|
3
|
|
edited May 5 2012 at 21:29
|
In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.
All distance-like functions seem to agree in being
- nonnegative: $d(x,y) \geq 0$
- symmetric: $d(x,y) = d(y,x)$
- fulfilling $d(x,x)=0$
—
– let me call such functions proto-metrics — – but differ in other respects:
a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.
an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$
a dissimilarity function — – conceived as some kind of distance, e.g. in a property space — – is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.
One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing – without any restriction – two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.
Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:
Which conditions does a function $F$ have to
fulfill such that
$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$
can be seen as a reasonable triangle inequality,
thus making the proto-metric $d$
distance-like?
Some conditions on $F$ seem to spring to mind:
- symmetry: $F(x,y)=F(y,x)$
- monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
- scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
- associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
- $F(x,0)=x$ (Thanks to David Feldman)
Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?
|
|
|
|
2
|
|
edited May 5 2012 at 21:19
|
In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.
All distance-like functions seem to agree in being
- nonnegative: $d(x,y) \geq 0$
- symmetric: $d(x,y) = d(y,x)$
- fulfilling $d(x,x)=0$
— let me call such functions proto-metrics — but differ in other respects:
a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.
an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$
a dissimilarity function — conceived as some kind of distance, e.g. in a property space — is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality at all.
One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.
Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:
Which conditions does a function $F$ have to
fulfill such that
$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$
can be seen as a reasonable triangle inequality,
thus making the proto-metric $d$
distance-like?
Some conditions on $F$ seem to spring to mind:
- symmetry: $F(x,y)=F(y,x)$
- monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
- scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
- associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
- $F(x,0)=x$ (Thanks to David Feldman)
Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?
|
|
|
|
1
|
|
asked May 5 2012 at 20:54
|
What makes a distance?
In the answers to my previous question, I learned that there are different concepts of distance, that is of distance-like functions with the usual metric being only the most popular and important one.
All distance-like functions seem to agree in being
- nonnegative: $d(x,y) \geq 0$
- symmetric: $d(x,y) = d(y,x)$
- fulfilling $d(x,x)=0$
— let me call such functions proto-metrics — but differ in other respects:
a pseudo-metric fulfills the usual triangle inequality, but allows $d(x,y)=0$ for $x \neq y$.
an ultra-metric has $d(x,y)=0$ only for $x = y$, but fulfills a stronger triangle inequality, namely $d(x,z) \leq \text{max}(d(x,y),d(y,z))$
a dissimilarity function — conceived as some kind of distance, e.g. in a property space — is just any proto-metric, i.e. fulfills nothing but the core conditions above, esp. no triangle inequality
One might ask whether such a dissimilarity function deserves its name, because it violates a strong intuition on dissimilarity (resp. distance) by allowing two objects which are very similar (= close) to a third object to be arbitrarily dissimilar (= far away) from each other.
Given that a distance-like function has to fulfill some kind of triangle inequality, my question boils down to:
Which conditions does a function $F$ have to
fulfill such that
$$d(x,z) \leq F(d(x,y),d(y,z))\qquad(*)$$
can be seen as a reasonable triangle inequality,
thus making the proto-metric $d$
distance-like?
Some conditions on $F$ seem to spring to mind:
- symmetry: $F(x,y)=F(y,x)$
- monotonicity: $F(x,y) \leq F(x',y)$ for $x\leq x'$
- scale-invariance: $\alpha F(x,y) \leq F(\alpha x, \alpha y)$ for $\alpha > 0$
- associativity: $F(x,F(y,z)) = F(F(x,y),z)$ (Did I understand this correctly, Will Sawin? Where does this intution come from?)
- $F(x,0)=x$ (Thanks to David Feldman)
Is there hope to fix a set of conditions $\mathcal{C}$, such that one will be willing to accept any proto-metric $d$ to be distance-like if it fulfills $(*)$ for an $F$ that fulfills $\mathcal{C}$?
|
|
|