Mixing
Rewrite $ int_{mathcal{S}}dP_X=1 $ as conditions on boxes in $mathbb{R}^d$
$begingroup$
Take $rin mathbb{N}$ and let $dequiv r+binom{r}{2}$.
Consider a d-dimensional random vector $Xequiv (X_1,...,X_d)$. Let $P_X$ be the probability distribution of $X$. Assume that
$$
int_{mathcal{S}}dP_X=1
$$
where
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,..., b_d)in mathbb{R}^{d}: text{ } & b_{r+1}=b_1-b_2, b_{r+2}=b_1-b_3, ...,b_{2r-1}=b_1-b_r, \
&b_{2r}=b_2-b_3, ..., b_{3r-3}=b_2-b_r,\
&...,\
& b_d=b_{r-1}-b_r}
end{aligned}
$$
For example, when $r=2$ ($d=3$) we have the surface
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_3=b_1-b_2}={(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_1=b_2+b_3}
end{aligned}
$$
When $r=3$ ($d=6$) we have
$$
begin{aligned}
mathcal{S}equiv {(b_1,..., b_6)in mathbb{R}^{6}: text{ } & b_4=b_1-b_2, b_5=b_1-b_3, b_6=b_2-b_3}
end{aligned}
$$
My final goal: I'm interested in rewriting the condition $int_{mathcal{S}}dP_X=1$ as a collection of zero probability measure conditions on d-dimensional "boxes" in $mathbb{R}^d$. The idea is that any box in $mathbb{R}^d$ not intersecting $mathcal{S}$ should have probability measure equal to zero. Therefore, if we consider enough of these boxes, we should be able to equivalently rewrite $int_{mathcal{S}}dP_X=1$.
When $r=2$ ($d=3$), my goal is achieved by the following claim
Claim: For any two real numbers $(b,c)in mathbb{R}^2$, define the boxes $$B(b,c)equiv {(x,y,z)text{ s.t. } x> b+c, yleq b, zleq c}$$ and $$Q(b,c)equiv {(x,y,z)text{ s.t. } xleq b+c, y>b, z>c}$$
If $P_{X}(B(b,c))=0$ and $P_{X}(Q(b,c))=0$ $forall(b,c)in mathbb{Q}^2$, then $int_{mathcal{S}}dP_{X}=1$.
The proof of the claim is provided here
I would like your help to generalise the claim (and possibly the proof) to any $r$. What I find challenging is defining the relevant boxes for any $r>2$. I really can't see how to generalise the box definitions from $r=2$ to any $r$.
probability geometry probability-theory measure-theory lebesgue-measure
asked Jan 20 at 21:20
STFSTF
431422
$endgroup$
add a comment |
$begingroup$
Take $rin mathbb{N}$ and let $dequiv r+binom{r}{2}$.
Consider a d-dimensional random vector $Xequiv (X_1,...,X_d)$. Let $P_X$ be the probability distribution of $X$. Assume that
$$
int_{mathcal{S}}dP_X=1
$$
where
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,..., b_d)in mathbb{R}^{d}: text{ } & b_{r+1}=b_1-b_2, b_{r+2}=b_1-b_3, ...,b_{2r-1}=b_1-b_r, \
&b_{2r}=b_2-b_3, ..., b_{3r-3}=b_2-b_r,\
&...,\
& b_d=b_{r-1}-b_r}
end{aligned}
$$
For example, when $r=2$ ($d=3$) we have the surface
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_3=b_1-b_2}={(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_1=b_2+b_3}
end{aligned}
$$
When $r=3$ ($d=6$) we have
$$
begin{aligned}
mathcal{S}equiv {(b_1,..., b_6)in mathbb{R}^{6}: text{ } & b_4=b_1-b_2, b_5=b_1-b_3, b_6=b_2-b_3}
end{aligned}
$$
My final goal: I'm interested in rewriting the condition $int_{mathcal{S}}dP_X=1$ as a collection of zero probability measure conditions on d-dimensional "boxes" in $mathbb{R}^d$. The idea is that any box in $mathbb{R}^d$ not intersecting $mathcal{S}$ should have probability measure equal to zero. Therefore, if we consider enough of these boxes, we should be able to equivalently rewrite $int_{mathcal{S}}dP_X=1$.
