Stretching the Brownian Motion in $[0,1]$ to get another Brownian Motion in $[0,t]$












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I'm running a simulation of a Standard Brownian Motion by limit of a Symmetric Standard Random Walk ${S_n ,ngeq 1}$ and $S_n=sum_{k=1}^n X_k$, where $$P(X_k =-1)=P(X_k =1)=frac{1}{2},$$ and interpolate linearly between integer points in this way: $$S(t)=S_{[t]}+(t-[t])(S_{[t]+1}-S_{[t]})$$ using the Donsker's Invariance Principle:
$$Z_n(t)=frac{S(nt)}{sqrt{n}},$$ then ${ Z_n : ngeq1}$ converges in distribution to a standard brownian motion ${B(t):tin[0,1]}$.




This is working fine for $tin[0,1]$ but, what if I want Brownian Motion in $[0,10]$, for example ${W_t :tin[0,10]}$?




I'm thinking about generating a vector $[B_0,B_{1/n}, B_{2/n},dots ,B_{n-1/n},B_{1}]$ that contains the position of BM in some points of [0,1] and just assignate $W_0=B_0,W_{10/n}=B_{1/n},dots,W_{10}=B_1$. My question: Is this correct? Theroretically, this is also a BM? It's some kind of stretched BM in $[0,1]$ to get a BM in $[0,10]$. If is not, then do you have any ideas of how could I get a SBM in $[0,10]$ using the same Random Walk?






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    I'm running a simulation of a Standard Brownian Motion by limit of a Symmetric Standard Random Walk ${S_n ,ngeq 1}$ and $S_n=sum_{k=1}^n X_k$, where $$P(X_k =-1)=P(X_k =1)=frac{1}{2},$$ and interpolate linearly between integer points in this way: $$S(t)=S_{[t]}+(t-[t])(S_{[t]+1}-S_{[t]})$$ using the Donsker's Invariance Principle:
    $$Z_n(t)=frac{S(nt)}{sqrt{n}},$$ then ${ Z_n : ngeq1}$ converges in distribution to a standard brownian motion ${B(t):tin[0,1]}$.




    This is working fine for $tin[0,1]$ but, what if I want Brownian Motion in $[0,10]$, for example ${W_t :tin[0,10]}$?




    I'm thinking about generating a vector $[B_0,B_{1/n}, B_{2/n},dots ,B_{n-1/n},B_{1}]$ that contains the position of BM in some points of [0,1] and just assignate $W_0=B_0,W_{10/n}=B_{1/n},dots,W_{10}=B_1$. My question: Is this correct? Theroretically, this is also a BM? It's some kind of stretched BM in $[0,1]$ to get a BM in $[0,10]$. If is not, then do you have any ideas of how could I get a SBM in $[0,10]$ using the same Random Walk?






    brownian-motion random-walk simulation







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      I'm running a simulation of a Standard Brownian Motion by limit of a Symmetric Standard Random Walk ${S_n ,ngeq 1}$ and $S_n=sum_{k=1}^n X_k$, where $$P(X_k =-1)=P(X_k =1)=frac{1}{2},$$ and interpolate linearly between integer points in this way: $$S(t)=S_{[t]}+(t-[t])(S_{[t]+1}-S_{[t]})$$ using the Donsker's Invariance Principle:
      $$Z_n(t)=frac{S(nt)}{sqrt{n}},$$ then ${ Z_n : ngeq1}$ converges in distribution to a standard brownian motion ${B(t):tin[0,1]}$.




      This is working fine for $tin[0,1]$ but, what if I want Brownian Motion in $[0,10]$, for example ${W_t :tin[0,10]}$?




