Return to Question

ASSUMPTION 1: there exists a continuous random vector $(X,Y,Z)$ such that $$ \begin{cases} p_1=\Pr(X\geq 0, Z\geq 0)\\ p_2=\Pr(Y\geq 0, Z< 0)\\ p_3=\Pr(X< 0, Y<0)\\ \end{cases} $$ where $(p_1,p_2,p_3)\in [0,1]^3$ and $p_1+p_2+p_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.

Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p_1,p_2,p_3)$. This is why here I assume that it exists.

QUESTION: Does Assumption 1 imply that we construct from $(X,Y,Z)$ a continuous random vector $(W,H,Q)$ such that:

1. it holds that $$ \begin{cases} \Pr(W\geq 0, Q\geq 0)=p_1\\ \Pr(H\geq 0, Q< 0)= p_2\\ \Pr(W< 0, H<0)= p_3\\ \end{cases} $$

2. the marginal distribution of each of $W,H,Q$ are symmetric around 0.

3. $Q=W-H$.

Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.

SOME DISCUSSION ON THE MOTIVATION BEHIND THE QUESTION: I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints 1-3. However, constraint 3 is computationally intractable to implement because infinite-dimensional. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.

ATTEMPTED ANSWER (with questions):

Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.

Define $(W,H,Q)$ as follows:

• For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$

• For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$

• Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p_3$. Hence, we will need to have: $-2X, -Y$ for $W$; $-X, -2X$ for $H$; $-X, Y$ for $Q$. I'm not sure we can always do this, though. Can we?

ASSUMPTION 1: there exists a continuous random vector $(X,Y,Z)$ such that $$ \begin{cases} p_1=\Pr(X\geq 0, Z\geq 0)\\ p_2=\Pr(Y\geq 0, Z< 0)\\ p_3=\Pr(X< 0, Y<0)\\ \end{cases} $$ where $(p_1,p_2,p_3)\in [0,1]^3$ and $p_1+p_2+p_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.

Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p_1,p_2,p_3)$. This is why here I assume that it exists.

QUESTION: Does Assumption 1 imply that there exists (i.e., we can construct from $(X,Y,Z)$) a continuous random vector $(W,H,Q)$ such that:

1. it holds that $$ \begin{cases} \Pr(W\geq 0, Q\geq 0)=p_1\\ \Pr(H\geq 0, Q< 0)= p_2\\ \Pr(W< 0, H<0)= p_3\\ \end{cases} $$

2. the marginal distribution of each of $W,H,Q$ are symmetric around 0.

3. $Q=W-H$.

Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.

SOME DISCUSSION ON THE MOTIVATION BEHIND THE QUESTION: I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints 1-3. However, constraint 3 is computationally intractable to implement because infinite-dimensional. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.

ATTEMPTED ANSWER (with questions):

Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.

Define $(W,H,Q)$ as follows:

• For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$

• For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$

• Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p_3$. Hence, we will need to have: $-2X, -Y$ for $W$; $-X, -2X$ for $H$; $-X, Y$ for $Q$. I'm not sure we can always do this, though. Can we?

ASSUMPTION 1: there exists a continuous random vector $(X,Y,Z)$ such that $$ \begin{cases} p_1=\Pr(X\geq 0, Z\geq 0)\\ p_2=\Pr(Y\geq 0, Z< 0)\\ p_3=\Pr(X< 0, Y<0)\\ \end{cases} $$ where $(p_1,p_2,p_3)\in [0,1]^3$ and $p_1+p_2+p_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.

Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p_1,p_2,p_3)$. This is why here I assume that it exists.

QUESTION: Does Assumption 1 imply that we construct from $(X,Y,Z)$ a continuous random vector $(W,H,Q)$ such that:

1. it holds that $$ \begin{cases} \Pr(W\geq 0, Q\geq 0)=p_1\\ \Pr(H\geq 0, Q< 0)= p_2\\ \Pr(W< 0, H<0)= p_3\\ \end{cases} $$

2. the marginal distribution of each of $W,H,Q$ are symmetric around 0.

3. $Q=W-H$.

Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.

SOME DISCUSSION ON THE MOTIVATION BEHIND THE QUESTION: I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints 1-3. However, constraint 3 is computationally intractable to implement because infinite-dimensional. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.

ATTEMPTED ANSWER (with questions):

Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.

Define $(W,H,Q)$ as follows:

• For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$

• For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$

• Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p_3$. Hence, we will need to have: $-2X, -Y$ for $W$; $-X, -2X$ for $H$; $X, Y$ for $Q$. I'm not sure we can always do this, though. Can we?

