Proving Kolmogorov's continuity condition holds for Brownian motion?
$begingroup$
Let $B(t)$ be $n$-dimensional Brownian motion.
I want to show that $E|B(t)-B(s)|^4 = n(n+2)|t-s|^2$ so that Kolmogorov's continuity theorem can be used to show that Brownian motion has a continuous version.
I tried the following:
$$E|B(t)-B(s)|^4 = Eleft[ sqrt{sum_{i=1}^n (B_i(t)-B_i(s))^2}^4 right]=Eleft[ left(sum_{i=1}^n (B_i(t)-B_i(s))^2right)^2 right]$$
$$= sum_{i,j=1}^n Eleft[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2right]$$
Since the individual components of Brownian motion are independent, if $i neq j$ we get
$$E[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2] = (t-s)^2$$
while if $i=j$ we get
$$E[(B_i(t)-B_i(s))^4]= 3(t-s)^2$$
Adding together we see the RHS is then equal to
$$3n (t-s)^2 + (n^2-n)(t-s)^2$$
and, since $3n + n^2 - n = n(n+2)$, the result follows by Kolmogorov's Continuity Theorem that a continuous version of Brownian motion exists.
Is this correct?
probability brownian-motion
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
$endgroup$
add a comment |
$begingroup$
Let $B(t)$ be $n$-dimensional Brownian motion.
I want to show that $E|B(t)-B(s)|^4 = n(n+2)|t-s|^2$ so that Kolmogorov's continuity theorem can be used to show that Brownian motion has a continuous version.
I tried the following:
$$E|B(t)-B(s)|^4 = Eleft[ sqrt{sum_{i=1}^n (B_i(t)-B_i(s))^2}^4 right]=Eleft[ left(sum_{i=1}^n (B_i(t)-B_i(s))^2right)^2 right]$$
$$= sum_{i,j=1}^n Eleft[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2right]$$
Since the individual components of Brownian motion are independent, if $i neq j$ we get
$$E[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2] = (t-s)^2$$
while if $i=j$ we get
$$E[(B_i(t)-B_i(s))^4]= 3(t-s)^2$$
Adding together we see the RHS is then equal to
$$3n (t-s)^2 + (n^2-n)(t-s)^2$$
and, since $3n + n^2 - n = n(n+2)$, the result follows by Kolmogorov's Continuity Theorem that a continuous version of Brownian motion exists.
Is this correct?
probability brownian-motion
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
$endgroup$
$begingroup$
To show that a continuous version of n-dimensional BM exists you just prove this for $n=1$ and then take $n$ independent copies (by going to to a product space).
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 5:29
1
$begingroup$
Your calculation seems to be fine.Altermatively you could use the scaling property, i.e. the fact that $B_t-B_s$ equals in distriution $sqrt{t-s} B_1$ which yields immediately that $$mathbb{E}(|B_t-B_s|^4) = |t-s|^2 mathbb{E}(|B_1|^4).$$
$endgroup$
– saz
Dec 11 '18 at 7:25
add a comment |
$begingroup$
Let $B(t)$ be $n$-dimensional Brownian motion.
I want to show that $E|B(t)-B(s)|^4 = n(n+2)|t-s|^2$ so that Kolmogorov's continuity theorem can be used to show that Brownian motion has a continuous version.
I tried the following:
$$E|B(t)-B(s)|^4 = Eleft[ sqrt{sum_{i=1}^n (B_i(t)-B_i(s))^2}^4 right]=Eleft[ left(sum_{i=1}^n (B_i(t)-B_i(s))^2right)^2 right]$$
$$= sum_{i,j=1}^n Eleft[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2right]$$
Since the individual components of Brownian motion are independent, if $i neq j$ we get
$$E[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2] = (t-s)^2$$
while if $i=j$ we get
$$E[(B_i(t)-B_i(s))^4]= 3(t-s)^2$$
Adding together we see the RHS is then equal to
$$3n (t-s)^2 + (n^2-n)(t-s)^2$$
and, since $3n + n^2 - n = n(n+2)$, the result follows by Kolmogorov's Continuity Theorem that a continuous version of Brownian motion exists.
Is this correct?
probability brownian-motion
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
$endgroup$
Let $B(t)$ be $n$-dimensional Brownian motion.
