Concentration of two independent sub-Gaussian random variables
$begingroup$
Suppose $X$ and $Y$ are independent sub-Gaussian random variables with 0 mean and $sigma^2$ sub-Gaussian parameter. More specifically, $mathbb E[exp(a^T X)]leq exp{|a|_2^2sigma^2/2}$ for all $a$, and the same holds for $Y$ as well.
I wish to upper bound the tail probability
$$
Prleft[|X^T Y|>tright]
$$
using $sigma^2$ and dimension $n$ (that is, both $X$ and $Y$ are $d$-dimensional random variables). How can I achieve this? $X^T Y$ does not seem to be either sub-Gaussian or sub-exponential.
probability concentration-of-measure
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
$endgroup$
add a comment |
$begingroup$
Suppose $X$ and $Y$ are independent sub-Gaussian random variables with 0 mean and $sigma^2$ sub-Gaussian parameter. More specifically, $mathbb E[exp(a^T X)]leq exp{|a|_2^2sigma^2/2}$ for all $a$, and the same holds for $Y$ as well.
I wish to upper bound the tail probability
$$
Prleft[|X^T Y|>tright]
$$
using $sigma^2$ and dimension $n$ (that is, both $X$ and $Y$ are $d$-dimensional random variables). How can I achieve this? $X^T Y$ does not seem to be either sub-Gaussian or sub-exponential.
probability concentration-of-measure
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
$endgroup$
add a comment |
$begingroup$
Suppose $X$ and $Y$ are independent sub-Gaussian random variables with 0 mean and $sigma^2$ sub-Gaussian parameter. More specifically, $mathbb E[exp(a^T X)]leq exp{|a|_2^2sigma^2/2}$ for all $a$, and the same holds for $Y$ as well.
I wish to upper bound the tail probability
$$
Prleft[|X^T Y|>tright]
$$
using $sigma^2$ and dimension $n$ (that is, both $X$ and $Y$ are $d$-dimensional random variables). How can I achieve this? $X^T Y$ does not seem to be either sub-Gaussian or sub-exponential.
probability concentration-of-measure
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
$endgroup$
Suppose $X$ and $Y$ are independent sub-Gaussian random variables with 0 mean and $sigma^2$ sub-Gaussian parameter. More specifically, $mathbb E[exp(a^T X)]leq exp{|a|_2^2sigma^2/2}$ for all $a$, and the same holds for $Y$ as well.
I wish to upper bound the tail probability
$$
Prleft[|X^T Y|>tright]
$$
using $sigma^2$ and dimension $n$ (that is, both $X$ and $Y$ are $d$-dimensional random variables). How can I achieve this? $X^T Y$ does not seem to be either sub-Gaussian or sub-exponential.
probability concentration-of-measure
probability concentration-of-measure
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
asked Jul 6 '17 at 12:36
Yining WangYining Wang
784315
784315
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Though you are right that the product is in general not sub-Gaussian, but it is still subexponential.
Note that for a sub-Gaussian vector $X$ in $mathbb{R}^n$ with parameter $sigma^2$ and for $r<sigma^2$,
$$
mathrm{E}bigg[expBig{frac{r||X||^2}2Big}bigg] = frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r}Big}, mathbb{E}[e^{a^T X}]da \
le frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r} + frac{||a^2||sigma^2}{2}Big}da = frac{1}{big(2pi (sigma^2-r)big)^{n/2}}.
$$
Therefore, thanks to independence, for any $lambda<1$,
$$
mathrm{E}big[exp{lambda X^T Y}big] le mathrm{E}bigg[expBig{frac{lambda^2sigma^2||X||^2}2Big}bigg]lefrac{1}{big(2pi sigma^2(1-lambda^2)big)^{n/2}}.
$$
Hence you can easily derive an exponential bound for $mathrm{P}(X^TY>t)$.
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
$endgroup$
$begingroup$
You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty.
$endgroup$
– Yining Wang
Aug 25 '17 at 17:06
add a comment |
$begingroup$
By Cauchy-Schwarz, the event $[ |X^TY|>t]$ is a subset of $[| X|>sqrt t] cup[|Y|>sqrt t]$, so its probability is bounded above by $P(| X|>sqrt t)+P(| Y|>sqrt t)$, for which $O(exp{-At})$ bounds can probably be derived, for some $sigma$-dependent $A$.
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
$endgroup$
$begingroup$
Thanks for your answer. Unfortunately, I think this upper bound is very loose because $|X|$ is far from 0 because $mathbb E|X|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $mathbb EX^T Y = 0$. In particular, in the $P(|X|>sqrt{t})$ bound the $t$ can never be arbitrarily small.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:36
$begingroup$
And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:37
add a comment |
$begingroup$
Moment generating functions of subgaussian vector can often be bounded from above by the same moment generating function, with the subgaussian vector replaced by a standard normal. This is equivalent to zhoraster's answer.
