Expected value of Brownian motion at a time decided by a rate one Poisson process.












1












$begingroup$


Situation:




We have that ${W_{t},tgeq 0}$ is a Brownian motion and
${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
Poisson distribution with parameter $t$. This process is independent
from our Brownian motion.




Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$



b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.



My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:



$mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.



Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.



b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.



Any help is appreciated.






probability-theory brownian-motion stopping-times poisson-process expected-value







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Situation:




    We have that ${W_{t},tgeq 0}$ is a Brownian motion and
    ${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
    Poisson distribution with parameter $t$. This process is independent
    from our Brownian motion.




    Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$



    b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.



    My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:



    $mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.



    Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.



    b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.



    Any help is appreciated.






    probability-theory brownian-motion stopping-times poisson-process expected-value







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Situation:




      We have that ${W_{t},tgeq 0}$ is a Brownian motion and
      ${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
      Poisson distribution with parameter $t$. This process is independent
      from our Brownian motion.




      Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$



      b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.



      My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:



      $mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.



      Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.



      b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.



      Any help is appreciated.






      probability-theory brownian-motion stopping-times poisson-process expected-value







      share|cite|improve this question











      $endgroup$




      Situation:




      We have that ${W_{t},tgeq 0}$ is a Brownian motion and
      ${N_{t},tgeq 0}$ is a Poisson process such that $N_{t}$ follows a
      Poisson distribution with parameter $t$. This process is independent
      from our Brownian motion.




      Question: a) Show that $mathbb{E}[W^{2}_{N_{t}}]=t$



      b) For $ageq 0$ we define the stopping time: $tau=inf{tgeq 0:|W_{t}|>a}$. Show that this stopping time is finite.



      My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:



      $mathbb{E}[W^{2}_{N_{t}}]=mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0leq s<N_{t}$.



      Using the Brownian motion properties a few more times, this reduces to $mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.



      b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.



      Any help is appreciated.






      probability-theory brownian-motion stopping-times poisson-process expected-value




      probability-theory brownian-motion stopping-times poisson-process expected-value






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 1 at 8:49









      saz

      81.3k861127




      81.3k861127










      asked Dec 31 '18 at 17:23









      S. CrimS. Crim

      389112




      389112






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.




          $N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$



          Alternatively, we can use the tower property of the conditional expectation to show that



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$




          To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
            $endgroup$
            – S. Crim
            Dec 31 '18 at 18:14








          • 1




            $begingroup$
            @S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
            $endgroup$
            – saz
            Dec 31 '18 at 18:17








          • 1




            $begingroup$
            @S.Crim Yeah, sorry for the typo.
            $endgroup$
            – saz
            Dec 31 '18 at 18:21






          • 1




            $begingroup$
            @S.Crim $min{t,tau} uparrow tau$ holds also on ${tau=infty}$; just note that $$lim_{t to infty} min{t,tau(omega)} = infty = tau(omega)$$ for all $omega in {tau=infty}$.
            $endgroup$
            – saz
            Jan 26 at 19:00






          • 1




            $begingroup$
            @S.Crim Not sure whether it's just a typo... we have $$lim_{t to infty} min{t,tau(omega)} = lim_{t to infty} t = infty = tau(omega)$$ on ${tau=infty}$. (You wrote $tau(omega)=t$; that's wrong, note that this equation doesn't even make sense here)
            $endgroup$
            – saz
            Jan 26 at 19:09













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          1 Answer
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          active

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          active

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          2












          $begingroup$

          Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.




          $N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$



          Alternatively, we can use the tower property of the conditional expectation to show that



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$




          To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
            $endgroup$
            – S. Crim
            Dec 31 '18 at 18:14








          • 1




            $begingroup$
            @S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
            $endgroup$
            – saz
            Dec 31 '18 at 18:17








          • 1




            $begingroup$
            @S.Crim Yeah, sorry for the typo.
            $endgroup$
            – saz
            Dec 31 '18 at 18:21






          • 1




            $begingroup$
            @S.Crim $min{t,tau} uparrow tau$ holds also on ${tau=infty}$; just note that $$lim_{t to infty} min{t,tau(omega)} = infty = tau(omega)$$ for all $omega in {tau=infty}$.
            $endgroup$
            – saz
            Jan 26 at 19:00






          • 1




            $begingroup$
            @S.Crim Not sure whether it's just a typo... we have $$lim_{t to infty} min{t,tau(omega)} = lim_{t to infty} t = infty = tau(omega)$$ on ${tau=infty}$. (You wrote $tau(omega)=t$; that's wrong, note that this equation doesn't even make sense here)
            $endgroup$
            – saz
            Jan 26 at 19:09


















          2












          $begingroup$

          Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.