ASSUMPTION 1: there exists a continuous random vector $(X,Y,Z)$ such that $$ \begin{cases} p_1=\Pr(X\geq 0, Z\geq 0)\\ p_2=\Pr(Y\geq 0, Z< 0)\\ p_3=\Pr(X< 0, Y<0)\\ \end{cases} $$ where $(p_1,p_2,p_3)\in [0,1]^3$ and $p_1+p_2+p_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.

Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p_1,p_2,p_3)$. This is why here I assume that it exists.

QUESTION: Does Assumption 1 imply that we construct from $(X,Y,Z)$ a continuous random vector $(W,H,Q)$ such that:

1. it holds that $$ \begin{cases} \Pr(W\geq 0, Q\geq 0)=p_1\\ \Pr(H\geq 0, Q< 0)= p_2\\ \Pr(W< 0, H<0)= p_3\\ \end{cases} $$

2. the marginal distribution of each of $W,H,Q$ are symmetric around 0.

3. $Q=W-H$.

Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.

SOME DISCUSSION ON THE MOTIVATION BEHIND THE QUESTION: I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints 1-3. However, constraint 3 is computationally intractable to implement because infinite-dimensional. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.

ATTEMPTED ANSWER (with questions):

Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.

Define $(W,H,Q)$ as follows:

• For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$

• For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$

• Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p_3$. Hence, we will need to have: $-2X, -Y$ for $W$; $-X, -2X$ for $H$; $-X, Y$ for $Q$. I'm not sure we can always do this, though. Can we?

ASSUMPTION 1: there exists a continuous random vector $(X,Y,Z)$ such that $$ \begin{cases} p_1=\Pr(X\geq 0, Z\geq 0)\\ p_2=\Pr(Y\geq 0, Z< 0)\\ p_3=\Pr(X< 0, Y<0)\\ \end{cases} $$ where $(p_1,p_2,p_3)\in [0,1]^3$ and $p_1+p_2+p_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.

Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p_1,p_2,p_3)$. This is why here I assume that it exists.

QUESTION: Does Assumption 1 imply that we construct from $(X,Y,Z)$ a continuous random vector $(W,H,Q)$ such that:

1. it holds that $$ \begin{cases} \Pr(W\geq 0, Q\geq 0)=p_1\\ \Pr(H\geq 0, Q< 0)= p_2\\ \Pr(W< 0, H<0)= p_3\\ \end{cases} $$

2. the marginal distribution of each of $W,H,Q$ are symmetric around 0.

3. $Q=W-H$.

Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.

SOME DISCUSSION ON THE MOTIVATION BEHIND THE QUESTION: I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints 1-3. However, constraint 3 is computationally intractable to implement because infinite-dimensional. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.

ATTEMPTED ANSWER (with questions):

Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.

Define $(W,H,Q)$ as follows:

• For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$

• For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$

• Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p_3$. I'm not sure we can always do this, though. Can we?

ASSUMPTION 1: there exists a continuous random vector $(X,Y,Z)$ such that $$ \begin{cases} p_1=\Pr(X\geq 0, Z\geq 0)\\ p_2=\Pr(Y\geq 0, Z< 0)\\ p_3=\Pr(X< 0, Y<0)\\ \end{cases} $$ where $(p_1,p_2,p_3)\in [0,1]^3$ and $p_1+p_2+p_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.

Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p_1,p_2,p_3)$. This is why here I assume that it exists.

QUESTION: Does Assumption 1 imply that we construct from $(X,Y,Z)$ a continuous random vector $(W,H,Q)$ such that:

1. it holds that $$ \begin{cases} \Pr(W\geq 0, Q\geq 0)=p_1\\ \Pr(H\geq 0, Q< 0)= p_2\\ \Pr(W< 0, H<0)= p_3\\ \end{cases} $$

2. the marginal distribution of each of $W,H,Q$ are symmetric around 0.

3. $Q=W-H$.

Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.

SOME DISCUSSION ON THE MOTIVATION BEHIND THE QUESTION: I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints 1-3. However, constraint 3 is computationally intractable to implement because infinite-dimensional. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.

ATTEMPTED ANSWER (with questions):

Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.

Define $(W,H,Q)$ as follows:

• For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$

• For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:

$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$

• Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p_3$. Hence, we will need to have: $-2X, -Y$ for $W$; $-X, -2X$ for $H$; $X, Y$ for $Q$. I'm not sure we can always do this, though. Can we?