Brownian motion and covariance
$begingroup$
Show that for $B = (B_t)$ Brownian motion, its covariance is
$cov(B_s, B_t) = min(s, t)$.
The solution I was given was:
For $s ≤ t$,
$B_t = B_s + (B_t − B_s)$,
$B_sB_t = B_s^2 + Bs(Bt − Bs)$
$cov(B_s,B_t)=E[B_sB_t]$(as $E(B_i)=0)$
so $cov(B_s,B_t)=E[B_s^2]+E[B_s(B_t-B_s)]$,
as all increments of Brown motion are independent the second term in the RHS
$=E[B_s]E[B_t-B_s]=0*0=0$. Now $cov(B_s,B_t)=E(B_s^2)=Var(B_s)=s$.
Similarly if $tleq s$ we get $=t$.
My question is, could I not have done this arguement the same way without assuming that $sleq t$ in the first place? I mean, if $sleq t$ why cant I say
$B_s = B_t + (B_s − B_t)$ and conitnue like this to get the answer with a max instead of a min?
Is it because $B_t-B_s$ is only an increment if $tgeq s$? I mean otherwise we dont have independency or something?
probability brownian-motion finance
asked May 8 '15 at 20:23
vounoovounoo
156112
$endgroup$
add a comment |
$begingroup$
Show that for $B = (B_t)$ Brownian motion, its covariance is
$cov(B_s, B_t) = min(s, t)$.
The solution I was given was:
For $s ≤ t$,
$B_t = B_s + (B_t − B_s)$,
$B_sB_t = B_s^2 + Bs(Bt − Bs)$
$cov(B_s,B_t)=E[B_sB_t]$(as $E(B_i)=0)$
so $cov(B_s,B_t)=E[B_s^2]+E[B_s(B_t-B_s)]$,
as all increments of Brown motion are independent the second term in the RHS
$=E[B_s]E[B_t-B_s]=0*0=0$. Now $cov(B_s,B_t)=E(B_s^2)=Var(B_s)=s$.
Similarly if $tleq s$ we get $=t$.
My question is, could I not have done this arguement the same way without assuming that $sleq t$ in the first place? I mean, if $sleq t$ why cant I say
$B_s = B_t + (B_s − B_t)$ and conitnue like this to get the answer with a max instead of a min?
Is it because $B_t-B_s$ is only an increment if $tgeq s$? I mean otherwise we dont have independency or something?
probability brownian-motion finance
asked May 8 '15 at 20:23
vounoovounoo
156112
$endgroup$
2
$begingroup$
Why don't you first assume $s leq t$ without loss of generality?
$endgroup$
– Calculon
May 8 '15 at 21:58
add a comment |
$begingroup$
Show that for $B = (B_t)$ Brownian motion, its covariance is
$cov(B_s, B_t) = min(s, t)$.
The solution I was given was:
For $s ≤ t$,
$B_t = B_s + (B_t − B_s)$,
$B_sB_t = B_s^2 + Bs(Bt − Bs)$
$cov(B_s,B_t)=E[B_sB_t]$(as $E(B_i)=0)$
so $cov(B_s,B_t)=E[B_s^2]+E[B_s(B_t-B_s)]$,
as all increments of Brown motion are independent the second term in the RHS
$=E[B_s]E[B_t-B_s]=0*0=0$. Now $cov(B_s,B_t)=E(B_s^2)=Var(B_s)=s$.
Similarly if $tleq s$ we get $=t$.
My question is, could I not have done this arguement the same way without assuming that $sleq t$ in the first place? I mean, if $sleq t$ why cant I say
$B_s = B_t + (B_s − B_t)$ and conitnue like this to get the answer with a max instead of a min?
Is it because $B_t-B_s$ is only an increment if $tgeq s$? I mean otherwise we dont have independency or something?
probability brownian-motion finance
asked May 8 '15 at 20:23
vounoovounoo
156112
$endgroup$
Show that for $B = (B_t)$ Brownian motion, its covariance is
$cov(B_s, B_t) = min(s, t)$.
