Distribution of $(langle X_i,X_jrangle)_{i,j=1}^n$ for $X_ksimoperatorname{Unif}(S^d)$
$begingroup$
For two independent random variables distributed uniformly on a $d$-sphere surface ($X_1,X_2simoperatorname{Unif}(S^d)$), it is obvious that
$$langle X_1,X_2ranglesim -langle X_1,X_2rangle$$
by symmetry of $X_1$ and linearity of the inner product. Can this notion be generalized to the joint distribution of the pairwise scalar combination of $n$ such variables by using symmetries in a clever way? Namely, does
$$((1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^nsim (-(1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^n$$
hold? By rotational invarince, one can derive the density of each such entry rather quick, but that does not seem helpful as the dependence/interaction is rather involved.
probability-distributions random-graphs random-matrices
$endgroup$
add a comment |
$begingroup$
For two independent random variables distributed uniformly on a $d$-sphere surface ($X_1,X_2simoperatorname{Unif}(S^d)$), it is obvious that
$$langle X_1,X_2ranglesim -langle X_1,X_2rangle$$
by symmetry of $X_1$ and linearity of the inner product. Can this notion be generalized to the joint distribution of the pairwise scalar combination of $n$ such variables by using symmetries in a clever way? Namely, does
$$((1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^nsim (-(1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^n$$
hold? By rotational invarince, one can derive the density of each such entry rather quick, but that does not seem helpful as the dependence/interaction is rather involved.
probability-distributions random-graphs random-matrices
$endgroup$
$begingroup$
What's the connection to random graphs? Is there one?
$endgroup$
– Misha Lavrov
Jan 15 at 14:45
$begingroup$
I have not stated the detailed relation to random graphs since it's not crucial for the question asked (and i didn't want to include unnecessary information), although some may recognize a connection. However, i came across this issue when facing a random graph problem, where random adjacency matrices arise by applying indicators on the entries above (symmetry and a vanishing diagonal are already provided).
$endgroup$
– Fiff
Jan 16 at 9:08
$begingroup$
That's interesting. If you place an edge $ij$ whenever $langle X_i, X_jrangle ge alpha$ for some threshold $alpha$, then you get a random geometric graph in the $d$-sphere, for instance.
$endgroup$
– Misha Lavrov
Jan 16 at 14:30
$begingroup$
That is exactly the context in which i am studying these type of matrices. I suspected that the total variation distance between the random geometric graph on the $d$-sphere und the unstructured Erdős–Rényi graph might be symmetric around single-edge probability $p=frac{1}{2}$ and the above would have shown that. But apparently, this approach is just a dead end.
$endgroup$
– Fiff
Jan 16 at 21:03
add a comment |
$begingroup$
For two independent random variables distributed uniformly on a $d$-sphere surface ($X_1,X_2simoperatorname{Unif}(S^d)$), it is obvious that
$$langle X_1,X_2ranglesim -langle X_1,X_2rangle$$
by symmetry of $X_1$ and linearity of the inner product. Can this notion be generalized to the joint distribution of the pairwise scalar combination of $n$ such variables by using symmetries in a clever way? Namely, does
$$((1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^nsim (-(1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^n$$
hold? By rotational invarince, one can derive the density of each such entry rather quick, but that does not seem helpful as the dependence/interaction is rather involved.
probability-distributions random-graphs random-matrices
$endgroup$
For two independent random variables distributed uniformly on a $d$-sphere surface ($X_1,X_2simoperatorname{Unif}(S^d)$), it is obvious that
$$langle X_1,X_2ranglesim -langle X_1,X_2rangle$$
by symmetry of $X_1$ and linearity of the inner product. Can this notion be generalized to the joint distribution of the pairwise scalar combination of $n$ such variables by using symmetries in a clever way? Namely, does
$$((1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^nsim (-(1-delta_{ij})langle X_i,X_jrangle)_{i,j=1}^n$$
hold? By rotational invarince, one can derive the density of each such entry rather quick, but that does not seem helpful as the dependence/interaction is rather involved.
probability-distributions random-graphs random-matrices
probability-distributions random-graphs random-matrices
asked Jan 13 at 16:11
FiffFiff
204
204
$begingroup$
What's the connection to random graphs? Is there one?
$endgroup$
– Misha Lavrov
Jan 15 at 14:45
$begingroup$
I have not stated the detailed relation to random graphs since it's not crucial for the question asked (and i didn't want to include unnecessary information), although some may recognize a connection. However, i came across this issue when facing a random graph problem, where random adjacency matrices arise by applying indicators on the entries above (symmetry and a vanishing diagonal are already provided).
