Variance of the Euclidean norm under finite moment assumptions
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Let $X = (X_1,X_2 cdots X_n)$ be random vector in $R^n$ with independent coordinate $X_i$ that satisfy $E[X_i^2]=1$ and $E[X_i^4] leq K^4$. Then show that $$operatorname{Var}(| X|_2) leq CK^4$$
where $C$ is a absolute constant and $| |_2$ denotes euclidian norm.
Here is my attempt:
$$begin{align*} E(|X|_2^2 -n)^2 &= E[(sum_{i=1}^n X_i^2)^2 ]-n^2 \
&=E[sum_{i=1}^n X_i^4]+E[sum_{i<j}X_i^2X_j^2] -n^2 \
&leq nK^4 + 2{{n}choose {2}}-n^2 \
&leq n(K^4-1) \
& leq nk^4
end{align*}$$
since $$ E(|X|_2^2 -n)^2 leq nk^4 rightarrow Eleft(frac{|X|_2^2}{n} -1right)^2 leq frac{K^4}{n}$$
and since
$$(forall z geq 0 |z-1|leq |z^2-1|) rightarrow
E(frac{|X|_2}{sqrt n} -1)^2leq E(frac{|X|_2^2}{n} -1)^2 $$
thus:
$$E(frac{|X|_2}{sqrt n} -1)^2 leq K^4/n rightarrow E(|X|_2-sqrt n)^2leq K^4$$
by Jensen inequality:
$$(E[|X|_2] - sqrt n)^2 leq K^4 $$
which is equivalence to
$$ |E[|X|_2] - sqrt n)| leq K^2$$
then when I am trying to bound $Var(| X|_2)$ I meet some problem :
$$operatorname{Var}(| X|_2)=E[|X|_2^2] -(E[|X|_2])^2 leq n- (K^2-sqrt n)^2 leq -K^4+2K^2sqrt n$$ which is not bound by constant , how can I bound that?
probability random-variables norm variance concentration-of-measure
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
asked Nov 17 at 21:10
ShaoyuPei
1537
add a comment |
up vote
4
down vote
favorite
Let $X = (X_1,X_2 cdots X_n)$ be random vector in $R^n$ with independent coordinate $X_i$ that satisfy $E[X_i^2]=1$ and $E[X_i^4] leq K^4$. Then show that $$operatorname{Var}(| X|_2) leq CK^4$$
where $C$ is a absolute constant and $| |_2$ denotes euclidian norm.
Here is my attempt:
$$begin{align*} E(|X|_2^2 -n)^2 &= E[(sum_{i=1}^n X_i^2)^2 ]-n^2 \
&=E[sum_{i=1}^n X_i^4]+E[sum_{i<j}X_i^2X_j^2] -n^2 \
&leq nK^4 + 2{{n}choose {2}}-n^2 \
&leq n(K^4-1) \
& leq nk^4
end{align*}$$
since $$ E(|X|_2^2 -n)^2 leq nk^4 rightarrow Eleft(frac{|X|_2^2}{n} -1right)^2 leq frac{K^4}{n}$$
and since
$$(forall z geq 0 |z-1|leq |z^2-1|) rightarrow
E(frac{|X|_2}{sqrt n} -1)^2leq E(frac{|X|_2^2}{n} -1)^2 $$
thus:
$$E(frac{|X|_2}{sqrt n} -1)^2 leq K^4/n rightarrow E(|X|_2-sqrt n)^2leq K^4$$
by Jensen inequality:
$$(E[|X|_2] - sqrt n)^2 leq K^4 $$
which is equivalence to
$$ |E[|X|_2] - sqrt n)| leq K^2$$
then when I am trying to bound $Var(| X|_2)$ I meet some problem :
$$operatorname{Var}(| X|_2)=E[|X|_2^2] -(E[|X|_2])^2 leq n- (K^2-sqrt n)^2 leq -K^4+2K^2sqrt n$$ which is not bound by constant , how can I bound that?
probability random-variables norm variance concentration-of-measure
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
asked Nov 17 at 21:10
ShaoyuPei
1537
Can you cite the source of your claim? My guess is that $text{Var}|X|$ will grow with $n$.
– nemo
yesterday
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $X = (X_1,X_2 cdots X_n)$ be random vector in $R^n$ with independent coordinate $X_i$ that satisfy $E[X_i^2]=1$ and $E[X_i^4] leq K^4$. Then show that $$operatorname{Var}(| X|_2) leq CK^4$$
where $C$ is a absolute constant and $| |_2$ denotes euclidian norm.
