Mixing
Show that $suplimits_tX_t-t^{p/2}$ and $suplimits_tleft(frac{X_t}{1+t^{p/2}}right)^q$ have the same law
$begingroup$
I have a question: Let $X=B^{+}$ or $X=|B|$ where $B$ is the standard Brownian motion.
Set
$$J_p=sup_{tgeq 0}(X_t-t^{frac{p}{2}})$$
where $p>1$ and $q$ its conjugate number($p^{-1}+q^{-1}=1$). Prove $J_p$ is a.s. strictly positive and has the same law as
$$sup_{tgeq 0}left(frac{X_t}{1+t^{frac{p}{2}}}right)^q$$
Thanks a lot for your help!
probability-theory stochastic-calculus brownian-motion
edited Jan 11 at 21:56
Did
248k23223460
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
$endgroup$
add a comment |
$begingroup$
I have a question: Let $X=B^{+}$ or $X=|B|$ where $B$ is the standard Brownian motion.
Set
$$J_p=sup_{tgeq 0}(X_t-t^{frac{p}{2}})$$
where $p>1$ and $q$ its conjugate number($p^{-1}+q^{-1}=1$). Prove $J_p$ is a.s. strictly positive and has the same law as
$$sup_{tgeq 0}left(frac{X_t}{1+t^{frac{p}{2}}}right)^q$$
Thanks a lot for your help!
probability-theory stochastic-calculus brownian-motion
edited Jan 11 at 21:56
Did
248k23223460
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
$endgroup$
add a comment |
$begingroup$
I have a question: Let $X=B^{+}$ or $X=|B|$ where $B$ is the standard Brownian motion.
Set
$$J_p=sup_{tgeq 0}(X_t-t^{frac{p}{2}})$$
where $p>1$ and $q$ its conjugate number($p^{-1}+q^{-1}=1$). Prove $J_p$ is a.s. strictly positive and has the same law as
$$sup_{tgeq 0}left(frac{X_t}{1+t^{frac{p}{2}}}right)^q$$
Thanks a lot for your help!
probability-theory stochastic-calculus brownian-motion
edited Jan 11 at 21:56
Did
248k23223460
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
$endgroup$
I have a question: Let $X=B^{+}$ or $X=|B|$ where $B$ is the standard Brownian motion.
Set
$$J_p=sup_{tgeq 0}(X_t-t^{frac{p}{2}})$$
where $p>1$ and $q$ its conjugate number($p^{-1}+q^{-1}=1$). Prove $J_p$ is a.s. strictly positive and has the same law as
$$sup_{tgeq 0}left(frac{X_t}{1+t^{frac{p}{2}}}right)^q$$
Thanks a lot for your help!
probability-theory stochastic-calculus brownian-motion
probability-theory stochastic-calculus brownian-motion
edited Jan 11 at 21:56
Did
248k23223460
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
edited Jan 11 at 21:56
Did
248k23223460
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
edited Jan 11 at 21:56
Did
248k23223460
edited Jan 11 at 21:56
Did
248k23223460
edited Jan 11 at 21:56
Did
248k23223460
248k23223460
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
asked Jan 27 '13 at 21:23
Higgs88Higgs88
427212
427212
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In both cases, $(X_t)$ and $(Y_t)$ coincide in law where $Y_t=a^{-1}X_{a^2t}$ for some $agt0$. Fix $xgt0$, then
$$
[J_pleqslant x]=[forall t,X_tleqslant x+t^{p/2}]=[forall t,Y_{t}leqslant a^{-1}x+a^{p-1}t^{p/2}].
$$
Choosing $a=x^{1/p}$, one gets
$$
[J_pleqslant x]=[forall t,Y_tleqslant x^{1/q}+x^{1/q}t^{p/2}]=[forall t,(1+t^{p/2})^{-q}Y_t^qleqslant x]=[K_pleqslant x],
$$
where
$$
K_p=suplimits_{tgeqslant0},(1+t^{p/2})^{-q}Y_t^q.
$$
This holds for every $x$ hence $J_p=K_p$ almost surely. Finally, $(X_t)$ and $(Y_t)$ coincide in distribution hence $J_p$ coincides in distribution with
$suplimits_{tgeqslant0}, Z_t^q$, where $Z_t=X_t/(1+t^{p/2})$.
answered Jan 30 '13 at 12:20
DidDid
248k23223460
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In both cases, $(X_t)$ and $(Y_t)$ coincide in law where $Y_t=a^{-1}X_{a^2t}$ for some $agt0$. Fix $xgt0$, then
$$
[J_pleqslant x]=[forall t,X_tleqslant x+t^{p/2}]=[forall t,Y_{t}leqslant a^{-1}x+a^{p-1}t^{p/2}].