      I'm thinking about generating a vector $[B_0,B_{1/n}, B_{2/n},dots ,B_{n-1/n},B_{1}]$ that contains the position of BM in some points of [0,1] and just assignate $W_0=B_0,W_{10/n}=B_{1/n},dots,W_{10}=B_1$. My question: Is this correct? Theroretically, this is also a BM? It's some kind of stretched BM in $[0,1]$ to get a BM in $[0,10]$. If is not, then do you have any ideas of how could I get a SBM in $[0,10]$ using the same Random Walk?






      brownian-motion random-walk simulation







      share|cite|improve this question















      I'm running a simulation of a Standard Brownian Motion by limit of a Symmetric Standard Random Walk ${S_n ,ngeq 1}$ and $S_n=sum_{k=1}^n X_k$, where $$P(X_k =-1)=P(X_k =1)=frac{1}{2},$$ and interpolate linearly between integer points in this way: $$S(t)=S_{[t]}+(t-[t])(S_{[t]+1}-S_{[t]})$$ using the Donsker's Invariance Principle:
      $$Z_n(t)=frac{S(nt)}{sqrt{n}},$$ then ${ Z_n : ngeq1}$ converges in distribution to a standard brownian motion ${B(t):tin[0,1]}$.




      This is working fine for $tin[0,1]$ but, what if I want Brownian Motion in $[0,10]$, for example ${W_t :tin[0,10]}$?




      I'm thinking about generating a vector $[B_0,B_{1/n}, B_{2/n},dots ,B_{n-1/n},B_{1}]$ that contains the position of BM in some points of [0,1] and just assignate $W_0=B_0,W_{10/n}=B_{1/n},dots,W_{10}=B_1$. My question: Is this correct? Theroretically, this is also a BM? It's some kind of stretched BM in $[0,1]$ to get a BM in $[0,10]$. If is not, then do you have any ideas of how could I get a SBM in $[0,10]$ using the same Random Walk?






      brownian-motion random-walk simulation




      brownian-motion random-walk simulation






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      edited Nov 21 '18 at 8:17









      saz

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      asked Nov 16 '18 at 15:56









      WienerFan

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          If you take a brownian motion $(B_t)_{tgeq 0}$ and $lambda > 0$, then the process $(W_t)_{tgeq 0}$ defined by $W_t := sqrt{lambda} B_{frac 1 {lambda} t} $ is again a Brownian Motion.



          So, lets suppose you want to sample $W_s$ where $sin [0,10]$. What can you do? You said you can sample from $(B_t)_{tin [0,1]}$. Therefore you could sample $B_{frac s {10}}$ and scale it by ${sqrt{10}}$. Thus you have a sample of $W_s$.






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            Yes, it is possible to strech a Brownian motion. There is the following result (which is not difficult to prove)




            Proposition Let $(B_t)_{t geq 0}$ be a one-dimensional Brownian motion. Then for any $a>0$ the process defined by $$W_t := frac{1}{sqrt{a}} B_{at}, qquad t geq 0, $$ is a Brownian motion.




            If you have a Brownian motion $(B_t)_{t in [0,1]}$ then you can use this result to strech the Brownian motion to a larger time interval $[0,T]$ by choosing small $a:=1/T$, i.e. $$W_t = sqrt{T} B_{t/T}, qquad t in [0,T].$$






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              2 Answers
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              2 Answers
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              If you take a brownian motion $(B_t)_{tgeq 0}$ and $lambda > 0$, then the process $(W_t)_{tgeq 0}$ defined by $W_t := sqrt{lambda} B_{frac 1 {lambda} t} $ is again a Brownian Motion.



              So, lets suppose you want to sample $W_s$ where $sin [0,10]$. What can you do? You said you can sample from $(B_t)_{tin [0,1]}$. Therefore you could sample $B_{frac s {10}}$ and scale it by ${sqrt{10}}$. Thus you have a sample of $W_s$.






              share|cite|improve this answer


























                1














                If you take a brownian motion $(B_t)_{tgeq 0}$ and $lambda > 0$, then the process $(W_t)_{tgeq 0}$ defined by $W_t := sqrt{lambda} B_{frac 1 {lambda} t} $ is again a Brownian Motion.



