Mixing
Convergence in probability of stopping times
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Let $(X^n)_n$ be a sequence of continuous real-valued stochastic processes in $[0,T]$, $T<infty$ such that for every $tin [0,T]$ $X^n_t$ converges in probability to $X_t$.
Let $tau^n=inf{tin[0,T], X^n<0}wedge T$ and $tau=inf{tin[0,T], X<0}wedge T$.
Why does $tau^n$ converge in probability to $tau$ ?
probability stochastic-processes stopping-times
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Al Bundy
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Let $(X^n)_n$ be a sequence of continuous real-valued stochastic processes in $[0,T]$, $T<infty$ such that for every $tin [0,T]$ $X^n_t$ converges in probability to $X_t$.
Let $tau^n=inf{tin[0,T], X^n<0}wedge T$ and $tau=inf{tin[0,T], X<0}wedge T$.
Why does $tau^n$ converge in probability to $tau$ ?
probability stochastic-processes stopping-times
asked yesterday
Al Bundy
12410
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X^n)_n$ be a sequence of continuous real-valued stochastic processes in $[0,T]$, $T<infty$ such that for every $tin [0,T]$ $X^n_t$ converges in probability to $X_t$.
Let $tau^n=inf{tin[0,T], X^n<0}wedge T$ and $tau=inf{tin[0,T], X<0}wedge T$.
Why does $tau^n$ converge in probability to $tau$ ?
probability stochastic-processes stopping-times
asked yesterday
Al Bundy
12410
Let $(X^n)_n$ be a sequence of continuous real-valued stochastic processes in $[0,T]$, $T<infty$ such that for every $tin [0,T]$ $X^n_t$ converges in probability to $X_t$.
Let $tau^n=inf{tin[0,T], X^n<0}wedge T$ and $tau=inf{tin[0,T], X<0}wedge T$.
Why does $tau^n$ converge in probability to $tau$ ?
probability stochastic-processes stopping-times
probability stochastic-processes stopping-times
asked yesterday
Al Bundy
12410
asked yesterday
Al Bundy
12410
edited yesterday
asked yesterday
Al Bundy
12410
asked yesterday
Al Bundy
12410
asked yesterday
Al Bundy
12410
12410
add a comment |
add a comment |
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