Extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore the probability $mathbb{P}$
$begingroup$
We have a probability space ($Omega, mathcal{F}, mathbb{P}$), and we denote $mathcal{V}$ the vector space of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P})$.
$mathcal{V}$ has some nice properties. For example it's an inner-product space since : $(X,Y) mapsto mathbb{E}[XY]$ is an inner product. Hence it gives a geometric intuition about variance and expectation.
The main problem with this vector space $mathcal{V}$ is that it completely ignores the probability $mathbb{P}$. Hence if we have an other probability $mathbb{P}'$ and we consider the vecor space $mathcal{V}'$ of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P}')$ then we have $mathcal{V} = mathcal{V}'$.
That's why I am wondering if it's possible to add extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore the probability $mathbb{P}$ ? For example is it possible with an extra structure (let's suppose it's a function $B : mathcal{V}^2 to mathbb{R}$) such that :($pi$ is the canonical map that send a random variable to it's vector representation in $mathcal{V}$)
$$forall v, v' in mathcal{V}, B(v,v') = 0 Leftrightarrowforall A_1, A_2 in mathbb{R}, mathbb{P}(pi^{-1}(v) in A_1, pi^{-1}(v') in A_2) = mathbb{P}(pi^{-1}(v) in A_1) mathbb{P}(pi^{-1}(v') in A_2)$$
?
Hence the function $B$ traduces the notion of independent vectors.
Some thoughts :
I know that it's easy to defined a probability on $mathcal{V}$ simply : $mathbb{P}(v = k) = mathbb{P}(pi^{-1}(v) = k)$. But this is not what I want. I am really looking for an extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore $mathbb{P}$ but without using the canonical map $pi$ (for me in this case we are not adding structure to $mathcal{V}$). For example expectation can be easily traduced on $mathcal{V}$ without using $pi$. This extra structure on $mathcal{V}$ transform $mathcal{V}$ in an inner product space. I am looking for a similar result but for the probability $mathbb{P}$.
We know that : $X, Y$ are independent random variables $Rightarrow mathbb{E}[XY] = mathbb{E}[X]mathbb{E}[Y]$. What is nice with this is that it can be easily traduce on the inner product space $mathcal{V}$. So maybe there is a geometric intuition in this case. We can find many answers here.
I am sorry if my question is strange/unclear.
Thank you.
linear-algebra probability abstract-algebra vector-spaces random-variables
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
$endgroup$
add a comment |
$begingroup$
We have a probability space ($Omega, mathcal{F}, mathbb{P}$), and we denote $mathcal{V}$ the vector space of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P})$.
$mathcal{V}$ has some nice properties. For example it's an inner-product space since : $(X,Y) mapsto mathbb{E}[XY]$ is an inner product. Hence it gives a geometric intuition about variance and expectation.
The main problem with this vector space $mathcal{V}$ is that it completely ignores the probability $mathbb{P}$. Hence if we have an other probability $mathbb{P}'$ and we consider the vecor space $mathcal{V}'$ of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P}')$ then we have $mathcal{V} = mathcal{V}'$.
That's why I am wondering if it's possible to add extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore the probability $mathbb{P}$ ? For example is it possible with an extra structure (let's suppose it's a function $B : mathcal{V}^2 to mathbb{R}$) such that :($pi$ is the canonical map that send a random variable to it's vector representation in $mathcal{V}$)
$$forall v, v' in mathcal{V}, B(v,v') = 0 Leftrightarrowforall A_1, A_2 in mathbb{R}, mathbb{P}(pi^{-1}(v) in A_1, pi^{-1}(v') in A_2) = mathbb{P}(pi^{-1}(v) in A_1) mathbb{P}(pi^{-1}(v') in A_2)$$
?
Hence the function $B$ traduces the notion of independent vectors.
Some thoughts :
I know that it's easy to defined a probability on $mathcal{V}$ simply : $mathbb{P}(v = k) = mathbb{P}(pi^{-1}(v) = k)$. But this is not what I want. I am really looking for an extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore $mathbb{P}$ but without using the canonical map $pi$ (for me in this case we are not adding structure to $mathcal{V}$). For example expectation can be easily traduced on $mathcal{V}$ without using $pi$. This extra structure on $mathcal{V}$ transform $mathcal{V}$ in an inner product space. I am looking for a similar result but for the probability $mathbb{P}$.