The solution I was given was:
For $s ≤ t$,
$B_t = B_s + (B_t − B_s)$,
$B_sB_t = B_s^2 + Bs(Bt − Bs)$
$cov(B_s,B_t)=E[B_sB_t]$(as $E(B_i)=0)$
so $cov(B_s,B_t)=E[B_s^2]+E[B_s(B_t-B_s)]$,
as all increments of Brown motion are independent the second term in the RHS
$=E[B_s]E[B_t-B_s]=0*0=0$. Now $cov(B_s,B_t)=E(B_s^2)=Var(B_s)=s$.
Similarly if $tleq s$ we get $=t$.
My question is, could I not have done this arguement the same way without assuming that $sleq t$ in the first place? I mean, if $sleq t$ why cant I say
$B_s = B_t + (B_s − B_t)$ and conitnue like this to get the answer with a max instead of a min?
Is it because $B_t-B_s$ is only an increment if $tgeq s$? I mean otherwise we dont have independency or something?
probability brownian-motion finance
probability brownian-motion finance
asked May 8 '15 at 20:23
vounoovounoo
156112
asked May 8 '15 at 20:23
vounoovounoo
156112
asked May 8 '15 at 20:23
vounoovounoo
156112
asked May 8 '15 at 20:23
vounoovounoo
156112
asked May 8 '15 at 20:23
vounoovounoo
156112
156112
2
$begingroup$
Why don't you first assume $s leq t$ without loss of generality?
$endgroup$
– Calculon
May 8 '15 at 21:58
add a comment |
2
$begingroup$
Why don't you first assume $s leq t$ without loss of generality?
$endgroup$
– Calculon
May 8 '15 at 21:58
2
2
$begingroup$
Why don't you first assume $s leq t$ without loss of generality?
$endgroup$
– Calculon
May 8 '15 at 21:58
$begingroup$
Why don't you first assume $s leq t$ without loss of generality?
$endgroup$
– Calculon
May 8 '15 at 21:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In your proof you use the fact that $B_s$ is independent of $B_t-B_s$ (indepedent increments) but this is only true if $sleq t$.
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
$endgroup$
$begingroup$
thank you thats what I thought too
$endgroup$
– vounoo
May 8 '15 at 21:56
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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$begingroup$
In your proof you use the fact that $B_s$ is independent of $B_t-B_s$ (indepedent increments) but this is only true if $sleq t$.
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
$endgroup$
$begingroup$
thank you thats what I thought too
$endgroup$
– vounoo
May 8 '15 at 21:56
add a comment |
$begingroup$
In your proof you use the fact that $B_s$ is independent of $B_t-B_s$ (indepedent increments) but this is only true if $sleq t$.
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
$endgroup$
$begingroup$
thank you thats what I thought too
$endgroup$
– vounoo
May 8 '15 at 21:56
add a comment |
$begingroup$
In your proof you use the fact that $B_s$ is independent of $B_t-B_s$ (indepedent increments) but this is only true if $sleq t$.
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
$endgroup$
In your proof you use the fact that $B_s$ is independent of $B_t-B_s$ (indepedent increments) but this is only true if $sleq t$.
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
answered May 8 '15 at 21:49
Tyr CurtisTyr Curtis
1,056411
1,056411
$begingroup$
thank you thats what I thought too
$endgroup$
– vounoo
May 8 '15 at 21:56
add a comment |
$begingroup$
thank you thats what I thought too
$endgroup$
– vounoo
May 8 '15 at 21:56
$begingroup$
thank you thats what I thought too
$endgroup$
– vounoo
May 8 '15 at 21:56
$begingroup$
thank you thats what I thought too
$endgroup$
– vounoo
May 8 '15 at 21:56
add a comment |
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$begingroup$
Why don't you first assume $s leq t$ without loss of generality?
$endgroup$
– Calculon
May 8 '15 at 21:58