Take $sigma=1$ without loss of generality (otherwise, consider $X/sigma$ and $Y/sigma$ and scale back afterwards).
Let $g,h$ be standard normal random vectors, independent of $X,Y$. Then by independence,
[
E[e^{uX^TY} | X ] = E[e^{u^2|X|^2/2} ] = E[e^{u X^top g}].
]
Now by the law of total expectation, and a similar argument for $Y$,
[
E[e^{uX^TY}] le E[ E[e^{u X^T g}| g] ] = E[E[ e^{u^2|g|^2/2} | g]] = E[e^{u g^Th}].
]
So the problem is reduced to bounding the moment generating function where both random vectors are standard normal.
answered Dec 1 '18 at 1:02
jlewkjlewk
765
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Though you are right that the product is in general not sub-Gaussian, but it is still subexponential.
Note that for a sub-Gaussian vector $X$ in $mathbb{R}^n$ with parameter $sigma^2$ and for $r<sigma^2$,
$$
mathrm{E}bigg[expBig{frac{r||X||^2}2Big}bigg] = frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r}Big}, mathbb{E}[e^{a^T X}]da \
le frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r} + frac{||a^2||sigma^2}{2}Big}da = frac{1}{big(2pi (sigma^2-r)big)^{n/2}}.
$$
Therefore, thanks to independence, for any $lambda<1$,
$$
mathrm{E}big[exp{lambda X^T Y}big] le mathrm{E}bigg[expBig{frac{lambda^2sigma^2||X||^2}2Big}bigg]lefrac{1}{big(2pi sigma^2(1-lambda^2)big)^{n/2}}.
$$
Hence you can easily derive an exponential bound for $mathrm{P}(X^TY>t)$.
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
$endgroup$
$begingroup$
You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty.
$endgroup$
– Yining Wang
Aug 25 '17 at 17:06
add a comment |
$begingroup$
Though you are right that the product is in general not sub-Gaussian, but it is still subexponential.
Note that for a sub-Gaussian vector $X$ in $mathbb{R}^n$ with parameter $sigma^2$ and for $r<sigma^2$,
$$
mathrm{E}bigg[expBig{frac{r||X||^2}2Big}bigg] = frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r}Big}, mathbb{E}[e^{a^T X}]da \
le frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r} + frac{||a^2||sigma^2}{2}Big}da = frac{1}{big(2pi (sigma^2-r)big)^{n/2}}.
$$
Therefore, thanks to independence, for any $lambda<1$,
$$
mathrm{E}big[exp{lambda X^T Y}big] le mathrm{E}bigg[expBig{frac{lambda^2sigma^2||X||^2}2Big}bigg]lefrac{1}{big(2pi sigma^2(1-lambda^2)big)^{n/2}}.
$$
Hence you can easily derive an exponential bound for $mathrm{P}(X^TY>t)$.
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
$endgroup$
$begingroup$
You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty.
$endgroup$
– Yining Wang
Aug 25 '17 at 17:06
add a comment |
$begingroup$
Though you are right that the product is in general not sub-Gaussian, but it is still subexponential.
Note that for a sub-Gaussian vector $X$ in $mathbb{R}^n$ with parameter $sigma^2$ and for $r<sigma^2$,
$$
mathrm{E}bigg[expBig{frac{r||X||^2}2Big}bigg] = frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r}Big}, mathbb{E}[e^{a^T X}]da \
le frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r} + frac{||a^2||sigma^2}{2}Big}da = frac{1}{big(2pi (sigma^2-r)big)^{n/2}}.
$$
Therefore, thanks to independence, for any $lambda<1$,
$$
mathrm{E}big[exp{lambda X^T Y}big] le mathrm{E}bigg[expBig{frac{lambda^2sigma^2||X||^2}2Big}bigg]lefrac{1}{big(2pi sigma^2(1-lambda^2)big)^{n/2}}.
$$
Hence you can easily derive an exponential bound for $mathrm{P}(X^TY>t)$.
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
$endgroup$
Though you are right that the product is in general not sub-Gaussian, but it is still subexponential.
