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Write some probabilities using a common cumulative distribution function
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Consider the random variables $Y,epsilon_1, epsilon_2$. $epsilonequiv (epsilon_1, epsilon_2)$ has cumulative distribution function (cdf) $F$.
Let $u_1, u_2$ be some known real-valued parameters.
Suppose $Y$ can take values $0,1,2$ with the following probabilities
$$
(diamond) text
begincases
mathbbP(Y=1)=mathbbP(epsilon_1geq -u_1, epsilon_1-epsilon_2geq u_2-u_1)\
mathbbP(Y=2)=mathbbP(epsilon_2geq -u_2, epsilon_1-epsilon_2leq u_2-u_1)\
mathbbP(Y=0)=mathbbP( epsilon_1leq -u_1, epsilon_2leq-u_2)\
endcases
$$
(I haven't done a precise distinction between weak and strict inequalities as I assume that $F$ is continuous)
To understand better I have pictured an example
with $u_1equiv 3$ and $u_2equiv -1$. $mathbbP(Y=1)$ is the probability of the pink area, $mathbbP(Y=2)$ is the probability of the green area, $mathbbP(Y=0)$ is the probability of the blue area.
Question: I want to re-write these probabilities using one joint cdf only (not necessarily $F$). Is there a way to do that?
My thoughts:
The tricky part is that
these probabilities depend on different random variables: $mathbbP(Y=1)$ on $epsilon_1, epsilon_1-epsilon_2$; $mathbbP(Y=2)$ on $epsilon_2, epsilon_1-epsilon_2$; $mathbbP(Y=0)$ on $epsilon_1, epsilon_2$
the cdf's of $beginpmatrix
epsilon_1\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_2\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_1\
epsilon_2
endpmatrix$ do not seem to me expressible as marginals of a common joint cdf
My idea is to define $$Vequiv beginpmatrix
epsilon_1\
epsilon_2\
epsilon_1-epsilon_2
endpmatrix
$$
with cdf $G$ and use $G$ as common cdf to re-write those probabilities. With the help of the picture above,
$$
mathbbP(Y=1)=
$$
$$mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2,epsilon_1-epsilon_2geq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1geq -u_1, epsilon_2leq -u_2 rightarrow epsilon_1-epsilon_2geq u_2-u_1$$
Similarly,
$$
mathbbP(Y=2)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1leq -u_1, epsilon_2geq -u_2 rightarrow epsilon_1-epsilon_2leq u_2-u_1$$
Lastly,
$$
mathbbP(Y=0)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq infty)
$$
All these probabilities can be re-written using $G$. Is this correct?
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 11:55
TEX
2419
 |Â
show 3 more comments
up vote
0
down vote
favorite
Consider the random variables $Y,epsilon_1, epsilon_2$. $epsilonequiv (epsilon_1, epsilon_2)$ has cumulative distribution function (cdf) $F$.
Let $u_1, u_2$ be some known real-valued parameters.
Suppose $Y$ can take values $0,1,2$ with the following probabilities
$$
(diamond) text
begincases
mathbbP(Y=1)=mathbbP(epsilon_1geq -u_1, epsilon_1-epsilon_2geq u_2-u_1)\
mathbbP(Y=2)=mathbbP(epsilon_2geq -u_2, epsilon_1-epsilon_2leq u_2-u_1)\
mathbbP(Y=0)=mathbbP( epsilon_1leq -u_1, epsilon_2leq-u_2)\
endcases
$$
(I haven't done a precise distinction between weak and strict inequalities as I assume that $F$ is continuous)
To understand better I have pictured an example with $u_1equiv 3$ and $u_2equiv -1$. $mathbbP(Y=1)$ is the probability of the pink area, $mathbbP(Y=2)$ is the probability of the green area, $mathbbP(Y=0)$ is the probability of the blue area.
Question: I want to re-write these probabilities using one joint cdf only (not necessarily $F$). Is there a way to do that?
