Statistics:minimizing an interval for a standard normal distribution

Click For Summary
SUMMARY

The discussion focuses on minimizing the length of an interval for a standard normal distribution, specifically for arbitrary values of 0 < p < 1. The method involves determining z-scores using the expressions (1-p)/2 and [1-(1-p)/2]. Participants emphasize the importance of local extrema and suggest using Lagrange multipliers to minimize the interval length defined by the endpoints A and B, where B-A = (F^-1)(p+a) - (F^-1)(a). The standard normal cumulative distribution function (CDF) F(z) is central to this analysis.

PREREQUISITES
  • Understanding of standard normal distribution and its properties
  • Familiarity with cumulative distribution functions (CDF)
  • Knowledge of inverse functions and their derivatives
  • Experience with optimization techniques, particularly Lagrange multipliers
NEXT STEPS
  • Study the application of Lagrange multipliers in optimization problems
  • Learn about the properties and applications of the standard normal cumulative distribution function (CDF)
  • Explore the concept of inverse functions and their derivatives in detail
  • Investigate methods for calculating z-scores in statistical analysis
USEFUL FOR

Students in statistics, mathematicians, and data analysts seeking to understand interval minimization in the context of standard normal distributions.

k3k3
Messages
76
Reaction score
0

Homework Statement


Generalize: For arbitrary 0 < p < 1, show that the method giving a and b produces the minimum length interval.

Hint: It might be helpful to use local extrema for the inverse function of the distribution function.


Homework Equations


The method is is talking about is locating the z scores using (1-p)/2 and [1-(1-p)/2]


The Attempt at a Solution


Let a be the area on the tail end of the distribution not included in p
Let b be the other end so that

a+b=1-p and b=1-p-a

Then the points A and B are the end points of the interval containing p.

B-A = (F^-1)(p+a)-(F^-1)(a)

This is where I am stuck. I know f(y). So d/dy(F(y))=f(y) and then (f^-1)'(y)=1/(f'(f^-1)(y))

I am not sure how to proceed.
 
Physics news on Phys.org
k3k3 said:

Homework Statement


Generalize: For arbitrary 0 < p < 1, show that the method giving a and b produces the minimum length interval.

Hint: It might be helpful to use local extrema for the inverse function of the distribution function.


Homework Equations


The method is is talking about is locating the z scores using (1-p)/2 and [1-(1-p)/2]


The Attempt at a Solution


Let a be the area on the tail end of the distribution not included in p
Let b be the other end so that

a+b=1-p and b=1-p-a

Then the points A and B are the end points of the interval containing p.

B-A = (F^-1)(p+a)-(F^-1)(a)

This is where I am stuck. I know f(y). So d/dy(F(y))=f(y) and then (f^-1)'(y)=1/(f'(f^-1)(y))

I am not sure how to proceed.

It is helpful to assume the normal distribution is standard (mean = 0, variance = 1). Do you see why you can do that without loss of generality? Now, if F(z) is the standard normal cdf, you want to minimize b-a, subject to the constraint F(b) - F(a) = p. Hint: Lagrange multipliers.

RGV
 

Similar threads

Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
1K
Replies
32
Views
4K