What is $mathbb P{Wleq x,Yleq y}$ ?Is it $mathbb P{omega in Omega : X(omega )leq x,Y(omega )leq y}$?

5

$begingroup$

Let $(Omega ,mathbb F,mathbb P)$. I'm confuse about something : Let $X,Y:Omega to mathbb R$ r.v. What is $$mathbb P{Xleq x,Yleq y} ?tag{1}$$

Is it, $$mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$$
or $$mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y} ?tag{2}$$

I say the notation $$mathbb P{Xleq x,Yleq y}=mathbb P{{omega in Omega mid X(omega )leq x}cap {omega in Omega mid Y(omega )leq x}},$$

but in otherhand, I know that $mathbb P{Xin A,Yin B}$ is a product measure on $mathbb R^2$, so may be we also have a product measure on $Omega ^2$ ? I'm a bit confuse... Espacially that if $(Omega ',mathcal F',mathbb P')$, and $X:Omega to mathbb R$, $Y:Omega 'to mathbb R$, then $$mathbb Potimes mathbb P'{Xleq x, Yleq y}=mathbb Potimesmathbb P'{(omega ,omega ')in Omega times Omega 'mid X(omega )leq x, Y(omega ' )leq y}=mathbb P{Xleq x}mathbb P'{Yleq y},$$ which is a measure on $mathbb R^2$. That's why I would think that $(é)$ is true but not $(1)$. What do you think ?

probability measure-theory

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asked Jan 17 at 23:47

user623855user623855

1457

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5

$begingroup$

Let $(Omega ,mathbb F,mathbb P)$. I'm confuse about something : Let $X,Y:Omega to mathbb R$ r.v. What is $$mathbb P{Xleq x,Yleq y} ?tag{1}$$

Is it, $$mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$$
or $$mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y} ?tag{2}$$

I say the notation $$mathbb P{Xleq x,Yleq y}=mathbb P{{omega in Omega mid X(omega )leq x}cap {omega in Omega mid Y(omega )leq x}},$$

but in otherhand, I know that $mathbb P{Xin A,Yin B}$ is a product measure on $mathbb R^2$, so may be we also have a product measure on $Omega ^2$ ? I'm a bit confuse... Espacially that if $(Omega ',mathcal F',mathbb P')$, and $X:Omega to mathbb R$, $Y:Omega 'to mathbb R$, then $$mathbb Potimes mathbb P'{Xleq x, Yleq y}=mathbb Potimesmathbb P'{(omega ,omega ')in Omega times Omega 'mid X(omega )leq x, Y(omega ' )leq y}=mathbb P{Xleq x}mathbb P'{Yleq y},$$ which is a measure on $mathbb R^2$. That's why I would think that $(é)$ is true but not $(1)$. What do you think ?

probability measure-theory

share|cite|improve this question

asked Jan 17 at 23:47

user623855user623855

1457

$endgroup$

add a comment | 

5

5

5

$begingroup$

Let $(Omega ,mathbb F,mathbb P)$. I'm confuse about something : Let $X,Y:Omega to mathbb R$ r.v. What is $$mathbb P{Xleq x,Yleq y} ?tag{1}$$

Is it, $$mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$$
or $$mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y} ?tag{2}$$

I say the notation $$mathbb P{Xleq x,Yleq y}=mathbb P{{omega in Omega mid X(omega )leq x}cap {omega in Omega mid Y(omega )leq x}},$$

but in otherhand, I know that $mathbb P{Xin A,Yin B}$ is a product measure on $mathbb R^2$, so may be we also have a product measure on $Omega ^2$ ? I'm a bit confuse... Espacially that if $(Omega ',mathcal F',mathbb P')$, and $X:Omega to mathbb R$, $Y:Omega 'to mathbb R$, then $$mathbb Potimes mathbb P'{Xleq x, Yleq y}=mathbb Potimesmathbb P'{(omega ,omega ')in Omega times Omega 'mid X(omega )leq x, Y(omega ' )leq y}=mathbb P{Xleq x}mathbb P'{Yleq y},$$ which is a measure on $mathbb R^2$. That's why I would think that $(é)$ is true but not $(1)$. What do you think ?

probability measure-theory

share|cite|improve this question

asked Jan 17 at 23:47

user623855user623855

1457

$endgroup$

Let $(Omega ,mathbb F,mathbb P)$. I'm confuse about something : Let $X,Y:Omega to mathbb R$ r.v. What is $$mathbb P{Xleq x,Yleq y} ?tag{1}$$

Is it, $$mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$$
or $$mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y} ?tag{2}$$

I say the notation $$mathbb P{Xleq x,Yleq y}=mathbb P{{omega in Omega mid X(omega )leq x}cap {omega in Omega mid Y(omega )leq x}},$$

but in otherhand, I know that $mathbb P{Xin A,Yin B}$ is a product measure on $mathbb R^2$, so may be we also have a product measure on $Omega ^2$ ? I'm a bit confuse... Espacially that if $(Omega ',mathcal F',mathbb P')$, and $X:Omega to mathbb R$, $Y:Omega 'to mathbb R$, then $$mathbb Potimes mathbb P'{Xleq x, Yleq y}=mathbb Potimesmathbb P'{(omega ,omega ')in Omega times Omega 'mid X(omega )leq x, Y(omega ' )leq y}=mathbb P{Xleq x}mathbb P'{Yleq y},$$ which is a measure on $mathbb R^2$. That's why I would think that $(é)$ is true but not $(1)$. What do you think ?

