1. (Risk minimization and a simplified binary classifier) In lecture, we learnt about the Neyman-Pearson framework: we seek to find a "best" critical region that maximizes the power of the test while main- taining a given significance level. Another major paradigm in the literature to define a "best" critical region is known as risk minimization that we now introduce. Let X ~ N(μ, 1). We are interested in testing Ho:}=0, H₁ = 1. As we only have a single sample X, we intend to use the test statistic T and critical region C to be respectively T=X, C={X> c}, where c>0 is a constant. Let be the standard normal cdf. (a) Prove that the probability of type I error is a(c) = 1 − (c). The bracket of c on the left hand side is to indicate the dependence on c of a, that is, a is a function of c. (b) Prove that the probability of type II error is B(c) = (c-1). The bracket of c on the left hand side is to indicate the dependence on c of ẞ, that is, ẞ is a function of c. (c) Define the risk R to be the sum of the probability of type I and type II error, that is, R(c) = a(c) + ẞ(c) = 1 − Þ(c) + Þ(c − 1). In risk minimization, we seek to find an optimal critical region by minimizing R(c) with respect to c. Let c* be the resulting minimizer, that is, Prove that c* = arg min R(c). C 1 2 and hence R(c*) = 2(−1/2).
1. (Risk minimization and a simplified binary classifier) In lecture, we learnt about the Neyman-Pearson framework: we seek to find a "best" critical region that maximizes the power of the test while main- taining a given significance level. Another major paradigm in the literature to define a "best" critical region is known as risk minimization that we now introduce. Let X ~ N(μ, 1). We are interested in testing Ho:}=0, H₁ = 1. As we only have a single sample X, we intend to use the test statistic T and critical region C to be respectively T=X, C={X> c}, where c>0 is a constant. Let be the standard normal cdf. (a) Prove that the probability of type I error is a(c) = 1 − (c). The bracket of c on the left hand side is to indicate the dependence on c of a, that is, a is a function of c. (b) Prove that the probability of type II error is B(c) = (c-1). The bracket of c on the left hand side is to indicate the dependence on c of ẞ, that is, ẞ is a function of c. (c) Define the risk R to be the sum of the probability of type I and type II error, that is, R(c) = a(c) + ẞ(c) = 1 − Þ(c) + Þ(c − 1). In risk minimization, we seek to find an optimal critical region by minimizing R(c) with respect to c. Let c* be the resulting minimizer, that is, Prove that c* = arg min R(c). C 1 2 and hence R(c*) = 2(−1/2).
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Can I get help with parts a), b) and c)
Transcribed Image Text:1. (Risk minimization and a simplified binary classifier) In lecture, we learnt about the Neyman-Pearson
framework: we seek to find a "best" critical region that maximizes the power of the test while main-
taining a given significance level. Another major paradigm in the literature to define a "best" critical
region is known as risk minimization that we now introduce. Let X ~ N(μ, 1). We are interested in
testing
Ho:}=0, H₁ = 1.
As we only have a single sample X, we intend to use the test statistic T and critical region C to be
respectively
T=X, C={X> c},
where c>0 is a constant. Let be the standard normal cdf.
(a) Prove that the probability of type I error is
a(c) = 1 − (c).
The bracket of c on the left hand side is to indicate the dependence on c of a, that is, a is a
function of c.
(b) Prove that the probability of type II error is
B(c) = (c-1).
The bracket of c on the left hand side is to indicate the dependence on c of ẞ, that is, ẞ is a
function of c.
(c) Define the risk R to be the sum of the probability of type I and type II error, that is,
R(c) = a(c) + ẞ(c) = 1 − Þ(c) + Þ(c − 1).
In risk minimization, we seek to find an optimal critical region by minimizing R(c) with respect
to c. Let c* be the resulting minimizer, that is,
Prove that
c* = arg min R(c).
C
1
2
and hence
R(c*) = 2(−1/2).
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