Expectation in spectral sparsification algorithms
$begingroup$
I am new to random matrices. I am studying the (Sampling) sparsification algorithms done by Daniel Spielman, Teng, Srivastava.
They used the concept of graph sampling to obtain a good spectral sparsifier $H $ for every graph $G $.
The sampling procedure involves assigning a probabilty $p_{u,v} $ to each edge $(u,v)$ in $G$ and then selecting edge $(u,v)$ to be in graph $H $ with probability $p_{u,v} $. When edge $(u,v) $ is chosen to be in $H $, we multiply its weight by $1/p_{u,v} $.
Apparently this procedure guarantees that the expectation of the Laplacian of $H$ is
$E (L_{H})= L_{G}$. Now I am not fully understanding how they figured out the expectation.please explain
probability algorithms random-matrices spectral-graph-theory algorithmic-randomness
asked Dec 10 '18 at 10:41
SalSal
14010
$endgroup$
add a comment |
$begingroup$
I am new to random matrices. I am studying the (Sampling) sparsification algorithms done by Daniel Spielman, Teng, Srivastava.
They used the concept of graph sampling to obtain a good spectral sparsifier $H $ for every graph $G $.
The sampling procedure involves assigning a probabilty $p_{u,v} $ to each edge $(u,v)$ in $G$ and then selecting edge $(u,v)$ to be in graph $H $ with probability $p_{u,v} $. When edge $(u,v) $ is chosen to be in $H $, we multiply its weight by $1/p_{u,v} $.
Apparently this procedure guarantees that the expectation of the Laplacian of $H$ is
$E (L_{H})= L_{G}$. Now I am not fully understanding how they figured out the expectation.please explain
probability algorithms random-matrices spectral-graph-theory algorithmic-randomness
asked Dec 10 '18 at 10:41
SalSal
14010
$endgroup$
add a comment |
$begingroup$
I am new to random matrices. I am studying the (Sampling) sparsification algorithms done by Daniel Spielman, Teng, Srivastava.
They used the concept of graph sampling to obtain a good spectral sparsifier $H $ for every graph $G $.
The sampling procedure involves assigning a probabilty $p_{u,v} $ to each edge $(u,v)$ in $G$ and then selecting edge $(u,v)$ to be in graph $H $ with probability $p_{u,v} $. When edge $(u,v) $ is chosen to be in $H $, we multiply its weight by $1/p_{u,v} $.
Apparently this procedure guarantees that the expectation of the Laplacian of $H$ is
$E (L_{H})= L_{G}$. Now I am not fully understanding how they figured out the expectation.please explain
probability algorithms random-matrices spectral-graph-theory algorithmic-randomness
asked Dec 10 '18 at 10:41
SalSal
14010
$endgroup$
I am new to random matrices. I am studying the (Sampling) sparsification algorithms done by Daniel Spielman, Teng, Srivastava.
They used the concept of graph sampling to obtain a good spectral sparsifier $H $ for every graph $G $.
The sampling procedure involves assigning a probabilty $p_{u,v} $ to each edge $(u,v)$ in $G$ and then selecting edge $(u,v)$ to be in graph $H $ with probability $p_{u,v} $. When edge $(u,v) $ is chosen to be in $H $, we multiply its weight by $1/p_{u,v} $.
Apparently this procedure guarantees that the expectation of the Laplacian of $H$ is
$E (L_{H})= L_{G}$. Now I am not fully understanding how they figured out the expectation.please explain
probability algorithms random-matrices spectral-graph-theory algorithmic-randomness
probability algorithms random-matrices spectral-graph-theory algorithmic-randomness
asked Dec 10 '18 at 10:41
SalSal
14010
asked Dec 10 '18 at 10:41
SalSal
14010
edited Dec 10 '18 at 13:11
Sal
asked Dec 10 '18 at 10:41
SalSal
14010
asked Dec 10 '18 at 10:41
SalSal
14010
asked Dec 10 '18 at 10:41
SalSal
14010
14010
add a comment |
add a comment |
1 Answer
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$begingroup$
To figure out the correct weights $tilde{w}_e$ of the randomly sampled graph, it is useful to write the Laplacian of a weighted graph $G$ as the sum of outer products
$$L_G = sum_{e in E} w_e b_e b_e^T,$$
where
$w_e$ is the weight of the edge $e$ and $b_e$ is the vector such that
$$b_e(w) =
begin{cases}
1 & w = mathrm{head}(e)\
-1 & w = mathrm{tail}(e)\
0 & text{otherwise},
end{cases}$$
after some arbitrary orientation of the edge $e$.
Now if we sample each edge $e$ with probability $p_e$, we are left with a graph $tilde{G}$ whose Laplacian is
$$L_tilde{G} = sum_{e in E} I_e tilde{w}_e b_e b_e^T,$$
where $I_e sim textsf{Bern}(p_e)$ is an indicator random variable for whether or not $e$ was chosen. To make the expectation line up, we just need $mathbb{E}[I_e tilde{w}_e] = w_e$. In other words, we want $tilde{w}_e = frac{w_e}{mathbb{E}[I_e]} = frac{w_e}{p_e}$.
