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Express $mathbbP(underliney_kleq Y_kleq bary_k text forall k) $ in terms of cdf
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Consider the k-variate random vector $Yequiv (Y_1, Y_2, ..., Y_K)$ with cumulative distribution function (cdf) $F$.
How can I express in terms of $F$ the following probability:
$$
mathbbP(a_kleq Y_kleq b_k text forall k)
$$
?
For example, if $K=2$, I know that
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2)= F(a_2, b_2)+F(a_1, b_1)- F(a_1, b_2)-F(a_2, b_1)
$$
(Sometimes this is also called rectangle formula)
Is there a way to generalise this to any $K>2$?
Thanks to the comment below, I now write the expression for $K=3$
1) I list all the $2^K=8$ vertices
$$
begincases
a_1,a_2, a_3\
b_1, a_2, a_3\
a_1, b_2, a_3 \
b_1, b_2, a_3\
a_1,a_2, b_3\
b_1, a_2, b_3\
a_1, b_2, b_3 \
b_1, b_2, b_3\
endcases
$$
2) Algebraic sum of cdf at each vertex with +1 if the number of $a$'s is even and -1 otherwise
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2text, a_3leq Y_3leq b_3)= -F(a_1,a_2, a_3)
+F(b_1, a_2, a_3)
+F(a_1, b_2, a_3)
-F(b_1, b_2, a_3)
+F(a_1,a_2, b_3)
-F(b_1, a_2, b_3)
-F(a_1, b_2, b_3)
+F(b_1, b_2, b_3)
$$
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 8:46
TEX
2419
 |Â
show 1 more comment
up vote
0
down vote
favorite
Consider the k-variate random vector $Yequiv (Y_1, Y_2, ..., Y_K)$ with cumulative distribution function (cdf) $F$.
How can I express in terms of $F$ the following probability:
$$
mathbbP(a_kleq Y_kleq b_k text forall k)
$$
?
For example, if $K=2$, I know that
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2)= F(a_2, b_2)+F(a_1, b_1)- F(a_1, b_2)-F(a_2, b_1)
$$
(Sometimes this is also called rectangle formula)
Is there a way to generalise this to any $K>2$?
Thanks to the comment below, I now write the expression for $K=3$
1) I list all the $2^K=8$ vertices
$$
begincases
a_1,a_2, a_3\
b_1, a_2, a_3\
a_1, b_2, a_3 \
b_1, b_2, a_3\
a_1,a_2, b_3\
b_1, a_2, b_3\
a_1, b_2, b_3 \
b_1, b_2, b_3\
endcases
$$
2) Algebraic sum of cdf at each vertex with +1 if the number of $a$'s is even and -1 otherwise
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2text, a_3leq Y_3leq b_3)= -F(a_1,a_2, a_3)
+F(b_1, a_2, a_3)
+F(a_1, b_2, a_3)
-F(b_1, b_2, a_3)
+F(a_1,a_2, b_3)
-F(b_1, a_2, b_3)
-F(a_1, b_2, b_3)
+F(b_1, b_2, b_3)
$$
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 8:46
TEX
2419
1
Billingsley's Probability and Measure, Section 12 has a thorough discussion of this.
– Kavi Rama Murthy
Jul 30 at 8:50
Thank you. It is a bit overcomplicated for me. Any easier source?
– TEX
Jul 30 at 8:56
1
Sorry to say that the expression for the probability in $mathbb R^k$ is a bit complicated and I cannot think of a better source.
– Kavi Rama Murthy
Jul 30 at 8:59
@KaviRamaMurthy Thank you, following the book I have written above the expression for $K=3$. Is it correct?
– TEX
Jul 30 at 9:20
1
I think you have got it right.
– Kavi Rama Murthy
Jul 30 at 9:23
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the k-variate random vector $Yequiv (Y_1, Y_2, ..., Y_K)$ with cumulative distribution function (cdf) $F$.
How can I express in terms of $F$ the following probability:
$$
mathbbP(a_kleq Y_kleq b_k text forall k)
$$
?
