Distribution of a squared standard Brownian motion
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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 '18 at 13:08
BazzanBazzan
83
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add a comment |
$begingroup$
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 '18 at 13:08
BazzanBazzan
83
$endgroup$
add a comment |
$begingroup$
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 '18 at 13:08
BazzanBazzan
83
$endgroup$
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 '18 at 13:08
BazzanBazzan
83
asked Dec 5 '18 at 13:08
BazzanBazzan
83
asked Dec 5 '18 at 13:08
BazzanBazzan
83
asked Dec 5 '18 at 13:08
BazzanBazzan
83
asked Dec 5 '18 at 13:08
BazzanBazzan
83
83
add a comment |
add a comment |
1 Answer
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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 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
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$begingroup$
Thank you! This was much easier than I thought
$endgroup$
– Bazzan
Dec 5 '18 at 13:16
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@Bazzan You're welcome!
$endgroup$
– Cm7F7Bb
Dec 5 '18 at 13:20
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
$endgroup$
$begingroup$
Thank you! This was much easier than I thought
$endgroup$
– Bazzan
Dec 5 '18 at 13:16
$begingroup$
@Bazzan You're welcome!
$endgroup$
– Cm7F7Bb
Dec 5 '18 at 13:20
add a comment |
$begingroup$
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 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
$endgroup$
$begingroup$
Thank you! This was much easier than I thought
$endgroup$
– Bazzan
Dec 5 '18 at 13:16
$begingroup$
@Bazzan You're welcome!
$endgroup$
– Cm7F7Bb
Dec 5 '18 at 13:20
add a comment |
$begingroup$
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 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
$endgroup$
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 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
edited Dec 5 '18 at 13:16
answered Dec 5 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
answered Dec 5 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
answered Dec 5 '18 at 13:14
Cm7F7BbCm7F7Bb
12.4k32242
12.4k32242
$begingroup$
Thank you! This was much easier than I thought
$endgroup$
– Bazzan
Dec 5 '18 at 13:16
$begingroup$
@Bazzan You're welcome!
$endgroup$
– Cm7F7Bb
Dec 5 '18 at 13:20
add a comment |
$begingroup$
Thank you! This was much easier than I thought
$endgroup$
– Bazzan
Dec 5 '18 at 13:16
$begingroup$
@Bazzan You're welcome!
$endgroup$
– Cm7F7Bb
Dec 5 '18 at 13:20
$begingroup$
Thank you! This was much easier than I thought
$endgroup$
– Bazzan
Dec 5 '18 at 13:16
$begingroup$
Thank you! This was much easier than I thought
$endgroup$
– Bazzan
Dec 5 '18 at 13:16
$begingroup$
@Bazzan You're welcome!
$endgroup$
– Cm7F7Bb
Dec 5 '18 at 13:20
$begingroup$
@Bazzan You're welcome!
$endgroup$
– Cm7F7Bb
Dec 5 '18 at 13:20
add a comment |
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