$endgroup$
– Fiff
Jan 16 at 9:08
$begingroup$
That's interesting. If you place an edge $ij$ whenever $langle X_i, X_jrangle ge alpha$ for some threshold $alpha$, then you get a random geometric graph in the $d$-sphere, for instance.
$endgroup$
– Misha Lavrov
Jan 16 at 14:30
$begingroup$
That is exactly the context in which i am studying these type of matrices. I suspected that the total variation distance between the random geometric graph on the $d$-sphere und the unstructured Erdős–Rényi graph might be symmetric around single-edge probability $p=frac{1}{2}$ and the above would have shown that. But apparently, this approach is just a dead end.
$endgroup$
– Fiff
Jan 16 at 21:03
add a comment |
$begingroup$
What's the connection to random graphs? Is there one?
$endgroup$
– Misha Lavrov
Jan 15 at 14:45
$begingroup$
I have not stated the detailed relation to random graphs since it's not crucial for the question asked (and i didn't want to include unnecessary information), although some may recognize a connection. However, i came across this issue when facing a random graph problem, where random adjacency matrices arise by applying indicators on the entries above (symmetry and a vanishing diagonal are already provided).
$endgroup$
– Fiff
Jan 16 at 9:08
$begingroup$
That's interesting. If you place an edge $ij$ whenever $langle X_i, X_jrangle ge alpha$ for some threshold $alpha$, then you get a random geometric graph in the $d$-sphere, for instance.
$endgroup$
– Misha Lavrov
Jan 16 at 14:30
$begingroup$
That is exactly the context in which i am studying these type of matrices. I suspected that the total variation distance between the random geometric graph on the $d$-sphere und the unstructured Erdős–Rényi graph might be symmetric around single-edge probability $p=frac{1}{2}$ and the above would have shown that. But apparently, this approach is just a dead end.
$endgroup$
– Fiff
Jan 16 at 21:03
$begingroup$
What's the connection to random graphs? Is there one?
$endgroup$
– Misha Lavrov
Jan 15 at 14:45
$begingroup$
What's the connection to random graphs? Is there one?
$endgroup$
– Misha Lavrov
Jan 15 at 14:45
$begingroup$
I have not stated the detailed relation to random graphs since it's not crucial for the question asked (and i didn't want to include unnecessary information), although some may recognize a connection. However, i came across this issue when facing a random graph problem, where random adjacency matrices arise by applying indicators on the entries above (symmetry and a vanishing diagonal are already provided).
$endgroup$
– Fiff
Jan 16 at 9:08
$begingroup$
I have not stated the detailed relation to random graphs since it's not crucial for the question asked (and i didn't want to include unnecessary information), although some may recognize a connection. However, i came across this issue when facing a random graph problem, where random adjacency matrices arise by applying indicators on the entries above (symmetry and a vanishing diagonal are already provided).
$endgroup$
– Fiff
Jan 16 at 9:08
$begingroup$
That's interesting. If you place an edge $ij$ whenever $langle X_i, X_jrangle ge alpha$ for some threshold $alpha$, then you get a random geometric graph in the $d$-sphere, for instance.
$endgroup$
– Misha Lavrov
Jan 16 at 14:30
$begingroup$
That's interesting. If you place an edge $ij$ whenever $langle X_i, X_jrangle ge alpha$ for some threshold $alpha$, then you get a random geometric graph in the $d$-sphere, for instance.
$endgroup$
– Misha Lavrov
Jan 16 at 14:30
$begingroup$
That is exactly the context in which i am studying these type of matrices. I suspected that the total variation distance between the random geometric graph on the $d$-sphere und the unstructured Erdős–Rényi graph might be symmetric around single-edge probability $p=frac{1}{2}$ and the above would have shown that. But apparently, this approach is just a dead end.
$endgroup$
– Fiff
Jan 16 at 21:03
$begingroup$
That is exactly the context in which i am studying these type of matrices. I suspected that the total variation distance between the random geometric graph on the $d$-sphere und the unstructured Erdős–Rényi graph might be symmetric around single-edge probability $p=frac{1}{2}$ and the above would have shown that. But apparently, this approach is just a dead end.
$endgroup$
– Fiff
Jan 16 at 21:03
add a comment |
1 Answer
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$begingroup$
The thing you want just isn't true: if you have $n=5$ random variables distributed uniformly on $S^1$, for instance, then it's impossible for $langle X_i, X_jrangle$ to be negative for all pairs $ine j$. However, it's perfectly possible (and happens with probability better than $frac{1}{4^4}$: the chances that all of $X_1, X_2, dots, X_5$ land in the same quadrant) that $langle X_i, X_jrangle$ is positive for all pairs $i ne j$.
(The same thing happens in any dimension if you take $n$ large enough as a function of $d$.)
