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Why is each component of a random vector a random variable?
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I am reading A Concise Introduction to Mathematical Statistics by Dragi Anevski and in chapter 5 about random vectors, he proves that each component of a random vector is a random variable. Unfortunately, I need some help with comprehending the authors proof:
"Suppose that $(Omega, F, P)$ is a probability space and $X : Omega to mathbbR^n$ is a random vector. We assume $(X_1,...,X_n)$ is a random vector and want to show that for any $1 leq i leq n$, $X_i$is a random variable. This follows since for every $x in mathbbR$, $$omega : X_i(omega) leq x = omega : X_i(omega) leq x cap Omega = omega : X_i(omega) leq x cap (cap_j neq iomega : X_j(omega) leq infty) = omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$$. And since this is in the sigma algebra, $F$, $X_i$ is a random variable.".
What I don't understand is the reformulation $Omega = cap_j neq iomega : X_j(omega) leq infty$. Are you supposed to read it as 'the outcome space is equal to the intersection of all events that each $X_j$ for $j neq i in 1,..,n$ gives rise to as $x to infty$'? Why is it an intersection and not a union?
I have not yet read about measure theory and looking at other answers it seems measure theory is a prerequisite.
I also dont understand the conclusion: why $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$.
I tried looking for an answer on the internet, on this website. Didn't find any other proof of it and has been searching for over two hours now.
Thank you in advance,
Isak
probability random-variables
asked Jul 25 at 11:30
iaenstrom
114
 |Â
show 1 more comment
up vote
1
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favorite
I am reading A Concise Introduction to Mathematical Statistics by Dragi Anevski and in chapter 5 about random vectors, he proves that each component of a random vector is a random variable. Unfortunately, I need some help with comprehending the authors proof:
"Suppose that $(Omega, F, P)$ is a probability space and $X : Omega to mathbbR^n$ is a random vector. We assume $(X_1,...,X_n)$ is a random vector and want to show that for any $1 leq i leq n$, $X_i$is a random variable. This follows since for every $x in mathbbR$, $$omega : X_i(omega) leq x = omega : X_i(omega) leq x cap Omega = omega : X_i(omega) leq x cap (cap_j neq iomega : X_j(omega) leq infty) = omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$$. And since this is in the sigma algebra, $F$, $X_i$ is a random variable.".
What I don't understand is the reformulation $Omega = cap_j neq iomega : X_j(omega) leq infty$. Are you supposed to read it as 'the outcome space is equal to the intersection of all events that each $X_j$ for $j neq i in 1,..,n$ gives rise to as $x to infty$'? Why is it an intersection and not a union?
I have not yet read about measure theory and looking at other answers it seems measure theory is a prerequisite.
I also dont understand the conclusion: why $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$.
I tried looking for an answer on the internet, on this website. Didn't find any other proof of it and has been searching for over two hours now.
Thank you in advance,
Isak
probability random-variables
asked Jul 25 at 11:30
iaenstrom
114
1
The last set is $X^-1((-infty,x]timesmathbbR^n-1)$. Since $(-infty,x]timesmathbbR^n-1$ is Borel and $X$ is a random variable then $X^-1((-infty,x]timesmathbbR^n-1)in F$. You can read $X^-1((-infty,x]timesmathbbR^n-1)$ as, the first component is not greater than $x$, while the others take any value, which is the same as just saying, the first component is not larger than $x$.
– user578878
Jul 25 at 11:41
1
More (tediously) precisely: $$X_i^-1((-infty,x])=X^-1(mathbb R^i-1times(-infty,x]timesmathbb R^n-i)$$
– Did
Jul 25 at 11:44
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:19
That would be wrong, but it does not happen.
– drhab
Jul 27 at 17:49
Ok. Where does my logic steer off?
– iaenstrom
Jul 27 at 18:29
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am reading A Concise Introduction to Mathematical Statistics by Dragi Anevski and in chapter 5 about random vectors, he proves that each component of a random vector is a random variable. Unfortunately, I need some help with comprehending the authors proof:
"Suppose that $(Omega, F, P)$ is a probability space and $X : Omega to mathbbR^n$ is a random vector. We assume $(X_1,...,X_n)$ is a random vector and want to show that for any $1 leq i leq n$, $X_i$is a random variable. This follows since for every $x in mathbbR$, $$omega : X_i(omega) leq x = omega : X_i(omega) leq x cap Omega = omega : X_i(omega) leq x cap (cap_j neq iomega : X_j(omega) leq infty) = omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$$. And since this is in the sigma algebra, $F$, $X_i$ is a random variable.".
