Mixing
X and Y random variables. How can I interpret $sqrt{E(Y-X)^{2}}$ as the distance between X and Y?
$begingroup$
X and Y random variables. How can I interpret $sqrt{E(Y-X)^{2}}$ as the distance between X and Y by showing (prooving) that i) and ii) below works:
if $sqrt{E(Y-X)^{2}}$ = 0 then $P(Y=X)=1$
The triangle inequality works: $sqrt{E(Y-X)^{2}} leq sqrt{E(Y^{2})} +sqrt{E(X^{2})}$
So, I have to use i) and ii) to show that $sqrt{E(Y-X)^{2}}$ can be interpretated as the distance between X and Y.
I am completely lost here.
Any help?
probability probability-theory probability-distributions
asked Jan 23 at 1:39
LauraLaura
3368
$endgroup$
add a comment |
$begingroup$
X and Y random variables. How can I interpret $sqrt{E(Y-X)^{2}}$ as the distance between X and Y by showing (prooving) that i) and ii) below works:
if $sqrt{E(Y-X)^{2}}$ = 0 then $P(Y=X)=1$
The triangle inequality works: $sqrt{E(Y-X)^{2}} leq sqrt{E(Y^{2})} +sqrt{E(X^{2})}$
So, I have to use i) and ii) to show that $sqrt{E(Y-X)^{2}}$ can be interpretated as the distance between X and Y.
I am completely lost here.
Any help?
probability probability-theory probability-distributions
asked Jan 23 at 1:39
LauraLaura
3368
$endgroup$
3
$begingroup$
You just need to show the facts listed above. I guess you haven't learned about distance, so this might help: en.wikipedia.org/wiki/Metric_(mathematics)#Definition
$endgroup$
– user549397
Jan 23 at 1:44
add a comment |
$begingroup$
X and Y random variables. How can I interpret $sqrt{E(Y-X)^{2}}$ as the distance between X and Y by showing (prooving) that i) and ii) below works:
if $sqrt{E(Y-X)^{2}}$ = 0 then $P(Y=X)=1$
The triangle inequality works: $sqrt{E(Y-X)^{2}} leq sqrt{E(Y^{2})} +sqrt{E(X^{2})}$
So, I have to use i) and ii) to show that $sqrt{E(Y-X)^{2}}$ can be interpretated as the distance between X and Y.
I am completely lost here.
Any help?
probability probability-theory probability-distributions
asked Jan 23 at 1:39
LauraLaura
3368
$endgroup$
X and Y random variables. How can I interpret $sqrt{E(Y-X)^{2}}$ as the distance between X and Y by showing (prooving) that i) and ii) below works:
if $sqrt{E(Y-X)^{2}}$ = 0 then $P(Y=X)=1$
The triangle inequality works: $sqrt{E(Y-X)^{2}} leq sqrt{E(Y^{2})} +sqrt{E(X^{2})}$
So, I have to use i) and ii) to show that $sqrt{E(Y-X)^{2}}$ can be interpretated as the distance between X and Y.
I am completely lost here.
Any help?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Jan 23 at 1:39
LauraLaura
3368
asked Jan 23 at 1:39
LauraLaura
3368
asked Jan 23 at 1:39
LauraLaura
3368
asked Jan 23 at 1:39
LauraLaura
3368
asked Jan 23 at 1:39
LauraLaura
3368
3368
3
$begingroup$
You just need to show the facts listed above. I guess you haven't learned about distance, so this might help: en.wikipedia.org/wiki/Metric_(mathematics)#Definition
$endgroup$
– user549397
Jan 23 at 1:44
add a comment |
3
$begingroup$
You just need to show the facts listed above. I guess you haven't learned about distance, so this might help: en.wikipedia.org/wiki/Metric_(mathematics)#Definition
$endgroup$
– user549397
Jan 23 at 1:44
3
3
$begingroup$
You just need to show the facts listed above. I guess you haven't learned about distance, so this might help: en.wikipedia.org/wiki/Metric_(mathematics)#Definition
$endgroup$
– user549397
Jan 23 at 1:44
$begingroup$
You just need to show the facts listed above. I guess you haven't learned about distance, so this might help: en.wikipedia.org/wiki/Metric_(mathematics)#Definition
$endgroup$
– user549397
Jan 23 at 1:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A distance function satisfies
$d(x,y)=0$ implies $x=y$
$d(x,y) leq d(x,c)+d(c,y)$.
