Mixing
Double conditional expectation of martingale
$begingroup$
Assume $X$ a martingale. A common exercise is showing that every martingale has uncorrelated increments. That is, with $a < b < c< d$,
$$ Cov( X_a - X_b, X_c - X_d) = 0 $$
While there are a few straightforward ways of showing this result, I was curious whether we could reach the same result through noting that, say, $X_a = mathbb{E}(X_n | mathcal{F}_a)$, for some $n$ such, that $mathcal{F}_a subset mathcal{F}_n$.
This way,
$$ Cov( X_a - X_b, X_c - X_d) = Cov( mathbb{E}(X_n | mathcal{F}_a) - mathbb{E}(X_n | mathcal{F}_b), mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d)) $$
And since $mathbb{E}(mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d))=0$, this should reduce to four sums, where I'm guessing every member should equate to zero.
Namely,
$$ mathbb{E}(mathbb{E}(X_n | mathcal{F}_i)mathbb{E}(X_n | mathcal{F}_j)) = 0$$
would seem to hold for every $i neq j$ such, that $mathcal{F}_{i,j} subset mathcal{F}_n$
Is this true, or am I messing things ups?
How should we be dealing with double expectation of a multiplication of moments? It's not clear that the two expectations are independent (since $mathcal{F}_i subset mathcal{F}_j$ might hold, or the other way round).
probability probability-theory martingales
asked Jan 14 at 22:44
NutleNutle
315110
$endgroup$
add a comment |
$begingroup$
Assume $X$ a martingale. A common exercise is showing that every martingale has uncorrelated increments. That is, with $a < b < c< d$,
$$ Cov( X_a - X_b, X_c - X_d) = 0 $$
While there are a few straightforward ways of showing this result, I was curious whether we could reach the same result through noting that, say, $X_a = mathbb{E}(X_n | mathcal{F}_a)$, for some $n$ such, that $mathcal{F}_a subset mathcal{F}_n$.
This way,
$$ Cov( X_a - X_b, X_c - X_d) = Cov( mathbb{E}(X_n | mathcal{F}_a) - mathbb{E}(X_n | mathcal{F}_b), mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d)) $$
And since $mathbb{E}(mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d))=0$, this should reduce to four sums, where I'm guessing every member should equate to zero.
Namely,
$$ mathbb{E}(mathbb{E}(X_n | mathcal{F}_i)mathbb{E}(X_n | mathcal{F}_j)) = 0$$
would seem to hold for every $i neq j$ such, that $mathcal{F}_{i,j} subset mathcal{F}_n$
Is this true, or am I messing things ups?
How should we be dealing with double expectation of a multiplication of moments? It's not clear that the two expectations are independent (since $mathcal{F}_i subset mathcal{F}_j$ might hold, or the other way round).
probability probability-theory martingales
asked Jan 14 at 22:44
NutleNutle
315110
$endgroup$
$begingroup$
No, the expectation of the product does not equal zero. Using the tower property (of conditional expectation) it is not difficult to show that $$mathbb{E}(mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j)) = mathbb{E}( mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2).$$ If this expression was equal for all $i neq j$ this would imply $X_i = mathbb{E}(X_n mid mathcal{F}_i)=0$ for all $i$.
$endgroup$
– saz
Jan 15 at 7:54
$begingroup$
@saz thanks, does the tower property hold for multiplication also? I thought about this too, but wasn't too sure whether $mathbb{E}(mathbb{E}(X|F_1)mathbb{E}(X|F_2)|F_1)$ reduces the multiplication in the same way as it does a single term?
$endgroup$
– Nutle
Jan 15 at 8:41
add a comment |
$begingroup$
Assume $X$ a martingale. A common exercise is showing that every martingale has uncorrelated increments. That is, with $a < b < c< d$,
$$ Cov( X_a - X_b, X_c - X_d) = 0 $$
While there are a few straightforward ways of showing this result, I was curious whether we could reach the same result through noting that, say, $X_a = mathbb{E}(X_n | mathcal{F}_a)$, for some $n$ such, that $mathcal{F}_a subset mathcal{F}_n$.
