Mixing
Expectation of the supremum of a sequence of random variables
$begingroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X_n(w)=
begin{cases}
0 quad frac{1}{n} < w leq 1 \
n-n^2w quad 0 leq w leq frac{1}{n}
end{cases}$
The first part of the exercise is proving that $X_{n}$ is a martingale, which I managed to do.
Now my problem is computing $E(sup_{n geq 1}|X_n|)$ and I don't know how to proceed. From the solutions I know that it must be equal to $+infty$.
probability probability-theory supremum-and-infimum expected-value
asked Jan 17 at 9:10
ank13ank13
255
$endgroup$
add a comment |
$begingroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X_n(w)=
begin{cases}
0 quad frac{1}{n} < w leq 1 \
n-n^2w quad 0 leq w leq frac{1}{n}
end{cases}$
The first part of the exercise is proving that $X_{n}$ is a martingale, which I managed to do.
Now my problem is computing $E(sup_{n geq 1}|X_n|)$ and I don't know how to proceed. From the solutions I know that it must be equal to $+infty$.
probability probability-theory supremum-and-infimum expected-value
asked Jan 17 at 9:10
ank13ank13
255
$endgroup$
add a comment |
$begingroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X_n(w)=
begin{cases}
0 quad frac{1}{n} < w leq 1 \
n-n^2w quad 0 leq w leq frac{1}{n}
end{cases}$
The first part of the exercise is proving that $X_{n}$ is a martingale, which I managed to do.
Now my problem is computing $E(sup_{n geq 1}|X_n|)$ and I don't know how to proceed. From the solutions I know that it must be equal to $+infty$.
probability probability-theory supremum-and-infimum expected-value
asked Jan 17 at 9:10
ank13ank13
255
$endgroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X_n(w)=
begin{cases}
0 quad frac{1}{n} < w leq 1 \
n-n^2w quad 0 leq w leq frac{1}{n}
end{cases}$
The first part of the exercise is proving that $X_{n}$ is a martingale, which I managed to do.
Now my problem is computing $E(sup_{n geq 1}|X_n|)$ and I don't know how to proceed. From the solutions I know that it must be equal to $+infty$.
probability probability-theory supremum-and-infimum expected-value
probability probability-theory supremum-and-infimum expected-value
asked Jan 17 at 9:10
ank13ank13
255
asked Jan 17 at 9:10
ank13ank13
255
asked Jan 17 at 9:10
ank13ank13
255
asked Jan 17 at 9:10
ank13ank13
255
asked Jan 17 at 9:10
ank13ank13
255
255
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Note that $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) to frac 1 2 -frac 1 4=frac 1 4$ as $omega to 0$. Hence for $omega$ sufficiently small $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) >frac 1 8$. Taking $k=[frac 1 {2omega}]$ we see that $sup_n |X_n(omega)| geq |X_k(omega)| geq frac 1 {4omega}$. Since $int frac 1 {4omega}, d omega =infty$ it follows that $Esup_n |X_n(omega)|=infty$.
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
$endgroup$
$begingroup$
sorry I don't understand, where does the first expression come from?
$endgroup$
– ank13
Jan 17 at 13:43
$begingroup$
Ok now i understand but i don’t get Why the value of that integral is plus infinity. Because that integral is equal to log(w)/4. Where does plus infinity comes from?
$endgroup$
– ank13
Jan 17 at 13:59
$begingroup$
@ank13 You have to evaluate the definite integral from $0$ to $1$ to find expectation. If $Y(omega) geq frac 1 {4omega}$ then $EY=int_0^{1} Y(omega), domega geq int_0^{1} frac 1 {4omega}, domega =infty$.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 23:15
$begingroup$
Ok now I understand! Thank you!
$endgroup$
– ank13
Jan 18 at 11:59
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Note that $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) to frac 1 2 -frac 1 4=frac 1 4$ as $omega to 0$. Hence for $omega$ sufficiently small $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) >frac 1 8$. Taking $k=[frac 1 {2omega}]$ we see that $sup_n |X_n(omega)| geq |X_k(omega)| geq frac 1 {4omega}$. Since $int frac 1 {4omega}, d omega =infty$ it follows that $Esup_n |X_n(omega)|=infty$.
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
$endgroup$
$begingroup$
sorry I don't understand, where does the first expression come from?
$endgroup$
– ank13
Jan 17 at 13:43
$begingroup$
Ok now i understand but i don’t get Why the value of that integral is plus infinity. Because that integral is equal to log(w)/4. Where does plus infinity comes from?
$endgroup$
– ank13
Jan 17 at 13:59
$begingroup$
@ank13 You have to evaluate the definite integral from $0$ to $1$ to find expectation. If $Y(omega) geq frac 1 {4omega}$ then $EY=int_0^{1} Y(omega), domega geq int_0^{1} frac 1 {4omega}, domega =infty$.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 23:15
$begingroup$
Ok now I understand! Thank you!
$endgroup$
– ank13
Jan 18 at 11:59
add a comment |
$begingroup$
Note that $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) to frac 1 2 -frac 1 4=frac 1 4$ as $omega to 0$. Hence for $omega$ sufficiently small $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) >frac 1 8$. Taking $k=[frac 1 {2omega}]$ we see that $sup_n |X_n(omega)| geq |X_k(omega)| geq frac 1 {4omega}$. Since $int frac 1 {4omega}, d omega =infty$ it follows that $Esup_n |X_n(omega)|=infty$.
