Distribution of a squared standard Brownian motion
up vote
1
down vote
favorite
During a self study I encountered the following issue:
I got a standard Brownian motion $B(t)$ on the interval $[0,1]$ which is $N(0,t)$ distributed by definition, but now I want to figure out the distribution of $B^2(t)$. I assume its a $chi^2$ - distribution, but im not sure how to show it.
One approach I see is using the fact that $B(1) sim N(0,1)$ and thus $B^2(1) sim chi^2(1)$, but I dont think its correct to rewrite $B^2(t) stackrel{d}{=} t B(1)$ and receiving a $chi^2$ - distribution this way.
Can anyone point me in the right direction? Thank you!
brownian-motion chi-squared
asked Dec 5 at 13:08
Bazzan
82
add a comment |
up vote
1
down vote
favorite
During a self study I encountered the following issue:
I got a standard Brownian motion $B(t)$ on the interval $[0,1]$ which is $N(0,t)$ distributed by definition, but now I want to figure out the distribution of $B^2(t)$. I assume its a $chi^2$ - distribution, but im not sure how to show it.
One approach I see is using the fact that $B(1) sim N(0,1)$ and thus $B^2(1) sim chi^2(1)$, but I dont think its correct to rewrite $B^2(t) stackrel{d}{=} t B(1)$ and receiving a $chi^2$ - distribution this way.
Can anyone point me in the right direction? Thank you!
brownian-motion chi-squared
asked Dec 5 at 13:08
Bazzan
82
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
During a self study I encountered the following issue:
I got a standard Brownian motion $B(t)$ on the interval $[0,1]$ which is $N(0,t)$ distributed by definition, but now I want to figure out the distribution of $B^2(t)$. I assume its a $chi^2$ - distribution, but im not sure how to show it.
One approach I see is using the fact that $B(1) sim N(0,1)$ and thus $B^2(1) sim chi^2(1)$, but I dont think its correct to rewrite $B^2(t) stackrel{d}{=} t B(1)$ and receiving a $chi^2$ - distribution this way.
Can anyone point me in the right direction? Thank you!
brownian-motion chi-squared
asked Dec 5 at 13:08
Bazzan
82
During a self study I encountered the following issue:
I got a standard Brownian motion $B(t)$ on the interval $[0,1]$ which is $N(0,t)$ distributed by definition, but now I want to figure out the distribution of $B^2(t)$. I assume its a $chi^2$ - distribution, but im not sure how to show it.
One approach I see is using the fact that $B(1) sim N(0,1)$ and thus $B^2(1) sim chi^2(1)$, but I dont think its correct to rewrite $B^2(t) stackrel{d}{=} t B(1)$ and receiving a $chi^2$ - distribution this way.
Can anyone point me in the right direction? Thank you!
brownian-motion chi-squared
brownian-motion chi-squared
asked Dec 5 at 13:08
Bazzan
82
asked Dec 5 at 13:08
Bazzan
82
asked Dec 5 at 13:08
Bazzan
82
asked Dec 5 at 13:08
Bazzan
82
asked Dec 5 at 13:08
Bazzan
82
82
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
Since $B(t)sim N(0,t)$, we have that $B(t)=t^{1/2}cdot t^{-1/2}B(t)$, where $t^{-1/2}B(t)sim N(0,1)$. Hence, $B^2(t)=tQ$, where $Q$ is the chi-squared distribution with $1$ degree of freedom. Indeed, it is correct to write $B^2(t)stackrel{d}{=}tB^2(1)$ since $B^2(1)$ has the chi-squared distribution with $1$ degree of freedom.
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
Thank you! This was much easier than I thought
– Bazzan
Dec 5 at 13:16
@Bazzan You're welcome!
– Cm7F7Bb
Dec 5 at 13:20
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',
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%2f3027044%2fdistribution-of-a-squared-standard-brownian-motion%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
up vote
0
down vote
accepted
Since $B(t)sim N(0,t)$, we have that $B(t)=t^{1/2}cdot t^{-1/2}B(t)$, where $t^{-1/2}B(t)sim N(0,1)$. Hence, $B^2(t)=tQ$, where $Q$ is the chi-squared distribution with $1$ degree of freedom. Indeed, it is correct to write $B^2(t)stackrel{d}{=}tB^2(1)$ since $B^2(1)$ has the chi-squared distribution with $1$ degree of freedom.
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
Thank you! This was much easier than I thought
– Bazzan
Dec 5 at 13:16
@Bazzan You're welcome!
– Cm7F7Bb
Dec 5 at 13:20
add a comment |
up vote
0
down vote
accepted
Since $B(t)sim N(0,t)$, we have that $B(t)=t^{1/2}cdot t^{-1/2}B(t)$, where $t^{-1/2}B(t)sim N(0,1)$. Hence, $B^2(t)=tQ$, where $Q$ is the chi-squared distribution with $1$ degree of freedom. Indeed, it is correct to write $B^2(t)stackrel{d}{=}tB^2(1)$ since $B^2(1)$ has the chi-squared distribution with $1$ degree of freedom.
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
Thank you! This was much easier than I thought
– Bazzan
Dec 5 at 13:16
@Bazzan You're welcome!
– Cm7F7Bb
Dec 5 at 13:20
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Since $B(t)sim N(0,t)$, we have that $B(t)=t^{1/2}cdot t^{-1/2}B(t)$, where $t^{-1/2}B(t)sim N(0,1)$. Hence, $B^2(t)=tQ$, where $Q$ is the chi-squared distribution with $1$ degree of freedom. Indeed, it is correct to write $B^2(t)stackrel{d}{=}tB^2(1)$ since $B^2(1)$ has the chi-squared distribution with $1$ degree of freedom.
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
Since $B(t)sim N(0,t)$, we have that $B(t)=t^{1/2}cdot t^{-1/2}B(t)$, where $t^{-1/2}B(t)sim N(0,1)$. Hence, $B^2(t)=tQ$, where $Q$ is the chi-squared distribution with $1$ degree of freedom. Indeed, it is correct to write $B^2(t)stackrel{d}{=}tB^2(1)$ since $B^2(1)$ has the chi-squared distribution with $1$ degree of freedom.
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
edited Dec 5 at 13:16
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
answered Dec 5 at 13:14
Cm7F7Bb
12.4k32142
12.4k32142
Thank you! This was much easier than I thought
– Bazzan
Dec 5 at 13:16
@Bazzan You're welcome!
– Cm7F7Bb
Dec 5 at 13:20
add a comment |
Thank you! This was much easier than I thought
– Bazzan
Dec 5 at 13:16
@Bazzan You're welcome!
– Cm7F7Bb
Dec 5 at 13:20
Thank you! This was much easier than I thought
– Bazzan
Dec 5 at 13:16
Thank you! This was much easier than I thought
– Bazzan
Dec 5 at 13:16
@Bazzan You're welcome!
– Cm7F7Bb
Dec 5 at 13:20
@Bazzan You're welcome!
– Cm7F7Bb
Dec 5 at 13:20
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3027044%2fdistribution-of-a-squared-standard-brownian-motion%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