My thoughts:
The tricky part is that
these probabilities depend on different random variables: $mathbbP(Y=1)$ on $epsilon_1, epsilon_1-epsilon_2$; $mathbbP(Y=2)$ on $epsilon_2, epsilon_1-epsilon_2$; $mathbbP(Y=0)$ on $epsilon_1, epsilon_2$
the cdf's of $beginpmatrix
epsilon_1\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_2\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_1\
epsilon_2
endpmatrix$ do not seem to me expressible as marginals of a common joint cdf
My idea is to define $$Vequiv beginpmatrix
epsilon_1\
epsilon_2\
epsilon_1-epsilon_2
endpmatrix
$$
with cdf $G$ and use $G$ as common cdf to re-write those probabilities. With the help of the picture above,
$$
mathbbP(Y=1)=
$$
$$mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2,epsilon_1-epsilon_2geq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1geq -u_1, epsilon_2leq -u_2 rightarrow epsilon_1-epsilon_2geq u_2-u_1$$
Similarly,
$$
mathbbP(Y=2)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1leq -u_1, epsilon_2geq -u_2 rightarrow epsilon_1-epsilon_2leq u_2-u_1$$
Lastly,
$$
mathbbP(Y=0)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq infty)
$$
All these probabilities can be re-written using $G$. Is this correct?
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 11:55
TEX
2419
At first sight: no. If e.g. $X,Y$ have a common distribution then there is no way to express the probability of sets like $X<Y$ or $X+Y<c$ by means of an expression in $F_X,Y$. We can write them as integrals $int[x<y]dF_X,Y(x,y)$ and $int[x+y<c]dF_X,Y(x,y)$ where $[textcondition on x,y]$ takes value $1$ if the condition is satisfied, and takes value $0$ otherwise. There it stops. Btw, CDF's are important as fully determining the distribution, but often not well-behaved as tools for the expression of probabilities. So don't bother too much.
– drhab
Jul 30 at 12:17
@drhab Thank you. I have tried to elaborate a bit more my thoughts. If you have time could you tell me where/why my derivations are wrong?
– TEX
Jul 30 at 12:53
Only the blue one.
– Did
Jul 30 at 13:51
@Did thanks, why the pink and green are wrong?
– TEX
Jul 30 at 14:01
Because the CDF only deals with corners such as the blue one, and combinations of those, and clearly the pink and green areas are not of this type.
– Did
Jul 30 at 14:03
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the random variables $Y,epsilon_1, epsilon_2$. $epsilonequiv (epsilon_1, epsilon_2)$ has cumulative distribution function (cdf) $F$.
Let $u_1, u_2$ be some known real-valued parameters.
Suppose $Y$ can take values $0,1,2$ with the following probabilities
$$
(diamond) text
begincases
mathbbP(Y=1)=mathbbP(epsilon_1geq -u_1, epsilon_1-epsilon_2geq u_2-u_1)\
mathbbP(Y=2)=mathbbP(epsilon_2geq -u_2, epsilon_1-epsilon_2leq u_2-u_1)\
mathbbP(Y=0)=mathbbP( epsilon_1leq -u_1, epsilon_2leq-u_2)\
endcases
$$
(I haven't done a precise distinction between weak and strict inequalities as I assume that $F$ is continuous)
To understand better I have pictured an example with $u_1equiv 3$ and $u_2equiv -1$. $mathbbP(Y=1)$ is the probability of the pink area, $mathbbP(Y=2)$ is the probability of the green area, $mathbbP(Y=0)$ is the probability of the blue area.
Question: I want to re-write these probabilities using one joint cdf only (not necessarily $F$). Is there a way to do that?
My thoughts:
The tricky part is that
these probabilities depend on different random variables: $mathbbP(Y=1)$ on $epsilon_1, epsilon_1-epsilon_2$; $mathbbP(Y=2)$ on $epsilon_2, epsilon_1-epsilon_2$; $mathbbP(Y=0)$ on $epsilon_1, epsilon_2$
the cdf's of $beginpmatrix
epsilon_1\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_2\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_1\
epsilon_2
endpmatrix$ do not seem to me expressible as marginals of a common joint cdf
My idea is to define $$Vequiv beginpmatrix
epsilon_1\
epsilon_2\
epsilon_1-epsilon_2
endpmatrix
$$
with cdf $G$ and use $G$ as common cdf to re-write those probabilities. With the help of the picture above,
$$
mathbbP(Y=1)=
$$
$$mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2,epsilon_1-epsilon_2geq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1geq -u_1, epsilon_2leq -u_2 rightarrow epsilon_1-epsilon_2geq u_2-u_1$$
Similarly,
$$
mathbbP(Y=2)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1leq -u_1, epsilon_2geq -u_2 rightarrow epsilon_1-epsilon_2leq u_2-u_1$$
Lastly,
$$
mathbbP(Y=0)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq infty)
$$
All these probabilities can be re-written using $G$. Is this correct?