                So, lets suppose you want to sample $W_s$ where $sin [0,10]$. What can you do? You said you can sample from $(B_t)_{tin [0,1]}$. Therefore you could sample $B_{frac s {10}}$ and scale it by ${sqrt{10}}$. Thus you have a sample of $W_s$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  If you take a brownian motion $(B_t)_{tgeq 0}$ and $lambda > 0$, then the process $(W_t)_{tgeq 0}$ defined by $W_t := sqrt{lambda} B_{frac 1 {lambda} t} $ is again a Brownian Motion.



                  So, lets suppose you want to sample $W_s$ where $sin [0,10]$. What can you do? You said you can sample from $(B_t)_{tin [0,1]}$. Therefore you could sample $B_{frac s {10}}$ and scale it by ${sqrt{10}}$. Thus you have a sample of $W_s$.






                  share|cite|improve this answer












                  If you take a brownian motion $(B_t)_{tgeq 0}$ and $lambda > 0$, then the process $(W_t)_{tgeq 0}$ defined by $W_t := sqrt{lambda} B_{frac 1 {lambda} t} $ is again a Brownian Motion.



                  So, lets suppose you want to sample $W_s$ where $sin [0,10]$. What can you do? You said you can sample from $(B_t)_{tin [0,1]}$. Therefore you could sample $B_{frac s {10}}$ and scale it by ${sqrt{10}}$. Thus you have a sample of $W_s$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 16 '18 at 16:32









                  Falrach

                  1,618223




                  1,618223























                      2














                      Yes, it is possible to strech a Brownian motion. There is the following result (which is not difficult to prove)




                      Proposition Let $(B_t)_{t geq 0}$ be a one-dimensional Brownian motion. Then for any $a>0$ the process defined by $$W_t := frac{1}{sqrt{a}} B_{at}, qquad t geq 0, $$ is a Brownian motion.




                      If you have a Brownian motion $(B_t)_{t in [0,1]}$ then you can use this result to strech the Brownian motion to a larger time interval $[0,T]$ by choosing small $a:=1/T$, i.e. $$W_t = sqrt{T} B_{t/T}, qquad t in [0,T].$$






                      share|cite|improve this answer


























                        2














                        Yes, it is possible to strech a Brownian motion. There is the following result (which is not difficult to prove)




                        Proposition Let $(B_t)_{t geq 0}$ be a one-dimensional Brownian motion. Then for any $a>0$ the process defined by $$W_t := frac{1}{sqrt{a}} B_{at}, qquad t geq 0, $$ is a Brownian motion.




                        If you have a Brownian motion $(B_t)_{t in [0,1]}$ then you can use this result to strech the Brownian motion to a larger time interval $[0,T]$ by choosing small $a:=1/T$, i.e. $$W_t = sqrt{T} B_{t/T}, qquad t in [0,T].$$






                        share|cite|improve this answer
























                          2












                          2








                          2






                          Yes, it is possible to strech a Brownian motion. There is the following result (which is not difficult to prove)




                          Proposition Let $(B_t)_{t geq 0}$ be a one-dimensional Brownian motion. Then for any $a>0$ the process defined by $$W_t := frac{1}{sqrt{a}} B_{at}, qquad t geq 0, $$ is a Brownian motion.




                          If you have a Brownian motion $(B_t)_{t in [0,1]}$ then you can use this result to strech the Brownian motion to a larger time interval $[0,T]$ by choosing small $a:=1/T$, i.e. $$W_t = sqrt{T} B_{t/T}, qquad t in [0,T].$$






                          share|cite|improve this answer












                          Yes, it is possible to strech a Brownian motion. There is the following result (which is not difficult to prove)




                          Proposition Let $(B_t)_{t geq 0}$ be a one-dimensional Brownian motion. Then for any $a>0$ the process defined by $$W_t := frac{1}{sqrt{a}} B_{at}, qquad t geq 0, $$ is a Brownian motion.




                          If you have a Brownian motion $(B_t)_{t in [0,1]}$ then you can use this result to strech the Brownian motion to a larger time interval $[0,T]$ by choosing small $a:=1/T$, i.e. $$W_t = sqrt{T} B_{t/T}, qquad t in [0,T].$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 16 '18 at 16:30









                          saz

                          78.3k758123




                          78.3k758123






























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