What I don't understand is the reformulation $Omega = cap_j neq iomega : X_j(omega) leq infty$. Are you supposed to read it as 'the outcome space is equal to the intersection of all events that each $X_j$ for $j neq i in 1,..,n$ gives rise to as $x to infty$'? Why is it an intersection and not a union?
I have not yet read about measure theory and looking at other answers it seems measure theory is a prerequisite.
I also dont understand the conclusion: why $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$.
I tried looking for an answer on the internet, on this website. Didn't find any other proof of it and has been searching for over two hours now.
Thank you in advance,
Isak
probability random-variables
asked Jul 25 at 11:30
iaenstrom
114
I am reading A Concise Introduction to Mathematical Statistics by Dragi Anevski and in chapter 5 about random vectors, he proves that each component of a random vector is a random variable. Unfortunately, I need some help with comprehending the authors proof:
"Suppose that $(Omega, F, P)$ is a probability space and $X : Omega to mathbbR^n$ is a random vector. We assume $(X_1,...,X_n)$ is a random vector and want to show that for any $1 leq i leq n$, $X_i$is a random variable. This follows since for every $x in mathbbR$, $$omega : X_i(omega) leq x = omega : X_i(omega) leq x cap Omega = omega : X_i(omega) leq x cap (cap_j neq iomega : X_j(omega) leq infty) = omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$$. And since this is in the sigma algebra, $F$, $X_i$ is a random variable.".
What I don't understand is the reformulation $Omega = cap_j neq iomega : X_j(omega) leq infty$. Are you supposed to read it as 'the outcome space is equal to the intersection of all events that each $X_j$ for $j neq i in 1,..,n$ gives rise to as $x to infty$'? Why is it an intersection and not a union?
I have not yet read about measure theory and looking at other answers it seems measure theory is a prerequisite.
I also dont understand the conclusion: why $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$.
I tried looking for an answer on the internet, on this website. Didn't find any other proof of it and has been searching for over two hours now.
Thank you in advance,
Isak
probability random-variables
asked Jul 25 at 11:30
iaenstrom
114
edited Jul 25 at 11:47
asked Jul 25 at 11:30
iaenstrom
114
asked Jul 25 at 11:30
iaenstrom
114
asked Jul 25 at 11:30
iaenstrom
114
114
1
The last set is $X^-1((-infty,x]timesmathbbR^n-1)$. Since $(-infty,x]timesmathbbR^n-1$ is Borel and $X$ is a random variable then $X^-1((-infty,x]timesmathbbR^n-1)in F$. You can read $X^-1((-infty,x]timesmathbbR^n-1)$ as, the first component is not greater than $x$, while the others take any value, which is the same as just saying, the first component is not larger than $x$.
– user578878
Jul 25 at 11:41
1
More (tediously) precisely: $$X_i^-1((-infty,x])=X^-1(mathbb R^i-1times(-infty,x]timesmathbb R^n-i)$$
– Did
Jul 25 at 11:44
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:19
That would be wrong, but it does not happen.
– drhab
Jul 27 at 17:49
Ok. Where does my logic steer off?
– iaenstrom
Jul 27 at 18:29
 |Â
show 1 more comment
1
The last set is $X^-1((-infty,x]timesmathbbR^n-1)$. Since $(-infty,x]timesmathbbR^n-1$ is Borel and $X$ is a random variable then $X^-1((-infty,x]timesmathbbR^n-1)in F$. You can read $X^-1((-infty,x]timesmathbbR^n-1)$ as, the first component is not greater than $x$, while the others take any value, which is the same as just saying, the first component is not larger than $x$.
– user578878
Jul 25 at 11:41
1
More (tediously) precisely: $$X_i^-1((-infty,x])=X^-1(mathbb R^i-1times(-infty,x]timesmathbb R^n-i)$$
– Did
Jul 25 at 11:44
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:19
That would be wrong, but it does not happen.
– drhab
Jul 27 at 17:49
Ok. Where does my logic steer off?
