Mixing
What is the PDF of $X_1X_2-X_3X_4$ conditionally on $sumlimits_{k=1}^4X_k=1$, for $(X_k)$ i.i.d. uniform on...
$begingroup$
Consider the following expression
$$D=X-(X+Y)(X+Z) text{under the constraint} X+Y+Z leq 1 tag{1}$$
(being understood that $X, Y$ and $Z$ are $geq 0$).
This expression can also be written under the form :
$$D=XT-YZ=begin{vmatrix}X&Y\Z&Tend{vmatrix} text{given that} X+Y+Z+T=1.tag{2}$$
where $(X,Y,Z,T)$ take their values in $[0,1]^4$ ($[0,1]$ being the set of real values between $0$ and $1$).
(all these constraints are due to the fact that we deal with a complete system of events ; see "Origin of the question" below with definitions (4)).
My question : what is the pdf (probability density function) of random variable D ?
My work :
I have attempted different computations, but they have failed to be conclusive ; I hit on the difficulty to assign a distribution to the random vector $(X,Y,Z,T)$ taking into account constraint $X+Y+Z+T=1$. I have in particular attempted to replace uniformly distributed $X,Y,Z,T$ on $mathbb{R}$ by random variables taking discrete values ${0/n,1/n,cdots (n-1)/n,1}$ without success.
I have made simulations displaying a rather "smooth" symmetrical behavior (see Fig. below displaying an histogram for a large number of simulations) with a relative spike around value $d=0$.
The origin of this question : A recent question about extreme values of what could be called a "degree of dependence':
$$D=P(A cap B) - P(A)P(B) tag{3}$$
(that can be transformed into (1)) and the answer I gave : https://math.stackexchange.com/q/3090278
has triggered (as often is the case) a different question : instead of restricting our attention to the extreme values of (1) (or the equivalent expression (2)), I asked myself the more general question of the pdf of the values taken by (1) with the following definitions :
$$X=P(A cap B), Y=P(A cap B^c), Z=P(A^c cap B), T=P(A^c cap B^c).tag{4}$$
as one can see globally on the following diagram :
Besides, if one drops constraint $X+Y+Z+T=1$ (yielding a very different issue), what would become the pdf of $D$ ?(I give below a histogram gathering results of simulation with this hypothesis).
probability probability-distributions
edited Jan 28 at 19:00
Did
249k23226466
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
$endgroup$
|
show 1 more comment
$begingroup$
Consider the following expression
$$D=X-(X+Y)(X+Z) text{under the constraint} X+Y+Z leq 1 tag{1}$$
(being understood that $X, Y$ and $Z$ are $geq 0$).
This expression can also be written under the form :
$$D=XT-YZ=begin{vmatrix}X&Y\Z&Tend{vmatrix} text{given that} X+Y+Z+T=1.tag{2}$$
where $(X,Y,Z,T)$ take their values in $[0,1]^4$ ($[0,1]$ being the set of real values between $0$ and $1$).
(all these constraints are due to the fact that we deal with a complete system of events ; see "Origin of the question" below with definitions (4)).
My question : what is the pdf (probability density function) of random variable D ?
My work :
I have attempted different computations, but they have failed to be conclusive ; I hit on the difficulty to assign a distribution to the random vector $(X,Y,Z,T)$ taking into account constraint $X+Y+Z+T=1$. I have in particular attempted to replace uniformly distributed $X,Y,Z,T$ on $mathbb{R}$ by random variables taking discrete values ${0/n,1/n,cdots (n-1)/n,1}$ without success.
I have made simulations displaying a rather "smooth" symmetrical behavior (see Fig. below displaying an histogram for a large number of simulations) with a relative spike around value $d=0$.
