Mixing
Martingale and indicator function
$begingroup$
For $tge 0$ define the stochastic process
$$Y_t:[0,1]rightarrowmathbb{R},quad Y_t(x)=begin{cases}0,&text{if }t-xnotinmathbb{Q}\
1,&text{if }t-xinmathbb{Q}
end{cases}$$
on the filtration $(F_t)_{tge0}$, $F_t=mathcal{B}(mathbb{R})$ for all $tge 0$. I want to show, that this is a martingale.
Of course $Y_tle1$, such that $Y_t$ is integrable. For $s,tinmathbb{Q}_+$ I find $Y_t=Y_s$. So we have the martingale property for this case. But how do you find the solution for the other cases? And what can you say about the continuity of the paths of this process? Thanks in advance for any help!
probability-theory stochastic-processes
asked Jan 8 at 22:15
user628255user628255
224
$endgroup$
add a comment |
$begingroup$
For $tge 0$ define the stochastic process
$$Y_t:[0,1]rightarrowmathbb{R},quad Y_t(x)=begin{cases}0,&text{if }t-xnotinmathbb{Q}\
1,&text{if }t-xinmathbb{Q}
end{cases}$$
on the filtration $(F_t)_{tge0}$, $F_t=mathcal{B}(mathbb{R})$ for all $tge 0$. I want to show, that this is a martingale.
Of course $Y_tle1$, such that $Y_t$ is integrable. For $s,tinmathbb{Q}_+$ I find $Y_t=Y_s$. So we have the martingale property for this case. But how do you find the solution for the other cases? And what can you say about the continuity of the paths of this process? Thanks in advance for any help!
probability-theory stochastic-processes
asked Jan 8 at 22:15
user628255user628255
224
$endgroup$
$begingroup$
For every $t$, $P(Y_t=0)=1$ and $Y_t$ is measurable with respect to $mathcal B(mathbb R)$ hence, for every $s<t$, $E(Y_tmidmathcal F_s)=Y_t=0=Y_s$ almost surely, which shows $(Y_t)$ is an $(mathcal F_t)$-martingale.
$endgroup$
– Did
Jan 8 at 22:30
add a comment |
$begingroup$
For $tge 0$ define the stochastic process
$$Y_t:[0,1]rightarrowmathbb{R},quad Y_t(x)=begin{cases}0,&text{if }t-xnotinmathbb{Q}\
1,&text{if }t-xinmathbb{Q}
end{cases}$$
on the filtration $(F_t)_{tge0}$, $F_t=mathcal{B}(mathbb{R})$ for all $tge 0$. I want to show, that this is a martingale.
Of course $Y_tle1$, such that $Y_t$ is integrable. For $s,tinmathbb{Q}_+$ I find $Y_t=Y_s$. So we have the martingale property for this case. But how do you find the solution for the other cases? And what can you say about the continuity of the paths of this process? Thanks in advance for any help!
probability-theory stochastic-processes
asked Jan 8 at 22:15
user628255user628255
224
$endgroup$
For $tge 0$ define the stochastic process
$$Y_t:[0,1]rightarrowmathbb{R},quad Y_t(x)=begin{cases}0,&text{if }t-xnotinmathbb{Q}\
1,&text{if }t-xinmathbb{Q}
end{cases}$$
on the filtration $(F_t)_{tge0}$, $F_t=mathcal{B}(mathbb{R})$ for all $tge 0$. I want to show, that this is a martingale.
Of course $Y_tle1$, such that $Y_t$ is integrable. For $s,tinmathbb{Q}_+$ I find $Y_t=Y_s$. So we have the martingale property for this case. But how do you find the solution for the other cases? And what can you say about the continuity of the paths of this process? Thanks in advance for any help!
probability-theory stochastic-processes
probability-theory stochastic-processes
asked Jan 8 at 22:15
user628255user628255
224
asked Jan 8 at 22:15
user628255user628255
224
edited Jan 8 at 22:24
user628255
asked Jan 8 at 22:15
user628255user628255
224
asked Jan 8 at 22:15
user628255user628255
224
asked Jan 8 at 22:15
user628255user628255
224
224
$begingroup$
For every $t$, $P(Y_t=0)=1$ and $Y_t$ is measurable with respect to $mathcal B(mathbb R)$ hence, for every $s<t$, $E(Y_tmidmathcal F_s)=Y_t=0=Y_s$ almost surely, which shows $(Y_t)$ is an $(mathcal F_t)$-martingale.
$endgroup$
– Did
Jan 8 at 22:30
add a comment |
$begingroup$
For every $t$, $P(Y_t=0)=1$ and $Y_t$ is measurable with respect to $mathcal B(mathbb R)$ hence, for every $s<t$, $E(Y_tmidmathcal F_s)=Y_t=0=Y_s$ almost surely, which shows $(Y_t)$ is an $(mathcal F_t)$-martingale.
$endgroup$
– Did
Jan 8 at 22:30
$begingroup$
For every $t$, $P(Y_t=0)=1$ and $Y_t$ is measurable with respect to $mathcal B(mathbb R)$ hence, for every $s<t$, $E(Y_tmidmathcal F_s)=Y_t=0=Y_s$ almost surely, which shows $(Y_t)$ is an $(mathcal F_t)$-martingale.