– iaenstrom
Jul 27 at 18:29
1
1
The last set is $X^-1((-infty,x]timesmathbbR^n-1)$. Since $(-infty,x]timesmathbbR^n-1$ is Borel and $X$ is a random variable then $X^-1((-infty,x]timesmathbbR^n-1)in F$. You can read $X^-1((-infty,x]timesmathbbR^n-1)$ as, the first component is not greater than $x$, while the others take any value, which is the same as just saying, the first component is not larger than $x$.
– user578878
Jul 25 at 11:41
The last set is $X^-1((-infty,x]timesmathbbR^n-1)$. Since $(-infty,x]timesmathbbR^n-1$ is Borel and $X$ is a random variable then $X^-1((-infty,x]timesmathbbR^n-1)in F$. You can read $X^-1((-infty,x]timesmathbbR^n-1)$ as, the first component is not greater than $x$, while the others take any value, which is the same as just saying, the first component is not larger than $x$.
– user578878
Jul 25 at 11:41
1
1
More (tediously) precisely: $$X_i^-1((-infty,x])=X^-1(mathbb R^i-1times(-infty,x]timesmathbb R^n-i)$$
– Did
Jul 25 at 11:44
More (tediously) precisely: $$X_i^-1((-infty,x])=X^-1(mathbb R^i-1times(-infty,x]timesmathbb R^n-i)$$
– Did
Jul 25 at 11:44
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:19
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:19
That would be wrong, but it does not happen.
– drhab
Jul 27 at 17:49
That would be wrong, but it does not happen.
– drhab
Jul 27 at 17:49
Ok. Where does my logic steer off?
– iaenstrom
Jul 27 at 18:29
Ok. Where does my logic steer off?
– iaenstrom
Jul 27 at 18:29
 |Â
show 1 more comment
1 Answer
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If $(X_1,dots,X_n)$ is a random vector then sets of the form $bigcap_i=1^nX_ileq x_i$ for $x_iin(-infty,infty]$ are measurable.
This fact can be used to prove that e.g. $X_1$ is a random variable.
This because the set $X_1leq x$ is exactly the same as the set $X_1leq x, X_2leqinfty,dots, X_nleqinfty$
so appears to be measurable if $(X_1,dots,X_n)$ is a random vector.
The last set can also be denoted as: $X_1leq xcap(bigcap_i=2^nX_ileqinfty)$
or as $bigcap_i=1^nX_ileq x_i$ where $x_1=x$ and $x_i=infty$ if $ineq 1$.
The "conclusion" $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$
makes no sense to me.
My conclusion is that $omega:X_1(omega)leq xin F$
answered Jul 25 at 11:45
drhab
86.2k541118
Thank you, your third line cleared up a lot of confusion but I am still confused about why the outcome space $Omega$ can be written as that intersection. Is it because it is assumed that each $X_i$ is a function mapping from the same outcome space?
– iaenstrom
Jul 25 at 11:57
Let $omega$ be an arbitrary element of $Omega$. For every $jneq i$ we have $X_j(omega)leqinfty$, or equivalently $omegainX_jleqinfty$. So we are justified to conclude that $omegainbigcap_jneq iX_jleqinfty$. So every element of $Omega$ is an element of $bigcap_jneq iX_jleqinftysubseteqOmega$.
– drhab
Jul 25 at 12:03
I agree about the "conclusion" stated in the book. The "set" notation $omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$ seems ill-formed to me. Perhaps they meant $omega : X_i(omega) leq x, X_1(omega) leq infty, ldots, X_i-1(omega) leq infty, X_i+1(omega) leq infty, ldots, X_n(omega) leq infty.$
– David K
Jul 25 at 12:18
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:17
$A_1=X_1leq xsubseteqOmegain F$ but the conclusion that $A_1in F$ is not based on that. For that it is used that $A_1=X_1leq x=X_1leq x,X_2leqinfty,dots,X_nleqinfty$. The set on RHS must be an element of $F$ because $(X_1,X_2,dots,X_n)$ is a random vector.
– drhab
Jul 27 at 12:34
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $(X_1,dots,X_n)$ is a random vector then sets of the form $bigcap_i=1^nX_ileq x_i$ for $x_iin(-infty,infty]$ are measurable.
This fact can be used to prove that e.g. $X_1$ is a random variable.
This because the set $X_1leq x$ is exactly the same as the set $X_1leq x, X_2leqinfty,dots, X_nleqinfty$
so appears to be measurable if $(X_1,dots,X_n)$ is a random vector.
