Covariance of Martingales Announcing the arrival of Valued Associate #679: Cesar Manara ...
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Covariance of Martingales
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Martingales, finite expectationDecomposition of local martingalesChebyshev Inequality for MartingalesResulting Covariance Matrix $Sigma$' after reducing space along the primary eigenvector?3-sigma Ellipse, why axis length scales with square root of eigenvalues of covariance-matrixHow can the variance of data be represented as this product of the principal direction and the covariance matrix?Is this a proof of $Eint^b_a f dZ = 0$?Martingales and stopping timesMartingales and dyadic intervalsConditional expectation of product of martingales
$begingroup$
I have proven that Martingales have orthogonal increments.
From this I need to show that $operatorname{Cov}[M(t),M(s)]$ relies only on $min{s,t}$.
I used the expected value definition of Covariance and eliminated one piece using the orthogonal increments bit. Also, I introduced conditioning on $F_s$.
$operatorname{Cov}[M(t),M(s)]= E[(M(t)-E[M(s)])(M(s)-E[M(s)])]$
$= .... $
$= E[E[((M(t)+E[M(t)])E[M(s)]|F_s)]]$
From here I am lost.
Should I....
$ = E[E[M(t)E[M(s)]|F_s] +E[E[E[M(t)]*E[M(s)]|F_s]]$
$ = E[E[M(t)M(s)|F_S] + E[E[E[M^2(s)]|F_s]]$
I know $E[M(t)] = E[M(s)]$. Thats how I got the second piece of the second line.
Is this correct? Where should I go from there?
stochastic-processes martingales conditional-expectation covariance
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
$endgroup$
add a comment |
$begingroup$
I have proven that Martingales have orthogonal increments.
From this I need to show that $operatorname{Cov}[M(t),M(s)]$ relies only on $min{s,t}$.
I used the expected value definition of Covariance and eliminated one piece using the orthogonal increments bit. Also, I introduced conditioning on $F_s$.
$operatorname{Cov}[M(t),M(s)]= E[(M(t)-E[M(s)])(M(s)-E[M(s)])]$
$= .... $
$= E[E[((M(t)+E[M(t)])E[M(s)]|F_s)]]$
From here I am lost.
Should I....
$ = E[E[M(t)E[M(s)]|F_s] +E[E[E[M(t)]*E[M(s)]|F_s]]$
$ = E[E[M(t)M(s)|F_S] + E[E[E[M^2(s)]|F_s]]$
I know $E[M(t)] = E[M(s)]$. Thats how I got the second piece of the second line.
Is this correct? Where should I go from there?
stochastic-processes martingales conditional-expectation covariance
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
$endgroup$
add a comment |
$begingroup$
I have proven that Martingales have orthogonal increments.
From this I need to show that $operatorname{Cov}[M(t),M(s)]$ relies only on $min{s,t}$.
I used the expected value definition of Covariance and eliminated one piece using the orthogonal increments bit. Also, I introduced conditioning on $F_s$.
$operatorname{Cov}[M(t),M(s)]= E[(M(t)-E[M(s)])(M(s)-E[M(s)])]$
$= .... $
$= E[E[((M(t)+E[M(t)])E[M(s)]|F_s)]]$
From here I am lost.
Should I....
$ = E[E[M(t)E[M(s)]|F_s] +E[E[E[M(t)]*E[M(s)]|F_s]]$
$ = E[E[M(t)M(s)|F_S] + E[E[E[M^2(s)]|F_s]]$
I know $E[M(t)] = E[M(s)]$. Thats how I got the second piece of the second line.
Is this correct? Where should I go from there?
stochastic-processes martingales conditional-expectation covariance
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
$endgroup$
I have proven that Martingales have orthogonal increments.
From this I need to show that $operatorname{Cov}[M(t),M(s)]$ relies only on $min{s,t}$.
I used the expected value definition of Covariance and eliminated one piece using the orthogonal increments bit. Also, I introduced conditioning on $F_s$.
$operatorname{Cov}[M(t),M(s)]= E[(M(t)-E[M(s)])(M(s)-E[M(s)])]$
$= .... $
$= E[E[((M(t)+E[M(t)])E[M(s)]|F_s)]]$
From here I am lost.
