MHB How to prove the normal distribution tail inequality for large x ?

Click For Summary
SUMMARY

This discussion focuses on proving the normal distribution tail inequality for large values of x, specifically addressing lemma (7.1) and problem 1. Key steps include analyzing derivatives of expressions in (1.8) and (1.9) and demonstrating that the left side of the inequality is slightly less than n(x) while the right side is slightly more. The discussion emphasizes the necessity of rewriting lemma (7.1) in the form of (1.8) and taking derivatives to establish the proof. Participants also seek clarification on specific computations related to the cumulative distribution function (CDF) of the standard normal distribution.

PREREQUISITES
  • Understanding of normal distribution and its properties
  • Familiarity with calculus, particularly differentiation
  • Knowledge of cumulative distribution functions (CDF)
  • Experience with mathematical proofs and inequalities
NEXT STEPS
  • Study the derivation of the cumulative distribution function (CDF) for the standard normal distribution
  • Learn about the properties of derivatives in the context of probability distributions
  • Explore advanced topics in mathematical proofs, focusing on inequalities
  • Investigate the implications of the normal distribution tail inequality in statistical applications
USEFUL FOR

Mathematicians, statisticians, and students studying probability theory, particularly those interested in the properties of the normal distribution and its applications in statistical inference.

WMDhamnekar
MHB
Messages
378
Reaction score
30
1635782991047.png

1635783023489.png

1635783043074.png


What is the meaning of this proof? What is the meaning of last statement of this proof? How to prove lemma (7.1)? or How to answer problem 1 given below?

1635783071928.png
1635783082068.png

1635784677943.png
 
Last edited:
Physics news on Phys.org
Dhamnekar Winod said:
View attachment 11392
View attachment 11393

What is the meaning of this proof? What is the meaning of last statement of this proof? How to prove lemma (7.1)? or How to answer problem 1 given below?

If $x>0$ then the inequality in (1.9) must be true, because the left side is then slightly less than n(x), and the right side is slightly more than n(x).

Let's consider the derivatives of the expressions in (1.8).
$$\frac d{dx}(1-\Re(x)) = -\Re'(x) = -n(x)$$

Let's first find $n'(x)$.
We have:
$$n'(x)=\frac d{dx} \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} = \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} \cdot -x =-xn(x)$$
Then we have for instance:
$$\frac d{dx}(x^{-1}n(x)) = -x^{-2}n(x) + x^{-1}n'(x) = -x^{-2}n(x)+x^{-1}\cdot -x n(x) = -(1+x^{-2})n(x)$$
So we see that the derivatives of the expressions in (1.8) are indeed the negatives of the expressions in (1.9).

We have that $1-\Re(x)$ is in between 2 expressions, so its integration must also be between the integrations of the those 2 expressions.
Qed.

To prove the more general formula, we need to repeat these steps for the additional terms.
 
Klaas van Aarsen said:
If $x>0$ then the inequality in (1.9) must be true, because the left side is then slightly less than n(x), and the right side is slightly more than n(x).

Let's consider the derivatives of the expressions in (1.8).
$$\frac d{dx}(1-\Re(x)) = -\Re'(x) = -n(x)$$

Let's first find $n'(x)$.
We have:
$$n'(x)=\frac d{dx} \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} = \frac 1{\sqrt{2\pi}} e^{-\frac 12x^2} \cdot -x =-xn(x)$$
Then we have for instance:
$$\frac d{dx}(x^{-1}n(x)) = -x^{-2}n(x) + x^{-1}n'(x) = -x^{-2}n(x)+x^{-1}\cdot -x n(x) = -(1+x^{-2})n(x)$$
So we see that the derivatives of the expressions in (1.8) are indeed the negatives of the expressions in (1.9).

We have that $1-\Re(x)$ is in between 2 expressions, so its integration must also be between the integrations of the those 2 expressions.
Qed.

To prove the more general formula, we need to repeat these steps for the additional terms.
But, how to use this information to solve the given problem 1 or lemma 7.1?
 
Dhamnekar Winod said:
But, how to use this information to solve the given problem 1 or lemma 7.1?
Write (7.1) in the same form as (1.8) with the series on the left and also on the right.
Take the derivatives to find an expression that is in the same form as (1.9).
Then the proof follows in the same fashion.
 
Klaas van Aarsen said:
Write (7.1) in the same form as (1.8) with the series on the left and also on the right.
Take the derivatives to find an expression that is in the same form as (1.9).
Then the proof follows in the same fashion.
Thanks for your answer. But sorry for not understanding it. I want to know your way of answering this problem.
But I got some math help from wikipedia.
1635875496626.png


How can we use the above two formulas of CDF of Normal distribution to prove lemma 7.1 in the original question?

I got the following proofs of expansion of CDF of standard normal distribution.

1635919963509.png


CDFNDINT.png

CDFNDINT2.png

CDFNDINT1.png

1635921860126.png


In the above expansions of CDF of standard normal distribution, I want to know how the highlighted or marked computations was performed. If any member of this MHB knows the method of these computations, may explain it in reply.

Following is the simple proof:
1635922488791.png

1635922506052.png

How the last omitted term was computed by the author?
 
Last edited:
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

Similar threads

  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K