When $r=2$ ($d=3$), my goal is achieved by the following claim
Claim: For any two real numbers $(b,c)in mathbb{R}^2$, define the boxes $$B(b,c)equiv {(x,y,z)text{ s.t. } x> b+c, yleq b, zleq c}$$ and $$Q(b,c)equiv {(x,y,z)text{ s.t. } xleq b+c, y>b, z>c}$$
If $P_{X}(B(b,c))=0$ and $P_{X}(Q(b,c))=0$ $forall(b,c)in mathbb{Q}^2$, then $int_{mathcal{S}}dP_{X}=1$.
The proof of the claim is provided here
I would like your help to generalise the claim (and possibly the proof) to any $r$. What I find challenging is defining the relevant boxes for any $r>2$. I really can't see how to generalise the box definitions from $r=2$ to any $r$.
probability geometry probability-theory measure-theory lebesgue-measure
asked Jan 20 at 21:20
STFSTF
431422
$endgroup$
add a comment |
$begingroup$
Take $rin mathbb{N}$ and let $dequiv r+binom{r}{2}$.
Consider a d-dimensional random vector $Xequiv (X_1,...,X_d)$. Let $P_X$ be the probability distribution of $X$. Assume that
$$
int_{mathcal{S}}dP_X=1
$$
where
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,..., b_d)in mathbb{R}^{d}: text{ } & b_{r+1}=b_1-b_2, b_{r+2}=b_1-b_3, ...,b_{2r-1}=b_1-b_r, \
&b_{2r}=b_2-b_3, ..., b_{3r-3}=b_2-b_r,\
&...,\
& b_d=b_{r-1}-b_r}
end{aligned}
$$
For example, when $r=2$ ($d=3$) we have the surface
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_3=b_1-b_2}={(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_1=b_2+b_3}
end{aligned}
$$
When $r=3$ ($d=6$) we have
$$
begin{aligned}
mathcal{S}equiv {(b_1,..., b_6)in mathbb{R}^{6}: text{ } & b_4=b_1-b_2, b_5=b_1-b_3, b_6=b_2-b_3}
end{aligned}
$$
My final goal: I'm interested in rewriting the condition $int_{mathcal{S}}dP_X=1$ as a collection of zero probability measure conditions on d-dimensional "boxes" in $mathbb{R}^d$. The idea is that any box in $mathbb{R}^d$ not intersecting $mathcal{S}$ should have probability measure equal to zero. Therefore, if we consider enough of these boxes, we should be able to equivalently rewrite $int_{mathcal{S}}dP_X=1$.
When $r=2$ ($d=3$), my goal is achieved by the following claim
Claim: For any two real numbers $(b,c)in mathbb{R}^2$, define the boxes $$B(b,c)equiv {(x,y,z)text{ s.t. } x> b+c, yleq b, zleq c}$$ and $$Q(b,c)equiv {(x,y,z)text{ s.t. } xleq b+c, y>b, z>c}$$
If $P_{X}(B(b,c))=0$ and $P_{X}(Q(b,c))=0$ $forall(b,c)in mathbb{Q}^2$, then $int_{mathcal{S}}dP_{X}=1$.
The proof of the claim is provided here
I would like your help to generalise the claim (and possibly the proof) to any $r$. What I find challenging is defining the relevant boxes for any $r>2$. I really can't see how to generalise the box definitions from $r=2$ to any $r$.
probability geometry probability-theory measure-theory lebesgue-measure
asked Jan 20 at 21:20
STFSTF
431422
$endgroup$
Take $rin mathbb{N}$ and let $dequiv r+binom{r}{2}$.