This way,
$$ Cov( X_a - X_b, X_c - X_d) = Cov( mathbb{E}(X_n | mathcal{F}_a) - mathbb{E}(X_n | mathcal{F}_b), mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d)) $$
And since $mathbb{E}(mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d))=0$, this should reduce to four sums, where I'm guessing every member should equate to zero.
Namely,
$$ mathbb{E}(mathbb{E}(X_n | mathcal{F}_i)mathbb{E}(X_n | mathcal{F}_j)) = 0$$
would seem to hold for every $i neq j$ such, that $mathcal{F}_{i,j} subset mathcal{F}_n$
Is this true, or am I messing things ups?
How should we be dealing with double expectation of a multiplication of moments? It's not clear that the two expectations are independent (since $mathcal{F}_i subset mathcal{F}_j$ might hold, or the other way round).
probability probability-theory martingales
asked Jan 14 at 22:44
NutleNutle
315110
$endgroup$
Assume $X$ a martingale. A common exercise is showing that every martingale has uncorrelated increments. That is, with $a < b < c< d$,
$$ Cov( X_a - X_b, X_c - X_d) = 0 $$
While there are a few straightforward ways of showing this result, I was curious whether we could reach the same result through noting that, say, $X_a = mathbb{E}(X_n | mathcal{F}_a)$, for some $n$ such, that $mathcal{F}_a subset mathcal{F}_n$.
This way,
$$ Cov( X_a - X_b, X_c - X_d) = Cov( mathbb{E}(X_n | mathcal{F}_a) - mathbb{E}(X_n | mathcal{F}_b), mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d)) $$
And since $mathbb{E}(mathbb{E}(X_n | mathcal{F}_c) - mathbb{E}(X_n | mathcal{F}_d))=0$, this should reduce to four sums, where I'm guessing every member should equate to zero.
Namely,
$$ mathbb{E}(mathbb{E}(X_n | mathcal{F}_i)mathbb{E}(X_n | mathcal{F}_j)) = 0$$
would seem to hold for every $i neq j$ such, that $mathcal{F}_{i,j} subset mathcal{F}_n$
Is this true, or am I messing things ups?
How should we be dealing with double expectation of a multiplication of moments? It's not clear that the two expectations are independent (since $mathcal{F}_i subset mathcal{F}_j$ might hold, or the other way round).
probability probability-theory martingales
probability probability-theory martingales
asked Jan 14 at 22:44
NutleNutle
315110
asked Jan 14 at 22:44
NutleNutle
315110
asked Jan 14 at 22:44
NutleNutle
315110
asked Jan 14 at 22:44
NutleNutle
315110
asked Jan 14 at 22:44
NutleNutle
315110
315110
$begingroup$
No, the expectation of the product does not equal zero. Using the tower property (of conditional expectation) it is not difficult to show that $$mathbb{E}(mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j)) = mathbb{E}( mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2).$$ If this expression was equal for all $i neq j$ this would imply $X_i = mathbb{E}(X_n mid mathcal{F}_i)=0$ for all $i$.
$endgroup$
– saz
Jan 15 at 7:54
$begingroup$
@saz thanks, does the tower property hold for multiplication also? I thought about this too, but wasn't too sure whether $mathbb{E}(mathbb{E}(X|F_1)mathbb{E}(X|F_2)|F_1)$ reduces the multiplication in the same way as it does a single term?
$endgroup$
– Nutle
Jan 15 at 8:41
add a comment |
$begingroup$
No, the expectation of the product does not equal zero. Using the tower property (of conditional expectation) it is not difficult to show that $$mathbb{E}(mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j)) = mathbb{E}( mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2).$$ If this expression was equal for all $i neq j$ this would imply $X_i = mathbb{E}(X_n mid mathcal{F}_i)=0$ for all $i$.
$endgroup$
– saz
Jan 15 at 7:54
$begingroup$
@saz thanks, does the tower property hold for multiplication also? I thought about this too, but wasn't too sure whether $mathbb{E}(mathbb{E}(X|F_1)mathbb{E}(X|F_2)|F_1)$ reduces the multiplication in the same way as it does a single term?