For example, if $K=2$, I know that
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2)= F(a_2, b_2)+F(a_1, b_1)- F(a_1, b_2)-F(a_2, b_1)
$$
(Sometimes this is also called rectangle formula)
Is there a way to generalise this to any $K>2$?
Thanks to the comment below, I now write the expression for $K=3$
1) I list all the $2^K=8$ vertices
$$
begincases
a_1,a_2, a_3\
b_1, a_2, a_3\
a_1, b_2, a_3 \
b_1, b_2, a_3\
a_1,a_2, b_3\
b_1, a_2, b_3\
a_1, b_2, b_3 \
b_1, b_2, b_3\
endcases
$$
2) Algebraic sum of cdf at each vertex with +1 if the number of $a$'s is even and -1 otherwise
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2text, a_3leq Y_3leq b_3)= -F(a_1,a_2, a_3)
+F(b_1, a_2, a_3)
+F(a_1, b_2, a_3)
-F(b_1, b_2, a_3)
+F(a_1,a_2, b_3)
-F(b_1, a_2, b_3)
-F(a_1, b_2, b_3)
+F(b_1, b_2, b_3)
$$
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 8:46
TEX
2419
Consider the k-variate random vector $Yequiv (Y_1, Y_2, ..., Y_K)$ with cumulative distribution function (cdf) $F$.
How can I express in terms of $F$ the following probability:
$$
mathbbP(a_kleq Y_kleq b_k text forall k)
$$
?
For example, if $K=2$, I know that
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2)= F(a_2, b_2)+F(a_1, b_1)- F(a_1, b_2)-F(a_2, b_1)
$$
(Sometimes this is also called rectangle formula)
Is there a way to generalise this to any $K>2$?
Thanks to the comment below, I now write the expression for $K=3$
1) I list all the $2^K=8$ vertices
$$
begincases
a_1,a_2, a_3\
b_1, a_2, a_3\
a_1, b_2, a_3 \
b_1, b_2, a_3\
a_1,a_2, b_3\
b_1, a_2, b_3\
a_1, b_2, b_3 \
b_1, b_2, b_3\
endcases
$$
2) Algebraic sum of cdf at each vertex with +1 if the number of $a$'s is even and -1 otherwise
$$
mathbbP(a_1leq Y_1leq b_1text, a_2leq Y_2leq b_2text, a_3leq Y_3leq b_3)= -F(a_1,a_2, a_3)
+F(b_1, a_2, a_3)
+F(a_1, b_2, a_3)
-F(b_1, b_2, a_3)
+F(a_1,a_2, b_3)
-F(b_1, a_2, b_3)
-F(a_1, b_2, b_3)
+F(b_1, b_2, b_3)
$$
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 8:46
TEX
2419
edited Jul 30 at 9:23
asked Jul 30 at 8:46
TEX
2419
asked Jul 30 at 8:46
TEX
2419
asked Jul 30 at 8:46
TEX
2419
2419
1
Billingsley's Probability and Measure, Section 12 has a thorough discussion of this.
– Kavi Rama Murthy
Jul 30 at 8:50
Thank you. It is a bit overcomplicated for me. Any easier source?
– TEX
Jul 30 at 8:56
1
Sorry to say that the expression for the probability in $mathbb R^k$ is a bit complicated and I cannot think of a better source.
– Kavi Rama Murthy
Jul 30 at 8:59
@KaviRamaMurthy Thank you, following the book I have written above the expression for $K=3$. Is it correct?
– TEX
Jul 30 at 9:20
1
I think you have got it right.
– Kavi Rama Murthy
Jul 30 at 9:23
 |Â
show 1 more comment
1
Billingsley's Probability and Measure, Section 12 has a thorough discussion of this.
– Kavi Rama Murthy
Jul 30 at 8:50
Thank you. It is a bit overcomplicated for me. Any easier source?
– TEX
Jul 30 at 8:56
1
Sorry to say that the expression for the probability in $mathbb R^k$ is a bit complicated and I cannot think of a better source.
