Mixing
Why stochastic dominance is “stochastic�
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I think the CDF is pretty much fixed, so the FOSD (first order stochastic dominance) is pretty much non-stochastic. Why does it have a "stochastic" in its name?
financial-economics decision-theory portfolio-theory
asked 4 hours ago
High GPA
1697
add a comment |Â
up vote
2
down vote
favorite
I think the CDF is pretty much fixed, so the FOSD (first order stochastic dominance) is pretty much non-stochastic. Why does it have a "stochastic" in its name?
financial-economics decision-theory portfolio-theory
asked 4 hours ago
High GPA
1697
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I think the CDF is pretty much fixed, so the FOSD (first order stochastic dominance) is pretty much non-stochastic. Why does it have a "stochastic" in its name?
financial-economics decision-theory portfolio-theory
asked 4 hours ago
High GPA
1697
I think the CDF is pretty much fixed, so the FOSD (first order stochastic dominance) is pretty much non-stochastic. Why does it have a "stochastic" in its name?
financial-economics decision-theory portfolio-theory
financial-economics decision-theory portfolio-theory
asked 4 hours ago
High GPA
1697
asked 4 hours ago
High GPA
1697
asked 4 hours ago
High GPA
1697
asked 4 hours ago
High GPA
1697
asked 4 hours ago
High GPA
1697
1697
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
In the below figure, CDF $F(cdot)$ is first-order stochastically dominated by $G(cdot)$. But $X_1$ and $X_2$ fall within the support of both distributions. So it would be possible to draw $X_1$ from $F$ and $X_2$ from $G$, or to draw $X_2$ from $F$ and $X_1$ from $G$.
More generally, if $X_G$ is a draw from $G$ and $X_F$ is a draw from $F$ then $X_F-X_G$ will sometimes be positive and sometimes negative. In this sense, the dominance is only stochastic: $G$ produces larger draws than $F$ on average, but not all of the time.
answered 2 hours ago
Ubiquitous♦
13.5k32363
Great explanation!
– High GPA
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In the below figure, CDF $F(cdot)$ is first-order stochastically dominated by $G(cdot)$. But $X_1$ and $X_2$ fall within the support of both distributions. So it would be possible to draw $X_1$ from $F$ and $X_2$ from $G$, or to draw $X_2$ from $F$ and $X_1$ from $G$.
More generally, if $X_G$ is a draw from $G$ and $X_F$ is a draw from $F$ then $X_F-X_G$ will sometimes be positive and sometimes negative. In this sense, the dominance is only stochastic: $G$ produces larger draws than $F$ on average, but not all of the time.
answered 2 hours ago
Ubiquitous♦
13.5k32363
Great explanation!
– High GPA
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
In the below figure, CDF $F(cdot)$ is first-order stochastically dominated by $G(cdot)$. But $X_1$ and $X_2$ fall within the support of both distributions. So it would be possible to draw $X_1$ from $F$ and $X_2$ from $G$, or to draw $X_2$ from $F$ and $X_1$ from $G$.
More generally, if $X_G$ is a draw from $G$ and $X_F$ is a draw from $F$ then $X_F-X_G$ will sometimes be positive and sometimes negative. In this sense, the dominance is only stochastic: $G$ produces larger draws than $F$ on average, but not all of the time.
answered 2 hours ago
Ubiquitous♦
13.5k32363
Great explanation!
– High GPA
2 hours ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In the below figure, CDF $F(cdot)$ is first-order stochastically dominated by $G(cdot)$. But $X_1$ and $X_2$ fall within the support of both distributions. So it would be possible to draw $X_1$ from $F$ and $X_2$ from $G$, or to draw $X_2$ from $F$ and $X_1$ from $G$.
More generally, if $X_G$ is a draw from $G$ and $X_F$ is a draw from $F$ then $X_F-X_G$ will sometimes be positive and sometimes negative. In this sense, the dominance is only stochastic: $G$ produces larger draws than $F$ on average, but not all of the time.
answered 2 hours ago
Ubiquitous♦
13.5k32363
In the below figure, CDF $F(cdot)$ is first-order stochastically dominated by $G(cdot)$. But $X_1$ and $X_2$ fall within the support of both distributions. So it would be possible to draw $X_1$ from $F$ and $X_2$ from $G$, or to draw $X_2$ from $F$ and $X_1$ from $G$.
More generally, if $X_G$ is a draw from $G$ and $X_F$ is a draw from $F$ then $X_F-X_G$ will sometimes be positive and sometimes negative. In this sense, the dominance is only stochastic: $G$ produces larger draws than $F$ on average, but not all of the time.
answered 2 hours ago
Ubiquitous♦
13.5k32363
answered 2 hours ago
Ubiquitous♦
13.5k32363
answered 2 hours ago
Ubiquitous♦
13.5k32363
answered 2 hours ago
Ubiquitous♦
13.5k32363
13.5k32363
Great explanation!
– High GPA
2 hours ago
add a comment |Â
Great explanation!
– High GPA
2 hours ago
Great explanation!
– High GPA
2 hours ago
Great explanation!
– High GPA
2 hours ago
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2feconomics.stackexchange.com%2fquestions%2f25204%2fwhy-stochastic-dominance-is-stochastic%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