We know that : $X, Y$ are independent random variables $Rightarrow mathbb{E}[XY] = mathbb{E}[X]mathbb{E}[Y]$. What is nice with this is that it can be easily traduce on the inner product space $mathcal{V}$. So maybe there is a geometric intuition in this case. We can find many answers here.
I am sorry if my question is strange/unclear.
Thank you.
linear-algebra probability abstract-algebra vector-spaces random-variables
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
$endgroup$
1
$begingroup$
That inner product isn't defined $mathcal{V}$, it's only defined on $L^2(Omega, mathbb{P}) subset $mathcal{V}$. Note that the expectation of any $L^2$-random variable already determines $mathbb{P}$ (consider indicator functions).
$endgroup$
– Rhys Steele
Dec 21 '18 at 9:54
1
$begingroup$
$EXY$ is not defined for arbitrary random variables $X$ and $Y$. If you restrict to square integrable random variables you get $L^{2}$ space and this space does not ignore $P$. If you restrict yourself to a vector space of jointly normal random variables the $B(X,Y)=cov (X,Y)$ meets your requirements.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 9:55
$begingroup$
@KaviRamaMurthy Thank you. I completely missed the fact that of course we need the assumption "square integrable" in order to get an inner product space. And what is nice is that as Rhys Steele said is that since the function $mathbb{1}_{omega}$ are in $L^2$, the inner product directly gives the probability. So the probability law can be thought as the length of the indicator functions of a single event. And you are right $cov(X,Y)$ completely meets my requirement. Thank you !
$endgroup$
– Thinking
Dec 21 '18 at 10:06
add a comment |
$begingroup$
We have a probability space ($Omega, mathcal{F}, mathbb{P}$), and we denote $mathcal{V}$ the vector space of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P})$.
$mathcal{V}$ has some nice properties. For example it's an inner-product space since : $(X,Y) mapsto mathbb{E}[XY]$ is an inner product. Hence it gives a geometric intuition about variance and expectation.
The main problem with this vector space $mathcal{V}$ is that it completely ignores the probability $mathbb{P}$. Hence if we have an other probability $mathbb{P}'$ and we consider the vecor space $mathcal{V}'$ of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P}')$ then we have $mathcal{V} = mathcal{V}'$.
That's why I am wondering if it's possible to add extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore the probability $mathbb{P}$ ? For example is it possible with an extra structure (let's suppose it's a function $B : mathcal{V}^2 to mathbb{R}$) such that :($pi$ is the canonical map that send a random variable to it's vector representation in $mathcal{V}$)
$$forall v, v' in mathcal{V}, B(v,v') = 0 Leftrightarrowforall A_1, A_2 in mathbb{R}, mathbb{P}(pi^{-1}(v) in A_1, pi^{-1}(v') in A_2) = mathbb{P}(pi^{-1}(v) in A_1) mathbb{P}(pi^{-1}(v') in A_2)$$
?
Hence the function $B$ traduces the notion of independent vectors.
Some thoughts :
I know that it's easy to defined a probability on $mathcal{V}$ simply : $mathbb{P}(v = k) = mathbb{P}(pi^{-1}(v) = k)$. But this is not what I want. I am really looking for an extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore $mathbb{P}$ but without using the canonical map $pi$ (for me in this case we are not adding structure to $mathcal{V}$). For example expectation can be easily traduced on $mathcal{V}$ without using $pi$. This extra structure on $mathcal{V}$ transform $mathcal{V}$ in an inner product space. I am looking for a similar result but for the probability $mathbb{P}$.
We know that : $X, Y$ are independent random variables $Rightarrow mathbb{E}[XY] = mathbb{E}[X]mathbb{E}[Y]$. What is nice with this is that it can be easily traduce on the inner product space $mathcal{V}$. So maybe there is a geometric intuition in this case. We can find many answers here.
I am sorry if my question is strange/unclear.
Thank you.
linear-algebra probability abstract-algebra vector-spaces random-variables
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
$endgroup$
We have a probability space ($Omega, mathcal{F}, mathbb{P}$), and we denote $mathcal{V}$ the vector space of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P})$.