Should I....
$ = E[E[M(t)E[M(s)]|F_s] +E[E[E[M(t)]*E[M(s)]|F_s]]$
$ = E[E[M(t)M(s)|F_S] + E[E[E[M^2(s)]|F_s]]$
I know $E[M(t)] = E[M(s)]$. Thats how I got the second piece of the second line.
Is this correct? Where should I go from there?
stochastic-processes martingales conditional-expectation covariance
stochastic-processes martingales conditional-expectation covariance
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
edited Apr 3 '16 at 13:30
Davide Giraudo
128k17156268
128k17156268
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
asked Apr 1 '16 at 18:31
Skuttle_ButtSkuttle_Butt
354213
354213
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are on the right track. Assuming $s<t$, by conditioning,
$$
E[M(s)M(t)]=E[M(s)E[M(t)|mathcal F_s]=E[M(s)^2].
$$
So,
$$
cov[M(s),M(t)]=E[M(s)M(t)]-E[M(s)]E[M(t)]=E[M(s)^2]-(E[M(s)])^2=var[M(s)],
$$
this depends on $t$ only through the fact that $tin(s,infty)$.
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
add a comment |
$begingroup$
What is the mathematical meaning of 'rely on'? Typically, to compute covariances of martingales you can do the following. Suppose $s<t$
begin{align}
mbox{Cov}[M(t),M(s)] &= mbox{Cov}[M(t)-M(s)+M(s),M(s)] \
&=mbox{Cov}[M(t)-M(s),M(s)] + mbox{Var}[M(s)] \
&= mathbb{E}[M(s)(M(t)-M(s))] - mathbb{E}[M(t)-M(s)]mathbb{E}[M(s)] + mbox{Var}[M(s)] \
end{align}
Using the orthogonal increment property we have that $mathbb{E}[M(s)(M(t)-M(s))] = 0$ and hence for $s<t$
$$mbox{Cov}[M(t),M(s)] = mbox{Var}[M(s)].$$
The proof goes similar for $t<s$.
answered Apr 1 '16 at 19:23
SironSiron
1,215614
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You are on the right track. Assuming $s<t$, by conditioning,
$$
E[M(s)M(t)]=E[M(s)E[M(t)|mathcal F_s]=E[M(s)^2].
$$
So,
$$
cov[M(s),M(t)]=E[M(s)M(t)]-E[M(s)]E[M(t)]=E[M(s)^2]-(E[M(s)])^2=var[M(s)],
$$
this depends on $t$ only through the fact that $tin(s,infty)$.
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
add a comment |
$begingroup$
You are on the right track. Assuming $s<t$, by conditioning,
$$
E[M(s)M(t)]=E[M(s)E[M(t)|mathcal F_s]=E[M(s)^2].
$$
So,
$$
cov[M(s),M(t)]=E[M(s)M(t)]-E[M(s)]E[M(t)]=E[M(s)^2]-(E[M(s)])^2=var[M(s)],
$$
this depends on $t$ only through the fact that $tin(s,infty)$.
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
add a comment |
$begingroup$
You are on the right track. Assuming $s<t$, by conditioning,
$$
E[M(s)M(t)]=E[M(s)E[M(t)|mathcal F_s]=E[M(s)^2].
$$
So,
$$
cov[M(s),M(t)]=E[M(s)M(t)]-E[M(s)]E[M(t)]=E[M(s)^2]-(E[M(s)])^2=var[M(s)],
$$
this depends on $t$ only through the fact that $tin(s,infty)$.
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
$endgroup$
You are on the right track. Assuming $s<t$, by conditioning,
$$
E[M(s)M(t)]=E[M(s)E[M(t)|mathcal F_s]=E[M(s)^2].
$$
So,
$$
cov[M(s),M(t)]=E[M(s)M(t)]-E[M(s)]E[M(t)]=E[M(s)^2]-(E[M(s)])^2=var[M(s)],
$$
this depends on $t$ only through the fact that $tin(s,infty)$.
