Two Brownian motions and stopping time

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X and Y are two independent Brownian motions both equal to $1$ at time $0$. Consider the first time, T, that Y process hits $0$. What is the probability that $X_T>0$?



My intuition: I think that at any time $t>0$, $P(X_t>Y_t)=0.5$. The question asks about the random time T and at T we know $Y_T=0$. So $P(X_T>0) = P(X_T>Y_T) = 0.5$.



I would like to know if my intuition is correct? I appreciate it if anyone can provide a hint for a more formal solution. Thanks, guys!






probability stochastic-processes brownian-motion stopping-times







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  • Your intuition does not work because $T$ is a random variable, and it is not independent of $Y$.
    – Mike Earnest
    Sep 10 at 20:13











  • running a sim, the probability is consistently less than 0.5
    – phdmba7of12
    Sep 10 at 20:18






  • 1




    @MikeEarnest: I see. So, the intuition fails because T is not independent of Y.
    – Xin Wei
    Sep 10 at 20:33






  • 1




    @phdmba7of12: Wow, running a simul to check the intuition is a brilliant idea. I guess you meant "larger than 0.5".
    – Xin Wei
    Sep 10 at 20:35










  • correct ... greater than 0.5
    – phdmba7of12
    Sep 11 at 18:08














up vote
1
down vote

favorite












X and Y are two independent Brownian motions both equal to $1$ at time $0$. Consider the first time, T, that Y process hits $0$. What is the probability that $X_T>0$?



My intuition: I think that at any time $t>0$, $P(X_t>Y_t)=0.5$. The question asks about the random time T and at T we know $Y_T=0$. So $P(X_T>0) = P(X_T>Y_T) = 0.5$.



I would like to know if my intuition is correct? I appreciate it if anyone can provide a hint for a more formal solution. Thanks, guys!






probability stochastic-processes brownian-motion stopping-times







share|cite|improve this question























  • Your intuition does not work because $T$ is a random variable, and it is not independent of $Y$.
    – Mike Earnest
    Sep 10 at 20:13











  • running a sim, the probability is consistently less than 0.5
    – phdmba7of12
    Sep 10 at 20:18






  • 1




    @MikeEarnest: I see. So, the intuition fails because T is not independent of Y.
    – Xin Wei
    Sep 10 at 20:33






  • 1




    @phdmba7of12: Wow, running a simul to check the intuition is a brilliant idea. I guess you meant "larger than 0.5".
    – Xin Wei
    Sep 10 at 20:35










  • correct ... greater than 0.5
    – phdmba7of12
    Sep 11 at 18:08












up vote
1
down vote

favorite









up vote
1
down vote

favorite











X and Y are two independent Brownian motions both equal to $1$ at time $0$. Consider the first time, T, that Y process hits $0$. What is the probability that $X_T>0$?



My intuition: I think that at any time $t>0$, $P(X_t>Y_t)=0.5$. The question asks about the random time T and at T we know $Y_T=0$. So $P(X_T>0) = P(X_T>Y_T) = 0.5$.



I would like to know if my intuition is correct? I appreciate it if anyone can provide a hint for a more formal solution. Thanks, guys!






probability stochastic-processes brownian-motion stopping-times







share|cite|improve this question















X and Y are two independent Brownian motions both equal to $1$ at time $0$. Consider the first time, T, that Y process hits $0$. What is the probability that $X_T>0$?



My intuition: I think that at any time $t>0$, $P(X_t>Y_t)=0.5$. The question asks about the random time T and at T we know $Y_T=0$. So $P(X_T>0) = P(X_T>Y_T) = 0.5$.



I would like to know if my intuition is correct? I appreciate it if anyone can provide a hint for a more formal solution. Thanks, guys!






probability stochastic-processes brownian-motion stopping-times




probability stochastic-processes brownian-motion stopping-times






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 10 at 20:47

























asked Sep 10 at 20:02









Xin Wei

154




154











  • Your intuition does not work because $T$ is a random variable, and it is not independent of $Y$.
    – Mike Earnest
    Sep 10 at 20:13











  • running a sim, the probability is consistently less than 0.5
    – phdmba7of12
    Sep 10 at 20:18






  • 1




    @MikeEarnest: I see. So, the intuition fails because T is not independent of Y.
    – Xin Wei
    Sep 10 at 20:33