– Kavi Rama Murthy
Jul 30 at 8:59
@KaviRamaMurthy Thank you, following the book I have written above the expression for $K=3$. Is it correct?
– TEX
Jul 30 at 9:20
1
I think you have got it right.
– Kavi Rama Murthy
Jul 30 at 9:23
1
1
Billingsley's Probability and Measure, Section 12 has a thorough discussion of this.
– Kavi Rama Murthy
Jul 30 at 8:50
Billingsley's Probability and Measure, Section 12 has a thorough discussion of this.
– Kavi Rama Murthy
Jul 30 at 8:50
Thank you. It is a bit overcomplicated for me. Any easier source?
– TEX
Jul 30 at 8:56
Thank you. It is a bit overcomplicated for me. Any easier source?
– TEX
Jul 30 at 8:56
1
1
Sorry to say that the expression for the probability in $mathbb R^k$ is a bit complicated and I cannot think of a better source.
– Kavi Rama Murthy
Jul 30 at 8:59
Sorry to say that the expression for the probability in $mathbb R^k$ is a bit complicated and I cannot think of a better source.
– Kavi Rama Murthy
Jul 30 at 8:59
@KaviRamaMurthy Thank you, following the book I have written above the expression for $K=3$. Is it correct?
– TEX
Jul 30 at 9:20
@KaviRamaMurthy Thank you, following the book I have written above the expression for $K=3$. Is it correct?
– TEX
Jul 30 at 9:20
1
1
I think you have got it right.
– Kavi Rama Murthy
Jul 30 at 9:23
I think you have got it right.
– Kavi Rama Murthy
Jul 30 at 9:23
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
In general $1_Acap B=1_Atimes 1_B$ and $1_(a_i,b_i]=1_(-infty,b_i]-1_(-infty,a_i]$.
This leads to:$$1_(a_1,b_1]timescdotstimes(a_n,b_n]=left(1_(-infty,b_1]-1_(-infty,a_1]right)timescdotstimesleft(1_(-infty,b_n]-1_(-infty,a_n]right)$$
Working out the RHS we get $2^n-1$ terms with sign $+$ and $2^n-1$ with sign $-$.
The sign is $+$ if and only if the number of $a_i$'s that appear in the term is even.
Taking the integral on both sides gives the equality:$$P(a_1<Y_1leq b_1,dots, a_n<Y_nleq b_n)=sum_(c_1,dots,c_n)inprod_i=1^na_i,b_i(-1)^e(c_1,dots,c_n)F(c_1,dots,c_n)$$where $e(c_1,dots,c_n)$ denotes the cardinality of the set $iin1,dots,nmid c_i=a_i$
answered Jul 30 at 9:47
drhab
85.9k540118
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
In general $1_Acap B=1_Atimes 1_B$ and $1_(a_i,b_i]=1_(-infty,b_i]-1_(-infty,a_i]$.
This leads to:$$1_(a_1,b_1]timescdotstimes(a_n,b_n]=left(1_(-infty,b_1]-1_(-infty,a_1]right)timescdotstimesleft(1_(-infty,b_n]-1_(-infty,a_n]right)$$
Working out the RHS we get $2^n-1$ terms with sign $+$ and $2^n-1$ with sign $-$.
The sign is $+$ if and only if the number of $a_i$'s that appear in the term is even.
Taking the integral on both sides gives the equality:$$P(a_1<Y_1leq b_1,dots, a_n<Y_nleq b_n)=sum_(c_1,dots,c_n)inprod_i=1^na_i,b_i(-1)^e(c_1,dots,c_n)F(c_1,dots,c_n)$$where $e(c_1,dots,c_n)$ denotes the cardinality of the set $iin1,dots,nmid c_i=a_i$
answered Jul 30 at 9:47
drhab
85.9k540118
add a comment |Â
up vote
1
down vote
accepted
In general $1_Acap B=1_Atimes 1_B$ and $1_(a_i,b_i]=1_(-infty,b_i]-1_(-infty,a_i]$.