I want to show that $E|B(t)-B(s)|^4 = n(n+2)|t-s|^2$ so that Kolmogorov's continuity theorem can be used to show that Brownian motion has a continuous version.
I tried the following:
$$E|B(t)-B(s)|^4 = Eleft[ sqrt{sum_{i=1}^n (B_i(t)-B_i(s))^2}^4 right]=Eleft[ left(sum_{i=1}^n (B_i(t)-B_i(s))^2right)^2 right]$$
$$= sum_{i,j=1}^n Eleft[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2right]$$
Since the individual components of Brownian motion are independent, if $i neq j$ we get
$$E[(B_i(t)-B_i(s))^2(B_j(t)-B_j(s))^2] = (t-s)^2$$
while if $i=j$ we get
$$E[(B_i(t)-B_i(s))^4]= 3(t-s)^2$$
Adding together we see the RHS is then equal to
$$3n (t-s)^2 + (n^2-n)(t-s)^2$$
and, since $3n + n^2 - n = n(n+2)$, the result follows by Kolmogorov's Continuity Theorem that a continuous version of Brownian motion exists.
Is this correct?
probability brownian-motion
probability brownian-motion
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
asked Dec 11 '18 at 4:44
XiaomiXiaomi
1,081115
1,081115
$begingroup$
To show that a continuous version of n-dimensional BM exists you just prove this for $n=1$ and then take $n$ independent copies (by going to to a product space).
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 5:29
1
$begingroup$
Your calculation seems to be fine.Altermatively you could use the scaling property, i.e. the fact that $B_t-B_s$ equals in distriution $sqrt{t-s} B_1$ which yields immediately that $$mathbb{E}(|B_t-B_s|^4) = |t-s|^2 mathbb{E}(|B_1|^4).$$
$endgroup$
– saz
Dec 11 '18 at 7:25
add a comment |
$begingroup$
To show that a continuous version of n-dimensional BM exists you just prove this for $n=1$ and then take $n$ independent copies (by going to to a product space).
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 5:29
1
$begingroup$
Your calculation seems to be fine.Altermatively you could use the scaling property, i.e. the fact that $B_t-B_s$ equals in distriution $sqrt{t-s} B_1$ which yields immediately that $$mathbb{E}(|B_t-B_s|^4) = |t-s|^2 mathbb{E}(|B_1|^4).$$
$endgroup$
– saz
Dec 11 '18 at 7:25
$begingroup$
To show that a continuous version of n-dimensional BM exists you just prove this for $n=1$ and then take $n$ independent copies (by going to to a product space).
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 5:29
$begingroup$
To show that a continuous version of n-dimensional BM exists you just prove this for $n=1$ and then take $n$ independent copies (by going to to a product space).
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 5:29
1
1
$begingroup$
Your calculation seems to be fine.Altermatively you could use the scaling property, i.e. the fact that $B_t-B_s$ equals in distriution $sqrt{t-s} B_1$ which yields immediately that $$mathbb{E}(|B_t-B_s|^4) = |t-s|^2 mathbb{E}(|B_1|^4).$$
$endgroup$
– saz
Dec 11 '18 at 7:25
$begingroup$
Your calculation seems to be fine.Altermatively you could use the scaling property, i.e. the fact that $B_t-B_s$ equals in distriution $sqrt{t-s} B_1$ which yields immediately that $$mathbb{E}(|B_t-B_s|^4) = |t-s|^2 mathbb{E}(|B_1|^4).$$
$endgroup$
– saz
Dec 11 '18 at 7:25
add a comment |
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$begingroup$
To show that a continuous version of n-dimensional BM exists you just prove this for $n=1$ and then take $n$ independent copies (by going to to a product space).
$endgroup$
– Kavi Rama Murthy
Dec 11 '18 at 5:29
1
$begingroup$
Your calculation seems to be fine.Altermatively you could use the scaling property, i.e. the fact that $B_t-B_s$ equals in distriution $sqrt{t-s} B_1$ which yields immediately that $$mathbb{E}(|B_t-B_s|^4) = |t-s|^2 mathbb{E}(|B_1|^4).$$
$endgroup$
– saz
Dec 11 '18 at 7:25