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
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add a comment |
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$begingroup$
To figure out the correct weights $tilde{w}_e$ of the randomly sampled graph, it is useful to write the Laplacian of a weighted graph $G$ as the sum of outer products
$$L_G = sum_{e in E} w_e b_e b_e^T,$$
where
$w_e$ is the weight of the edge $e$ and $b_e$ is the vector such that
$$b_e(w) =
begin{cases}
1 & w = mathrm{head}(e)\
-1 & w = mathrm{tail}(e)\
0 & text{otherwise},
end{cases}$$
after some arbitrary orientation of the edge $e$.
Now if we sample each edge $e$ with probability $p_e$, we are left with a graph $tilde{G}$ whose Laplacian is
$$L_tilde{G} = sum_{e in E} I_e tilde{w}_e b_e b_e^T,$$
where $I_e sim textsf{Bern}(p_e)$ is an indicator random variable for whether or not $e$ was chosen. To make the expectation line up, we just need $mathbb{E}[I_e tilde{w}_e] = w_e$. In other words, we want $tilde{w}_e = frac{w_e}{mathbb{E}[I_e]} = frac{w_e}{p_e}$.
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
$endgroup$
add a comment |
$begingroup$
To figure out the correct weights $tilde{w}_e$ of the randomly sampled graph, it is useful to write the Laplacian of a weighted graph $G$ as the sum of outer products
$$L_G = sum_{e in E} w_e b_e b_e^T,$$
where
$w_e$ is the weight of the edge $e$ and $b_e$ is the vector such that
$$b_e(w) =
begin{cases}
1 & w = mathrm{head}(e)\
-1 & w = mathrm{tail}(e)\
0 & text{otherwise},
end{cases}$$
after some arbitrary orientation of the edge $e$.
Now if we sample each edge $e$ with probability $p_e$, we are left with a graph $tilde{G}$ whose Laplacian is
$$L_tilde{G} = sum_{e in E} I_e tilde{w}_e b_e b_e^T,$$
where $I_e sim textsf{Bern}(p_e)$ is an indicator random variable for whether or not $e$ was chosen. To make the expectation line up, we just need $mathbb{E}[I_e tilde{w}_e] = w_e$. In other words, we want $tilde{w}_e = frac{w_e}{mathbb{E}[I_e]} = frac{w_e}{p_e}$.
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
$endgroup$
add a comment |
$begingroup$
To figure out the correct weights $tilde{w}_e$ of the randomly sampled graph, it is useful to write the Laplacian of a weighted graph $G$ as the sum of outer products
$$L_G = sum_{e in E} w_e b_e b_e^T,$$
where
$w_e$ is the weight of the edge $e$ and $b_e$ is the vector such that
$$b_e(w) =
begin{cases}
1 & w = mathrm{head}(e)\
-1 & w = mathrm{tail}(e)\
0 & text{otherwise},
end{cases}$$
after some arbitrary orientation of the edge $e$.
Now if we sample each edge $e$ with probability $p_e$, we are left with a graph $tilde{G}$ whose Laplacian is
$$L_tilde{G} = sum_{e in E} I_e tilde{w}_e b_e b_e^T,$$
where $I_e sim textsf{Bern}(p_e)$ is an indicator random variable for whether or not $e$ was chosen. To make the expectation line up, we just need $mathbb{E}[I_e tilde{w}_e] = w_e$. In other words, we want $tilde{w}_e = frac{w_e}{mathbb{E}[I_e]} = frac{w_e}{p_e}$.
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
$endgroup$
To figure out the correct weights $tilde{w}_e$ of the randomly sampled graph, it is useful to write the Laplacian of a weighted graph $G$ as the sum of outer products
$$L_G = sum_{e in E} w_e b_e b_e^T,$$
where
$w_e$ is the weight of the edge $e$ and $b_e$ is the vector such that
$$b_e(w) =
begin{cases}
1 & w = mathrm{head}(e)\
-1 & w = mathrm{tail}(e)\
0 & text{otherwise},
end{cases}$$
after some arbitrary orientation of the edge $e$.
Now if we sample each edge $e$ with probability $p_e$, we are left with a graph $tilde{G}$ whose Laplacian is
$$L_tilde{G} = sum_{e in E} I_e tilde{w}_e b_e b_e^T,$$
where $I_e sim textsf{Bern}(p_e)$ is an indicator random variable for whether or not $e$ was chosen. To make the expectation line up, we just need $mathbb{E}[I_e tilde{w}_e] = w_e$. In other words, we want $tilde{w}_e = frac{w_e}{mathbb{E}[I_e]} = frac{w_e}{p_e}$.
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
answered Dec 11 '18 at 0:58
Zach LangleyZach Langley
9731019
9731019
add a comment |
add a comment |
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