The last set can also be denoted as: $X_1leq xcap(bigcap_i=2^nX_ileqinfty)$
or as $bigcap_i=1^nX_ileq x_i$ where $x_1=x$ and $x_i=infty$ if $ineq 1$.
The "conclusion" $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$
makes no sense to me.
My conclusion is that $omega:X_1(omega)leq xin F$
answered Jul 25 at 11:45
drhab
86.2k541118
Thank you, your third line cleared up a lot of confusion but I am still confused about why the outcome space $Omega$ can be written as that intersection. Is it because it is assumed that each $X_i$ is a function mapping from the same outcome space?
– iaenstrom
Jul 25 at 11:57
Let $omega$ be an arbitrary element of $Omega$. For every $jneq i$ we have $X_j(omega)leqinfty$, or equivalently $omegainX_jleqinfty$. So we are justified to conclude that $omegainbigcap_jneq iX_jleqinfty$. So every element of $Omega$ is an element of $bigcap_jneq iX_jleqinftysubseteqOmega$.
– drhab
Jul 25 at 12:03
I agree about the "conclusion" stated in the book. The "set" notation $omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$ seems ill-formed to me. Perhaps they meant $omega : X_i(omega) leq x, X_1(omega) leq infty, ldots, X_i-1(omega) leq infty, X_i+1(omega) leq infty, ldots, X_n(omega) leq infty.$
– David K
Jul 25 at 12:18
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:17
$A_1=X_1leq xsubseteqOmegain F$ but the conclusion that $A_1in F$ is not based on that. For that it is used that $A_1=X_1leq x=X_1leq x,X_2leqinfty,dots,X_nleqinfty$. The set on RHS must be an element of $F$ because $(X_1,X_2,dots,X_n)$ is a random vector.
– drhab
Jul 27 at 12:34
add a comment |Â
up vote
0
down vote
If $(X_1,dots,X_n)$ is a random vector then sets of the form $bigcap_i=1^nX_ileq x_i$ for $x_iin(-infty,infty]$ are measurable.
This fact can be used to prove that e.g. $X_1$ is a random variable.
This because the set $X_1leq x$ is exactly the same as the set $X_1leq x, X_2leqinfty,dots, X_nleqinfty$
so appears to be measurable if $(X_1,dots,X_n)$ is a random vector.
The last set can also be denoted as: $X_1leq xcap(bigcap_i=2^nX_ileqinfty)$
or as $bigcap_i=1^nX_ileq x_i$ where $x_1=x$ and $x_i=infty$ if $ineq 1$.
The "conclusion" $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$
makes no sense to me.
My conclusion is that $omega:X_1(omega)leq xin F$
answered Jul 25 at 11:45
drhab
86.2k541118
Thank you, your third line cleared up a lot of confusion but I am still confused about why the outcome space $Omega$ can be written as that intersection. Is it because it is assumed that each $X_i$ is a function mapping from the same outcome space?
– iaenstrom
Jul 25 at 11:57
Let $omega$ be an arbitrary element of $Omega$. For every $jneq i$ we have $X_j(omega)leqinfty$, or equivalently $omegainX_jleqinfty$. So we are justified to conclude that $omegainbigcap_jneq iX_jleqinfty$. So every element of $Omega$ is an element of $bigcap_jneq iX_jleqinftysubseteqOmega$.
– drhab
Jul 25 at 12:03
I agree about the "conclusion" stated in the book. The "set" notation $omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$ seems ill-formed to me. Perhaps they meant $omega : X_i(omega) leq x, X_1(omega) leq infty, ldots, X_i-1(omega) leq infty, X_i+1(omega) leq infty, ldots, X_n(omega) leq infty.$
– David K
Jul 25 at 12:18
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:17
$A_1=X_1leq xsubseteqOmegain F$ but the conclusion that $A_1in F$ is not based on that. For that it is used that $A_1=X_1leq x=X_1leq x,X_2leqinfty,dots,X_nleqinfty$. The set on RHS must be an element of $F$ because $(X_1,X_2,dots,X_n)$ is a random vector.
– drhab
Jul 27 at 12:34
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $(X_1,dots,X_n)$ is a random vector then sets of the form $bigcap_i=1^nX_ileq x_i$ for $x_iin(-infty,infty]$ are measurable.