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
answered Apr 1 '16 at 19:22
John DawkinsJohn Dawkins
13.4k11017
13.4k11017
add a comment |
add a comment |
$begingroup$
What is the mathematical meaning of 'rely on'? Typically, to compute covariances of martingales you can do the following. Suppose $s<t$
begin{align}
mbox{Cov}[M(t),M(s)] &= mbox{Cov}[M(t)-M(s)+M(s),M(s)] \
&=mbox{Cov}[M(t)-M(s),M(s)] + mbox{Var}[M(s)] \
&= mathbb{E}[M(s)(M(t)-M(s))] - mathbb{E}[M(t)-M(s)]mathbb{E}[M(s)] + mbox{Var}[M(s)] \
end{align}
Using the orthogonal increment property we have that $mathbb{E}[M(s)(M(t)-M(s))] = 0$ and hence for $s<t$
$$mbox{Cov}[M(t),M(s)] = mbox{Var}[M(s)].$$
The proof goes similar for $t<s$.
answered Apr 1 '16 at 19:23
SironSiron
1,215614
$endgroup$
add a comment |
$begingroup$
What is the mathematical meaning of 'rely on'? Typically, to compute covariances of martingales you can do the following. Suppose $s<t$
begin{align}
mbox{Cov}[M(t),M(s)] &= mbox{Cov}[M(t)-M(s)+M(s),M(s)] \
&=mbox{Cov}[M(t)-M(s),M(s)] + mbox{Var}[M(s)] \
&= mathbb{E}[M(s)(M(t)-M(s))] - mathbb{E}[M(t)-M(s)]mathbb{E}[M(s)] + mbox{Var}[M(s)] \
end{align}
Using the orthogonal increment property we have that $mathbb{E}[M(s)(M(t)-M(s))] = 0$ and hence for $s<t$
$$mbox{Cov}[M(t),M(s)] = mbox{Var}[M(s)].$$
The proof goes similar for $t<s$.
answered Apr 1 '16 at 19:23
SironSiron
1,215614
$endgroup$
add a comment |
$begingroup$
What is the mathematical meaning of 'rely on'? Typically, to compute covariances of martingales you can do the following. Suppose $s<t$
begin{align}
mbox{Cov}[M(t),M(s)] &= mbox{Cov}[M(t)-M(s)+M(s),M(s)] \
&=mbox{Cov}[M(t)-M(s),M(s)] + mbox{Var}[M(s)] \
&= mathbb{E}[M(s)(M(t)-M(s))] - mathbb{E}[M(t)-M(s)]mathbb{E}[M(s)] + mbox{Var}[M(s)] \
end{align}
Using the orthogonal increment property we have that $mathbb{E}[M(s)(M(t)-M(s))] = 0$ and hence for $s<t$
$$mbox{Cov}[M(t),M(s)] = mbox{Var}[M(s)].$$
The proof goes similar for $t<s$.
answered Apr 1 '16 at 19:23
SironSiron
1,215614
$endgroup$
What is the mathematical meaning of 'rely on'? Typically, to compute covariances of martingales you can do the following. Suppose $s<t$
begin{align}
mbox{Cov}[M(t),M(s)] &= mbox{Cov}[M(t)-M(s)+M(s),M(s)] \
&=mbox{Cov}[M(t)-M(s),M(s)] + mbox{Var}[M(s)] \
&= mathbb{E}[M(s)(M(t)-M(s))] - mathbb{E}[M(t)-M(s)]mathbb{E}[M(s)] + mbox{Var}[M(s)] \
end{align}
Using the orthogonal increment property we have that $mathbb{E}[M(s)(M(t)-M(s))] = 0$ and hence for $s<t$
$$mbox{Cov}[M(t),M(s)] = mbox{Var}[M(s)].$$
The proof goes similar for $t<s$.
answered Apr 1 '16 at 19:23
SironSiron
1,215614
answered Apr 1 '16 at 19:23
SironSiron
1,215614
answered Apr 1 '16 at 19:23
SironSiron
1,215614
answered Apr 1 '16 at 19:23
SironSiron
1,215614
1,215614
add a comment |
add a comment |
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