Does partial averaging allow moving increments in and out of an expectation?Calculation of characteristic...
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Does partial averaging allow moving increments in and out of an expectation?
Calculation of characteristic functions of Levy processesJumping times of a Lévy ProcessSubordination of a Levy process when the “subordinator” is not nondecreasingMinimal value of probability according to the difference of a Levy-processTime Changing a Levy process to get rid of multipleWhy is $T_n-T_{n-1}$ independent of $mathcal{F}_{T_{n-1}}$ where $T_{n}=inf {t>T_{n-1}: |X_t-X_{T_{n-1}}| geq C}$?Characteristic Exponent of Levy ProcessShow that $ (e^{alpha X_t} int^ t_ 0 e ^{-alpha X_u}du, t geq 0) $ is a Markov processDefinition of Lévy processDynkin formula and expectation
$begingroup$
Given a Levy process $X$ at different points in time $s$ and $t$, and if I have an expression like this:
$$mathbb{E}[X_t cdot mathbb{E}[X_s]]$$
I want to know if I can use partial averaging to say that this is equal to
$$mathbb{E}[X_tcdot X_s].$$
Is that correct?
probability-theory levy-processes
edited Mar 13 at 10:03
Cettt
1,888622
asked Mar 12 at 21:57
jajajaja
8711
$endgroup$
add a comment |
$begingroup$
Given a Levy process $X$ at different points in time $s$ and $t$, and if I have an expression like this:
$$mathbb{E}[X_t cdot mathbb{E}[X_s]]$$
I want to know if I can use partial averaging to say that this is equal to
$$mathbb{E}[X_tcdot X_s].$$
Is that correct?
probability-theory levy-processes
edited Mar 13 at 10:03
Cettt
1,888622
asked Mar 12 at 21:57
jajajaja
8711
$endgroup$
4
$begingroup$
The first expression is equal to $mathbb E[X_t]mathbb E[X_s]$. Can you see why this is not equal to $mathbb E[X_tX_s]$, when $X_t$ is (say) Brownian motion?
$endgroup$
– Mike Earnest
Mar 12 at 22:59
$begingroup$
okay, yes I see that! But can you please tell me why $mathbb{E}[X_tmathbb{E}[X_s]]=mathbb{E}[X_t]mathbb{E}[X_s]$? I'm sorry if I'm not seeing obviously. $X_t$ is not necessarily independent of $X_s$. The increments are independent but not the specific points.
$endgroup$
– jaja
Mar 13 at 8:36
1
$begingroup$
For any constant, $k$, $mathbb E[kX_t]=kmathbb E[X_t]$. Let $k$ be the constant $E[X_s]$.
$endgroup$
– Mike Earnest
Mar 13 at 18:16
add a comment |
$begingroup$
Given a Levy process $X$ at different points in time $s$ and $t$, and if I have an expression like this:
$$mathbb{E}[X_t cdot mathbb{E}[X_s]]$$
I want to know if I can use partial averaging to say that this is equal to
$$mathbb{E}[X_tcdot X_s].$$
Is that correct?
probability-theory levy-processes
edited Mar 13 at 10:03
Cettt
1,888622
asked Mar 12 at 21:57
jajajaja
8711
$endgroup$
Given a Levy process $X$ at different points in time $s$ and $t$, and if I have an expression like this:
$$mathbb{E}[X_t cdot mathbb{E}[X_s]]$$
I want to know if I can use partial averaging to say that this is equal to
$$mathbb{E}[X_tcdot X_s].$$
Is that correct?
probability-theory levy-processes
probability-theory levy-processes
edited Mar 13 at 10:03
Cettt
1,888622
asked Mar 12 at 21:57
jajajaja
8711
edited Mar 13 at 10:03
Cettt
1,888622
asked Mar 12 at 21:57
jajajaja
8711
edited Mar 13 at 10:03
Cettt
1,888622
edited Mar 13 at 10:03
Cettt
1,888622
edited Mar 13 at 10:03
Cettt
1,888622
1,888622
asked Mar 12 at 21:57
jajajaja
8711
asked Mar 12 at 21:57
jajajaja
8711
asked Mar 12 at 21:57
jajajaja
8711
8711
4
$begingroup$
The first expression is equal to $mathbb E[X_t]mathbb E[X_s]$. Can you see why this is not equal to $mathbb E[X_tX_s]$, when $X_t$ is (say) Brownian motion?