Consider a d-dimensional random vector $Xequiv (X_1,...,X_d)$. Let $P_X$ be the probability distribution of $X$. Assume that
$$
int_{mathcal{S}}dP_X=1
$$
where
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,..., b_d)in mathbb{R}^{d}: text{ } & b_{r+1}=b_1-b_2, b_{r+2}=b_1-b_3, ...,b_{2r-1}=b_1-b_r, \
&b_{2r}=b_2-b_3, ..., b_{3r-3}=b_2-b_r,\
&...,\
& b_d=b_{r-1}-b_r}
end{aligned}
$$
For example, when $r=2$ ($d=3$) we have the surface
$$
begin{aligned}
mathcal{S}equiv {(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_3=b_1-b_2}={(b_1,b_2,b_3)in mathbb{R}^{3}: text{ } & b_1=b_2+b_3}
end{aligned}
$$
When $r=3$ ($d=6$) we have
$$
begin{aligned}
mathcal{S}equiv {(b_1,..., b_6)in mathbb{R}^{6}: text{ } & b_4=b_1-b_2, b_5=b_1-b_3, b_6=b_2-b_3}
end{aligned}
$$
My final goal: I'm interested in rewriting the condition $int_{mathcal{S}}dP_X=1$ as a collection of zero probability measure conditions on d-dimensional "boxes" in $mathbb{R}^d$. The idea is that any box in $mathbb{R}^d$ not intersecting $mathcal{S}$ should have probability measure equal to zero. Therefore, if we consider enough of these boxes, we should be able to equivalently rewrite $int_{mathcal{S}}dP_X=1$.
When $r=2$ ($d=3$), my goal is achieved by the following claim
Claim: For any two real numbers $(b,c)in mathbb{R}^2$, define the boxes $$B(b,c)equiv {(x,y,z)text{ s.t. } x> b+c, yleq b, zleq c}$$ and $$Q(b,c)equiv {(x,y,z)text{ s.t. } xleq b+c, y>b, z>c}$$
If $P_{X}(B(b,c))=0$ and $P_{X}(Q(b,c))=0$ $forall(b,c)in mathbb{Q}^2$, then $int_{mathcal{S}}dP_{X}=1$.
The proof of the claim is provided here
I would like your help to generalise the claim (and possibly the proof) to any $r$. What I find challenging is defining the relevant boxes for any $r>2$. I really can't see how to generalise the box definitions from $r=2$ to any $r$.
probability geometry probability-theory measure-theory lebesgue-measure
probability geometry probability-theory measure-theory lebesgue-measure
asked Jan 20 at 21:20
STFSTF
431422
asked Jan 20 at 21:20
STFSTF
431422
edited Jan 28 at 14:11
STF
asked Jan 20 at 21:20
STFSTF
431422
asked Jan 20 at 21:20
STFSTF
431422
asked Jan 20 at 21:20
STFSTF
431422
431422
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I try to post an answer. It is just an attempt, aiming to encourage comments from your side. The attempt mimics for any $r$ the claim and the proof provided here for the case $r=2$ ($d=3$).
Claim
For any $(bar{b}, tilde{b})in mathbb{R}^2$, consider the $d$-dimensional boxes in $mathbb{R}^d$
$$
B_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p leq bar{b}, text{ } z_qleq tilde{b}, text{ } z_t>bar{b}+tilde{b} }
$$
and
$$
Q_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p >bar{b}, text{ } z_q> tilde{b}, text{ } z_tleq bar{b}+tilde{b} }
$$
$forall t in {1,...,r-1}$ and $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$.
Let $mathbb{Q}$ denote the set of rational numbers.
If
begin{equation}
label{integral}
begin{aligned}
P_{X}(B_{t,p,q}(bar{b}, tilde{b}))=& P_{X}(Q_{t,p,q}(bar{b}, tilde{b}))=0 \
& text{$forall t in {1,...,r-1}$, $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$}
end{aligned}
end{equation}
$forall (b, tilde{b})in mathbb{Q}^{2}$, then $P_{X}(mathcal{S})=1$.
Proof
Step 1: Using the fact that $mathbb{Q}$ is dense in $mathbb{R}$, we can show that
$$
mathcal{S}^c= overbrace{bigcup_{(bar{b}, tilde{b})in mathbb{Q}^2} Big{bigcup_{substack{text{$tin{1,...,r_1}$} \ text{$(p,q)in {(t+1,r),...,(r,d)}$}}}{B_{t,q,p}(bar{b}, tilde{b}) cup Q_{t,q,p}(bar{b}, tilde{b})}Big}}^{equiv A}
$$
where $mathcal{S}^c$ denotes the complement of $mathcal{S}$.
Step 2: Therefore
$$
mathbb{P}(A)=0 Leftrightarrow mathbb{P}(mathcal{S})=1
$$
from which the conclusion of the claim follows.
answered Jan 28 at 14:35
STFSTF
431422
$endgroup$
add a comment |
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$begingroup$
I try to post an answer. It is just an attempt, aiming to encourage comments from your side. The attempt mimics for any $r$ the claim and the proof provided here for the case $r=2$ ($d=3$).