$endgroup$
– Did
Jan 8 at 22:30
$begingroup$
For every $t$, $P(Y_t=0)=1$ and $Y_t$ is measurable with respect to $mathcal B(mathbb R)$ hence, for every $s<t$, $E(Y_tmidmathcal F_s)=Y_t=0=Y_s$ almost surely, which shows $(Y_t)$ is an $(mathcal F_t)$-martingale.
$endgroup$
– Did
Jan 8 at 22:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We may write $$
Y_t(x) = 1_{(t+mathbb{Q})cap [0,1]}(x)
$$ and it follows $Y_t$ is $F_t=mathcal{B}(mathbb{R})$ measurable. And also, we can see that $
Y_t(x)= 0
$ holds almost surely, thus giving
$$
E[Y_t|F_s]=E[0|F_s]=0=Y_s
$$ holds almost surely for all $s<t$. This establishes that ${Y_t,F_t}$ is a martingale. Finally, we can observe that for all $xin [0,1]$, the path
$$
tmapsto Y_t(x)=1_{x+mathbb{Q}}(t)
$$ of $Y_t$ is everywhere discontinuous.
answered Jan 8 at 22:53
SongSong
11.1k628
$endgroup$
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
We may write $$
Y_t(x) = 1_{(t+mathbb{Q})cap [0,1]}(x)
$$ and it follows $Y_t$ is $F_t=mathcal{B}(mathbb{R})$ measurable. And also, we can see that $
Y_t(x)= 0
$ holds almost surely, thus giving
$$
E[Y_t|F_s]=E[0|F_s]=0=Y_s
$$ holds almost surely for all $s<t$. This establishes that ${Y_t,F_t}$ is a martingale. Finally, we can observe that for all $xin [0,1]$, the path
$$
tmapsto Y_t(x)=1_{x+mathbb{Q}}(t)
$$ of $Y_t$ is everywhere discontinuous.
answered Jan 8 at 22:53
SongSong
11.1k628
$endgroup$
add a comment |
$begingroup$
We may write $$
Y_t(x) = 1_{(t+mathbb{Q})cap [0,1]}(x)
$$ and it follows $Y_t$ is $F_t=mathcal{B}(mathbb{R})$ measurable. And also, we can see that $
Y_t(x)= 0
$ holds almost surely, thus giving
$$
E[Y_t|F_s]=E[0|F_s]=0=Y_s
$$ holds almost surely for all $s<t$. This establishes that ${Y_t,F_t}$ is a martingale. Finally, we can observe that for all $xin [0,1]$, the path
$$
tmapsto Y_t(x)=1_{x+mathbb{Q}}(t)
$$ of $Y_t$ is everywhere discontinuous.
answered Jan 8 at 22:53
SongSong
11.1k628
$endgroup$
add a comment |
$begingroup$
We may write $$
Y_t(x) = 1_{(t+mathbb{Q})cap [0,1]}(x)
$$ and it follows $Y_t$ is $F_t=mathcal{B}(mathbb{R})$ measurable. And also, we can see that $
Y_t(x)= 0
$ holds almost surely, thus giving
$$
E[Y_t|F_s]=E[0|F_s]=0=Y_s
$$ holds almost surely for all $s<t$. This establishes that ${Y_t,F_t}$ is a martingale. Finally, we can observe that for all $xin [0,1]$, the path
$$
tmapsto Y_t(x)=1_{x+mathbb{Q}}(t)
$$ of $Y_t$ is everywhere discontinuous.
answered Jan 8 at 22:53
SongSong
11.1k628
$endgroup$
We may write $$
Y_t(x) = 1_{(t+mathbb{Q})cap [0,1]}(x)
$$ and it follows $Y_t$ is $F_t=mathcal{B}(mathbb{R})$ measurable. And also, we can see that $
Y_t(x)= 0
$ holds almost surely, thus giving
$$
E[Y_t|F_s]=E[0|F_s]=0=Y_s
$$ holds almost surely for all $s<t$. This establishes that ${Y_t,F_t}$ is a martingale. Finally, we can observe that for all $xin [0,1]$, the path
$$
tmapsto Y_t(x)=1_{x+mathbb{Q}}(t)
$$ of $Y_t$ is everywhere discontinuous.
answered Jan 8 at 22:53
SongSong
11.1k628
answered Jan 8 at 22:53
SongSong
11.1k628
answered Jan 8 at 22:53
SongSong
11.1k628
answered Jan 8 at 22:53
SongSong
11.1k628
11.1k628
add a comment |
add a comment |
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$begingroup$
For every $t$, $P(Y_t=0)=1$ and $Y_t$ is measurable with respect to $mathcal B(mathbb R)$ hence, for every $s<t$, $E(Y_tmidmathcal F_s)=Y_t=0=Y_s$ almost surely, which shows $(Y_t)$ is an $(mathcal F_t)$-martingale.
$endgroup$
– Did
Jan 8 at 22:30