$endgroup$
– Nutle
Jan 15 at 8:41
$begingroup$
No, the expectation of the product does not equal zero. Using the tower property (of conditional expectation) it is not difficult to show that $$mathbb{E}(mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j)) = mathbb{E}( mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2).$$ If this expression was equal for all $i neq j$ this would imply $X_i = mathbb{E}(X_n mid mathcal{F}_i)=0$ for all $i$.
$endgroup$
– saz
Jan 15 at 7:54
$begingroup$
No, the expectation of the product does not equal zero. Using the tower property (of conditional expectation) it is not difficult to show that $$mathbb{E}(mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j)) = mathbb{E}( mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2).$$ If this expression was equal for all $i neq j$ this would imply $X_i = mathbb{E}(X_n mid mathcal{F}_i)=0$ for all $i$.
$endgroup$
– saz
Jan 15 at 7:54
$begingroup$
@saz thanks, does the tower property hold for multiplication also? I thought about this too, but wasn't too sure whether $mathbb{E}(mathbb{E}(X|F_1)mathbb{E}(X|F_2)|F_1)$ reduces the multiplication in the same way as it does a single term?
$endgroup$
– Nutle
Jan 15 at 8:41
$begingroup$
@saz thanks, does the tower property hold for multiplication also? I thought about this too, but wasn't too sure whether $mathbb{E}(mathbb{E}(X|F_1)mathbb{E}(X|F_2)|F_1)$ reduces the multiplication in the same way as it does a single term?
$endgroup$
– Nutle
Jan 15 at 8:41
add a comment |
1 Answer
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$begingroup$
No, the expectation of the product does, in general, not equal zero.
Lemma: Let $(X_n)_{n in mathbb{N}}$ be a square-integrable martingale. Then $$mathbb{E}(X_i X_j) = mathbb{E}(X_{min{i,j}}^2) tag{1}$$ for any $i,j in mathbb{N}$.
Since $mathbb{E}(X_n mid mathcal{F}_i) = X_i$ for all $i leq n$, we can $(1)$ rewrite as
$$mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j) big] = mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2 big]$$
for $i,j leq n$. Hence, the expression equals zero if, and only if, $ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})=0$ almost surely (i.e. $X_{min{i,j}}=0$ almost surely). However, it is possible to use the above lemma to prove that $X$ has uncorrelated increments, see below.
Proof of the lemma: The assertion is obvious for $i=j$. Now if $i,j in mathbb{N}$ are such that $i<j$, then $$mathbb{E}(X_j mid mathcal{F}_i) = X_i,$$ and therefore it follows from the tower property that
$$mathbb{E}(X_i X_j) = mathbb{E} big[ mathbb{E}(X_i X_j mid mathcal{F}_i big] = mathbb{E} big[ X_i mathbb{E}(X_j mid mathcal{F}_i big] = mathbb{E}(X_i^2)$$ which proves the assertion.
Corollary Any square integrable martingale $(X_n)_{n in mathbb{N}}$ has uncorrelated increments.
Proof: Since a martingale has constant expectation, we have
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}((X_a-X_b)(X_c-X_d))$$
for any $a<b<c<d$. Hence,
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a X_c - X_a X_d - X_b X_c+ X_b X_d).$$
Applying the above lemma (four times), we get
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a^2) - mathbb{E}(X_a^2) - mathbb{E}(X_b^2)+ mathbb{E}(X_b^2)=0.$$
answered Jan 15 at 15:53
sazsaz
80.8k860127
$endgroup$
add a comment |
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$begingroup$
No, the expectation of the product does, in general, not equal zero.
Lemma: Let $(X_n)_{n in mathbb{N}}$ be a square-integrable martingale. Then $$mathbb{E}(X_i X_j) = mathbb{E}(X_{min{i,j}}^2) tag{1}$$ for any $i,j in mathbb{N}$.
Since $mathbb{E}(X_n mid mathcal{F}_i) = X_i$ for all $i leq n$, we can $(1)$ rewrite as
$$mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j) big] = mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2 big]$$
for $i,j leq n$. Hence, the expression equals zero if, and only if, $ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})=0$ almost surely (i.e. $X_{min{i,j}}=0$ almost surely). However, it is possible to use the above lemma to prove that $X$ has uncorrelated increments, see below.