Claim
For any $(bar{b}, tilde{b})in mathbb{R}^2$, consider the $d$-dimensional boxes in $mathbb{R}^d$
$$
B_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p leq bar{b}, text{ } z_qleq tilde{b}, text{ } z_t>bar{b}+tilde{b} }
$$
and
$$
Q_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p >bar{b}, text{ } z_q> tilde{b}, text{ } z_tleq bar{b}+tilde{b} }
$$
$forall t in {1,...,r-1}$ and $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$.
Let $mathbb{Q}$ denote the set of rational numbers.
If
begin{equation}
label{integral}
begin{aligned}
P_{X}(B_{t,p,q}(bar{b}, tilde{b}))=& P_{X}(Q_{t,p,q}(bar{b}, tilde{b}))=0 \
& text{$forall t in {1,...,r-1}$, $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$}
end{aligned}
end{equation}
$forall (b, tilde{b})in mathbb{Q}^{2}$, then $P_{X}(mathcal{S})=1$.
Proof
Step 1: Using the fact that $mathbb{Q}$ is dense in $mathbb{R}$, we can show that
$$
mathcal{S}^c= overbrace{bigcup_{(bar{b}, tilde{b})in mathbb{Q}^2} Big{bigcup_{substack{text{$tin{1,...,r_1}$} \ text{$(p,q)in {(t+1,r),...,(r,d)}$}}}{B_{t,q,p}(bar{b}, tilde{b}) cup Q_{t,q,p}(bar{b}, tilde{b})}Big}}^{equiv A}
$$
where $mathcal{S}^c$ denotes the complement of $mathcal{S}$.
Step 2: Therefore
$$
mathbb{P}(A)=0 Leftrightarrow mathbb{P}(mathcal{S})=1
$$
from which the conclusion of the claim follows.
answered Jan 28 at 14:35
STFSTF
431422
$endgroup$
add a comment |
$begingroup$
I try to post an answer. It is just an attempt, aiming to encourage comments from your side. The attempt mimics for any $r$ the claim and the proof provided here for the case $r=2$ ($d=3$).
Claim
For any $(bar{b}, tilde{b})in mathbb{R}^2$, consider the $d$-dimensional boxes in $mathbb{R}^d$
$$
B_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p leq bar{b}, text{ } z_qleq tilde{b}, text{ } z_t>bar{b}+tilde{b} }
$$
and
$$
Q_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p >bar{b}, text{ } z_q> tilde{b}, text{ } z_tleq bar{b}+tilde{b} }
$$
$forall t in {1,...,r-1}$ and $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$.
Let $mathbb{Q}$ denote the set of rational numbers.
If
begin{equation}
label{integral}
begin{aligned}
P_{X}(B_{t,p,q}(bar{b}, tilde{b}))=& P_{X}(Q_{t,p,q}(bar{b}, tilde{b}))=0 \
& text{$forall t in {1,...,r-1}$, $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$}
end{aligned}
end{equation}
$forall (b, tilde{b})in mathbb{Q}^{2}$, then $P_{X}(mathcal{S})=1$.
Proof
Step 1: Using the fact that $mathbb{Q}$ is dense in $mathbb{R}$, we can show that
$$
mathcal{S}^c= overbrace{bigcup_{(bar{b}, tilde{b})in mathbb{Q}^2} Big{bigcup_{substack{text{$tin{1,...,r_1}$} \ text{$(p,q)in {(t+1,r),...,(r,d)}$}}}{B_{t,q,p}(bar{b}, tilde{b}) cup Q_{t,q,p}(bar{b}, tilde{b})}Big}}^{equiv A}
$$
where $mathcal{S}^c$ denotes the complement of $mathcal{S}$.
Step 2: Therefore
$$
mathbb{P}(A)=0 Leftrightarrow mathbb{P}(mathcal{S})=1
$$
from which the conclusion of the claim follows.
answered Jan 28 at 14:35
STFSTF
431422
$endgroup$
add a comment |
$begingroup$
I try to post an answer. It is just an attempt, aiming to encourage comments from your side. The attempt mimics for any $r$ the claim and the proof provided here for the case $r=2$ ($d=3$).