          $N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$



          Alternatively, we can use the tower property of the conditional expectation to show that



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$




          To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
            $endgroup$
            – S. Crim
            Dec 31 '18 at 18:14








          • 1




            $begingroup$
            @S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
            $endgroup$
            – saz
            Dec 31 '18 at 18:17








          • 1




            $begingroup$
            @S.Crim Yeah, sorry for the typo.
            $endgroup$
            – saz
            Dec 31 '18 at 18:21






          • 1




            $begingroup$
            @S.Crim $min{t,tau} uparrow tau$ holds also on ${tau=infty}$; just note that $$lim_{t to infty} min{t,tau(omega)} = infty = tau(omega)$$ for all $omega in {tau=infty}$.
            $endgroup$
            – saz
            Jan 26 at 19:00






          • 1




            $begingroup$
            @S.Crim Not sure whether it's just a typo... we have $$lim_{t to infty} min{t,tau(omega)} = lim_{t to infty} t = infty = tau(omega)$$ on ${tau=infty}$. (You wrote $tau(omega)=t$; that's wrong, note that this equation doesn't even make sense here)
            $endgroup$
            – saz
            Jan 26 at 19:09
















          2












          2








          2





          $begingroup$

          Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.




          $N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$



          Alternatively, we can use the tower property of the conditional expectation to show that



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$




          To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.






          share|cite|improve this answer











          $endgroup$



          Note that $N_t$ is a random variable and therefore the statement "where $0 leq s < N_t$" doesn't make sense.




          $N_t$ takes values in $mathbb{N}_0$ and because of the independence from $(W_t)_{t geq 0}$ we have



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} left( sum_{k geq 0} W_k^2 1_{{N_t = k}} right) = sum_{k geq 0} underbrace{mathbb{E}(W_k^2)}_{=k} underbrace{mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} sum_{k geq 1} frac{t^{k-1}}{(k-1)!} =t.$$



          Alternatively, we can use the tower property of the conditional expectation to show that



          $$mathbb{E}(W_{N_t}^2) = mathbb{E} big[ mathbb{E}(W_{N_t}^2 mid N_t) big] = mathbb{E} big[ underbrace{mathbb{E}(W_k^2)}_{=k} big|_{k=N_t} big] = mathbb{E}(N_t)=t.$$




          To prove that the stopping time $tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$mathbb{E}(W_{t wedge tau}^2) = mathbb{E}(t wedge tau).$$ Since $|W_{t wedge tau}| leq a$ this gives $$mathbb{E}(t wedge tau) leq a^2,$$ and now the monotone convergence theorem yields $$mathbb{E}(tau) leq a^2$$ which shows, in particular, that $tau<infty$ almost surely.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 31 '18 at 18:18

























          answered Dec 31 '18 at 17:45









          sazsaz

          81.3k861127




          81.3k861127












          • $begingroup$
            Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
            $endgroup$
            – S. Crim
            Dec 31 '18 at 18:14








          • 1




            $begingroup$
            @S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
            $endgroup$
            – saz
            Dec 31 '18 at 18:17








          • 1




            $begingroup$
            @S.Crim Yeah, sorry for the typo.
            $endgroup$
            – saz
            Dec 31 '18 at 18:21






          • 1




            $begingroup$
            @S.Crim $min{t,tau} uparrow tau$ holds also on ${tau=infty}$; just note that $$lim_{t to infty} min{t,tau(omega)} = infty = tau(omega)$$ for all $omega in {tau=infty}$.
            $endgroup$
            – saz
            Jan 26 at 19:00