  • 1




    @phdmba7of12: Wow, running a simul to check the intuition is a brilliant idea. I guess you meant "larger than 0.5".
    – Xin Wei
    Sep 10 at 20:35










  • correct ... greater than 0.5
    – phdmba7of12
    Sep 11 at 18:08
















  • Your intuition does not work because $T$ is a random variable, and it is not independent of $Y$.
    – Mike Earnest
    Sep 10 at 20:13











  • running a sim, the probability is consistently less than 0.5
    – phdmba7of12
    Sep 10 at 20:18






  • 1




    @MikeEarnest: I see. So, the intuition fails because T is not independent of Y.
    – Xin Wei
    Sep 10 at 20:33






  • 1




    @phdmba7of12: Wow, running a simul to check the intuition is a brilliant idea. I guess you meant "larger than 0.5".
    – Xin Wei
    Sep 10 at 20:35










  • correct ... greater than 0.5
    – phdmba7of12
    Sep 11 at 18:08















Your intuition does not work because $T$ is a random variable, and it is not independent of $Y$.
– Mike Earnest
Sep 10 at 20:13





Your intuition does not work because $T$ is a random variable, and it is not independent of $Y$.
– Mike Earnest
Sep 10 at 20:13













running a sim, the probability is consistently less than 0.5
– phdmba7of12
Sep 10 at 20:18




running a sim, the probability is consistently less than 0.5
– phdmba7of12
Sep 10 at 20:18




1




1




@MikeEarnest: I see. So, the intuition fails because T is not independent of Y.
– Xin Wei
Sep 10 at 20:33




@MikeEarnest: I see. So, the intuition fails because T is not independent of Y.
– Xin Wei
Sep 10 at 20:33




1




1




@phdmba7of12: Wow, running a simul to check the intuition is a brilliant idea. I guess you meant "larger than 0.5".
– Xin Wei
Sep 10 at 20:35




@phdmba7of12: Wow, running a simul to check the intuition is a brilliant idea. I guess you meant "larger than 0.5".
– Xin Wei
Sep 10 at 20:35












correct ... greater than 0.5
– phdmba7of12
Sep 11 at 18:08




correct ... greater than 0.5
– phdmba7of12
Sep 11 at 18:08










1 Answer
1






active

oldest

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up vote
4
down vote



accepted










Hint: Let $S$ be the first time that $X$ hits $0$.



  • If $S>T$, then certainly $X_T>0$.


  • If $S<T$, then $Z_r:= X_S+r$ is a Brownian motion independent of $Y$ with $Z_0=0$, so $P(Z_T-S>0mid S<T)=frac12$.


Furthermore, $P(T>S)=frac12$.






share|cite|improve this answer






















  • but that isn't what the question is asking ... it's asking $P(X_T>0)$
    – phdmba7of12
    Sep 10 at 20:27






  • 1




    @Mike: Thanks for the hint! So, P(X(T)>0) = P(S>T)+P(S<T)*P(Z_T-S>0)=1/2+1/2*1/2=3/4.
    – Xin Wei
    Sep 10 at 20:37










  • Yes, exactly. Perhaps it takes a little more work to make these arguments rigorous, but I think you now have a good road map to find a rigorous proof. @XinWei
    – Mike Earnest
    Sep 10 at 20:40










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










Hint: Let $S$ be the first time that $X$ hits $0$.



  • If $S>T$, then certainly $X_T>0$.


  • If $S<T$, then $Z_r:= X_S+r$ is a Brownian motion independent of $Y$ with $Z_0=0$, so $P(Z_T-S>0mid S<T)=frac12$.


Furthermore, $P(T>S)=frac12$.






share|cite|improve this answer






















  • but that isn't what the question is asking ... it's asking $P(X_T>0)$
    – phdmba7of12
    Sep 10 at 20:27






  • 1




    @Mike: Thanks for the hint! So, P(X(T)>0) = P(S>T)+P(S<T)*P(Z_T-S>0)=1/2+1/2*1/2=3/4.
    – Xin Wei
    Sep 10 at 20:37










  • Yes, exactly. Perhaps it takes a little more work to make these arguments rigorous, but I think you now have a good road map to find a rigorous proof. @XinWei
    – Mike Earnest
    Sep 10 at 20:40














up vote
4
down vote



accepted










Hint: Let $S$ be the first time that $X$ hits $0$.