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 11:55
TEX
2419
Consider the random variables $Y,epsilon_1, epsilon_2$. $epsilonequiv (epsilon_1, epsilon_2)$ has cumulative distribution function (cdf) $F$.
Let $u_1, u_2$ be some known real-valued parameters.
Suppose $Y$ can take values $0,1,2$ with the following probabilities
$$
(diamond) text
begincases
mathbbP(Y=1)=mathbbP(epsilon_1geq -u_1, epsilon_1-epsilon_2geq u_2-u_1)\
mathbbP(Y=2)=mathbbP(epsilon_2geq -u_2, epsilon_1-epsilon_2leq u_2-u_1)\
mathbbP(Y=0)=mathbbP( epsilon_1leq -u_1, epsilon_2leq-u_2)\
endcases
$$
(I haven't done a precise distinction between weak and strict inequalities as I assume that $F$ is continuous)
To understand better I have pictured an example with $u_1equiv 3$ and $u_2equiv -1$. $mathbbP(Y=1)$ is the probability of the pink area, $mathbbP(Y=2)$ is the probability of the green area, $mathbbP(Y=0)$ is the probability of the blue area.
Question: I want to re-write these probabilities using one joint cdf only (not necessarily $F$). Is there a way to do that?
My thoughts:
The tricky part is that
these probabilities depend on different random variables: $mathbbP(Y=1)$ on $epsilon_1, epsilon_1-epsilon_2$; $mathbbP(Y=2)$ on $epsilon_2, epsilon_1-epsilon_2$; $mathbbP(Y=0)$ on $epsilon_1, epsilon_2$
the cdf's of $beginpmatrix
epsilon_1\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_2\
epsilon_1-epsilon_2
endpmatrix$, $beginpmatrix
epsilon_1\
epsilon_2
endpmatrix$ do not seem to me expressible as marginals of a common joint cdf
My idea is to define $$Vequiv beginpmatrix
epsilon_1\
epsilon_2\
epsilon_1-epsilon_2
endpmatrix
$$
with cdf $G$ and use $G$ as common cdf to re-write those probabilities. With the help of the picture above,
$$
mathbbP(Y=1)=
$$
$$mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1geq -u_1, epsilon_2leq -u_2,epsilon_1-epsilon_2geq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1geq -u_1, epsilon_2leq -u_2 rightarrow epsilon_1-epsilon_2geq u_2-u_1$$
Similarly,
$$
mathbbP(Y=2)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP(epsilon_1geq -u_1, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)
$$
where the second equality comes from observing that $$epsilon_1leq -u_1, epsilon_2geq -u_2 rightarrow epsilon_1-epsilon_2leq u_2-u_1$$
Lastly,
$$
mathbbP(Y=0)=
$$
$$mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq u_2-u_1)+ mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2geq u_2-u_1)
$$
$$
=
mathbbP( epsilon_1leq -u_1, epsilon_2leq -u_2, epsilon_2geq -u_2,epsilon_1-epsilon_2leq infty)
$$
All these probabilities can be re-written using $G$. Is this correct?
probability probability-theory probability-distributions random-variables random
asked Jul 30 at 11:55
TEX
2419
edited Aug 2 at 8:47
asked Jul 30 at 11:55
TEX
2419
asked Jul 30 at 11:55
TEX
2419
asked Jul 30 at 11:55
TEX
2419
2419
At first sight: no. If e.g. $X,Y$ have a common distribution then there is no way to express the probability of sets like $X<Y$ or $X+Y<c$ by means of an expression in $F_X,Y$. We can write them as integrals $int[x<y]dF_X,Y(x,y)$ and $int[x+y<c]dF_X,Y(x,y)$ where $[textcondition on x,y]$ takes value $1$ if the condition is satisfied, and takes value $0$ otherwise. There it stops. Btw, CDF's are important as fully determining the distribution, but often not well-behaved as tools for the expression of probabilities. So don't bother too much.
– drhab
Jul 30 at 12:17
@drhab Thank you. I have tried to elaborate a bit more my thoughts. If you have time could you tell me where/why my derivations are wrong?
– TEX
Jul 30 at 12:53
Only the blue one.
– Did
Jul 30 at 13:51
@Did thanks, why the pink and green are wrong?
– TEX
Jul 30 at 14:01
Because the CDF only deals with corners such as the blue one, and combinations of those, and clearly the pink and green areas are not of this type.