$mathcal{V}$ has some nice properties. For example it's an inner-product space since : $(X,Y) mapsto mathbb{E}[XY]$ is an inner product. Hence it gives a geometric intuition about variance and expectation.
The main problem with this vector space $mathcal{V}$ is that it completely ignores the probability $mathbb{P}$. Hence if we have an other probability $mathbb{P}'$ and we consider the vecor space $mathcal{V}'$ of all real-valued random variables defined on $(Omega, mathcal{F}, mathbb{P}')$ then we have $mathcal{V} = mathcal{V}'$.
That's why I am wondering if it's possible to add extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore the probability $mathbb{P}$ ? For example is it possible with an extra structure (let's suppose it's a function $B : mathcal{V}^2 to mathbb{R}$) such that :($pi$ is the canonical map that send a random variable to it's vector representation in $mathcal{V}$)
$$forall v, v' in mathcal{V}, B(v,v') = 0 Leftrightarrowforall A_1, A_2 in mathbb{R}, mathbb{P}(pi^{-1}(v) in A_1, pi^{-1}(v') in A_2) = mathbb{P}(pi^{-1}(v) in A_1) mathbb{P}(pi^{-1}(v') in A_2)$$
?
Hence the function $B$ traduces the notion of independent vectors.
Some thoughts :
I know that it's easy to defined a probability on $mathcal{V}$ simply : $mathbb{P}(v = k) = mathbb{P}(pi^{-1}(v) = k)$. But this is not what I want. I am really looking for an extra structure on $mathcal{V}$ so that $mathcal{V}$ doesn't ignore $mathbb{P}$ but without using the canonical map $pi$ (for me in this case we are not adding structure to $mathcal{V}$). For example expectation can be easily traduced on $mathcal{V}$ without using $pi$. This extra structure on $mathcal{V}$ transform $mathcal{V}$ in an inner product space. I am looking for a similar result but for the probability $mathbb{P}$.
We know that : $X, Y$ are independent random variables $Rightarrow mathbb{E}[XY] = mathbb{E}[X]mathbb{E}[Y]$. What is nice with this is that it can be easily traduce on the inner product space $mathcal{V}$. So maybe there is a geometric intuition in this case. We can find many answers here.
I am sorry if my question is strange/unclear.
Thank you.
linear-algebra probability abstract-algebra vector-spaces random-variables
linear-algebra probability abstract-algebra vector-spaces random-variables
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
asked Dec 21 '18 at 9:47
ThinkingThinking
1,07416
1,07416
1
$begingroup$
That inner product isn't defined $mathcal{V}$, it's only defined on $L^2(Omega, mathbb{P}) subset $mathcal{V}$. Note that the expectation of any $L^2$-random variable already determines $mathbb{P}$ (consider indicator functions).
$endgroup$
– Rhys Steele
Dec 21 '18 at 9:54
1
$begingroup$
$EXY$ is not defined for arbitrary random variables $X$ and $Y$. If you restrict to square integrable random variables you get $L^{2}$ space and this space does not ignore $P$. If you restrict yourself to a vector space of jointly normal random variables the $B(X,Y)=cov (X,Y)$ meets your requirements.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 9:55
$begingroup$
@KaviRamaMurthy Thank you. I completely missed the fact that of course we need the assumption "square integrable" in order to get an inner product space. And what is nice is that as Rhys Steele said is that since the function $mathbb{1}_{omega}$ are in $L^2$, the inner product directly gives the probability. So the probability law can be thought as the length of the indicator functions of a single event. And you are right $cov(X,Y)$ completely meets my requirement. Thank you !
$endgroup$
– Thinking
Dec 21 '18 at 10:06
add a comment |
1
$begingroup$
That inner product isn't defined $mathcal{V}$, it's only defined on $L^2(Omega, mathbb{P}) subset $mathcal{V}$. Note that the expectation of any $L^2$-random variable already determines $mathbb{P}$ (consider indicator functions).
