Brownian motion and covariance












4












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Why don't you first assume $s leq t$ without loss of generality?
    $endgroup$
    – Calculon
    May 8 '15 at 21:58
















4












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Why don't you first assume $s leq t$ without loss of generality?
    $endgroup$
    – Calculon
    May 8 '15 at 21:58














4












4








4





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






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oldest

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2












$begingroup$

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 $sleq t$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you thats what I thought too
    $endgroup$
    – vounoo
    May 8 '15 at 21:56












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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $sleq t$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you thats what I thought too
    $endgroup$
    – vounoo
    May 8 '15 at 21:56
















2












$begingroup$

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 $sleq t$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you thats what I thought too
    $endgroup$
    – vounoo
    May 8 '15 at 21:56














2












2








2





$begingroup$

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 $sleq t$.






share|cite|improve this answer









$endgroup$



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 $sleq t$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • $begingroup$
    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


















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