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Sub-Gaussian distributions have a close relationship to the concentration of measure phenomenon [link] . To illustrate this, we note that we can combine Lemma 2 and Theorem 1 from "Sub-Gaussian random variables" to obtain deviation bounds for weighted sums of sub-Gaussian random variables. For our purposes, however, it will be more enlightening to study the norm of a vector of sub-Gaussian random variables. In particular, if is a vector where each is i.i.d. with , then we would like to know how deviates from its mean.
In order to establish the result, we will make use of Markov's inequality for nonnegative random variables.
For any nonnegative random variable and ,
Let denote the probability density function for .
In addition, we will require the following bound on the exponential moment of a sub-Gaussian random variable.
Suppose . Then
for any .
First, observe that if , then the lemma holds trivially. Thus, suppose that . Let denote the probability density function for . Since is sub-Gaussian, we have by definition that
for any . If we multiply by , then we obtain
Now, integrating both sides with respect to , we obtain
which reduces to
This simplifies to prove the lemma.
We now state our main theorem, which generalizes the results of [link] and uses substantially the same proof technique.
Suppose that , where each is i.i.d. with and . Then
Moreover, for any and for any , there exists a constant depending only on and the ratio such that
and
Since the are independent, we obtain
and hence [link] holds. We now turn to [link] and [link] . Let us first consider [link] . We begin by applying Markov's inequality:
Since , we have from [link] that
Thus,
and hence
By setting the derivative to zero and solving for , one can show that the optimal is
Plugging this in we obtain
Similarly,
In order to combine and simplify these inequalities, note that if we define
then we have that for any we have the bound
and hence
By setting , [link] reduces to yield [link] . Similarly, setting establishes [link] .
This result tells us that given a vector with entries drawn according to a sub-Gaussian distribution, we can expect the norm of the vector to concentrate around its expected value of with exponentially high probability as grows. Note, however, that the range of allowable choices for in [link] is limited to . Thus, for a general sub-Gaussian distribution, we may be unable to achieve an arbitrarily tight concentration. However, recall that for strictly sub-Gaussian distributions we have that , in which there is no such restriction. Moreover, for strictly sub-Gaussian distributions we also have the following useful corollary. [link] exploits the strictness in the strictly sub-Gaussian distribution twice — first to ensure that is an admissible range for and then to simplify the computation of . One could easily establish a different version of this corollary for non-strictly sub-Gaussian vectors but for which we consider a more restricted range of provided that . However, since most of the distributions of interest in this thesis are indeed strictly sub-Gaussian, we do not pursue this route. Note also that if one is interested in very small , then there is considerable room for improvement in the constant .
Suppose that , where each is i.i.d. with . Then
and for any ,
with .
Since each , we have that and , in which case we may apply [link] with and . This allows us to simplify and combine the bounds in [link] and [link] to obtain [link] . The value of follows from the observation that so that we can set .
Finally, from [link] we also have the following additional useful corollary. This result generalizes the main results of [link] to the broader family of general strictly sub-Gaussian distributions via a much simpler proof.
Suppose that is an matrix whose entries are i.i.d. with . Let for . Then for any , and any ,
and
with .
Let denote the row of . Observe that if denotes the first element of , then , and thus by Lemma 2 from "Sub-Gaussian random variables" , . Applying [link] to the -dimensional random vector , we obtain [link] .
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