$endgroup$
– Mike Earnest
Mar 12 at 22:59
$begingroup$
okay, yes I see that! But can you please tell me why $mathbb{E}[X_tmathbb{E}[X_s]]=mathbb{E}[X_t]mathbb{E}[X_s]$? I'm sorry if I'm not seeing obviously. $X_t$ is not necessarily independent of $X_s$. The increments are independent but not the specific points.
$endgroup$
– jaja
Mar 13 at 8:36
1
$begingroup$
For any constant, $k$, $mathbb E[kX_t]=kmathbb E[X_t]$. Let $k$ be the constant $E[X_s]$.
$endgroup$
– Mike Earnest
Mar 13 at 18:16
add a comment |
4
$begingroup$
The first expression is equal to $mathbb E[X_t]mathbb E[X_s]$. Can you see why this is not equal to $mathbb E[X_tX_s]$, when $X_t$ is (say) Brownian motion?
$endgroup$
– Mike Earnest
Mar 12 at 22:59
$begingroup$
okay, yes I see that! But can you please tell me why $mathbb{E}[X_tmathbb{E}[X_s]]=mathbb{E}[X_t]mathbb{E}[X_s]$? I'm sorry if I'm not seeing obviously. $X_t$ is not necessarily independent of $X_s$. The increments are independent but not the specific points.
$endgroup$
– jaja
Mar 13 at 8:36
1
$begingroup$
For any constant, $k$, $mathbb E[kX_t]=kmathbb E[X_t]$. Let $k$ be the constant $E[X_s]$.
$endgroup$
– Mike Earnest
Mar 13 at 18:16
4
4
$begingroup$
The first expression is equal to $mathbb E[X_t]mathbb E[X_s]$. Can you see why this is not equal to $mathbb E[X_tX_s]$, when $X_t$ is (say) Brownian motion?
$endgroup$
– Mike Earnest
Mar 12 at 22:59
$begingroup$
The first expression is equal to $mathbb E[X_t]mathbb E[X_s]$. Can you see why this is not equal to $mathbb E[X_tX_s]$, when $X_t$ is (say) Brownian motion?
$endgroup$
– Mike Earnest
Mar 12 at 22:59
$begingroup$
okay, yes I see that! But can you please tell me why $mathbb{E}[X_tmathbb{E}[X_s]]=mathbb{E}[X_t]mathbb{E}[X_s]$? I'm sorry if I'm not seeing obviously. $X_t$ is not necessarily independent of $X_s$. The increments are independent but not the specific points.
$endgroup$
– jaja
Mar 13 at 8:36
$begingroup$
okay, yes I see that! But can you please tell me why $mathbb{E}[X_tmathbb{E}[X_s]]=mathbb{E}[X_t]mathbb{E}[X_s]$? I'm sorry if I'm not seeing obviously. $X_t$ is not necessarily independent of $X_s$. The increments are independent but not the specific points.
$endgroup$
– jaja
Mar 13 at 8:36
1
1
$begingroup$
For any constant, $k$, $mathbb E[kX_t]=kmathbb E[X_t]$. Let $k$ be the constant $E[X_s]$.
$endgroup$
– Mike Earnest
Mar 13 at 18:16
$begingroup$
For any constant, $k$, $mathbb E[kX_t]=kmathbb E[X_t]$. Let $k$ be the constant $E[X_s]$.
$endgroup$
– Mike Earnest
Mar 13 at 18:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
let $X$ be a Levy process. Note that $Bbb E[X_s]$ is then a real number.
Therefore
$$
Bbb E[Bbb E[X_s] cdot X_t] = Bbb E[X_s] cdot Bbb E[X_t].
$$
As mentioned in the comments, this is (in general) not equal to $Bbb E[X_s cdot X_t]$.
answered Mar 13 at 9:22
CetttCettt
1,888622
$endgroup$
$begingroup$
Why does it matterthat it is a real number?
$endgroup$
– jaja
Mar 13 at 9:59
2
$begingroup$
One of the most basic properties for expectations states that for any real number $c$ we have that $Bbb E[ccdot X] = c cdot Bbb E[X]$.
$endgroup$
– Cettt
Mar 13 at 10:09
add a comment |
Your Answer
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1 Answer
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$begingroup$
let $X$ be a Levy process. Note that $Bbb E[X_s]$ is then a real number.
Therefore
$$
Bbb E[Bbb E[X_s] cdot X_t] = Bbb E[X_s] cdot Bbb E[X_t].
$$
As mentioned in the comments, this is (in general) not equal to $Bbb E[X_s cdot X_t]$.
answered Mar 13 at 9:22
CetttCettt
1,888622
$endgroup$
$begingroup$
Why does it matterthat it is a real number?
