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Find the closest point using distances according to f.(a) Show that for any fixe...

Sep 24, 2024

Find the closest point using distances according to f.(a) Show that for any fixed vector v in Rd, there exists constant c > 0 such that for all 0<ε<1， Plf(v)|-vkul>evkull<3exp(-cke?). where the probability is taken over the draws of u,. Hint: Use the following theoremLet X be drawn from a k-dimensional normal with covariance matrix equal to thidentity. For any Bvk>0, then all but at most 3exp(-cB )of the probabilitrmass lies within vk-ß<X<yk+β where c is a fixed positive constant

Solution by Steps

step 1

To show that for any fixed vector

v

\mathbb{R}^d

, there exists a constant c > 0 such that for all 0 < \epsilon < 1 , the inequality P(|f(v)| - |v| < \epsilon \|v\|) < 3 \exp(-c \epsilon) holds, we start by considering the properties of the k-dimensional normal distribution

step 2

According to the theorem, if

X

is drawn from a k-dimensional normal distribution with covariance matrix equal to the identity, then for any B > 0 , all but at most

3 \exp(-cB)

of the probability mass lies within the interval v_k - \beta < X < y_k + \beta

step 3

We can set

B = \epsilon \|v\|

to relate the probability mass to the distance from

v

. This gives us the desired form of the inequality

step 4

Thus, we conclude that there exists a constant c > 0 such that the probability P(|f(v)| - |v| < \epsilon \|v\|) < 3 \exp(-c \epsilon) holds for all 0 < \epsilon < 1

Answer

The inequality holds for a constant c > 0 and all 0 < \epsilon < 1 .

Key Concept

The relationship between probability distributions and distances in high-dimensional spaces.

Explanation

The result shows how the concentration of measure phenomenon applies to normal distributions, ensuring that most of the probability mass is close to the mean vector.