The origin of this question : A recent question about extreme values of what could be called a "degree of dependence':
$$D=P(A cap B) - P(A)P(B) tag{3}$$
(that can be transformed into (1)) and the answer I gave : https://math.stackexchange.com/q/3090278
has triggered (as often is the case) a different question : instead of restricting our attention to the extreme values of (1) (or the equivalent expression (2)), I asked myself the more general question of the pdf of the values taken by (1) with the following definitions :
$$X=P(A cap B), Y=P(A cap B^c), Z=P(A^c cap B), T=P(A^c cap B^c).tag{4}$$
as one can see globally on the following diagram :
Besides, if one drops constraint $X+Y+Z+T=1$ (yielding a very different issue), what would become the pdf of $D$ ?(I give below a histogram gathering results of simulation with this hypothesis).
probability probability-distributions
edited Jan 28 at 19:00
Did
249k23226466
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
$endgroup$
1
$begingroup$
First, note that you can eliminate the condition by doing a few operations on the determinant to transform it into: $D=X-(X+Y)(X+Z)$. (Just add the first column to the last column and then the first row to the last row.) So that's already one less constraint.
$endgroup$
– Raskolnikov
Jan 28 at 16:47
1
$begingroup$
Then the condition $Dleq d$ determines a section of space bounded by a hyperboloid. Adding the conditions that $X,Y,Z in [0,1]$ means that you have to find the volume of that section of space within the unit cube to find the probability.
$endgroup$
– Raskolnikov
Jan 28 at 16:52
$begingroup$
@Raskolnikov In fact expression $D=X-(X+Y)(X+Z)=P(A cap B)-P(A)P(B)$ is the initial expression ... and you are right that, although less symmetrical, this expression avoids to use constraint (2).
$endgroup$
– Jean Marie
Jan 28 at 17:24
$begingroup$
@Raskolnikov ... but it will remain the constraint $X+Y+Z leq 1$.
$endgroup$
– Jean Marie
Jan 28 at 17:34
$begingroup$
@Raskolnikov I have slightly modified my first sentences in order to take into account your remarks.
$endgroup$
– Jean Marie
Jan 28 at 18:50
|
show 1 more comment
$begingroup$
Consider the following expression
$$D=X-(X+Y)(X+Z) text{under the constraint} X+Y+Z leq 1 tag{1}$$
(being understood that $X, Y$ and $Z$ are $geq 0$).
This expression can also be written under the form :
$$D=XT-YZ=begin{vmatrix}X&Y\Z&Tend{vmatrix} text{given that} X+Y+Z+T=1.tag{2}$$
where $(X,Y,Z,T)$ take their values in $[0,1]^4$ ($[0,1]$ being the set of real values between $0$ and $1$).
(all these constraints are due to the fact that we deal with a complete system of events ; see "Origin of the question" below with definitions (4)).
My question : what is the pdf (probability density function) of random variable D ?
My work :
I have attempted different computations, but they have failed to be conclusive ; I hit on the difficulty to assign a distribution to the random vector $(X,Y,Z,T)$ taking into account constraint $X+Y+Z+T=1$. I have in particular attempted to replace uniformly distributed $X,Y,Z,T$ on $mathbb{R}$ by random variables taking discrete values ${0/n,1/n,cdots (n-1)/n,1}$ without success.
I have made simulations displaying a rather "smooth" symmetrical behavior (see Fig. below displaying an histogram for a large number of simulations) with a relative spike around value $d=0$.
The origin of this question : A recent question about extreme values of what could be called a "degree of dependence':
$$D=P(A cap B) - P(A)P(B) tag{3}$$
(that can be transformed into (1)) and the answer I gave : https://math.stackexchange.com/q/3090278
has triggered (as often is the case) a different question : instead of restricting our attention to the extreme values of (1) (or the equivalent expression (2)), I asked myself the more general question of the pdf of the values taken by (1) with the following definitions :
$$X=P(A cap B), Y=P(A cap B^c), Z=P(A^c cap B), T=P(A^c cap B^c).tag{4}$$
as one can see globally on the following diagram :
Besides, if one drops constraint $X+Y+Z+T=1$ (yielding a very different issue), what would become the pdf of $D$ ?(I give below a histogram gathering results of simulation with this hypothesis).
probability probability-distributions
edited Jan 28 at 19:00
Did
249k23226466
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
$endgroup$
Consider the following expression
$$D=X-(X+Y)(X+Z) text{under the constraint} X+Y+Z leq 1 tag{1}$$
(being understood that $X, Y$ and $Z$ are $geq 0$).