Here is my attempt:
$$begin{align*} E(|X|_2^2 -n)^2 &= E[(sum_{i=1}^n X_i^2)^2 ]-n^2 \
&=E[sum_{i=1}^n X_i^4]+E[sum_{i<j}X_i^2X_j^2] -n^2 \
&leq nK^4 + 2{{n}choose {2}}-n^2 \
&leq n(K^4-1) \
& leq nk^4
end{align*}$$
since $$ E(|X|_2^2 -n)^2 leq nk^4 rightarrow Eleft(frac{|X|_2^2}{n} -1right)^2 leq frac{K^4}{n}$$
and since
$$(forall z geq 0 |z-1|leq |z^2-1|) rightarrow
E(frac{|X|_2}{sqrt n} -1)^2leq E(frac{|X|_2^2}{n} -1)^2 $$
thus:
$$E(frac{|X|_2}{sqrt n} -1)^2 leq K^4/n rightarrow E(|X|_2-sqrt n)^2leq K^4$$
by Jensen inequality:
$$(E[|X|_2] - sqrt n)^2 leq K^4 $$
which is equivalence to
$$ |E[|X|_2] - sqrt n)| leq K^2$$
then when I am trying to bound $Var(| X|_2)$ I meet some problem :
$$operatorname{Var}(| X|_2)=E[|X|_2^2] -(E[|X|_2])^2 leq n- (K^2-sqrt n)^2 leq -K^4+2K^2sqrt n$$ which is not bound by constant , how can I bound that?
probability random-variables norm variance concentration-of-measure
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
asked Nov 17 at 21:10
ShaoyuPei
1537
Let $X = (X_1,X_2 cdots X_n)$ be random vector in $R^n$ with independent coordinate $X_i$ that satisfy $E[X_i^2]=1$ and $E[X_i^4] leq K^4$. Then show that $$operatorname{Var}(| X|_2) leq CK^4$$
where $C$ is a absolute constant and $| |_2$ denotes euclidian norm.
Here is my attempt:
$$begin{align*} E(|X|_2^2 -n)^2 &= E[(sum_{i=1}^n X_i^2)^2 ]-n^2 \
&=E[sum_{i=1}^n X_i^4]+E[sum_{i<j}X_i^2X_j^2] -n^2 \
&leq nK^4 + 2{{n}choose {2}}-n^2 \
&leq n(K^4-1) \
& leq nk^4
end{align*}$$
since $$ E(|X|_2^2 -n)^2 leq nk^4 rightarrow Eleft(frac{|X|_2^2}{n} -1right)^2 leq frac{K^4}{n}$$
and since
$$(forall z geq 0 |z-1|leq |z^2-1|) rightarrow
E(frac{|X|_2}{sqrt n} -1)^2leq E(frac{|X|_2^2}{n} -1)^2 $$
thus:
$$E(frac{|X|_2}{sqrt n} -1)^2 leq K^4/n rightarrow E(|X|_2-sqrt n)^2leq K^4$$
by Jensen inequality:
$$(E[|X|_2] - sqrt n)^2 leq K^4 $$
which is equivalence to
$$ |E[|X|_2] - sqrt n)| leq K^2$$
then when I am trying to bound $Var(| X|_2)$ I meet some problem :
$$operatorname{Var}(| X|_2)=E[|X|_2^2] -(E[|X|_2])^2 leq n- (K^2-sqrt n)^2 leq -K^4+2K^2sqrt n$$ which is not bound by constant , how can I bound that?
probability random-variables norm variance concentration-of-measure
probability random-variables norm variance concentration-of-measure
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
asked Nov 17 at 21:10
ShaoyuPei
1537
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
asked Nov 17 at 21:10
ShaoyuPei
1537
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
edited Nov 26 at 14:29
Davide Giraudo
124k16150256
124k16150256
asked Nov 17 at 21:10
ShaoyuPei
1537
asked Nov 17 at 21:10
ShaoyuPei
1537
asked Nov 17 at 21:10
ShaoyuPei
1537
1537
Can you cite the source of your claim? My guess is that $text{Var}|X|$ will grow with $n$.
– nemo
yesterday
add a comment |
Can you cite the source of your claim? My guess is that $text{Var}|X|$ will grow with $n$.
– nemo
yesterday
Can you cite the source of your claim? My guess is that $text{Var}|X|$ will grow with $n$.
– nemo
yesterday
Can you cite the source of your claim? My guess is that $text{Var}|X|$ will grow with $n$.
– nemo
yesterday
add a comment |
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Can you cite the source of your claim? My guess is that $text{Var}|X|$ will grow with $n$.
– nemo
yesterday