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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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3 Rules For Clinical Trials

{\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. .