Claim
For any $(bar{b}, tilde{b})in mathbb{R}^2$, consider the $d$-dimensional boxes in $mathbb{R}^d$
$$
B_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p leq bar{b}, text{ } z_qleq tilde{b}, text{ } z_t>bar{b}+tilde{b} }
$$
and
$$
Q_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p >bar{b}, text{ } z_q> tilde{b}, text{ } z_tleq bar{b}+tilde{b} }
$$
$forall t in {1,...,r-1}$ and $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$.
Let $mathbb{Q}$ denote the set of rational numbers.
If
begin{equation}
label{integral}
begin{aligned}
P_{X}(B_{t,p,q}(bar{b}, tilde{b}))=& P_{X}(Q_{t,p,q}(bar{b}, tilde{b}))=0 \
& text{$forall t in {1,...,r-1}$, $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$}
end{aligned}
end{equation}
$forall (b, tilde{b})in mathbb{Q}^{2}$, then $P_{X}(mathcal{S})=1$.
Proof
Step 1: Using the fact that $mathbb{Q}$ is dense in $mathbb{R}$, we can show that
$$
mathcal{S}^c= overbrace{bigcup_{(bar{b}, tilde{b})in mathbb{Q}^2} Big{bigcup_{substack{text{$tin{1,...,r_1}$} \ text{$(p,q)in {(t+1,r),...,(r,d)}$}}}{B_{t,q,p}(bar{b}, tilde{b}) cup Q_{t,q,p}(bar{b}, tilde{b})}Big}}^{equiv A}
$$
where $mathcal{S}^c$ denotes the complement of $mathcal{S}$.
Step 2: Therefore
$$
mathbb{P}(A)=0 Leftrightarrow mathbb{P}(mathcal{S})=1
$$
from which the conclusion of the claim follows.
answered Jan 28 at 14:35
STFSTF
431422
$endgroup$
I try to post an answer. It is just an attempt, aiming to encourage comments from your side. The attempt mimics for any $r$ the claim and the proof provided here for the case $r=2$ ($d=3$).
Claim
For any $(bar{b}, tilde{b})in mathbb{R}^2$, consider the $d$-dimensional boxes in $mathbb{R}^d$
$$
B_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p leq bar{b}, text{ } z_qleq tilde{b}, text{ } z_t>bar{b}+tilde{b} }
$$
and
$$
Q_{t,p,q}(bar{b}, tilde{b})equiv {(z_1,...,z_d)in mathbb{R}^d text{: } z_p >bar{b}, text{ } z_q> tilde{b}, text{ } z_tleq bar{b}+tilde{b} }
$$
$forall t in {1,...,r-1}$ and $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$.
Let $mathbb{Q}$ denote the set of rational numbers.
If
begin{equation}
label{integral}
begin{aligned}
P_{X}(B_{t,p,q}(bar{b}, tilde{b}))=& P_{X}(Q_{t,p,q}(bar{b}, tilde{b}))=0 \
& text{$forall t in {1,...,r-1}$, $forall (p,q)in {(t+1,r), (t+2, r+1),...,(r, d)}$}
end{aligned}
end{equation}
$forall (b, tilde{b})in mathbb{Q}^{2}$, then $P_{X}(mathcal{S})=1$.
Proof
Step 1: Using the fact that $mathbb{Q}$ is dense in $mathbb{R}$, we can show that
$$
mathcal{S}^c= overbrace{bigcup_{(bar{b}, tilde{b})in mathbb{Q}^2} Big{bigcup_{substack{text{$tin{1,...,r_1}$} \ text{$(p,q)in {(t+1,r),...,(r,d)}$}}}{B_{t,q,p}(bar{b}, tilde{b}) cup Q_{t,q,p}(bar{b}, tilde{b})}Big}}^{equiv A}
$$
where $mathcal{S}^c$ denotes the complement of $mathcal{S}$.
Step 2: Therefore
$$
mathbb{P}(A)=0 Leftrightarrow mathbb{P}(mathcal{S})=1
$$
from which the conclusion of the claim follows.
answered Jan 28 at 14:35
STFSTF
431422
edited Jan 28 at 15:44
answered Jan 28 at 14:35
STFSTF
431422
answered Jan 28 at 14:35
STFSTF
431422
answered Jan 28 at 14:35
STFSTF
431422
431422
add a comment |
add a comment |
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