$endgroup$
$begingroup$
Thanks a lot. Embarrassingly, it did not cross my mind to simply check for counter examples.
$endgroup$
– Fiff
Jan 16 at 20:54
add a comment |
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1 Answer
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$begingroup$
The thing you want just isn't true: if you have $n=5$ random variables distributed uniformly on $S^1$, for instance, then it's impossible for $langle X_i, X_jrangle$ to be negative for all pairs $ine j$. However, it's perfectly possible (and happens with probability better than $frac{1}{4^4}$: the chances that all of $X_1, X_2, dots, X_5$ land in the same quadrant) that $langle X_i, X_jrangle$ is positive for all pairs $i ne j$.
(The same thing happens in any dimension if you take $n$ large enough as a function of $d$.)
$endgroup$
$begingroup$
Thanks a lot. Embarrassingly, it did not cross my mind to simply check for counter examples.
$endgroup$
– Fiff
Jan 16 at 20:54
add a comment |
$begingroup$
The thing you want just isn't true: if you have $n=5$ random variables distributed uniformly on $S^1$, for instance, then it's impossible for $langle X_i, X_jrangle$ to be negative for all pairs $ine j$. However, it's perfectly possible (and happens with probability better than $frac{1}{4^4}$: the chances that all of $X_1, X_2, dots, X_5$ land in the same quadrant) that $langle X_i, X_jrangle$ is positive for all pairs $i ne j$.
(The same thing happens in any dimension if you take $n$ large enough as a function of $d$.)
$endgroup$
$begingroup$
Thanks a lot. Embarrassingly, it did not cross my mind to simply check for counter examples.
$endgroup$
– Fiff
Jan 16 at 20:54
add a comment |
$begingroup$
The thing you want just isn't true: if you have $n=5$ random variables distributed uniformly on $S^1$, for instance, then it's impossible for $langle X_i, X_jrangle$ to be negative for all pairs $ine j$. However, it's perfectly possible (and happens with probability better than $frac{1}{4^4}$: the chances that all of $X_1, X_2, dots, X_5$ land in the same quadrant) that $langle X_i, X_jrangle$ is positive for all pairs $i ne j$.
(The same thing happens in any dimension if you take $n$ large enough as a function of $d$.)
$endgroup$
The thing you want just isn't true: if you have $n=5$ random variables distributed uniformly on $S^1$, for instance, then it's impossible for $langle X_i, X_jrangle$ to be negative for all pairs $ine j$. However, it's perfectly possible (and happens with probability better than $frac{1}{4^4}$: the chances that all of $X_1, X_2, dots, X_5$ land in the same quadrant) that $langle X_i, X_jrangle$ is positive for all pairs $i ne j$.
(The same thing happens in any dimension if you take $n$ large enough as a function of $d$.)
answered Jan 15 at 14:42
Misha LavrovMisha Lavrov
46.2k656107
46.2k656107
$begingroup$
Thanks a lot. Embarrassingly, it did not cross my mind to simply check for counter examples.
$endgroup$
– Fiff
Jan 16 at 20:54
add a comment |
$begingroup$
Thanks a lot. Embarrassingly, it did not cross my mind to simply check for counter examples.
$endgroup$
– Fiff
Jan 16 at 20:54
$begingroup$
Thanks a lot. Embarrassingly, it did not cross my mind to simply check for counter examples.
$endgroup$
– Fiff
Jan 16 at 20:54
$begingroup$
Thanks a lot. Embarrassingly, it did not cross my mind to simply check for counter examples.
$endgroup$
– Fiff
Jan 16 at 20:54
add a comment |
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$begingroup$
What's the connection to random graphs? Is there one?
$endgroup$
– Misha Lavrov
Jan 15 at 14:45
$begingroup$
I have not stated the detailed relation to random graphs since it's not crucial for the question asked (and i didn't want to include unnecessary information), although some may recognize a connection. However, i came across this issue when facing a random graph problem, where random adjacency matrices arise by applying indicators on the entries above (symmetry and a vanishing diagonal are already provided).
$endgroup$
– Fiff
Jan 16 at 9:08
$begingroup$
That's interesting. If you place an edge $ij$ whenever $langle X_i, X_jrangle ge alpha$ for some threshold $alpha$, then you get a random geometric graph in the $d$-sphere, for instance.
$endgroup$
– Misha Lavrov
Jan 16 at 14:30
$begingroup$
That is exactly the context in which i am studying these type of matrices. I suspected that the total variation distance between the random geometric graph on the $d$-sphere und the unstructured Erdős–Rényi graph might be symmetric around single-edge probability $p=frac{1}{2}$ and the above would have shown that. But apparently, this approach is just a dead end.
$endgroup$
– Fiff
Jan 16 at 21:03