Here $d(x,y)=sqrt{mathbb{E}(X-Y)^2}$ and $c=0$. In words, if the distance function is zero, the points are the same (distance from a point to itself is zero, right?). The triangle inequality just means that the distance from $x$ to $y$ is less than the sum of the distances from $x$ to $c$ and the distance from $c$ to $y$ where $c$ is some intermediate point. They ask you to prove $sqrt{mathbb{E}(cdot)}$ satisfies these two properties, so that it can be interpreted as a distance function.
They seemed to have omitted symmetry though...
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
$begingroup$
So simple. Many many thanks!
$endgroup$
– Laura
Jan 23 at 1:55
$begingroup$
One comment: I could have problem sugesting that $c=0$? There will be a number c more general?
$endgroup$
– Laura
Jan 23 at 1:57
1
$begingroup$
Indeed. They only ask you to prove property 2) for the specific case $c=0$. In general, for this specific distance function (there are many!) you would show instead $sqrt{mathbb{E}(X-Y)^2}leq sqrt{mathbb{E}(X-Z)^2}+sqrt{mathbb{E}(Z-Y)^2}$.
$endgroup$
– LoveTooNap29
Jan 23 at 1:58
1
$begingroup$
Note, what I write here above and in my comment—is not a proof—you still need to show the two properties hold to finish the exercise. I was just trying to elucidate it what it meant to for a function $d(x,y)$ to be a distance function, since that is what seemed unfamiliar to you.
$endgroup$
– LoveTooNap29
Jan 23 at 2:01
$begingroup$
Yes, yes @LoveTooNap29 I understand. My question was not clear, but I understand your point.
$endgroup$
– Laura
Jan 23 at 2:06
add a comment |
Your Answer
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1 Answer
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active
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1 Answer
1
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oldest
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$begingroup$
A distance function satisfies
$d(x,y)=0$ implies $x=y$
$d(x,y) leq d(x,c)+d(c,y)$.
Here $d(x,y)=sqrt{mathbb{E}(X-Y)^2}$ and $c=0$. In words, if the distance function is zero, the points are the same (distance from a point to itself is zero, right?). The triangle inequality just means that the distance from $x$ to $y$ is less than the sum of the distances from $x$ to $c$ and the distance from $c$ to $y$ where $c$ is some intermediate point. They ask you to prove $sqrt{mathbb{E}(cdot)}$ satisfies these two properties, so that it can be interpreted as a distance function.
They seemed to have omitted symmetry though...
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
$begingroup$
So simple. Many many thanks!
$endgroup$
– Laura
Jan 23 at 1:55
$begingroup$
One comment: I could have problem sugesting that $c=0$? There will be a number c more general?
$endgroup$
– Laura
Jan 23 at 1:57
1
$begingroup$
Indeed. They only ask you to prove property 2) for the specific case $c=0$. In general, for this specific distance function (there are many!) you would show instead $sqrt{mathbb{E}(X-Y)^2}leq sqrt{mathbb{E}(X-Z)^2}+sqrt{mathbb{E}(Z-Y)^2}$.
$endgroup$
– LoveTooNap29
Jan 23 at 1:58
1
$begingroup$
Note, what I write here above and in my comment—is not a proof—you still need to show the two properties hold to finish the exercise. I was just trying to elucidate it what it meant to for a function $d(x,y)$ to be a distance function, since that is what seemed unfamiliar to you.
$endgroup$
– LoveTooNap29
Jan 23 at 2:01
$begingroup$
Yes, yes @LoveTooNap29 I understand. My question was not clear, but I understand your point.
$endgroup$
– Laura
Jan 23 at 2:06
add a comment |
$begingroup$
A distance function satisfies
$d(x,y)=0$ implies $x=y$
$d(x,y) leq d(x,c)+d(c,y)$.