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
$endgroup$
$begingroup$
sorry I don't understand, where does the first expression come from?
$endgroup$
– ank13
Jan 17 at 13:43
$begingroup$
Ok now i understand but i don’t get Why the value of that integral is plus infinity. Because that integral is equal to log(w)/4. Where does plus infinity comes from?
$endgroup$
– ank13
Jan 17 at 13:59
$begingroup$
@ank13 You have to evaluate the definite integral from $0$ to $1$ to find expectation. If $Y(omega) geq frac 1 {4omega}$ then $EY=int_0^{1} Y(omega), domega geq int_0^{1} frac 1 {4omega}, domega =infty$.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 23:15
$begingroup$
Ok now I understand! Thank you!
$endgroup$
– ank13
Jan 18 at 11:59
add a comment |
$begingroup$
Note that $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) to frac 1 2 -frac 1 4=frac 1 4$ as $omega to 0$. Hence for $omega$ sufficiently small $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) >frac 1 8$. Taking $k=[frac 1 {2omega}]$ we see that $sup_n |X_n(omega)| geq |X_k(omega)| geq frac 1 {4omega}$. Since $int frac 1 {4omega}, d omega =infty$ it follows that $Esup_n |X_n(omega)|=infty$.
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
$endgroup$
Note that $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) to frac 1 2 -frac 1 4=frac 1 4$ as $omega to 0$. Hence for $omega$ sufficiently small $omega ([frac 1 {2omega}] -omega [frac 1 {2omega}]^{2}) >frac 1 8$. Taking $k=[frac 1 {2omega}]$ we see that $sup_n |X_n(omega)| geq |X_k(omega)| geq frac 1 {4omega}$. Since $int frac 1 {4omega}, d omega =infty$ it follows that $Esup_n |X_n(omega)|=infty$.
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
answered Jan 17 at 9:38
Kavi Rama MurthyKavi Rama Murthy
62.4k42262
62.4k42262
$begingroup$
sorry I don't understand, where does the first expression come from?
$endgroup$
– ank13
Jan 17 at 13:43
$begingroup$
Ok now i understand but i don’t get Why the value of that integral is plus infinity. Because that integral is equal to log(w)/4. Where does plus infinity comes from?
$endgroup$
– ank13
Jan 17 at 13:59
$begingroup$
@ank13 You have to evaluate the definite integral from $0$ to $1$ to find expectation. If $Y(omega) geq frac 1 {4omega}$ then $EY=int_0^{1} Y(omega), domega geq int_0^{1} frac 1 {4omega}, domega =infty$.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 23:15
$begingroup$
Ok now I understand! Thank you!
$endgroup$
– ank13
Jan 18 at 11:59
add a comment |
$begingroup$
sorry I don't understand, where does the first expression come from?
$endgroup$
– ank13
Jan 17 at 13:43
$begingroup$
Ok now i understand but i don’t get Why the value of that integral is plus infinity. Because that integral is equal to log(w)/4. Where does plus infinity comes from?
$endgroup$
– ank13
Jan 17 at 13:59
$begingroup$
@ank13 You have to evaluate the definite integral from $0$ to $1$ to find expectation. If $Y(omega) geq frac 1 {4omega}$ then $EY=int_0^{1} Y(omega), domega geq int_0^{1} frac 1 {4omega}, domega =infty$.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 23:15
$begingroup$
Ok now I understand! Thank you!
$endgroup$
– ank13
Jan 18 at 11:59
$begingroup$
sorry I don't understand, where does the first expression come from?
$endgroup$
– ank13
Jan 17 at 13:43
$begingroup$
sorry I don't understand, where does the first expression come from?
$endgroup$
– ank13
Jan 17 at 13:43
$begingroup$
Ok now i understand but i don’t get Why the value of that integral is plus infinity. Because that integral is equal to log(w)/4. Where does plus infinity comes from?
$endgroup$
– ank13
Jan 17 at 13:59
$begingroup$
Ok now i understand but i don’t get Why the value of that integral is plus infinity. Because that integral is equal to log(w)/4. Where does plus infinity comes from?
$endgroup$
– ank13
Jan 17 at 13:59
$begingroup$
@ank13 You have to evaluate the definite integral from $0$ to $1$ to find expectation. If $Y(omega) geq frac 1 {4omega}$ then $EY=int_0^{1} Y(omega), domega geq int_0^{1} frac 1 {4omega}, domega =infty$.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 23:15
$begingroup$
@ank13 You have to evaluate the definite integral from $0$ to $1$ to find expectation. If $Y(omega) geq frac 1 {4omega}$ then $EY=int_0^{1} Y(omega), domega geq int_0^{1} frac 1 {4omega}, domega =infty$.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 23:15
$begingroup$
Ok now I understand! Thank you!
$endgroup$
– ank13
Jan 18 at 11:59
$begingroup$
Ok now I understand! Thank you!
$endgroup$
– ank13
Jan 18 at 11:59
add a comment |
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