Mixing
Proving Brownian motions have bounded quadratic variation
$begingroup$
I am looking into arguments which yield the result $(dB_t)^2=dt$ and authors first start by letting P be a partition $P={t_0, t_1, ..., t_n}$ of the interval $ [0,T]$ where $t_i=frac{i}{n}T$ and then use the approximation of $$int_{0}^{T}(dB_t^2)=lim_{nto infty}sum_{i=1}^{n} (B_{t_i}-B_{t_{i-1}})^2.$$
Then they use properties of Brownian motions to achieve that $$lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=lim_{n to infty}Tsum_{i=1}^{n}frac{Z_i^2}{n}$$ where $Z_isim mathcal{N}(0, 1)$.
Now where I start to get unsure is that texts opt to use the Weak Law of Large Numbers to obtain convergnce in probability but my question is, why do they avoid using the strong law when we can obtain a stronger result, that $$P(lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=T)=1.$$
probability-theory brownian-motion probability-limit-theorems
asked Jan 9 at 19:42
DLBDLB
548
$endgroup$
add a comment |
$begingroup$
I am looking into arguments which yield the result $(dB_t)^2=dt$ and authors first start by letting P be a partition $P={t_0, t_1, ..., t_n}$ of the interval $ [0,T]$ where $t_i=frac{i}{n}T$ and then use the approximation of $$int_{0}^{T}(dB_t^2)=lim_{nto infty}sum_{i=1}^{n} (B_{t_i}-B_{t_{i-1}})^2.$$
Then they use properties of Brownian motions to achieve that $$lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=lim_{n to infty}Tsum_{i=1}^{n}frac{Z_i^2}{n}$$ where $Z_isim mathcal{N}(0, 1)$.
Now where I start to get unsure is that texts opt to use the Weak Law of Large Numbers to obtain convergnce in probability but my question is, why do they avoid using the strong law when we can obtain a stronger result, that $$P(lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=T)=1.$$
probability-theory brownian-motion probability-limit-theorems
asked Jan 9 at 19:42
DLBDLB
548
$endgroup$
$begingroup$
What is $T$ here?
$endgroup$
– Math1000
Jan 9 at 21:15
$begingroup$
@Math1000 I have edited my question to hopefully make it more clear.
$endgroup$
– DLB
Jan 9 at 21:36
add a comment |
$begingroup$
I am looking into arguments which yield the result $(dB_t)^2=dt$ and authors first start by letting P be a partition $P={t_0, t_1, ..., t_n}$ of the interval $ [0,T]$ where $t_i=frac{i}{n}T$ and then use the approximation of $$int_{0}^{T}(dB_t^2)=lim_{nto infty}sum_{i=1}^{n} (B_{t_i}-B_{t_{i-1}})^2.$$
Then they use properties of Brownian motions to achieve that $$lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=lim_{n to infty}Tsum_{i=1}^{n}frac{Z_i^2}{n}$$ where $Z_isim mathcal{N}(0, 1)$.
Now where I start to get unsure is that texts opt to use the Weak Law of Large Numbers to obtain convergnce in probability but my question is, why do they avoid using the strong law when we can obtain a stronger result, that $$P(lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=T)=1.$$
probability-theory brownian-motion probability-limit-theorems
asked Jan 9 at 19:42
DLBDLB
548
$endgroup$
I am looking into arguments which yield the result $(dB_t)^2=dt$ and authors first start by letting P be a partition $P={t_0, t_1, ..., t_n}$ of the interval $ [0,T]$ where $t_i=frac{i}{n}T$ and then use the approximation of $$int_{0}^{T}(dB_t^2)=lim_{nto infty}sum_{i=1}^{n} (B_{t_i}-B_{t_{i-1}})^2.$$
Then they use properties of Brownian motions to achieve that $$lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=lim_{n to infty}Tsum_{i=1}^{n}frac{Z_i^2}{n}$$ where $Z_isim mathcal{N}(0, 1)$.
