What Your Can Reveal About Your Blumenthals 0 1 law
{\displaystyle \mathbb {P} (\Lambda )=1. }[/math]
Suppose that [math]\displaystyle{ X=(X_t:t\geq 0) }[/math] is an adapted stochastic process on a probability space [math]\displaystyle{ (\Omega,\mathcal{F},\{\mathcal{F}_{t}\}_{t\geq 0},\mathbb{P}) }[/math] such that [math]\displaystyle{ X_0 }[/math] is constant with probability one. Then any event in the germ sigma algebra
F
0
+
find here
X
{\displaystyle \Lambda \in {\mathcal {F}}_{0+}^{X}}
has either
P
(
)
=
0
{\displaystyle \mathbb {P} (\Lambda )=0}
or
P
(
)
=
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{\displaystyle \mathbb {P} (\Lambda )=1. Let [math]\displaystyle{ \mathcal{F}^X_t:=\sigma(X_s; s\leq t), \mathcal{F}^X_{t^+}:=\bigcap_{s\gt t}\mathcal{F}^X_s }[/math]. Then any event in the germ sigma algebra [math]\displaystyle{ \Lambda \in \mathcal{F}^X_{0+} }[/math] has either [math]\displaystyle{ \mathbb{P}(\Lambda)=0 }[/math] or [math]\displaystyle{ \mathbb{P}(\Lambda)=1. }
Suppose that
X
=
(
X
t
:
t
0
)
{\displaystyle X=(X_{t}:t\geq 0)}
is an adapted stochastic process on a probability space
(
,
F
,
{
F
t
}
t
0
,
P
)
{\displaystyle (\Omega ,{\mathcal {F}},\{{\mathcal {F}}_{t}\}_{t\geq 0},\mathbb {P} )}
such that
X
0
{\displaystyle X_{0}}
is constant with probability one. .