convergence in law of an exponential brownian motion
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I have a queston about the convergence in law of the following stochastic processe:
$$leftI_t=left(int_0^te^B_sdsright)^1/sqrttright_tgeq 0$$
with $B_t_tgeq 0$ is a standrad brownian motion.
Prove that $I_trightarrow e^$ in law, where $N$ has the gaussian distribution $N(0,1)$.
I have tried by scaling property of brownian motion, but it does not work. I try also with the Laplace tranformation, but it is really difficult to discrible $I_t$'s transformation. Does someone have an idea? Thanks a lot!
stochastic-processes
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
asked Oct 12 '12 at 17:21
Higgs88
422212
add a comment |Â
up vote
2
down vote
favorite
I have a queston about the convergence in law of the following stochastic processe:
$$leftI_t=left(int_0^te^B_sdsright)^1/sqrttright_tgeq 0$$
with $B_t_tgeq 0$ is a standrad brownian motion.
Prove that $I_trightarrow e^$ in law, where $N$ has the gaussian distribution $N(0,1)$.
I have tried by scaling property of brownian motion, but it does not work. I try also with the Laplace tranformation, but it is really difficult to discrible $I_t$'s transformation. Does someone have an idea? Thanks a lot!
stochastic-processes
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
asked Oct 12 '12 at 17:21
Higgs88
422212
1
@ Higgs88 : You should think about Laplace Method (the one that talk about $L^P$ convergence when $ptoinfty$ of a suitable random variable (or measurable function), by the way scaling property is clearly a step in the process. Best regards
– TheBridge
Oct 13 '12 at 21:30
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a queston about the convergence in law of the following stochastic processe:
$$leftI_t=left(int_0^te^B_sdsright)^1/sqrttright_tgeq 0$$
with $B_t_tgeq 0$ is a standrad brownian motion.
Prove that $I_trightarrow e^$ in law, where $N$ has the gaussian distribution $N(0,1)$.
I have tried by scaling property of brownian motion, but it does not work. I try also with the Laplace tranformation, but it is really difficult to discrible $I_t$'s transformation. Does someone have an idea? Thanks a lot!
stochastic-processes
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
asked Oct 12 '12 at 17:21
Higgs88
422212
I have a queston about the convergence in law of the following stochastic processe:
$$leftI_t=left(int_0^te^B_sdsright)^1/sqrttright_tgeq 0$$
with $B_t_tgeq 0$ is a standrad brownian motion.
Prove that $I_trightarrow e^$ in law, where $N$ has the gaussian distribution $N(0,1)$.
I have tried by scaling property of brownian motion, but it does not work. I try also with the Laplace tranformation, but it is really difficult to discrible $I_t$'s transformation. Does someone have an idea? Thanks a lot!
stochastic-processes
stochastic-processes
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
asked Oct 12 '12 at 17:21
Higgs88
422212
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
asked Oct 12 '12 at 17:21
Higgs88
422212
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
edited Oct 12 '12 at 17:27
MJD
46.2k28199378
46.2k28199378
asked Oct 12 '12 at 17:21
Higgs88
422212
asked Oct 12 '12 at 17:21
Higgs88
422212
asked Oct 12 '12 at 17:21
Higgs88
422212
422212
1
@ Higgs88 : You should think about Laplace Method (the one that talk about $L^P$ convergence when $ptoinfty$ of a suitable random variable (or measurable function), by the way scaling property is clearly a step in the process. Best regards
– TheBridge
Oct 13 '12 at 21:30
add a comment |Â
1
@ Higgs88 : You should think about Laplace Method (the one that talk about $L^P$ convergence when $ptoinfty$ of a suitable random variable (or measurable function), by the way scaling property is clearly a step in the process. Best regards
– TheBridge
Oct 13 '12 at 21:30
1
1
@ Higgs88 : You should think about Laplace Method (the one that talk about $L^P$ convergence when $ptoinfty$ of a suitable random variable (or measurable function), by the way scaling property is clearly a step in the process. Best regards
– TheBridge
Oct 13 '12 at 21:30
@ Higgs88 : You should think about Laplace Method (the one that talk about $L^P$ convergence when $ptoinfty$ of a suitable random variable (or measurable function), by the way scaling property is clearly a step in the process. Best regards
– TheBridge
Oct 13 '12 at 21:30
add a comment |Â
1 Answer
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This is an answer based on the comment given by TheBridge above.
Basically we note by the scaling property that $(B_s)_sin[0,t] stackreld= t^1/2 (B_s/t)_s in [0,t]$. Therefore $$int_0^t e^B_sds stackreld= int_0^t e^t^1/2 B_s/tds = tint_0^1 e^t^1/2B_udu,$$ where we make a change of variable $s=tu$ in the final equality. Now using the fact that $L^p$ norms converge to the $L^infty$ norm as $p to infty$, and setting $p = sqrt t$ we see that $$lim_t to infty bigg(tint_0^1 e^t^1/2B_udu bigg)^1/sqrtt = 1 cdot e^max_t in [0,1] B_t.$$ We used the fact that $t^1/sqrtt to 1$.
