Martingale-like properties of extreme value process?
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Imagine I have a sequence of i.i.d. random variables $x_1,x_2,...,x_n,...$ and let $M_n = max(x_1,...,x_n)$. Is there something I can say about the expectation $mathbbE[M_n | M_n-1]$? What if $x_1,...,x_n$ have a known distribution, such as the exponential distribution or an extreme value distribution?
probability probability-distributions random-variables
asked Aug 15 at 3:45
Asterix
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favorite
Imagine I have a sequence of i.i.d. random variables $x_1,x_2,...,x_n,...$ and let $M_n = max(x_1,...,x_n)$. Is there something I can say about the expectation $mathbbE[M_n | M_n-1]$? What if $x_1,...,x_n$ have a known distribution, such as the exponential distribution or an extreme value distribution?
probability probability-distributions random-variables
asked Aug 15 at 3:45
Asterix
1827
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Imagine I have a sequence of i.i.d. random variables $x_1,x_2,...,x_n,...$ and let $M_n = max(x_1,...,x_n)$. Is there something I can say about the expectation $mathbbE[M_n | M_n-1]$? What if $x_1,...,x_n$ have a known distribution, such as the exponential distribution or an extreme value distribution?
probability probability-distributions random-variables
asked Aug 15 at 3:45
Asterix
1827
Imagine I have a sequence of i.i.d. random variables $x_1,x_2,...,x_n,...$ and let $M_n = max(x_1,...,x_n)$. Is there something I can say about the expectation $mathbbE[M_n | M_n-1]$? What if $x_1,...,x_n$ have a known distribution, such as the exponential distribution or an extreme value distribution?
probability probability-distributions random-variables
asked Aug 15 at 3:45
Asterix
1827
asked Aug 15 at 3:45
Asterix
1827
asked Aug 15 at 3:45
Asterix
1827
asked Aug 15 at 3:45
Asterix
1827
1827
add a comment |Â
add a comment |Â
2 Answers
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accepted
Since the cumulative maximum will only increase if the next $X_n$ is greater than the current value of the maximum, we have $$E(M_nmid M_n-1) = P(Xle M_n-1)M_n-1 + P(X> M_n-1)E(Xmid X> M_n-1).$$
In particular, for an exponential with mean $theta$, this works out to $$ (1-e^-M_n-1/theta)M_n-1 + e^-M_n-1/theta(M_n-1+theta)=M_n-1 +theta e^-M_n-1/theta$$ where we used the memorylessness property to get the conditional expectation.
answered Aug 15 at 3:56
spaceisdarkgreen
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We also know $M_n$ is a sub-martingale since $M_n+1 geq M_n$ almost surely, so
$$mathbbE[M_n+1|M_1,ldots,M_n] = mathbbE[M_n+1 | M_n] geq M_n.$$
answered Aug 15 at 4:29
cdipaolo
677311
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Since the cumulative maximum will only increase if the next $X_n$ is greater than the current value of the maximum, we have $$E(M_nmid M_n-1) = P(Xle M_n-1)M_n-1 + P(X> M_n-1)E(Xmid X> M_n-1).$$
In particular, for an exponential with mean $theta$, this works out to $$ (1-e^-M_n-1/theta)M_n-1 + e^-M_n-1/theta(M_n-1+theta)=M_n-1 +theta e^-M_n-1/theta$$ where we used the memorylessness property to get the conditional expectation.
answered Aug 15 at 3:56
spaceisdarkgreen
28k21548
add a comment |Â
up vote
2
down vote
accepted
Since the cumulative maximum will only increase if the next $X_n$ is greater than the current value of the maximum, we have $$E(M_nmid M_n-1) = P(Xle M_n-1)M_n-1 + P(X> M_n-1)E(Xmid X> M_n-1).$$
In particular, for an exponential with mean $theta$, this works out to $$ (1-e^-M_n-1/theta)M_n-1 + e^-M_n-1/theta(M_n-1+theta)=M_n-1 +theta e^-M_n-1/theta$$ where we used the memorylessness property to get the conditional expectation.
answered Aug 15 at 3:56
spaceisdarkgreen
28k21548
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Since the cumulative maximum will only increase if the next $X_n$ is greater than the current value of the maximum, we have $$E(M_nmid M_n-1) = P(Xle M_n-1)M_n-1 + P(X> M_n-1)E(Xmid X> M_n-1).$$
In particular, for an exponential with mean $theta$, this works out to $$ (1-e^-M_n-1/theta)M_n-1 + e^-M_n-1/theta(M_n-1+theta)=M_n-1 +theta e^-M_n-1/theta$$ where we used the memorylessness property to get the conditional expectation.
answered Aug 15 at 3:56
spaceisdarkgreen
28k21548
Since the cumulative maximum will only increase if the next $X_n$ is greater than the current value of the maximum, we have $$E(M_nmid M_n-1) = P(Xle M_n-1)M_n-1 + P(X> M_n-1)E(Xmid X> M_n-1).$$
In particular, for an exponential with mean $theta$, this works out to $$ (1-e^-M_n-1/theta)M_n-1 + e^-M_n-1/theta(M_n-1+theta)=M_n-1 +theta e^-M_n-1/theta$$ where we used the memorylessness property to get the conditional expectation.
answered Aug 15 at 3:56
spaceisdarkgreen
28k21548
answered Aug 15 at 3:56
spaceisdarkgreen
28k21548
answered Aug 15 at 3:56
spaceisdarkgreen
28k21548
answered Aug 15 at 3:56
spaceisdarkgreen
28k21548
28k21548
add a comment |Â
add a comment |Â
up vote
0
down vote
We also know $M_n$ is a sub-martingale since $M_n+1 geq M_n$ almost surely, so
$$mathbbE[M_n+1|M_1,ldots,M_n] = mathbbE[M_n+1 | M_n] geq M_n.$$
answered Aug 15 at 4:29
cdipaolo
677311
add a comment |Â
up vote
0
down vote
We also know $M_n$ is a sub-martingale since $M_n+1 geq M_n$ almost surely, so
$$mathbbE[M_n+1|M_1,ldots,M_n] = mathbbE[M_n+1 | M_n] geq M_n.$$
answered Aug 15 at 4:29
cdipaolo
677311
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We also know $M_n$ is a sub-martingale since $M_n+1 geq M_n$ almost surely, so
$$mathbbE[M_n+1|M_1,ldots,M_n] = mathbbE[M_n+1 | M_n] geq M_n.$$
answered Aug 15 at 4:29
cdipaolo
677311
We also know $M_n$ is a sub-martingale since $M_n+1 geq M_n$ almost surely, so
$$mathbbE[M_n+1|M_1,ldots,M_n] = mathbbE[M_n+1 | M_n] geq M_n.$$
answered Aug 15 at 4:29
cdipaolo
677311
answered Aug 15 at 4:29
cdipaolo
677311
answered Aug 15 at 4:29
cdipaolo
677311
answered Aug 15 at 4:29
cdipaolo
677311
677311
add a comment |Â
add a comment |Â
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