Here $d(x,y)=sqrt{mathbb{E}(X-Y)^2}$ and $c=0$. In words, if the distance function is zero, the points are the same (distance from a point to itself is zero, right?). The triangle inequality just means that the distance from $x$ to $y$ is less than the sum of the distances from $x$ to $c$ and the distance from $c$ to $y$ where $c$ is some intermediate point. They ask you to prove $sqrt{mathbb{E}(cdot)}$ satisfies these two properties, so that it can be interpreted as a distance function.
They seemed to have omitted symmetry though...
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
$begingroup$
So simple. Many many thanks!
$endgroup$
– Laura
Jan 23 at 1:55
$begingroup$
One comment: I could have problem sugesting that $c=0$? There will be a number c more general?
$endgroup$
– Laura
Jan 23 at 1:57
1
$begingroup$
Indeed. They only ask you to prove property 2) for the specific case $c=0$. In general, for this specific distance function (there are many!) you would show instead $sqrt{mathbb{E}(X-Y)^2}leq sqrt{mathbb{E}(X-Z)^2}+sqrt{mathbb{E}(Z-Y)^2}$.
$endgroup$
– LoveTooNap29
Jan 23 at 1:58
1
$begingroup$
Note, what I write here above and in my comment—is not a proof—you still need to show the two properties hold to finish the exercise. I was just trying to elucidate it what it meant to for a function $d(x,y)$ to be a distance function, since that is what seemed unfamiliar to you.
$endgroup$
– LoveTooNap29
Jan 23 at 2:01
$begingroup$
Yes, yes @LoveTooNap29 I understand. My question was not clear, but I understand your point.
$endgroup$
– Laura
Jan 23 at 2:06
add a comment |
$begingroup$
A distance function satisfies
$d(x,y)=0$ implies $x=y$
$d(x,y) leq d(x,c)+d(c,y)$.
Here $d(x,y)=sqrt{mathbb{E}(X-Y)^2}$ and $c=0$. In words, if the distance function is zero, the points are the same (distance from a point to itself is zero, right?). The triangle inequality just means that the distance from $x$ to $y$ is less than the sum of the distances from $x$ to $c$ and the distance from $c$ to $y$ where $c$ is some intermediate point. They ask you to prove $sqrt{mathbb{E}(cdot)}$ satisfies these two properties, so that it can be interpreted as a distance function.
They seemed to have omitted symmetry though...
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
$endgroup$
A distance function satisfies
$d(x,y)=0$ implies $x=y$
$d(x,y) leq d(x,c)+d(c,y)$.
Here $d(x,y)=sqrt{mathbb{E}(X-Y)^2}$ and $c=0$. In words, if the distance function is zero, the points are the same (distance from a point to itself is zero, right?). The triangle inequality just means that the distance from $x$ to $y$ is less than the sum of the distances from $x$ to $c$ and the distance from $c$ to $y$ where $c$ is some intermediate point. They ask you to prove $sqrt{mathbb{E}(cdot)}$ satisfies these two properties, so that it can be interpreted as a distance function.
They seemed to have omitted symmetry though...
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
edited Jan 23 at 1:53
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
answered Jan 23 at 1:47
LoveTooNap29LoveTooNap29
1,1331614
1,1331614
$begingroup$
So simple. Many many thanks!
$endgroup$
– Laura
Jan 23 at 1:55
$begingroup$
One comment: I could have problem sugesting that $c=0$? There will be a number c more general?
$endgroup$
– Laura
Jan 23 at 1:57
1
$begingroup$
Indeed. They only ask you to prove property 2) for the specific case $c=0$. In general, for this specific distance function (there are many!) you would show instead $sqrt{mathbb{E}(X-Y)^2}leq sqrt{mathbb{E}(X-Z)^2}+sqrt{mathbb{E}(Z-Y)^2}$.
$endgroup$
– LoveTooNap29
Jan 23 at 1:58
1
$begingroup$
Note, what I write here above and in my comment—is not a proof—you still need to show the two properties hold to finish the exercise. I was just trying to elucidate it what it meant to for a function $d(x,y)$ to be a distance function, since that is what seemed unfamiliar to you.
