<< Chapter < Page Chapter >> Page >
This module establishes concentration bounds for sub-Gaussian vectors and matrices.

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 X is a vector where each X i is i.i.d. with X i Sub ( c ) , then we would like to know how X 2 deviates from its mean.

In order to establish the result, we will make use of Markov's inequality for nonnegative random variables.

(markov's inequality)

For any nonnegative random variable X and t > 0 ,

P ( X t ) E ( X ) t .

Let f ( x ) denote the probability density function for X .

E ( X ) = 0 x f ( x ) d x t x f ( x ) d x t t f ( x ) d x = t P ( X t ) .

In addition, we will require the following bound on the exponential moment of a sub-Gaussian random variable.

Suppose X Sub ( c 2 ) . Then

E exp ( λ X 2 / 2 c 2 ) 1 1 - λ ,

for any λ [ 0 , 1 ) .

First, observe that if λ = 0 , then the lemma holds trivially. Thus, suppose that λ ( 0 , 1 ) . Let f ( x ) denote the probability density function for X . Since X is sub-Gaussian, we have by definition that

- exp ( t x ) f ( x ) d x exp ( c 2 t 2 / 2 )

for any t R . If we multiply by exp ( - c 2 t 2 / 2 λ ) , then we obtain

- exp ( t x - c 2 t 2 / 2 λ ) f ( x ) d x exp ( c 2 t 2 ( λ - 1 ) / 2 λ ) .

Now, integrating both sides with respect to t , we obtain

- - exp ( t x - c 2 t 2 / 2 λ ) d t f ( x ) d x - exp ( c 2 t 2 ( λ - 1 ) / 2 λ ) d t ,

which reduces to

1 c 2 π λ - exp ( λ x 2 / 2 c 2 ) f ( x ) d x 1 c 2 π λ 1 - λ .

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 X = [ X 1 , X 2 , ... , X M ] , where each X i is i.i.d. with X i Sub ( c 2 ) and E ( X i 2 ) = σ 2 . Then

E X 2 2 = M σ 2 .

Moreover, for any α ( 0 , 1 ) and for any β [ c 2 / σ 2 , β max ] , there exists a constant κ * 4 depending only on β max and the ratio σ 2 / c 2 such that

P X 2 2 α M σ 2 exp - M ( 1 - α ) 2 / κ *

and

P X 2 2 β M σ 2 exp - M ( β - 1 ) 2 / κ * .

Since the X i are independent, we obtain

E X 2 2 = i = 1 M E X i 2 = i = 1 M σ 2 = M σ 2

and hence [link] holds. We now turn to [link] and [link] . Let us first consider [link] . We begin by applying Markov's inequality:

P X 2 2 β M σ 2 = P exp ( λ X 2 2 ) exp λ β M σ 2 E exp ( λ X 2 2 ) exp λ β M σ 2 = i = 1 M E exp ( λ X i 2 ) exp λ β M σ 2 .

Since X i Sub ( c 2 ) , we have from [link] that

E exp ( λ X i 2 ) = E exp ( 2 c 2 λ X i 2 / 2 c 2 ) 1 1 - 2 c 2 λ .

Thus,

i = 1 M E exp λ X i 2 1 1 - 2 c 2 λ M / 2

and hence

P X 2 2 β M σ 2 exp - 2 λ β σ 2 1 - 2 c 2 λ M / 2 .

By setting the derivative to zero and solving for λ , one can show that the optimal λ is

λ = β σ 2 - c 2 2 c 2 σ 2 ( 1 + β ) .

Plugging this in we obtain

P X 2 2 β M σ 2 β σ 2 c 2 exp 1 - β σ 2 c 2 M / 2 .

Similarly,

P X 2 2 α M σ 2 α σ 2 c 2 exp 1 - α σ 2 c 2 M / 2 .

In order to combine and simplify these inequalities, note that if we define

κ * = max 4 , 2 ( β max σ 2 / c - 1 ) 2 ( β max σ 2 / c - 1 ) - log ( β max σ 2 / c )

then we have that for any γ [ 0 , β max σ 2 / c ] we have the bound

log ( γ ) ( γ - 1 ) - 2 ( γ - 1 ) 2 κ * ,

and hence

γ exp ( γ - 1 ) - 2 ( γ - 1 ) 2 κ * .

By setting γ = α σ 2 / c 2 , [link] reduces to yield [link] . Similarly, setting γ = β σ 2 / c 2 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 M σ 2 with exponentially high probability as M grows. Note, however, that the range of allowable choices for β in [link] is limited to β c 2 / σ 2 1 . 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 c 2 = σ 2 , 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 β ( 1 , 2 ] is an admissible range for β and then to simplify the computation of κ * . 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 ϵ provided that c 2 / σ 2 < 2 . 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 ϵ , then there is considerable room for improvement in the constant C * .

Suppose that X = [ X 1 , X 2 , ... , X M ] , where each X i is i.i.d. with X i SSub ( σ 2 ) . Then

E X 2 2 = M σ 2

and for any ϵ > 0 ,

P X 2 2 - M σ 2 ϵ M σ 2 2 exp - M ϵ 2 κ *

with κ * = 2 / ( 1 - log ( 2 ) ) 6 . 52 .

Since each X i SSub ( σ 2 ) , we have that X i Sub ( σ 2 ) and E ( X i 2 ) = σ 2 , in which case we may apply [link] with α = 1 - ϵ and β = 1 + ϵ . This allows us to simplify and combine the bounds in [link] and [link] to obtain [link] . The value of κ * follows from the observation that 1 + ϵ 2 so that we can set β max = 2 .

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 Φ is an M × N matrix whose entries φ i j are i.i.d. with φ i j SSub ( 1 / M ) . Let Y = Φ x for x R N . Then for any ϵ > 0 , and any x R N ,

E Y 2 2 = x 2 2

and

P Y 2 2 - x 2 2 ϵ x 2 2 2 exp - M ϵ 2 κ *

with κ * = 2 / ( 1 - log ( 2 ) ) 6 . 52 .

Let φ i denote the i th row of Φ . Observe that if Y i denotes the first element of Y , then Y i = φ i , x , and thus by Lemma 2 from "Sub-Gaussian random variables" , Y i SSub x 2 2 / M . Applying [link] to the M -dimensional random vector Y , we obtain [link] .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, An introduction to compressive sensing. OpenStax CNX. Apr 02, 2011 Download for free at http://legacy.cnx.org/content/col11133/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'An introduction to compressive sensing' conversation and receive update notifications?

Ask