This fact can be used to prove that e.g. $X_1$ is a random variable.
This because the set $X_1leq x$ is exactly the same as the set $X_1leq x, X_2leqinfty,dots, X_nleqinfty$
so appears to be measurable if $(X_1,dots,X_n)$ is a random vector.
The last set can also be denoted as: $X_1leq xcap(bigcap_i=2^nX_ileqinfty)$
or as $bigcap_i=1^nX_ileq x_i$ where $x_1=x$ and $x_i=infty$ if $ineq 1$.
The "conclusion" $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$
makes no sense to me.
My conclusion is that $omega:X_1(omega)leq xin F$
answered Jul 25 at 11:45
drhab
86.2k541118
If $(X_1,dots,X_n)$ is a random vector then sets of the form $bigcap_i=1^nX_ileq x_i$ for $x_iin(-infty,infty]$ are measurable.
This fact can be used to prove that e.g. $X_1$ is a random variable.
This because the set $X_1leq x$ is exactly the same as the set $X_1leq x, X_2leqinfty,dots, X_nleqinfty$
so appears to be measurable if $(X_1,dots,X_n)$ is a random vector.
The last set can also be denoted as: $X_1leq xcap(bigcap_i=2^nX_ileqinfty)$
or as $bigcap_i=1^nX_ileq x_i$ where $x_1=x$ and $x_i=infty$ if $ineq 1$.
The "conclusion" $$omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty in F$$
makes no sense to me.
My conclusion is that $omega:X_1(omega)leq xin F$
answered Jul 25 at 11:45
drhab
86.2k541118
edited Jul 25 at 11:52
answered Jul 25 at 11:45
drhab
86.2k541118
answered Jul 25 at 11:45
drhab
86.2k541118
answered Jul 25 at 11:45
drhab
86.2k541118
86.2k541118
Thank you, your third line cleared up a lot of confusion but I am still confused about why the outcome space $Omega$ can be written as that intersection. Is it because it is assumed that each $X_i$ is a function mapping from the same outcome space?
– iaenstrom
Jul 25 at 11:57
Let $omega$ be an arbitrary element of $Omega$. For every $jneq i$ we have $X_j(omega)leqinfty$, or equivalently $omegainX_jleqinfty$. So we are justified to conclude that $omegainbigcap_jneq iX_jleqinfty$. So every element of $Omega$ is an element of $bigcap_jneq iX_jleqinftysubseteqOmega$.
– drhab
Jul 25 at 12:03
I agree about the "conclusion" stated in the book. The "set" notation $omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$ seems ill-formed to me. Perhaps they meant $omega : X_i(omega) leq x, X_1(omega) leq infty, ldots, X_i-1(omega) leq infty, X_i+1(omega) leq infty, ldots, X_n(omega) leq infty.$
– David K
Jul 25 at 12:18
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:17
$A_1=X_1leq xsubseteqOmegain F$ but the conclusion that $A_1in F$ is not based on that. For that it is used that $A_1=X_1leq x=X_1leq x,X_2leqinfty,dots,X_nleqinfty$. The set on RHS must be an element of $F$ because $(X_1,X_2,dots,X_n)$ is a random vector.
– drhab
Jul 27 at 12:34
add a comment |Â
Thank you, your third line cleared up a lot of confusion but I am still confused about why the outcome space $Omega$ can be written as that intersection. Is it because it is assumed that each $X_i$ is a function mapping from the same outcome space?
– iaenstrom
Jul 25 at 11:57
Let $omega$ be an arbitrary element of $Omega$. For every $jneq i$ we have $X_j(omega)leqinfty$, or equivalently $omegainX_jleqinfty$. So we are justified to conclude that $omegainbigcap_jneq iX_jleqinfty$. So every element of $Omega$ is an element of $bigcap_jneq iX_jleqinftysubseteqOmega$.
– drhab
Jul 25 at 12:03
I agree about the "conclusion" stated in the book. The "set" notation $omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$ seems ill-formed to me. Perhaps they meant $omega : X_i(omega) leq x, X_1(omega) leq infty, ldots, X_i-1(omega) leq infty, X_i+1(omega) leq infty, ldots, X_n(omega) leq infty.$
– David K
Jul 25 at 12:18
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:17
$A_1=X_1leq xsubseteqOmegain F$ but the conclusion that $A_1in F$ is not based on that. For that it is used that $A_1=X_1leq x=X_1leq x,X_2leqinfty,dots,X_nleqinfty$. The set on RHS must be an element of $F$ because $(X_1,X_2,dots,X_n)$ is a random vector.