  • If $S>T$, then certainly $X_T>0$.


  • If $S<T$, then $Z_r:= X_S+r$ is a Brownian motion independent of $Y$ with $Z_0=0$, so $P(Z_T-S>0mid S<T)=frac12$.


Furthermore, $P(T>S)=frac12$.






share|cite|improve this answer






















  • but that isn't what the question is asking ... it's asking $P(X_T>0)$
    – phdmba7of12
    Sep 10 at 20:27






  • 1




    @Mike: Thanks for the hint! So, P(X(T)>0) = P(S>T)+P(S<T)*P(Z_T-S>0)=1/2+1/2*1/2=3/4.
    – Xin Wei
    Sep 10 at 20:37










  • Yes, exactly. Perhaps it takes a little more work to make these arguments rigorous, but I think you now have a good road map to find a rigorous proof. @XinWei
    – Mike Earnest
    Sep 10 at 20:40












up vote
4
down vote



accepted







up vote
4
down vote



accepted






Hint: Let $S$ be the first time that $X$ hits $0$.



  • If $S>T$, then certainly $X_T>0$.


  • If $S<T$, then $Z_r:= X_S+r$ is a Brownian motion independent of $Y$ with $Z_0=0$, so $P(Z_T-S>0mid S<T)=frac12$.


Furthermore, $P(T>S)=frac12$.






share|cite|improve this answer














Hint: Let $S$ be the first time that $X$ hits $0$.



  • If $S>T$, then certainly $X_T>0$.


  • If $S<T$, then $Z_r:= X_S+r$ is a Brownian motion independent of $Y$ with $Z_0=0$, so $P(Z_T-S>0mid S<T)=frac12$.


Furthermore, $P(T>S)=frac12$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 10 at 20:44









Did

243k23209445




243k23209445










answered Sep 10 at 20:22









Mike Earnest

18.6k11950




18.6k11950











  • but that isn't what the question is asking ... it's asking $P(X_T>0)$
    – phdmba7of12
    Sep 10 at 20:27






  • 1




    @Mike: Thanks for the hint! So, P(X(T)>0) = P(S>T)+P(S<T)*P(Z_T-S>0)=1/2+1/2*1/2=3/4.
    – Xin Wei
    Sep 10 at 20:37










  • Yes, exactly. Perhaps it takes a little more work to make these arguments rigorous, but I think you now have a good road map to find a rigorous proof. @XinWei
    – Mike Earnest
    Sep 10 at 20:40
















  • but that isn't what the question is asking ... it's asking $P(X_T>0)$
    – phdmba7of12
    Sep 10 at 20:27






  • 1




    @Mike: Thanks for the hint! So, P(X(T)>0) = P(S>T)+P(S<T)*P(Z_T-S>0)=1/2+1/2*1/2=3/4.
    – Xin Wei
    Sep 10 at 20:37










  • Yes, exactly. Perhaps it takes a little more work to make these arguments rigorous, but I think you now have a good road map to find a rigorous proof. @XinWei
    – Mike Earnest
    Sep 10 at 20:40















but that isn't what the question is asking ... it's asking $P(X_T>0)$
– phdmba7of12
Sep 10 at 20:27




but that isn't what the question is asking ... it's asking $P(X_T>0)$
– phdmba7of12
Sep 10 at 20:27




1




1




@Mike: Thanks for the hint! So, P(X(T)>0) = P(S>T)+P(S<T)*P(Z_T-S>0)=1/2+1/2*1/2=3/4.
– Xin Wei
Sep 10 at 20:37




@Mike: Thanks for the hint! So, P(X(T)>0) = P(S>T)+P(S<T)*P(Z_T-S>0)=1/2+1/2*1/2=3/4.
– Xin Wei
Sep 10 at 20:37












Yes, exactly. Perhaps it takes a little more work to make these arguments rigorous, but I think you now have a good road map to find a rigorous proof. @XinWei
– Mike Earnest
Sep 10 at 20:40




Yes, exactly. Perhaps it takes a little more work to make these arguments rigorous, but I think you now have a good road map to find a rigorous proof. @XinWei
– Mike Earnest
Sep 10 at 20:40

















 

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