Showing $text{cov} (x_d-x_c, x_b-x_a)=0$
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Suppose that $(x_k, mathcal{D}_k), k=1,...,n,$ is a martingale define on finite probability space. Show that $$text{cov} (x_d-x_c, x_b-x_a)=0$$ for arbitrary integers $a<b<c<d$.
How can I show it? Do I need to show that $x_d-x_c, x_b-x_a$ are independent? If yes, than how can I do it?
I know that $text{cov}(X,Y)=mathbb{E}(X-mathbb{E}X)(Y-mathbb{E}Y)=mathbb{E}XY-mathbb{E}Xmathbb{E}Y$ so when $mathbb{E}XY=mathbb{E}Xmathbb{E}Y$ than $text{cov}(X,Y)=0$
probability probability-theory random-variables martingales covariance
asked Jan 15 at 16:24
AtstovasAtstovas
1089
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add a comment |
$begingroup$
Suppose that $(x_k, mathcal{D}_k), k=1,...,n,$ is a martingale define on finite probability space. Show that $$text{cov} (x_d-x_c, x_b-x_a)=0$$ for arbitrary integers $a<b<c<d$.
How can I show it? Do I need to show that $x_d-x_c, x_b-x_a$ are independent? If yes, than how can I do it?
I know that $text{cov}(X,Y)=mathbb{E}(X-mathbb{E}X)(Y-mathbb{E}Y)=mathbb{E}XY-mathbb{E}Xmathbb{E}Y$ so when $mathbb{E}XY=mathbb{E}Xmathbb{E}Y$ than $text{cov}(X,Y)=0$
probability probability-theory random-variables martingales covariance
asked Jan 15 at 16:24
AtstovasAtstovas
1089
$endgroup$
$begingroup$
Take e.g. a look at this question (....and no, the increments are in general not independent).
$endgroup$
– saz
Jan 15 at 17:58
add a comment |
$begingroup$
Suppose that $(x_k, mathcal{D}_k), k=1,...,n,$ is a martingale define on finite probability space. Show that $$text{cov} (x_d-x_c, x_b-x_a)=0$$ for arbitrary integers $a<b<c<d$.
How can I show it? Do I need to show that $x_d-x_c, x_b-x_a$ are independent? If yes, than how can I do it?
I know that $text{cov}(X,Y)=mathbb{E}(X-mathbb{E}X)(Y-mathbb{E}Y)=mathbb{E}XY-mathbb{E}Xmathbb{E}Y$ so when $mathbb{E}XY=mathbb{E}Xmathbb{E}Y$ than $text{cov}(X,Y)=0$
probability probability-theory random-variables martingales covariance
asked Jan 15 at 16:24
AtstovasAtstovas
1089
$endgroup$
Suppose that $(x_k, mathcal{D}_k), k=1,...,n,$ is a martingale define on finite probability space. Show that $$text{cov} (x_d-x_c, x_b-x_a)=0$$ for arbitrary integers $a<b<c<d$.
How can I show it? Do I need to show that $x_d-x_c, x_b-x_a$ are independent? If yes, than how can I do it?
I know that $text{cov}(X,Y)=mathbb{E}(X-mathbb{E}X)(Y-mathbb{E}Y)=mathbb{E}XY-mathbb{E}Xmathbb{E}Y$ so when $mathbb{E}XY=mathbb{E}Xmathbb{E}Y$ than $text{cov}(X,Y)=0$
probability probability-theory random-variables martingales covariance
probability probability-theory random-variables martingales covariance
asked Jan 15 at 16:24
AtstovasAtstovas
1089
asked Jan 15 at 16:24
AtstovasAtstovas
1089
asked Jan 15 at 16:24
AtstovasAtstovas
1089
asked Jan 15 at 16:24
AtstovasAtstovas
1089
asked Jan 15 at 16:24
AtstovasAtstovas
1089
1089
$begingroup$
Take e.g. a look at this question (....and no, the increments are in general not independent).
$endgroup$
– saz
Jan 15 at 17:58
add a comment |
$begingroup$
Take e.g. a look at this question (....and no, the increments are in general not independent).
$endgroup$
– saz
Jan 15 at 17:58
$begingroup$
Take e.g. a look at this question (....and no, the increments are in general not independent).
$endgroup$
– saz
Jan 15 at 17:58
$begingroup$
Take e.g. a look at this question (....and no, the increments are in general not independent).
$endgroup$
– saz
Jan 15 at 17:58
add a comment |
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$begingroup$
Take e.g. a look at this question (....and no, the increments are in general not independent).
$endgroup$
– saz
Jan 15 at 17:58