probability measure-theory

probability measure-theory

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asked Jan 17 at 23:47

user623855user623855

1457

share|cite|improve this question

asked Jan 17 at 23:47

user623855user623855

1457

share|cite|improve this question

share|cite|improve this question

asked Jan 17 at 23:47

user623855user623855

1457

asked Jan 17 at 23:47

user623855user623855

1457

asked Jan 17 at 23:47

user623855user623855

1457

1457

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2 Answers
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5

$begingroup$

The correct interpretation is the first one: $mathbb P{Xle x,Yle y}=mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$. You say

I know that $P(Xin A,Yin B)$ is a product measure on $mathbb R^2$,

but this is untrue in general. This is only true when $X$ and $Y$ are independent.

The interpretation $mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y}$ cannot be correct, since $mathbb P(A)$ is defined for measurable subsets of $Omega$, not subsets of $Omega^2$.

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answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

$endgroup$

add a comment | 

3

$begingroup$

The product measure is in no way involved here. Product measure makes its appearance only when $X$ and $Y$ are independent. The correct expression is $P{omega in Omega: X(omega) leq x,Y(omega) leq y}$.

share|cite|improve this answer

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

$endgroup$

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

5

$begingroup$

The correct interpretation is the first one: $mathbb P{Xle x,Yle y}=mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$. You say

I know that $P(Xin A,Yin B)$ is a product measure on $mathbb R^2$,

but this is untrue in general. This is only true when $X$ and $Y$ are independent.

The interpretation $mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y}$ cannot be correct, since $mathbb P(A)$ is defined for measurable subsets of $Omega$, not subsets of $Omega^2$.

share|cite|improve this answer

answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

$endgroup$

add a comment | 

5

$begingroup$

The correct interpretation is the first one: $mathbb P{Xle x,Yle y}=mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$. You say

I know that $P(Xin A,Yin B)$ is a product measure on $mathbb R^2$,

but this is untrue in general. This is only true when $X$ and $Y$ are independent.

The interpretation $mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y}$ cannot be correct, since $mathbb P(A)$ is defined for measurable subsets of $Omega$, not subsets of $Omega^2$.

share|cite|improve this answer

answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

$endgroup$

add a comment | 

5

5

5

$begingroup$

The correct interpretation is the first one: $mathbb P{Xle x,Yle y}=mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$. You say

I know that $P(Xin A,Yin B)$ is a product measure on $mathbb R^2$,

but this is untrue in general. This is only true when $X$ and $Y$ are independent.

The interpretation $mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y}$ cannot be correct, since $mathbb P(A)$ is defined for measurable subsets of $Omega$, not subsets of $Omega^2$.

share|cite|improve this answer

answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

$endgroup$

The correct interpretation is the first one: $mathbb P{Xle x,Yle y}=mathbb P{omega in Omega mid X(omega )leq x,Y(omega )leq y}$. You say

I know that $P(Xin A,Yin B)$ is a product measure on $mathbb R^2$,

but this is untrue in general. This is only true when $X$ and $Y$ are independent.

The interpretation $mathbb P{(omega ,omega ')in Omega ^2mid X(omega )leq x,Y(omega ')leq y}$ cannot be correct, since $mathbb P(A)$ is defined for measurable subsets of $Omega$, not subsets of $Omega^2$.

share|cite|improve this answer

answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

share|cite|improve this answer

share|cite|improve this answer

answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

answered Jan 18 at 0:02

Mike EarnestMike Earnest

23.8k12051

23.8k12051

add a comment | 

add a comment | 

3

$begingroup$

The product measure is in no way involved here. Product measure makes its appearance only when $X$ and $Y$ are independent. The correct expression is $P{omega in Omega: X(omega) leq x,Y(omega) leq y}$.

share|cite|improve this answer

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

$endgroup$

add a comment | 

3

$begingroup$

The product measure is in no way involved here. Product measure makes its appearance only when $X$ and $Y$ are independent. The correct expression is $P{omega in Omega: X(omega) leq x,Y(omega) leq y}$.

share|cite|improve this answer

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

$endgroup$

add a comment | 

3

3

3

$begingroup$

The product measure is in no way involved here. Product measure makes its appearance only when $X$ and $Y$ are independent. The correct expression is $P{omega in Omega: X(omega) leq x,Y(omega) leq y}$.

share|cite|improve this answer

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

$endgroup$

The product measure is in no way involved here. Product measure makes its appearance only when $X$ and $Y$ are independent. The correct expression is $P{omega in Omega: X(omega) leq x,Y(omega) leq y}$.

share|cite|improve this answer

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

share|cite|improve this answer

share|cite|improve this answer

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

answered Jan 18 at 0:01

Kavi Rama MurthyKavi Rama Murthy

62.8k42262

62.8k42262

add a comment | 

add a comment | 

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