$endgroup$
– Rhys Steele
Dec 21 '18 at 9:54
1
$begingroup$
$EXY$ is not defined for arbitrary random variables $X$ and $Y$. If you restrict to square integrable random variables you get $L^{2}$ space and this space does not ignore $P$. If you restrict yourself to a vector space of jointly normal random variables the $B(X,Y)=cov (X,Y)$ meets your requirements.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 9:55
$begingroup$
@KaviRamaMurthy Thank you. I completely missed the fact that of course we need the assumption "square integrable" in order to get an inner product space. And what is nice is that as Rhys Steele said is that since the function $mathbb{1}_{omega}$ are in $L^2$, the inner product directly gives the probability. So the probability law can be thought as the length of the indicator functions of a single event. And you are right $cov(X,Y)$ completely meets my requirement. Thank you !
$endgroup$
– Thinking
Dec 21 '18 at 10:06
1
1
$begingroup$
That inner product isn't defined $mathcal{V}$, it's only defined on $L^2(Omega, mathbb{P}) subset $mathcal{V}$. Note that the expectation of any $L^2$-random variable already determines $mathbb{P}$ (consider indicator functions).
$endgroup$
– Rhys Steele
Dec 21 '18 at 9:54
$begingroup$
That inner product isn't defined $mathcal{V}$, it's only defined on $L^2(Omega, mathbb{P}) subset $mathcal{V}$. Note that the expectation of any $L^2$-random variable already determines $mathbb{P}$ (consider indicator functions).
$endgroup$
– Rhys Steele
Dec 21 '18 at 9:54
1
1
$begingroup$
$EXY$ is not defined for arbitrary random variables $X$ and $Y$. If you restrict to square integrable random variables you get $L^{2}$ space and this space does not ignore $P$. If you restrict yourself to a vector space of jointly normal random variables the $B(X,Y)=cov (X,Y)$ meets your requirements.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 9:55
$begingroup$
$EXY$ is not defined for arbitrary random variables $X$ and $Y$. If you restrict to square integrable random variables you get $L^{2}$ space and this space does not ignore $P$. If you restrict yourself to a vector space of jointly normal random variables the $B(X,Y)=cov (X,Y)$ meets your requirements.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 9:55
$begingroup$
@KaviRamaMurthy Thank you. I completely missed the fact that of course we need the assumption "square integrable" in order to get an inner product space. And what is nice is that as Rhys Steele said is that since the function $mathbb{1}_{omega}$ are in $L^2$, the inner product directly gives the probability. So the probability law can be thought as the length of the indicator functions of a single event. And you are right $cov(X,Y)$ completely meets my requirement. Thank you !
$endgroup$
– Thinking
Dec 21 '18 at 10:06
$begingroup$
@KaviRamaMurthy Thank you. I completely missed the fact that of course we need the assumption "square integrable" in order to get an inner product space. And what is nice is that as Rhys Steele said is that since the function $mathbb{1}_{omega}$ are in $L^2$, the inner product directly gives the probability. So the probability law can be thought as the length of the indicator functions of a single event. And you are right $cov(X,Y)$ completely meets my requirement. Thank you !
$endgroup$
– Thinking
Dec 21 '18 at 10:06
add a comment |
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$begingroup$
That inner product isn't defined $mathcal{V}$, it's only defined on $L^2(Omega, mathbb{P}) subset $mathcal{V}$. Note that the expectation of any $L^2$-random variable already determines $mathbb{P}$ (consider indicator functions).
$endgroup$
– Rhys Steele
Dec 21 '18 at 9:54
1
$begingroup$
$EXY$ is not defined for arbitrary random variables $X$ and $Y$. If you restrict to square integrable random variables you get $L^{2}$ space and this space does not ignore $P$. If you restrict yourself to a vector space of jointly normal random variables the $B(X,Y)=cov (X,Y)$ meets your requirements.
$endgroup$
– Kavi Rama Murthy
Dec 21 '18 at 9:55
$begingroup$
@KaviRamaMurthy Thank you. I completely missed the fact that of course we need the assumption "square integrable" in order to get an inner product space. And what is nice is that as Rhys Steele said is that since the function $mathbb{1}_{omega}$ are in $L^2$, the inner product directly gives the probability. So the probability law can be thought as the length of the indicator functions of a single event. And you are right $cov(X,Y)$ completely meets my requirement. Thank you !
$endgroup$
– Thinking
Dec 21 '18 at 10:06