Now where I start to get unsure is that texts opt to use the Weak Law of Large Numbers to obtain convergnce in probability but my question is, why do they avoid using the strong law when we can obtain a stronger result, that $$P(lim_{n to infty}sum_{i=1}^{n} (B_{t_{i}}-B_{t_i-1})^2=T)=1.$$
probability-theory brownian-motion probability-limit-theorems
probability-theory brownian-motion probability-limit-theorems
asked Jan 9 at 19:42
DLBDLB
548
asked Jan 9 at 19:42
DLBDLB
548
edited Jan 9 at 21:36
DLB
asked Jan 9 at 19:42
DLBDLB
548
asked Jan 9 at 19:42
DLBDLB
548
asked Jan 9 at 19:42
DLBDLB
548
548
$begingroup$
What is $T$ here?
$endgroup$
– Math1000
Jan 9 at 21:15
$begingroup$
@Math1000 I have edited my question to hopefully make it more clear.
$endgroup$
– DLB
Jan 9 at 21:36
add a comment |
$begingroup$
What is $T$ here?
$endgroup$
– Math1000
Jan 9 at 21:15
$begingroup$
@Math1000 I have edited my question to hopefully make it more clear.
$endgroup$
– DLB
Jan 9 at 21:36
$begingroup$
What is $T$ here?
$endgroup$
– Math1000
Jan 9 at 21:15
$begingroup$
What is $T$ here?
$endgroup$
– Math1000
Jan 9 at 21:15
$begingroup$
@Math1000 I have edited my question to hopefully make it more clear.
$endgroup$
– DLB
Jan 9 at 21:36
$begingroup$
@Math1000 I have edited my question to hopefully make it more clear.
$endgroup$
– DLB
Jan 9 at 21:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Strong Law requires a single sequence ${X_i}$ of ii.d. random variables. Here $Z_i$ is actually dependent on $n$, so there is no single i.i.d. sequence ${X_i}$. You cannot apply Weak Law directly but a simple argument using Chebychef's Inequality gives convergence in probability.
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
$begingroup$
thank you for this, could you please point me to a source which shows the proof?
$endgroup$
– DLB
Jan 10 at 11:16
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067876%2fproving-brownian-motions-have-bounded-quadratic-variation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Strong Law requires a single sequence ${X_i}$ of ii.d. random variables. Here $Z_i$ is actually dependent on $n$, so there is no single i.i.d. sequence ${X_i}$. You cannot apply Weak Law directly but a simple argument using Chebychef's Inequality gives convergence in probability.
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
$begingroup$
thank you for this, could you please point me to a source which shows the proof?
$endgroup$
– DLB
Jan 10 at 11:16
add a comment |
$begingroup$
Strong Law requires a single sequence ${X_i}$ of ii.d. random variables. Here $Z_i$ is actually dependent on $n$, so there is no single i.i.d. sequence ${X_i}$. You cannot apply Weak Law directly but a simple argument using Chebychef's Inequality gives convergence in probability.
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
$begingroup$
thank you for this, could you please point me to a source which shows the proof?
$endgroup$
– DLB
Jan 10 at 11:16
add a comment |
$begingroup$
Strong Law requires a single sequence ${X_i}$ of ii.d. random variables. Here $Z_i$ is actually dependent on $n$, so there is no single i.i.d. sequence ${X_i}$. You cannot apply Weak Law directly but a simple argument using Chebychef's Inequality gives convergence in probability.
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
$endgroup$
Strong Law requires a single sequence ${X_i}$ of ii.d. random variables. Here $Z_i$ is actually dependent on $n$, so there is no single i.i.d. sequence ${X_i}$. You cannot apply Weak Law directly but a simple argument using Chebychef's Inequality gives convergence in probability.
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
answered Jan 10 at 0:03
Kavi Rama MurthyKavi Rama Murthy
57.6k42160
57.6k42160
$begingroup$
thank you for this, could you please point me to a source which shows the proof?
$endgroup$
– DLB
Jan 10 at 11:16
add a comment |
$begingroup$
thank you for this, could you please point me to a source which shows the proof?
$endgroup$
– DLB
Jan 10 at 11:16
$begingroup$
thank you for this, could you please point me to a source which shows the proof?
$endgroup$
– DLB
Jan 10 at 11:16
$begingroup$
thank you for this, could you please point me to a source which shows the proof?
$endgroup$
– DLB
Jan 10 at 11:16
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067876%2fproving-brownian-motions-have-bounded-quadratic-variation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
What is $T$ here?
$endgroup$
– Math1000
Jan 9 at 21:15
$begingroup$
@Math1000 I have edited my question to hopefully make it more clear.
$endgroup$
– DLB
Jan 9 at 21:36