Proving that a process is a brownian motion: How do I show independent increments?
$begingroup$
$W_t$ is a brownian motion and I want to show that $W^*_t=(-W_t)$ is also a brownian motion. I can easily show the distribution the new variable: $W^*_t sim N(0,t)$.
But one of the properties of brownian motion is that the process has independent increments; how do I argue for that?
stochastic-processes brownian-motion
edited Dec 8 '18 at 13:07
Bernard
119k740113
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
$endgroup$
add a comment |
$begingroup$
$W_t$ is a brownian motion and I want to show that $W^*_t=(-W_t)$ is also a brownian motion. I can easily show the distribution the new variable: $W^*_t sim N(0,t)$.
But one of the properties of brownian motion is that the process has independent increments; how do I argue for that?
stochastic-processes brownian-motion
edited Dec 8 '18 at 13:07
Bernard
119k740113
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
$endgroup$
$begingroup$
Maybe it helps to show that if $X$ and $Y$ are independent, then $-X$ and $-Y$ are independent as well.
$endgroup$
– Shashi
Dec 8 '18 at 14:30
add a comment |
$begingroup$
$W_t$ is a brownian motion and I want to show that $W^*_t=(-W_t)$ is also a brownian motion. I can easily show the distribution the new variable: $W^*_t sim N(0,t)$.
But one of the properties of brownian motion is that the process has independent increments; how do I argue for that?
stochastic-processes brownian-motion
edited Dec 8 '18 at 13:07
Bernard
119k740113
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
$endgroup$
$W_t$ is a brownian motion and I want to show that $W^*_t=(-W_t)$ is also a brownian motion. I can easily show the distribution the new variable: $W^*_t sim N(0,t)$.
But one of the properties of brownian motion is that the process has independent increments; how do I argue for that?
stochastic-processes brownian-motion
stochastic-processes brownian-motion
edited Dec 8 '18 at 13:07
Bernard
119k740113
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
edited Dec 8 '18 at 13:07
Bernard
119k740113
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
edited Dec 8 '18 at 13:07
Bernard
119k740113
edited Dec 8 '18 at 13:07
Bernard
119k740113
edited Dec 8 '18 at 13:07
Bernard
119k740113
119k740113
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
asked Dec 8 '18 at 13:04
k.dkhkk.dkhk
16410
16410
$begingroup$
Maybe it helps to show that if $X$ and $Y$ are independent, then $-X$ and $-Y$ are independent as well.
$endgroup$
– Shashi
Dec 8 '18 at 14:30
add a comment |
$begingroup$
Maybe it helps to show that if $X$ and $Y$ are independent, then $-X$ and $-Y$ are independent as well.
$endgroup$
– Shashi
Dec 8 '18 at 14:30
$begingroup$
Maybe it helps to show that if $X$ and $Y$ are independent, then $-X$ and $-Y$ are independent as well.
$endgroup$
– Shashi
Dec 8 '18 at 14:30
$begingroup$
Maybe it helps to show that if $X$ and $Y$ are independent, then $-X$ and $-Y$ are independent as well.
$endgroup$
– Shashi
Dec 8 '18 at 14:30
add a comment |
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$begingroup$
Maybe it helps to show that if $X$ and $Y$ are independent, then $-X$ and $-Y$ are independent as well.
$endgroup$
– Shashi
Dec 8 '18 at 14:30