$endgroup$
– jaja
Mar 13 at 9:59
2
$begingroup$
One of the most basic properties for expectations states that for any real number $c$ we have that $Bbb E[ccdot X] = c cdot Bbb E[X]$.
$endgroup$
– Cettt
Mar 13 at 10:09
add a comment |
$begingroup$
let $X$ be a Levy process. Note that $Bbb E[X_s]$ is then a real number.
Therefore
$$
Bbb E[Bbb E[X_s] cdot X_t] = Bbb E[X_s] cdot Bbb E[X_t].
$$
As mentioned in the comments, this is (in general) not equal to $Bbb E[X_s cdot X_t]$.
answered Mar 13 at 9:22
CetttCettt
1,888622
$endgroup$
$begingroup$
Why does it matterthat it is a real number?
$endgroup$
– jaja
Mar 13 at 9:59
2
$begingroup$
One of the most basic properties for expectations states that for any real number $c$ we have that $Bbb E[ccdot X] = c cdot Bbb E[X]$.
$endgroup$
– Cettt
Mar 13 at 10:09
add a comment |
$begingroup$
let $X$ be a Levy process. Note that $Bbb E[X_s]$ is then a real number.
Therefore
$$
Bbb E[Bbb E[X_s] cdot X_t] = Bbb E[X_s] cdot Bbb E[X_t].
$$
As mentioned in the comments, this is (in general) not equal to $Bbb E[X_s cdot X_t]$.
answered Mar 13 at 9:22
CetttCettt
1,888622
$endgroup$
let $X$ be a Levy process. Note that $Bbb E[X_s]$ is then a real number.
Therefore
$$
Bbb E[Bbb E[X_s] cdot X_t] = Bbb E[X_s] cdot Bbb E[X_t].
$$
As mentioned in the comments, this is (in general) not equal to $Bbb E[X_s cdot X_t]$.
answered Mar 13 at 9:22
CetttCettt
1,888622
answered Mar 13 at 9:22
CetttCettt
1,888622
answered Mar 13 at 9:22
CetttCettt
1,888622
answered Mar 13 at 9:22
CetttCettt
1,888622
1,888622
$begingroup$
Why does it matterthat it is a real number?
$endgroup$
– jaja
Mar 13 at 9:59
2
$begingroup$
One of the most basic properties for expectations states that for any real number $c$ we have that $Bbb E[ccdot X] = c cdot Bbb E[X]$.
$endgroup$
– Cettt
Mar 13 at 10:09
add a comment |
$begingroup$
Why does it matterthat it is a real number?
$endgroup$
– jaja
Mar 13 at 9:59
2
$begingroup$
One of the most basic properties for expectations states that for any real number $c$ we have that $Bbb E[ccdot X] = c cdot Bbb E[X]$.
$endgroup$
– Cettt
Mar 13 at 10:09
$begingroup$
Why does it matterthat it is a real number?
$endgroup$
– jaja
Mar 13 at 9:59
$begingroup$
Why does it matterthat it is a real number?
$endgroup$
– jaja
Mar 13 at 9:59
2
2
$begingroup$
One of the most basic properties for expectations states that for any real number $c$ we have that $Bbb E[ccdot X] = c cdot Bbb E[X]$.
$endgroup$
– Cettt
Mar 13 at 10:09
$begingroup$
One of the most basic properties for expectations states that for any real number $c$ we have that $Bbb E[ccdot X] = c cdot Bbb E[X]$.
$endgroup$
– Cettt
Mar 13 at 10:09
add a comment |
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4
$begingroup$
The first expression is equal to $mathbb E[X_t]mathbb E[X_s]$. Can you see why this is not equal to $mathbb E[X_tX_s]$, when $X_t$ is (say) Brownian motion?
$endgroup$
– Mike Earnest
Mar 12 at 22:59
$begingroup$
okay, yes I see that! But can you please tell me why $mathbb{E}[X_tmathbb{E}[X_s]]=mathbb{E}[X_t]mathbb{E}[X_s]$? I'm sorry if I'm not seeing obviously. $X_t$ is not necessarily independent of $X_s$. The increments are independent but not the specific points.
$endgroup$
– jaja
Mar 13 at 8:36
1
$begingroup$
For any constant, $k$, $mathbb E[kX_t]=kmathbb E[X_t]$. Let $k$ be the constant $E[X_s]$.
$endgroup$
– Mike Earnest
Mar 13 at 18:16