          • 1




            $begingroup$
            @S.Crim Not sure whether it's just a typo... we have $$lim_{t to infty} min{t,tau(omega)} = lim_{t to infty} t = infty = tau(omega)$$ on ${tau=infty}$. (You wrote $tau(omega)=t$; that's wrong, note that this equation doesn't even make sense here)
            $endgroup$
            – saz
            Jan 26 at 19:09




















          • $begingroup$
            Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
            $endgroup$
            – S. Crim
            Dec 31 '18 at 18:14








          • 1




            $begingroup$
            @S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
            $endgroup$
            – saz
            Dec 31 '18 at 18:17








          • 1




            $begingroup$
            @S.Crim Yeah, sorry for the typo.
            $endgroup$
            – saz
            Dec 31 '18 at 18:21






          • 1




            $begingroup$
            @S.Crim $min{t,tau} uparrow tau$ holds also on ${tau=infty}$; just note that $$lim_{t to infty} min{t,tau(omega)} = infty = tau(omega)$$ for all $omega in {tau=infty}$.
            $endgroup$
            – saz
            Jan 26 at 19:00






          • 1




            $begingroup$
            @S.Crim Not sure whether it's just a typo... we have $$lim_{t to infty} min{t,tau(omega)} = lim_{t to infty} t = infty = tau(omega)$$ on ${tau=infty}$. (You wrote $tau(omega)=t$; that's wrong, note that this equation doesn't even make sense here)
            $endgroup$
            – saz
            Jan 26 at 19:09


















          $begingroup$
          Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
          $endgroup$
          – S. Crim
          Dec 31 '18 at 18:14






          $begingroup$
          Thanks alot, once again. Your solution for a) using the tower property makes perfect sense. I have a question about your solution for b), though. How do you conclude from $|W_{t wedge tau}| leq a$ that $mathbb{E}(t wedge tau) leq a$? I'm struggling to relate the fact that $|W_{t wedge tau}| leq a$ to $mathbb{E}(W_{t wedge tau}^2)$.
          $endgroup$
          – S. Crim
          Dec 31 '18 at 18:14






          1




          1




          $begingroup$
          @S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
          $endgroup$
          – saz
          Dec 31 '18 at 18:17






          $begingroup$
          @S.Crim If $X$ is a random variable such that $0 leq X leq b$, then $mathbb{E}(X) leq b$, right? That's what I used to estimate $mathbb{E}(W_{t wedge tau}^2)$. (Sorry, there was a square missing in my estimate; should be fine now... perhaps this was the reason for your confusion.)
          $endgroup$
          – saz
          Dec 31 '18 at 18:17






          1




          1




          $begingroup$
          @S.Crim Yeah, sorry for the typo.
          $endgroup$
          – saz
          Dec 31 '18 at 18:21




          $begingroup$
          @S.Crim Yeah, sorry for the typo.
          $endgroup$
          – saz
          Dec 31 '18 at 18:21




          1




          1




          $begingroup$
          @S.Crim $min{t,tau} uparrow tau$ holds also on ${tau=infty}$; just note that $$lim_{t to infty} min{t,tau(omega)} = infty = tau(omega)$$ for all $omega in {tau=infty}$.
          $endgroup$
          – saz
          Jan 26 at 19:00




          $begingroup$
          @S.Crim $min{t,tau} uparrow tau$ holds also on ${tau=infty}$; just note that $$lim_{t to infty} min{t,tau(omega)} = infty = tau(omega)$$ for all $omega in {tau=infty}$.
          $endgroup$
          – saz
          Jan 26 at 19:00




          1




          1




          $begingroup$
          @S.Crim Not sure whether it's just a typo... we have $$lim_{t to infty} min{t,tau(omega)} = lim_{t to infty} t = infty = tau(omega)$$ on ${tau=infty}$. (You wrote $tau(omega)=t$; that's wrong, note that this equation doesn't even make sense here)
          $endgroup$
          – saz
          Jan 26 at 19:09






          $begingroup$
          @S.Crim Not sure whether it's just a typo... we have $$lim_{t to infty} min{t,tau(omega)} = lim_{t to infty} t = infty = tau(omega)$$ on ${tau=infty}$. (You wrote $tau(omega)=t$; that's wrong, note that this equation doesn't even make sense here)
          $endgroup$
          – saz
          Jan 26 at 19:09




















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