– Did
Jul 30 at 14:03
 |Â
show 3 more comments
At first sight: no. If e.g. $X,Y$ have a common distribution then there is no way to express the probability of sets like $X<Y$ or $X+Y<c$ by means of an expression in $F_X,Y$. We can write them as integrals $int[x<y]dF_X,Y(x,y)$ and $int[x+y<c]dF_X,Y(x,y)$ where $[textcondition on x,y]$ takes value $1$ if the condition is satisfied, and takes value $0$ otherwise. There it stops. Btw, CDF's are important as fully determining the distribution, but often not well-behaved as tools for the expression of probabilities. So don't bother too much.
– drhab
Jul 30 at 12:17
@drhab Thank you. I have tried to elaborate a bit more my thoughts. If you have time could you tell me where/why my derivations are wrong?
– TEX
Jul 30 at 12:53
Only the blue one.
– Did
Jul 30 at 13:51
@Did thanks, why the pink and green are wrong?
– TEX
Jul 30 at 14:01
Because the CDF only deals with corners such as the blue one, and combinations of those, and clearly the pink and green areas are not of this type.
– Did
Jul 30 at 14:03
At first sight: no. If e.g. $X,Y$ have a common distribution then there is no way to express the probability of sets like $X<Y$ or $X+Y<c$ by means of an expression in $F_X,Y$. We can write them as integrals $int[x<y]dF_X,Y(x,y)$ and $int[x+y<c]dF_X,Y(x,y)$ where $[textcondition on x,y]$ takes value $1$ if the condition is satisfied, and takes value $0$ otherwise. There it stops. Btw, CDF's are important as fully determining the distribution, but often not well-behaved as tools for the expression of probabilities. So don't bother too much.
– drhab
Jul 30 at 12:17
At first sight: no. If e.g. $X,Y$ have a common distribution then there is no way to express the probability of sets like $X<Y$ or $X+Y<c$ by means of an expression in $F_X,Y$. We can write them as integrals $int[x<y]dF_X,Y(x,y)$ and $int[x+y<c]dF_X,Y(x,y)$ where $[textcondition on x,y]$ takes value $1$ if the condition is satisfied, and takes value $0$ otherwise. There it stops. Btw, CDF's are important as fully determining the distribution, but often not well-behaved as tools for the expression of probabilities. So don't bother too much.
– drhab
Jul 30 at 12:17
@drhab Thank you. I have tried to elaborate a bit more my thoughts. If you have time could you tell me where/why my derivations are wrong?
– TEX
Jul 30 at 12:53
@drhab Thank you. I have tried to elaborate a bit more my thoughts. If you have time could you tell me where/why my derivations are wrong?
– TEX
Jul 30 at 12:53
Only the blue one.
– Did
Jul 30 at 13:51
Only the blue one.
– Did
Jul 30 at 13:51
@Did thanks, why the pink and green are wrong?
– TEX
Jul 30 at 14:01
@Did thanks, why the pink and green are wrong?
– TEX
Jul 30 at 14:01
Because the CDF only deals with corners such as the blue one, and combinations of those, and clearly the pink and green areas are not of this type.
– Did
Jul 30 at 14:03
Because the CDF only deals with corners such as the blue one, and combinations of those, and clearly the pink and green areas are not of this type.
– Did
Jul 30 at 14:03
 |Â
show 3 more comments
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At first sight: no. If e.g. $X,Y$ have a common distribution then there is no way to express the probability of sets like $X<Y$ or $X+Y<c$ by means of an expression in $F_X,Y$. We can write them as integrals $int[x<y]dF_X,Y(x,y)$ and $int[x+y<c]dF_X,Y(x,y)$ where $[textcondition on x,y]$ takes value $1$ if the condition is satisfied, and takes value $0$ otherwise. There it stops. Btw, CDF's are important as fully determining the distribution, but often not well-behaved as tools for the expression of probabilities. So don't bother too much.
– drhab
Jul 30 at 12:17
@drhab Thank you. I have tried to elaborate a bit more my thoughts. If you have time could you tell me where/why my derivations are wrong?
– TEX
Jul 30 at 12:53
Only the blue one.
– Did
Jul 30 at 13:51
@Did thanks, why the pink and green are wrong?
– TEX
Jul 30 at 14:01
Because the CDF only deals with corners such as the blue one, and combinations of those, and clearly the pink and green areas are not of this type.
– Did
Jul 30 at 14:03