What does that mean support of a distribution is contained in unit ball?
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In page 4, part 2 of assumption I in On the Regret Minimization of Nonconvex Online Gradient Ascent for Online PCA, the papers says "the support of $D$ is contained in a Euclidean ball of radius $R$ centered at the origin, i.e.,"
$$
sup_{textbf{q} in text{support}(D)} |textbf{q}|_2 leq R
$$
where $D$ is a distribution and $textbf{q}$'s are sampled i.i.d from $D$.
What is the meaning of that assumption? Could you elaborate it intuitively for me? What is the significance of such an assumption?
probability statistics probability-distributions definition
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
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add a comment |
$begingroup$
In page 4, part 2 of assumption I in On the Regret Minimization of Nonconvex Online Gradient Ascent for Online PCA, the papers says "the support of $D$ is contained in a Euclidean ball of radius $R$ centered at the origin, i.e.,"
$$
sup_{textbf{q} in text{support}(D)} |textbf{q}|_2 leq R
$$
where $D$ is a distribution and $textbf{q}$'s are sampled i.i.d from $D$.
What is the meaning of that assumption? Could you elaborate it intuitively for me? What is the significance of such an assumption?
probability statistics probability-distributions definition
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
In page 4, part 2 of assumption I in On the Regret Minimization of Nonconvex Online Gradient Ascent for Online PCA, the papers says "the support of $D$ is contained in a Euclidean ball of radius $R$ centered at the origin, i.e.,"
$$
sup_{textbf{q} in text{support}(D)} |textbf{q}|_2 leq R
$$
where $D$ is a distribution and $textbf{q}$'s are sampled i.i.d from $D$.
What is the meaning of that assumption? Could you elaborate it intuitively for me? What is the significance of such an assumption?
probability statistics probability-distributions definition
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
$endgroup$
In page 4, part 2 of assumption I in On the Regret Minimization of Nonconvex Online Gradient Ascent for Online PCA, the papers says "the support of $D$ is contained in a Euclidean ball of radius $R$ centered at the origin, i.e.,"
$$
sup_{textbf{q} in text{support}(D)} |textbf{q}|_2 leq R
$$
where $D$ is a distribution and $textbf{q}$'s are sampled i.i.d from $D$.
What is the meaning of that assumption? Could you elaborate it intuitively for me? What is the significance of such an assumption?
probability statistics probability-distributions definition
probability statistics probability-distributions definition
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
asked Dec 27 '18 at 0:40
SaeedSaeed
1,036310
1,036310
add a comment |
add a comment |
1 Answer
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Roughly speaking, the support of a distribution is a closed set that contains [as subsets] all the sets with positive probability. For example, the uniform distribution on $[0,1]$ has support $[0,1]$, the exponential distribution has support $[0, infty)$, and the Gaussian distribution has support $mathbb{R}$. See the Wikipedia link for a more rigorous definition.
So $text{support}(D)$ is a set, and the statement is simply saying $text{support}(D) subseteq B_R$ where $B_R$ is the Euclidean ball of radius $R$ centered at the origin. The inequality written in your question is equivalent: it simply states that all elements in the set $text{support}(D)$ have Euclidean norm $le R$.
Long story short, the assumption is saying that the distribution $D$ gives zero probability to subsets lying outside of the Euclidean ball of radius $R$.
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
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$begingroup$
Thank you so much. It was very intuitive, specially the conclusion. Is this assumption something as we assume the norm of decision variable is less than $M$ in optimization?
$endgroup$
– Saeed
Dec 27 '18 at 1:00
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Roughly speaking, the support of a distribution is a closed set that contains [as subsets] all the sets with positive probability. For example, the uniform distribution on $[0,1]$ has support $[0,1]$, the exponential distribution has support $[0, infty)$, and the Gaussian distribution has support $mathbb{R}$. See the Wikipedia link for a more rigorous definition.
So $text{support}(D)$ is a set, and the statement is simply saying $text{support}(D) subseteq B_R$ where $B_R$ is the Euclidean ball of radius $R$ centered at the origin. The inequality written in your question is equivalent: it simply states that all elements in the set $text{support}(D)$ have Euclidean norm $le R$.
