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 $tau$ ?










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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 $tau$ ?










    share|cite|improve this question


























      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 $tau$ ?










      share|cite|improve this question















      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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      share|cite|improve this question













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      edited Nov 19 at 16:38

























      asked Nov 19 at 16:31









      Al Bundy

      13410




      13410



























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