Proof of the lemma: The assertion is obvious for $i=j$. Now if $i,j in mathbb{N}$ are such that $i<j$, then $$mathbb{E}(X_j mid mathcal{F}_i) = X_i,$$ and therefore it follows from the tower property that
$$mathbb{E}(X_i X_j) = mathbb{E} big[ mathbb{E}(X_i X_j mid mathcal{F}_i big] = mathbb{E} big[ X_i mathbb{E}(X_j mid mathcal{F}_i big] = mathbb{E}(X_i^2)$$ which proves the assertion.
Corollary Any square integrable martingale $(X_n)_{n in mathbb{N}}$ has uncorrelated increments.
Proof: Since a martingale has constant expectation, we have
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}((X_a-X_b)(X_c-X_d))$$
for any $a<b<c<d$. Hence,
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a X_c - X_a X_d - X_b X_c+ X_b X_d).$$
Applying the above lemma (four times), we get
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a^2) - mathbb{E}(X_a^2) - mathbb{E}(X_b^2)+ mathbb{E}(X_b^2)=0.$$
answered Jan 15 at 15:53
sazsaz
80.8k860127
$endgroup$
add a comment |
$begingroup$
No, the expectation of the product does, in general, not equal zero.
Lemma: Let $(X_n)_{n in mathbb{N}}$ be a square-integrable martingale. Then $$mathbb{E}(X_i X_j) = mathbb{E}(X_{min{i,j}}^2) tag{1}$$ for any $i,j in mathbb{N}$.
Since $mathbb{E}(X_n mid mathcal{F}_i) = X_i$ for all $i leq n$, we can $(1)$ rewrite as
$$mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j) big] = mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2 big]$$
for $i,j leq n$. Hence, the expression equals zero if, and only if, $ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})=0$ almost surely (i.e. $X_{min{i,j}}=0$ almost surely). However, it is possible to use the above lemma to prove that $X$ has uncorrelated increments, see below.
Proof of the lemma: The assertion is obvious for $i=j$. Now if $i,j in mathbb{N}$ are such that $i<j$, then $$mathbb{E}(X_j mid mathcal{F}_i) = X_i,$$ and therefore it follows from the tower property that
$$mathbb{E}(X_i X_j) = mathbb{E} big[ mathbb{E}(X_i X_j mid mathcal{F}_i big] = mathbb{E} big[ X_i mathbb{E}(X_j mid mathcal{F}_i big] = mathbb{E}(X_i^2)$$ which proves the assertion.
Corollary Any square integrable martingale $(X_n)_{n in mathbb{N}}$ has uncorrelated increments.
Proof: Since a martingale has constant expectation, we have
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}((X_a-X_b)(X_c-X_d))$$
for any $a<b<c<d$. Hence,
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a X_c - X_a X_d - X_b X_c+ X_b X_d).$$
Applying the above lemma (four times), we get
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a^2) - mathbb{E}(X_a^2) - mathbb{E}(X_b^2)+ mathbb{E}(X_b^2)=0.$$
answered Jan 15 at 15:53
sazsaz
80.8k860127
$endgroup$
add a comment |
$begingroup$
No, the expectation of the product does, in general, not equal zero.
Lemma: Let $(X_n)_{n in mathbb{N}}$ be a square-integrable martingale. Then $$mathbb{E}(X_i X_j) = mathbb{E}(X_{min{i,j}}^2) tag{1}$$ for any $i,j in mathbb{N}$.
Since $mathbb{E}(X_n mid mathcal{F}_i) = X_i$ for all $i leq n$, we can $(1)$ rewrite as
$$mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j) big] = mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2 big]$$
for $i,j leq n$. Hence, the expression equals zero if, and only if, $ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})=0$ almost surely (i.e. $X_{min{i,j}}=0$ almost surely). However, it is possible to use the above lemma to prove that $X$ has uncorrelated increments, see below.