Note that for a sub-Gaussian vector $X$ in $mathbb{R}^n$ with parameter $sigma^2$ and for $r<sigma^2$,
$$
mathrm{E}bigg[expBig{frac{r||X||^2}2Big}bigg] = frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r}Big}, mathbb{E}[e^{a^T X}]da \
le frac{1}{(2pi r)^{n/2}}int_{mathrm{R}^n} expBig{-frac{||a||^2}{2r} + frac{||a^2||sigma^2}{2}Big}da = frac{1}{big(2pi (sigma^2-r)big)^{n/2}}.
$$
Therefore, thanks to independence, for any $lambda<1$,
$$
mathrm{E}big[exp{lambda X^T Y}big] le mathrm{E}bigg[expBig{frac{lambda^2sigma^2||X||^2}2Big}bigg]lefrac{1}{big(2pi sigma^2(1-lambda^2)big)^{n/2}}.
$$
Hence you can easily derive an exponential bound for $mathrm{P}(X^TY>t)$.
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
answered Aug 25 '17 at 6:14
zhorasterzhoraster
15.7k21752
15.7k21752
$begingroup$
You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty.
$endgroup$
– Yining Wang
Aug 25 '17 at 17:06
add a comment |
$begingroup$
You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty.
$endgroup$
– Yining Wang
Aug 25 '17 at 17:06
$begingroup$
You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty.
$endgroup$
– Yining Wang
Aug 25 '17 at 17:06
$begingroup$
You're definitely right! I obtained the same results (sub-exponentiality of $X^T Y$) by bounding $mathbb E|X^T Y|^p$ and invoking the Bernstein condition. Anyway I accepted your answer and awarded the bounty.
$endgroup$
– Yining Wang
Aug 25 '17 at 17:06
add a comment |
$begingroup$
By Cauchy-Schwarz, the event $[ |X^TY|>t]$ is a subset of $[| X|>sqrt t] cup[|Y|>sqrt t]$, so its probability is bounded above by $P(| X|>sqrt t)+P(| Y|>sqrt t)$, for which $O(exp{-At})$ bounds can probably be derived, for some $sigma$-dependent $A$.
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
$endgroup$
$begingroup$
Thanks for your answer. Unfortunately, I think this upper bound is very loose because $|X|$ is far from 0 because $mathbb E|X|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $mathbb EX^T Y = 0$. In particular, in the $P(|X|>sqrt{t})$ bound the $t$ can never be arbitrarily small.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:36
$begingroup$
And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:37
add a comment |
$begingroup$
By Cauchy-Schwarz, the event $[ |X^TY|>t]$ is a subset of $[| X|>sqrt t] cup[|Y|>sqrt t]$, so its probability is bounded above by $P(| X|>sqrt t)+P(| Y|>sqrt t)$, for which $O(exp{-At})$ bounds can probably be derived, for some $sigma$-dependent $A$.
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
$endgroup$
$begingroup$
Thanks for your answer. Unfortunately, I think this upper bound is very loose because $|X|$ is far from 0 because $mathbb E|X|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $mathbb EX^T Y = 0$. In particular, in the $P(|X|>sqrt{t})$ bound the $t$ can never be arbitrarily small.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:36
$begingroup$
And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:37
add a comment |
$begingroup$
By Cauchy-Schwarz, the event $[ |X^TY|>t]$ is a subset of $[| X|>sqrt t] cup[|Y|>sqrt t]$, so its probability is bounded above by $P(| X|>sqrt t)+P(| Y|>sqrt t)$, for which $O(exp{-At})$ bounds can probably be derived, for some $sigma$-dependent $A$.
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
$endgroup$
By Cauchy-Schwarz, the event $[ |X^TY|>t]$ is a subset of $[| X|>sqrt t] cup[|Y|>sqrt t]$, so its probability is bounded above by $P(| X|>sqrt t)+P(| Y|>sqrt t)$, for which $O(exp{-At})$ bounds can probably be derived, for some $sigma$-dependent $A$.
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
answered Jul 6 '17 at 17:39
kimchi loverkimchi lover
9,70631128
9,70631128
$begingroup$
Thanks for your answer. Unfortunately, I think this upper bound is very loose because $|X|$ is far from 0 because $mathbb E|X|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $mathbb EX^T Y = 0$. In particular, in the $P(|X|>sqrt{t})$ bound the $t$ can never be arbitrarily small.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:36
$begingroup$
And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:37
add a comment |
$begingroup$
Thanks for your answer. Unfortunately, I think this upper bound is very loose because $|X|$ is far from 0 because $mathbb E|X|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $mathbb EX^T Y = 0$. In particular, in the $P(|X|>sqrt{t})$ bound the $t$ can never be arbitrarily small.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:36
$begingroup$
And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:37
$begingroup$
Thanks for your answer. Unfortunately, I think this upper bound is very loose because $|X|$ is far from 0 because $mathbb E|X|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $mathbb EX^T Y = 0$. In particular, in the $P(|X|>sqrt{t})$ bound the $t$ can never be arbitrarily small.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:36
$begingroup$
Thanks for your answer. Unfortunately, I think this upper bound is very loose because $|X|$ is far from 0 because $mathbb E|X|_2^2>0$, and $X^T Y$ is known to concentrate to 0 because $mathbb EX^T Y = 0$. In particular, in the $P(|X|>sqrt{t})$ bound the $t$ can never be arbitrarily small.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:36
$begingroup$
And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:37
$begingroup$
And also, this argument applies to dependent $X$ and $Y$: which is not good because we know that when $X$ and $Y$ are independent, $X^T Y$ should be much smaller and close to 0.