This expression can also be written under the form :
$$D=XT-YZ=begin{vmatrix}X&Y\Z&Tend{vmatrix} text{given that} X+Y+Z+T=1.tag{2}$$
where $(X,Y,Z,T)$ take their values in $[0,1]^4$ ($[0,1]$ being the set of real values between $0$ and $1$).
(all these constraints are due to the fact that we deal with a complete system of events ; see "Origin of the question" below with definitions (4)).
My question : what is the pdf (probability density function) of random variable D ?
My work :
I have attempted different computations, but they have failed to be conclusive ; I hit on the difficulty to assign a distribution to the random vector $(X,Y,Z,T)$ taking into account constraint $X+Y+Z+T=1$. I have in particular attempted to replace uniformly distributed $X,Y,Z,T$ on $mathbb{R}$ by random variables taking discrete values ${0/n,1/n,cdots (n-1)/n,1}$ without success.
I have made simulations displaying a rather "smooth" symmetrical behavior (see Fig. below displaying an histogram for a large number of simulations) with a relative spike around value $d=0$.
The origin of this question : A recent question about extreme values of what could be called a "degree of dependence':
$$D=P(A cap B) - P(A)P(B) tag{3}$$
(that can be transformed into (1)) and the answer I gave : https://math.stackexchange.com/q/3090278
has triggered (as often is the case) a different question : instead of restricting our attention to the extreme values of (1) (or the equivalent expression (2)), I asked myself the more general question of the pdf of the values taken by (1) with the following definitions :
$$X=P(A cap B), Y=P(A cap B^c), Z=P(A^c cap B), T=P(A^c cap B^c).tag{4}$$
as one can see globally on the following diagram :
Besides, if one drops constraint $X+Y+Z+T=1$ (yielding a very different issue), what would become the pdf of $D$ ?(I give below a histogram gathering results of simulation with this hypothesis).
probability probability-distributions
probability probability-distributions
edited Jan 28 at 19:00
Did
249k23226466
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
edited Jan 28 at 19:00
Did
249k23226466
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
edited Jan 28 at 19:00
Did
249k23226466
edited Jan 28 at 19:00
Did
249k23226466
edited Jan 28 at 19:00
Did
249k23226466
249k23226466
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
asked Jan 28 at 10:41
Jean MarieJean Marie
31k42255
31k42255
1
$begingroup$
First, note that you can eliminate the condition by doing a few operations on the determinant to transform it into: $D=X-(X+Y)(X+Z)$. (Just add the first column to the last column and then the first row to the last row.) So that's already one less constraint.
$endgroup$
– Raskolnikov
Jan 28 at 16:47
1
$begingroup$
Then the condition $Dleq d$ determines a section of space bounded by a hyperboloid. Adding the conditions that $X,Y,Z in [0,1]$ means that you have to find the volume of that section of space within the unit cube to find the probability.
$endgroup$
– Raskolnikov
Jan 28 at 16:52
$begingroup$
@Raskolnikov In fact expression $D=X-(X+Y)(X+Z)=P(A cap B)-P(A)P(B)$ is the initial expression ... and you are right that, although less symmetrical, this expression avoids to use constraint (2).
$endgroup$
– Jean Marie
Jan 28 at 17:24
$begingroup$
@Raskolnikov ... but it will remain the constraint $X+Y+Z leq 1$.
$endgroup$
– Jean Marie
Jan 28 at 17:34
$begingroup$
@Raskolnikov I have slightly modified my first sentences in order to take into account your remarks.
$endgroup$
– Jean Marie
Jan 28 at 18:50
|
show 1 more comment
1
$begingroup$
First, note that you can eliminate the condition by doing a few operations on the determinant to transform it into: $D=X-(X+Y)(X+Z)$. (Just add the first column to the last column and then the first row to the last row.) So that's already one less constraint.
$endgroup$
– Raskolnikov
Jan 28 at 16:47
1
$begingroup$
Then the condition $Dleq d$ determines a section of space bounded by a hyperboloid. Adding the conditions that $X,Y,Z in [0,1]$ means that you have to find the volume of that section of space within the unit cube to find the probability.
$endgroup$
– Raskolnikov
Jan 28 at 16:52
$begingroup$
@Raskolnikov In fact expression $D=X-(X+Y)(X+Z)=P(A cap B)-P(A)P(B)$ is the initial expression ... and you are right that, although less symmetrical, this expression avoids to use constraint (2).