Proof of the lemma: The assertion is obvious for $i=j$. Now if $i,j in mathbb{N}$ are such that $i<j$, then $$mathbb{E}(X_j mid mathcal{F}_i) = X_i,$$ and therefore it follows from the tower property that
$$mathbb{E}(X_i X_j) = mathbb{E} big[ mathbb{E}(X_i X_j mid mathcal{F}_i big] = mathbb{E} big[ X_i mathbb{E}(X_j mid mathcal{F}_i big] = mathbb{E}(X_i^2)$$ which proves the assertion.
Corollary Any square integrable martingale $(X_n)_{n in mathbb{N}}$ has uncorrelated increments.
Proof: Since a martingale has constant expectation, we have
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}((X_a-X_b)(X_c-X_d))$$
for any $a<b<c<d$. Hence,
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a X_c - X_a X_d - X_b X_c+ X_b X_d).$$
Applying the above lemma (four times), we get
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a^2) - mathbb{E}(X_a^2) - mathbb{E}(X_b^2)+ mathbb{E}(X_b^2)=0.$$
answered Jan 15 at 15:53
sazsaz
80.8k860127
$endgroup$
No, the expectation of the product does, in general, not equal zero.
Lemma: Let $(X_n)_{n in mathbb{N}}$ be a square-integrable martingale. Then $$mathbb{E}(X_i X_j) = mathbb{E}(X_{min{i,j}}^2) tag{1}$$ for any $i,j in mathbb{N}$.
Since $mathbb{E}(X_n mid mathcal{F}_i) = X_i$ for all $i leq n$, we can $(1)$ rewrite as
$$mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j) big] = mathbb{E} big[ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2 big]$$
for $i,j leq n$. Hence, the expression equals zero if, and only if, $ mathbb{E}(X_n mid mathcal{F}_{min{i,j}})=0$ almost surely (i.e. $X_{min{i,j}}=0$ almost surely). However, it is possible to use the above lemma to prove that $X$ has uncorrelated increments, see below.
Proof of the lemma: The assertion is obvious for $i=j$. Now if $i,j in mathbb{N}$ are such that $i<j$, then $$mathbb{E}(X_j mid mathcal{F}_i) = X_i,$$ and therefore it follows from the tower property that
$$mathbb{E}(X_i X_j) = mathbb{E} big[ mathbb{E}(X_i X_j mid mathcal{F}_i big] = mathbb{E} big[ X_i mathbb{E}(X_j mid mathcal{F}_i big] = mathbb{E}(X_i^2)$$ which proves the assertion.
Corollary Any square integrable martingale $(X_n)_{n in mathbb{N}}$ has uncorrelated increments.
Proof: Since a martingale has constant expectation, we have
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}((X_a-X_b)(X_c-X_d))$$
for any $a<b<c<d$. Hence,
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a X_c - X_a X_d - X_b X_c+ X_b X_d).$$
Applying the above lemma (four times), we get
$$text{cov}(X_a -X_b, X_c-X_d) = mathbb{E}(X_a^2) - mathbb{E}(X_a^2) - mathbb{E}(X_b^2)+ mathbb{E}(X_b^2)=0.$$
answered Jan 15 at 15:53
sazsaz
80.8k860127
edited Jan 15 at 17:55
answered Jan 15 at 15:53
sazsaz
80.8k860127
answered Jan 15 at 15:53
sazsaz
80.8k860127
answered Jan 15 at 15:53
sazsaz
80.8k860127
80.8k860127
add a comment |
add a comment |
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$begingroup$
No, the expectation of the product does not equal zero. Using the tower property (of conditional expectation) it is not difficult to show that $$mathbb{E}(mathbb{E}(X_n mid mathcal{F}_i) mathbb{E}(X_n mid mathcal{F}_j)) = mathbb{E}( mathbb{E}(X_n mid mathcal{F}_{min{i,j}})^2).$$ If this expression was equal for all $i neq j$ this would imply $X_i = mathbb{E}(X_n mid mathcal{F}_i)=0$ for all $i$.
$endgroup$
– saz
Jan 15 at 7:54
$begingroup$
@saz thanks, does the tower property hold for multiplication also? I thought about this too, but wasn't too sure whether $mathbb{E}(mathbb{E}(X|F_1)mathbb{E}(X|F_2)|F_1)$ reduces the multiplication in the same way as it does a single term?
$endgroup$
– Nutle
Jan 15 at 8:41