Long story short, the assumption is saying that the distribution $D$ gives zero probability to subsets lying outside of the Euclidean ball of radius $R$.
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Thank you so much. It was very intuitive, specially the conclusion. Is this assumption something as we assume the norm of decision variable is less than $M$ in optimization?
$endgroup$
– Saeed
Dec 27 '18 at 1:00
add a comment |
$begingroup$
Roughly speaking, the support of a distribution is a closed set that contains [as subsets] all the sets with positive probability. For example, the uniform distribution on $[0,1]$ has support $[0,1]$, the exponential distribution has support $[0, infty)$, and the Gaussian distribution has support $mathbb{R}$. See the Wikipedia link for a more rigorous definition.
So $text{support}(D)$ is a set, and the statement is simply saying $text{support}(D) subseteq B_R$ where $B_R$ is the Euclidean ball of radius $R$ centered at the origin. The inequality written in your question is equivalent: it simply states that all elements in the set $text{support}(D)$ have Euclidean norm $le R$.
Long story short, the assumption is saying that the distribution $D$ gives zero probability to subsets lying outside of the Euclidean ball of radius $R$.
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
$endgroup$
$begingroup$
Thank you so much. It was very intuitive, specially the conclusion. Is this assumption something as we assume the norm of decision variable is less than $M$ in optimization?
$endgroup$
– Saeed
Dec 27 '18 at 1:00
add a comment |
$begingroup$
Roughly speaking, the support of a distribution is a closed set that contains [as subsets] all the sets with positive probability. For example, the uniform distribution on $[0,1]$ has support $[0,1]$, the exponential distribution has support $[0, infty)$, and the Gaussian distribution has support $mathbb{R}$. See the Wikipedia link for a more rigorous definition.
So $text{support}(D)$ is a set, and the statement is simply saying $text{support}(D) subseteq B_R$ where $B_R$ is the Euclidean ball of radius $R$ centered at the origin. The inequality written in your question is equivalent: it simply states that all elements in the set $text{support}(D)$ have Euclidean norm $le R$.
Long story short, the assumption is saying that the distribution $D$ gives zero probability to subsets lying outside of the Euclidean ball of radius $R$.
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
$endgroup$
Roughly speaking, the support of a distribution is a closed set that contains [as subsets] all the sets with positive probability. For example, the uniform distribution on $[0,1]$ has support $[0,1]$, the exponential distribution has support $[0, infty)$, and the Gaussian distribution has support $mathbb{R}$. See the Wikipedia link for a more rigorous definition.
So $text{support}(D)$ is a set, and the statement is simply saying $text{support}(D) subseteq B_R$ where $B_R$ is the Euclidean ball of radius $R$ centered at the origin. The inequality written in your question is equivalent: it simply states that all elements in the set $text{support}(D)$ have Euclidean norm $le R$.
Long story short, the assumption is saying that the distribution $D$ gives zero probability to subsets lying outside of the Euclidean ball of radius $R$.
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
answered Dec 27 '18 at 0:45
angryavianangryavian
41.1k23380
41.1k23380
$begingroup$
Thank you so much. It was very intuitive, specially the conclusion. Is this assumption something as we assume the norm of decision variable is less than $M$ in optimization?
$endgroup$
– Saeed
Dec 27 '18 at 1:00
add a comment |
$begingroup$
Thank you so much. It was very intuitive, specially the conclusion. Is this assumption something as we assume the norm of decision variable is less than $M$ in optimization?
$endgroup$
– Saeed
Dec 27 '18 at 1:00
$begingroup$
Thank you so much. It was very intuitive, specially the conclusion. Is this assumption something as we assume the norm of decision variable is less than $M$ in optimization?
$endgroup$
– Saeed
Dec 27 '18 at 1:00
$begingroup$
Thank you so much. It was very intuitive, specially the conclusion. Is this assumption something as we assume the norm of decision variable is less than $M$ in optimization?
$endgroup$
– Saeed
Dec 27 '18 at 1:00
add a comment |
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