$endgroup$
– Jean Marie
Jan 28 at 17:24
$begingroup$
@Raskolnikov ... but it will remain the constraint $X+Y+Z leq 1$.
$endgroup$
– Jean Marie
Jan 28 at 17:34
$begingroup$
@Raskolnikov I have slightly modified my first sentences in order to take into account your remarks.
$endgroup$
– Jean Marie
Jan 28 at 18:50
1
1
$begingroup$
First, note that you can eliminate the condition by doing a few operations on the determinant to transform it into: $D=X-(X+Y)(X+Z)$. (Just add the first column to the last column and then the first row to the last row.) So that's already one less constraint.
$endgroup$
– Raskolnikov
Jan 28 at 16:47
$begingroup$
First, note that you can eliminate the condition by doing a few operations on the determinant to transform it into: $D=X-(X+Y)(X+Z)$. (Just add the first column to the last column and then the first row to the last row.) So that's already one less constraint.
$endgroup$
– Raskolnikov
Jan 28 at 16:47
1
1
$begingroup$
Then the condition $Dleq d$ determines a section of space bounded by a hyperboloid. Adding the conditions that $X,Y,Z in [0,1]$ means that you have to find the volume of that section of space within the unit cube to find the probability.
$endgroup$
– Raskolnikov
Jan 28 at 16:52
$begingroup$
Then the condition $Dleq d$ determines a section of space bounded by a hyperboloid. Adding the conditions that $X,Y,Z in [0,1]$ means that you have to find the volume of that section of space within the unit cube to find the probability.
$endgroup$
– Raskolnikov
Jan 28 at 16:52
$begingroup$
@Raskolnikov In fact expression $D=X-(X+Y)(X+Z)=P(A cap B)-P(A)P(B)$ is the initial expression ... and you are right that, although less symmetrical, this expression avoids to use constraint (2).
$endgroup$
– Jean Marie
Jan 28 at 17:24
$begingroup$
@Raskolnikov In fact expression $D=X-(X+Y)(X+Z)=P(A cap B)-P(A)P(B)$ is the initial expression ... and you are right that, although less symmetrical, this expression avoids to use constraint (2).
$endgroup$
– Jean Marie
Jan 28 at 17:24
$begingroup$
@Raskolnikov ... but it will remain the constraint $X+Y+Z leq 1$.
$endgroup$
– Jean Marie
Jan 28 at 17:34
$begingroup$
@Raskolnikov ... but it will remain the constraint $X+Y+Z leq 1$.
$endgroup$
– Jean Marie
Jan 28 at 17:34
$begingroup$
@Raskolnikov I have slightly modified my first sentences in order to take into account your remarks.
$endgroup$
– Jean Marie
Jan 28 at 18:50
$begingroup$
@Raskolnikov I have slightly modified my first sentences in order to take into account your remarks.
$endgroup$
– Jean Marie
Jan 28 at 18:50
|
show 1 more comment
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1
$begingroup$
First, note that you can eliminate the condition by doing a few operations on the determinant to transform it into: $D=X-(X+Y)(X+Z)$. (Just add the first column to the last column and then the first row to the last row.) So that's already one less constraint.
$endgroup$
– Raskolnikov
Jan 28 at 16:47
1
$begingroup$
Then the condition $Dleq d$ determines a section of space bounded by a hyperboloid. Adding the conditions that $X,Y,Z in [0,1]$ means that you have to find the volume of that section of space within the unit cube to find the probability.
$endgroup$
– Raskolnikov
Jan 28 at 16:52
$begingroup$
@Raskolnikov In fact expression $D=X-(X+Y)(X+Z)=P(A cap B)-P(A)P(B)$ is the initial expression ... and you are right that, although less symmetrical, this expression avoids to use constraint (2).
$endgroup$
– Jean Marie
Jan 28 at 17:24
$begingroup$
@Raskolnikov ... but it will remain the constraint $X+Y+Z leq 1$.
$endgroup$
– Jean Marie
Jan 28 at 17:34
$begingroup$
@Raskolnikov I have slightly modified my first sentences in order to take into account your remarks.
$endgroup$
– Jean Marie
Jan 28 at 18:50