But (see Revuz & Yor Chapter 3) it is well known that $max_[0,1] B_t stackreld= |N|$ where $N sim N(0,1)$. This completes the proof.
answered Sep 2 at 4:09
Shalop
9,04411029
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This is an answer based on the comment given by TheBridge above.
Basically we note by the scaling property that $(B_s)_sin[0,t] stackreld= t^1/2 (B_s/t)_s in [0,t]$. Therefore $$int_0^t e^B_sds stackreld= int_0^t e^t^1/2 B_s/tds = tint_0^1 e^t^1/2B_udu,$$ where we make a change of variable $s=tu$ in the final equality. Now using the fact that $L^p$ norms converge to the $L^infty$ norm as $p to infty$, and setting $p = sqrt t$ we see that $$lim_t to infty bigg(tint_0^1 e^t^1/2B_udu bigg)^1/sqrtt = 1 cdot e^max_t in [0,1] B_t.$$ We used the fact that $t^1/sqrtt to 1$.
But (see Revuz & Yor Chapter 3) it is well known that $max_[0,1] B_t stackreld= |N|$ where $N sim N(0,1)$. This completes the proof.
answered Sep 2 at 4:09
Shalop
9,04411029
add a comment |Â
up vote
0
down vote
This is an answer based on the comment given by TheBridge above.
Basically we note by the scaling property that $(B_s)_sin[0,t] stackreld= t^1/2 (B_s/t)_s in [0,t]$. Therefore $$int_0^t e^B_sds stackreld= int_0^t e^t^1/2 B_s/tds = tint_0^1 e^t^1/2B_udu,$$ where we make a change of variable $s=tu$ in the final equality. Now using the fact that $L^p$ norms converge to the $L^infty$ norm as $p to infty$, and setting $p = sqrt t$ we see that $$lim_t to infty bigg(tint_0^1 e^t^1/2B_udu bigg)^1/sqrtt = 1 cdot e^max_t in [0,1] B_t.$$ We used the fact that $t^1/sqrtt to 1$.
But (see Revuz & Yor Chapter 3) it is well known that $max_[0,1] B_t stackreld= |N|$ where $N sim N(0,1)$. This completes the proof.
answered Sep 2 at 4:09
Shalop
9,04411029
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is an answer based on the comment given by TheBridge above.
Basically we note by the scaling property that $(B_s)_sin[0,t] stackreld= t^1/2 (B_s/t)_s in [0,t]$. Therefore $$int_0^t e^B_sds stackreld= int_0^t e^t^1/2 B_s/tds = tint_0^1 e^t^1/2B_udu,$$ where we make a change of variable $s=tu$ in the final equality. Now using the fact that $L^p$ norms converge to the $L^infty$ norm as $p to infty$, and setting $p = sqrt t$ we see that $$lim_t to infty bigg(tint_0^1 e^t^1/2B_udu bigg)^1/sqrtt = 1 cdot e^max_t in [0,1] B_t.$$ We used the fact that $t^1/sqrtt to 1$.
But (see Revuz & Yor Chapter 3) it is well known that $max_[0,1] B_t stackreld= |N|$ where $N sim N(0,1)$. This completes the proof.
answered Sep 2 at 4:09
Shalop
9,04411029
This is an answer based on the comment given by TheBridge above.
Basically we note by the scaling property that $(B_s)_sin[0,t] stackreld= t^1/2 (B_s/t)_s in [0,t]$. Therefore $$int_0^t e^B_sds stackreld= int_0^t e^t^1/2 B_s/tds = tint_0^1 e^t^1/2B_udu,$$ where we make a change of variable $s=tu$ in the final equality. Now using the fact that $L^p$ norms converge to the $L^infty$ norm as $p to infty$, and setting $p = sqrt t$ we see that $$lim_t to infty bigg(tint_0^1 e^t^1/2B_udu bigg)^1/sqrtt = 1 cdot e^max_t in [0,1] B_t.$$ We used the fact that $t^1/sqrtt to 1$.
But (see Revuz & Yor Chapter 3) it is well known that $max_[0,1] B_t stackreld= |N|$ where $N sim N(0,1)$. This completes the proof.
answered Sep 2 at 4:09
Shalop
9,04411029
answered Sep 2 at 4:09
Shalop
9,04411029
answered Sep 2 at 4:09
Shalop
9,04411029
answered Sep 2 at 4:09
Shalop
9,04411029
9,04411029
add a comment |Â
add a comment |Â
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1
@ Higgs88 : You should think about Laplace Method (the one that talk about $L^P$ convergence when $ptoinfty$ of a suitable random variable (or measurable function), by the way scaling property is clearly a step in the process. Best regards
– TheBridge
Oct 13 '12 at 21:30