$endgroup$
– Yining Wang
Jul 8 '17 at 1:37
add a comment |
$begingroup$
Moment generating functions of subgaussian vector can often be bounded from above by the same moment generating function, with the subgaussian vector replaced by a standard normal. This is equivalent to zhoraster's answer.
Take $sigma=1$ without loss of generality (otherwise, consider $X/sigma$ and $Y/sigma$ and scale back afterwards).
Let $g,h$ be standard normal random vectors, independent of $X,Y$. Then by independence,
[
E[e^{uX^TY} | X ] = E[e^{u^2|X|^2/2} ] = E[e^{u X^top g}].
]
Now by the law of total expectation, and a similar argument for $Y$,
[
E[e^{uX^TY}] le E[ E[e^{u X^T g}| g] ] = E[E[ e^{u^2|g|^2/2} | g]] = E[e^{u g^Th}].
]
So the problem is reduced to bounding the moment generating function where both random vectors are standard normal.
answered Dec 1 '18 at 1:02
jlewkjlewk
765
$endgroup$
add a comment |
$begingroup$
Moment generating functions of subgaussian vector can often be bounded from above by the same moment generating function, with the subgaussian vector replaced by a standard normal. This is equivalent to zhoraster's answer.
Take $sigma=1$ without loss of generality (otherwise, consider $X/sigma$ and $Y/sigma$ and scale back afterwards).
Let $g,h$ be standard normal random vectors, independent of $X,Y$. Then by independence,
[
E[e^{uX^TY} | X ] = E[e^{u^2|X|^2/2} ] = E[e^{u X^top g}].
]
Now by the law of total expectation, and a similar argument for $Y$,
[
E[e^{uX^TY}] le E[ E[e^{u X^T g}| g] ] = E[E[ e^{u^2|g|^2/2} | g]] = E[e^{u g^Th}].
]
So the problem is reduced to bounding the moment generating function where both random vectors are standard normal.
answered Dec 1 '18 at 1:02
jlewkjlewk
765
$endgroup$
add a comment |
$begingroup$
Moment generating functions of subgaussian vector can often be bounded from above by the same moment generating function, with the subgaussian vector replaced by a standard normal. This is equivalent to zhoraster's answer.
Take $sigma=1$ without loss of generality (otherwise, consider $X/sigma$ and $Y/sigma$ and scale back afterwards).
Let $g,h$ be standard normal random vectors, independent of $X,Y$. Then by independence,
[
E[e^{uX^TY} | X ] = E[e^{u^2|X|^2/2} ] = E[e^{u X^top g}].
]
Now by the law of total expectation, and a similar argument for $Y$,
[
E[e^{uX^TY}] le E[ E[e^{u X^T g}| g] ] = E[E[ e^{u^2|g|^2/2} | g]] = E[e^{u g^Th}].
]
So the problem is reduced to bounding the moment generating function where both random vectors are standard normal.
answered Dec 1 '18 at 1:02
jlewkjlewk
765
$endgroup$
Moment generating functions of subgaussian vector can often be bounded from above by the same moment generating function, with the subgaussian vector replaced by a standard normal. This is equivalent to zhoraster's answer.
Take $sigma=1$ without loss of generality (otherwise, consider $X/sigma$ and $Y/sigma$ and scale back afterwards).
Let $g,h$ be standard normal random vectors, independent of $X,Y$. Then by independence,
[
E[e^{uX^TY} | X ] = E[e^{u^2|X|^2/2} ] = E[e^{u X^top g}].
]
Now by the law of total expectation, and a similar argument for $Y$,
[
E[e^{uX^TY}] le E[ E[e^{u X^T g}| g] ] = E[E[ e^{u^2|g|^2/2} | g]] = E[e^{u g^Th}].
]
So the problem is reduced to bounding the moment generating function where both random vectors are standard normal.
answered Dec 1 '18 at 1:02
jlewkjlewk
765
answered Dec 1 '18 at 1:02
jlewkjlewk
765
answered Dec 1 '18 at 1:02
jlewkjlewk
765
answered Dec 1 '18 at 1:02
jlewkjlewk
765
765
add a comment |
add a comment |
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