This leads to:$$1_(a_1,b_1]timescdotstimes(a_n,b_n]=left(1_(-infty,b_1]-1_(-infty,a_1]right)timescdotstimesleft(1_(-infty,b_n]-1_(-infty,a_n]right)$$
Working out the RHS we get $2^n-1$ terms with sign $+$ and $2^n-1$ with sign $-$.
The sign is $+$ if and only if the number of $a_i$'s that appear in the term is even.
Taking the integral on both sides gives the equality:$$P(a_1<Y_1leq b_1,dots, a_n<Y_nleq b_n)=sum_(c_1,dots,c_n)inprod_i=1^na_i,b_i(-1)^e(c_1,dots,c_n)F(c_1,dots,c_n)$$where $e(c_1,dots,c_n)$ denotes the cardinality of the set $iin1,dots,nmid c_i=a_i$
answered Jul 30 at 9:47
drhab
85.9k540118
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
In general $1_Acap B=1_Atimes 1_B$ and $1_(a_i,b_i]=1_(-infty,b_i]-1_(-infty,a_i]$.
This leads to:$$1_(a_1,b_1]timescdotstimes(a_n,b_n]=left(1_(-infty,b_1]-1_(-infty,a_1]right)timescdotstimesleft(1_(-infty,b_n]-1_(-infty,a_n]right)$$
Working out the RHS we get $2^n-1$ terms with sign $+$ and $2^n-1$ with sign $-$.
The sign is $+$ if and only if the number of $a_i$'s that appear in the term is even.
Taking the integral on both sides gives the equality:$$P(a_1<Y_1leq b_1,dots, a_n<Y_nleq b_n)=sum_(c_1,dots,c_n)inprod_i=1^na_i,b_i(-1)^e(c_1,dots,c_n)F(c_1,dots,c_n)$$where $e(c_1,dots,c_n)$ denotes the cardinality of the set $iin1,dots,nmid c_i=a_i$
answered Jul 30 at 9:47
drhab
85.9k540118
In general $1_Acap B=1_Atimes 1_B$ and $1_(a_i,b_i]=1_(-infty,b_i]-1_(-infty,a_i]$.
This leads to:$$1_(a_1,b_1]timescdotstimes(a_n,b_n]=left(1_(-infty,b_1]-1_(-infty,a_1]right)timescdotstimesleft(1_(-infty,b_n]-1_(-infty,a_n]right)$$
Working out the RHS we get $2^n-1$ terms with sign $+$ and $2^n-1$ with sign $-$.
The sign is $+$ if and only if the number of $a_i$'s that appear in the term is even.
Taking the integral on both sides gives the equality:$$P(a_1<Y_1leq b_1,dots, a_n<Y_nleq b_n)=sum_(c_1,dots,c_n)inprod_i=1^na_i,b_i(-1)^e(c_1,dots,c_n)F(c_1,dots,c_n)$$where $e(c_1,dots,c_n)$ denotes the cardinality of the set $iin1,dots,nmid c_i=a_i$
answered Jul 30 at 9:47
drhab
85.9k540118
edited Jul 30 at 10:04
answered Jul 30 at 9:47
drhab
85.9k540118
answered Jul 30 at 9:47
drhab
85.9k540118
answered Jul 30 at 9:47
drhab
85.9k540118
85.9k540118
add a comment |Â
add a comment |Â
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1
Billingsley's Probability and Measure, Section 12 has a thorough discussion of this.
– Kavi Rama Murthy
Jul 30 at 8:50
Thank you. It is a bit overcomplicated for me. Any easier source?
– TEX
Jul 30 at 8:56
1
Sorry to say that the expression for the probability in $mathbb R^k$ is a bit complicated and I cannot think of a better source.
– Kavi Rama Murthy
Jul 30 at 8:59
@KaviRamaMurthy Thank you, following the book I have written above the expression for $K=3$. Is it correct?
– TEX
Jul 30 at 9:20
1
I think you have got it right.
– Kavi Rama Murthy
Jul 30 at 9:23