Let $\{X(t),t\geq 0\}$ be a Brownian motion so that $X(t)\sim \mathcal{N}(0,\sigma^2 t)$.
Then for any $s,t>0$,
$$X(s+t)|(X(s)= x) \sim \mathcal{N}(x,\sigma^2 t)
$$
I get this intuitively (after the time unit $s$, we go another $t$ seconds. We pretend we started at $x$ at time unit $0$ and go another $t$ units which must be Brownian motion by independent and stationary increments).
But how do I prove this mathematically?
I tried this but it doesn't make sense:
Let $Y = X(s+t) - X(s) \sim X(t) \sim \mathcal{N}(0,\sigma^2 t).$
Then $Y + X(s) = X(s+t) - X(s)+X(s) \sim X(s+t) - x + x | X(s) = x \sim X(s+t) | X(s) = x \sim ???$
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Mr. Bromwich I
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what do you mean by $X(t+s)\mid X(s)=x$ ? – Surb May 30 '18 at 09:32
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$X(t)$ is brownian motion and it's the random variable $X(t+s)$ given $X(s) = x$. – Mr. Bromwich I May 30 '18 at 09:33
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but $\mathbb P{X(t+s)\mid X(s)=x}$ is not defined since $s\longmapsto X(s)$ is continuous. Maybe you have that $X(s)=x$ a.s. ? – Surb May 30 '18 at 09:35
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You should consider finding the moments of your conditional distribution, then work from the definition of the process ${X(t),t\geq 0}$ – Tony Hellmuth May 30 '18 at 09:38
1 Answers
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You can write $X(s+t)=Y+X(s)$ where $Y:=X(s+t)-X(s)$.
Here $Y$ and $X(s)$ are independent and $Y\sim\mathsf{Norm}(0,\sigma^2t)$.
Then under condition $X(s)=x$ the distribution of $Y+X(s)$ is the same as the distribution of $Y+x$ which is $\mathsf{Norm}(x,\sigma^2t)$.
drhab
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How would you show $Y+X(s) \sim Y+x$?
Is it by doing
$Y+X(s) \sim Y+X(s) | X(s) = x \sim Y+x |X(s) = x$ – Mr. Bromwich I May 30 '18 at 09:57 -
1In general if $X$ and $Y$ are independent then condition $X=x$ has no influence on $Y$ whatsoever (independence) and it makes $X$ equal to $x$ so under condition $X=x$ the rv $Y+X$ has the same distribution as $Y+x$. Is that enough? – drhab May 30 '18 at 10:01