Proof that supremum of an almost surely continuous random function is random variable
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Let ${X_t, tin[0,T]}$ on {$R, mathfrak B(R) $} be random, almost surely continuous, function. Show that $X^+=sup_{t in[0,T]} X_t$ is random variable.
My proof:
Let $P$ be set points of no continuity of $X_t$.
And let $widetilde X_t =left{
begin{array}{c}
X_t(omega), omegain mathbb Rsetminus P \
0, omegain P
end{array}
right. $
sup$widetilde X_t$ will be finite and: $
{omega: sup_{t in[0,1]}widetilde X_t>x}=bigcup_{tin [0,T]bigcap Q}{omega:widetilde X_t >x}$
Since ${omega:widetilde X_t >x}$ is in $mathfrak B(R)$ then ${omega : sup_{t in[0,1]}widetilde X_t>x}$ in $mathfrak B(R)$.
And since the intervals $(x, +infty)$ form $mathfrak B(R)$,$quad$$widetilde X_t$ is random variable.
Because sup$_{t in[0,T]}widetilde X_t$=sup $_{t in[0,T]}X_t$ $quad$ $forall omegain mathbb Rbackslash P$,$quad$ sup $_{t in[0,T]}X_t$ is random variable too.
probability-theory proof-verification continuity stochastic-processes
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up vote
1
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favorite
Let ${X_t, tin[0,T]}$ on {$R, mathfrak B(R) $} be random, almost surely continuous, function. Show that $X^+=sup_{t in[0,T]} X_t$ is random variable.
My proof:
Let $P$ be set points of no continuity of $X_t$.
And let $widetilde X_t =left{
begin{array}{c}
X_t(omega), omegain mathbb Rsetminus P \
0, omegain P
end{array}
right. $
sup$widetilde X_t$ will be finite and: $
{omega: sup_{t in[0,1]}widetilde X_t>x}=bigcup_{tin [0,T]bigcap Q}{omega:widetilde X_t >x}$
Since ${omega:widetilde X_t >x}$ is in $mathfrak B(R)$ then ${omega : sup_{t in[0,1]}widetilde X_t>x}$ in $mathfrak B(R)$.
And since the intervals $(x, +infty)$ form $mathfrak B(R)$,$quad$$widetilde X_t$ is random variable.
Because sup$_{t in[0,T]}widetilde X_t$=sup $_{t in[0,T]}X_t$ $quad$ $forall omegain mathbb Rbackslash P$,$quad$ sup $_{t in[0,T]}X_t$ is random variable too.
probability-theory proof-verification continuity stochastic-processes
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let ${X_t, tin[0,T]}$ on {$R, mathfrak B(R) $} be random, almost surely continuous, function. Show that $X^+=sup_{t in[0,T]} X_t$ is random variable.
My proof:
Let $P$ be set points of no continuity of $X_t$.
And let $widetilde X_t =left{
begin{array}{c}
X_t(omega), omegain mathbb Rsetminus P \
0, omegain P
end{array}
right. $
sup$widetilde X_t$ will be finite and: $
{omega: sup_{t in[0,1]}widetilde X_t>x}=bigcup_{tin [0,T]bigcap Q}{omega:widetilde X_t >x}$
Since ${omega:widetilde X_t >x}$ is in $mathfrak B(R)$ then ${omega : sup_{t in[0,1]}widetilde X_t>x}$ in $mathfrak B(R)$.
And since the intervals $(x, +infty)$ form $mathfrak B(R)$,$quad$$widetilde X_t$ is random variable.
Because sup$_{t in[0,T]}widetilde X_t$=sup $_{t in[0,T]}X_t$ $quad$ $forall omegain mathbb Rbackslash P$,$quad$ sup $_{t in[0,T]}X_t$ is random variable too.
probability-theory proof-verification continuity stochastic-processes
Let ${X_t, tin[0,T]}$ on {$R, mathfrak B(R) $} be random, almost surely continuous, function. Show that $X^+=sup_{t in[0,T]} X_t$ is random variable.
My proof:
Let $P$ be set points of no continuity of $X_t$.
And let $widetilde X_t =left{
begin{array}{c}
X_t(omega), omegain mathbb Rsetminus P \
0, omegain P
end{array}
right. $
sup$widetilde X_t$ will be finite and: $
{omega: sup_{t in[0,1]}widetilde X_t>x}=bigcup_{tin [0,T]bigcap Q}{omega:widetilde X_t >x}$
Since ${omega:widetilde X_t >x}$ is in $mathfrak B(R)$ then ${omega : sup_{t in[0,1]}widetilde X_t>x}$ in $mathfrak B(R)$.
And since the intervals $(x, +infty)$ form $mathfrak B(R)$,$quad$$widetilde X_t$ is random variable.
Because sup$_{t in[0,T]}widetilde X_t$=sup $_{t in[0,T]}X_t$ $quad$ $forall omegain mathbb Rbackslash P$,$quad$ sup $_{t in[0,T]}X_t$ is random variable too.
probability-theory proof-verification continuity stochastic-processes
probability-theory proof-verification continuity stochastic-processes
edited Nov 26 at 22:30
Davide Giraudo
124k16150256
124k16150256
asked Nov 18 at 15:44
Emerald
308
308
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