$$
Choosing $a=x^{1/p}$, one gets
$$
[J_pleqslant x]=[forall t,Y_tleqslant x^{1/q}+x^{1/q}t^{p/2}]=[forall t,(1+t^{p/2})^{-q}Y_t^qleqslant x]=[K_pleqslant x],
$$
where
$$
K_p=suplimits_{tgeqslant0},(1+t^{p/2})^{-q}Y_t^q.
$$
This holds for every $x$ hence $J_p=K_p$ almost surely. Finally, $(X_t)$ and $(Y_t)$ coincide in distribution hence $J_p$ coincides in distribution with
$suplimits_{tgeqslant0}, Z_t^q$, where $Z_t=X_t/(1+t^{p/2})$.
answered Jan 30 '13 at 12:20
DidDid
248k23223460
$endgroup$
add a comment |
$begingroup$
In both cases, $(X_t)$ and $(Y_t)$ coincide in law where $Y_t=a^{-1}X_{a^2t}$ for some $agt0$. Fix $xgt0$, then
$$
[J_pleqslant x]=[forall t,X_tleqslant x+t^{p/2}]=[forall t,Y_{t}leqslant a^{-1}x+a^{p-1}t^{p/2}].
$$
Choosing $a=x^{1/p}$, one gets
$$
[J_pleqslant x]=[forall t,Y_tleqslant x^{1/q}+x^{1/q}t^{p/2}]=[forall t,(1+t^{p/2})^{-q}Y_t^qleqslant x]=[K_pleqslant x],
$$
where
$$
K_p=suplimits_{tgeqslant0},(1+t^{p/2})^{-q}Y_t^q.
$$
This holds for every $x$ hence $J_p=K_p$ almost surely. Finally, $(X_t)$ and $(Y_t)$ coincide in distribution hence $J_p$ coincides in distribution with
$suplimits_{tgeqslant0}, Z_t^q$, where $Z_t=X_t/(1+t^{p/2})$.
answered Jan 30 '13 at 12:20
DidDid
248k23223460
$endgroup$
add a comment |
$begingroup$
In both cases, $(X_t)$ and $(Y_t)$ coincide in law where $Y_t=a^{-1}X_{a^2t}$ for some $agt0$. Fix $xgt0$, then
$$
[J_pleqslant x]=[forall t,X_tleqslant x+t^{p/2}]=[forall t,Y_{t}leqslant a^{-1}x+a^{p-1}t^{p/2}].
$$
Choosing $a=x^{1/p}$, one gets
$$
[J_pleqslant x]=[forall t,Y_tleqslant x^{1/q}+x^{1/q}t^{p/2}]=[forall t,(1+t^{p/2})^{-q}Y_t^qleqslant x]=[K_pleqslant x],
$$
where
$$
K_p=suplimits_{tgeqslant0},(1+t^{p/2})^{-q}Y_t^q.
$$
This holds for every $x$ hence $J_p=K_p$ almost surely. Finally, $(X_t)$ and $(Y_t)$ coincide in distribution hence $J_p$ coincides in distribution with
$suplimits_{tgeqslant0}, Z_t^q$, where $Z_t=X_t/(1+t^{p/2})$.
answered Jan 30 '13 at 12:20
DidDid
248k23223460
$endgroup$
In both cases, $(X_t)$ and $(Y_t)$ coincide in law where $Y_t=a^{-1}X_{a^2t}$ for some $agt0$. Fix $xgt0$, then
$$
[J_pleqslant x]=[forall t,X_tleqslant x+t^{p/2}]=[forall t,Y_{t}leqslant a^{-1}x+a^{p-1}t^{p/2}].
$$
Choosing $a=x^{1/p}$, one gets
$$
[J_pleqslant x]=[forall t,Y_tleqslant x^{1/q}+x^{1/q}t^{p/2}]=[forall t,(1+t^{p/2})^{-q}Y_t^qleqslant x]=[K_pleqslant x],
$$
where
$$
K_p=suplimits_{tgeqslant0},(1+t^{p/2})^{-q}Y_t^q.
$$
This holds for every $x$ hence $J_p=K_p$ almost surely. Finally, $(X_t)$ and $(Y_t)$ coincide in distribution hence $J_p$ coincides in distribution with
$suplimits_{tgeqslant0}, Z_t^q$, where $Z_t=X_t/(1+t^{p/2})$.
answered Jan 30 '13 at 12:20
DidDid
248k23223460
answered Jan 30 '13 at 12:20
DidDid
248k23223460
answered Jan 30 '13 at 12:20
DidDid
248k23223460
answered Jan 30 '13 at 12:20
DidDid
248k23223460
248k23223460
add a comment |
add a comment |
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