$endgroup$
– LoveTooNap29
Jan 23 at 2:01
$begingroup$
Yes, yes @LoveTooNap29 I understand. My question was not clear, but I understand your point.
$endgroup$
– Laura
Jan 23 at 2:06
add a comment |
$begingroup$
So simple. Many many thanks!
$endgroup$
– Laura
Jan 23 at 1:55
$begingroup$
One comment: I could have problem sugesting that $c=0$? There will be a number c more general?
$endgroup$
– Laura
Jan 23 at 1:57
1
$begingroup$
Indeed. They only ask you to prove property 2) for the specific case $c=0$. In general, for this specific distance function (there are many!) you would show instead $sqrt{mathbb{E}(X-Y)^2}leq sqrt{mathbb{E}(X-Z)^2}+sqrt{mathbb{E}(Z-Y)^2}$.
$endgroup$
– LoveTooNap29
Jan 23 at 1:58
1
$begingroup$
Note, what I write here above and in my comment—is not a proof—you still need to show the two properties hold to finish the exercise. I was just trying to elucidate it what it meant to for a function $d(x,y)$ to be a distance function, since that is what seemed unfamiliar to you.
$endgroup$
– LoveTooNap29
Jan 23 at 2:01
$begingroup$
Yes, yes @LoveTooNap29 I understand. My question was not clear, but I understand your point.
$endgroup$
– Laura
Jan 23 at 2:06
$begingroup$
So simple. Many many thanks!
$endgroup$
– Laura
Jan 23 at 1:55
$begingroup$
So simple. Many many thanks!
$endgroup$
– Laura
Jan 23 at 1:55
$begingroup$
One comment: I could have problem sugesting that $c=0$? There will be a number c more general?
$endgroup$
– Laura
Jan 23 at 1:57
$begingroup$
One comment: I could have problem sugesting that $c=0$? There will be a number c more general?
$endgroup$
– Laura
Jan 23 at 1:57
1
1
$begingroup$
Indeed. They only ask you to prove property 2) for the specific case $c=0$. In general, for this specific distance function (there are many!) you would show instead $sqrt{mathbb{E}(X-Y)^2}leq sqrt{mathbb{E}(X-Z)^2}+sqrt{mathbb{E}(Z-Y)^2}$.
$endgroup$
– LoveTooNap29
Jan 23 at 1:58
$begingroup$
Indeed. They only ask you to prove property 2) for the specific case $c=0$. In general, for this specific distance function (there are many!) you would show instead $sqrt{mathbb{E}(X-Y)^2}leq sqrt{mathbb{E}(X-Z)^2}+sqrt{mathbb{E}(Z-Y)^2}$.
$endgroup$
– LoveTooNap29
Jan 23 at 1:58
1
1
$begingroup$
Note, what I write here above and in my comment—is not a proof—you still need to show the two properties hold to finish the exercise. I was just trying to elucidate it what it meant to for a function $d(x,y)$ to be a distance function, since that is what seemed unfamiliar to you.
$endgroup$
– LoveTooNap29
Jan 23 at 2:01
$begingroup$
Note, what I write here above and in my comment—is not a proof—you still need to show the two properties hold to finish the exercise. I was just trying to elucidate it what it meant to for a function $d(x,y)$ to be a distance function, since that is what seemed unfamiliar to you.
$endgroup$
– LoveTooNap29
Jan 23 at 2:01
$begingroup$
Yes, yes @LoveTooNap29 I understand. My question was not clear, but I understand your point.
$endgroup$
– Laura
Jan 23 at 2:06
$begingroup$
Yes, yes @LoveTooNap29 I understand. My question was not clear, but I understand your point.
$endgroup$
– Laura
Jan 23 at 2:06
add a comment |
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$begingroup$
You just need to show the facts listed above. I guess you haven't learned about distance, so this might help: en.wikipedia.org/wiki/Metric_(mathematics)#Definition
$endgroup$
– user549397
Jan 23 at 1:44