– drhab
Jul 27 at 12:34
Thank you, your third line cleared up a lot of confusion but I am still confused about why the outcome space $Omega$ can be written as that intersection. Is it because it is assumed that each $X_i$ is a function mapping from the same outcome space?
– iaenstrom
Jul 25 at 11:57
Thank you, your third line cleared up a lot of confusion but I am still confused about why the outcome space $Omega$ can be written as that intersection. Is it because it is assumed that each $X_i$ is a function mapping from the same outcome space?
– iaenstrom
Jul 25 at 11:57
Let $omega$ be an arbitrary element of $Omega$. For every $jneq i$ we have $X_j(omega)leqinfty$, or equivalently $omegainX_jleqinfty$. So we are justified to conclude that $omegainbigcap_jneq iX_jleqinfty$. So every element of $Omega$ is an element of $bigcap_jneq iX_jleqinftysubseteqOmega$.
– drhab
Jul 25 at 12:03
Let $omega$ be an arbitrary element of $Omega$. For every $jneq i$ we have $X_j(omega)leqinfty$, or equivalently $omegainX_jleqinfty$. So we are justified to conclude that $omegainbigcap_jneq iX_jleqinfty$. So every element of $Omega$ is an element of $bigcap_jneq iX_jleqinftysubseteqOmega$.
– drhab
Jul 25 at 12:03
I agree about the "conclusion" stated in the book. The "set" notation $omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$ seems ill-formed to me. Perhaps they meant $omega : X_i(omega) leq x, X_1(omega) leq infty, ldots, X_i-1(omega) leq infty, X_i+1(omega) leq infty, ldots, X_n(omega) leq infty.$
– David K
Jul 25 at 12:18
I agree about the "conclusion" stated in the book. The "set" notation $omega : X_i(omega) leq x,cap_j neq iomega : X_j(omega) leq infty$ seems ill-formed to me. Perhaps they meant $omega : X_i(omega) leq x, X_1(omega) leq infty, ldots, X_i-1(omega) leq infty, X_i+1(omega) leq infty, ldots, X_n(omega) leq infty.$
– David K
Jul 25 at 12:18
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:17
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:17
$A_1=X_1leq xsubseteqOmegain F$ but the conclusion that $A_1in F$ is not based on that. For that it is used that $A_1=X_1leq x=X_1leq x,X_2leqinfty,dots,X_nleqinfty$. The set on RHS must be an element of $F$ because $(X_1,X_2,dots,X_n)$ is a random vector.
– drhab
Jul 27 at 12:34
$A_1=X_1leq xsubseteqOmegain F$ but the conclusion that $A_1in F$ is not based on that. For that it is used that $A_1=X_1leq x=X_1leq x,X_2leqinfty,dots,X_nleqinfty$. The set on RHS must be an element of $F$ because $(X_1,X_2,dots,X_n)$ is a random vector.
– drhab
Jul 27 at 12:34
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1
The last set is $X^-1((-infty,x]timesmathbbR^n-1)$. Since $(-infty,x]timesmathbbR^n-1$ is Borel and $X$ is a random variable then $X^-1((-infty,x]timesmathbbR^n-1)in F$. You can read $X^-1((-infty,x]timesmathbbR^n-1)$ as, the first component is not greater than $x$, while the others take any value, which is the same as just saying, the first component is not larger than $x$.
– user578878
Jul 25 at 11:41
1
More (tediously) precisely: $$X_i^-1((-infty,x])=X^-1(mathbb R^i-1times(-infty,x]timesmathbb R^n-i)$$
– Did
Jul 25 at 11:44
I just went over the arguments one more time, and I couldn’t help but to think that in the first equality in the proof imply that $A=omega : X_i(omega) leq x subseteq Omega$. And since $Omega in F$ this implies that $A in F$? I mean, isn’t it wrong to assume that from the beginning when that is in fact what we want to show?
– iaenstrom
Jul 27 at 12:19
That would be wrong, but it does not happen.
– drhab
Jul 27 at 17:49
Ok. Where does my logic steer off?
– iaenstrom
Jul 27 at 18:29