceisdarkgreenspaceisdarkgreen
32.8k21753
32.8k21753
add a comment |
add a comment |
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$begingroup$
There is a very elementary proof of the strong law of large numbers under the assumption of finite fourth moments (as you seem to have assumed). However, your argument isn't intelligible to me... too many $implies$'s and $iff$'s and very few words, and no clear statement of the theorem and the assumptions. The fact that only $sigma$ appears in the final sum shows that there is at least a minor error here. $E(X^4)$ is a fourth moment quantity, not a second moment quantity (perhaps you confused $var(x^2)$ for $var(x)^2$?)
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:18
1
$begingroup$
Squinting a little harder, assuming you used $sigma^4$ as a notation for the fourth moment, and reading between the lines and typos a bit, it resembles a correct argument.
$endgroup$
– spaceisdarkgreen
Jan 10 at 4:50