What is the value of this term and why is it not included in the given passage?

Click For Summary
SUMMARY

The discussion centers on the term derived from integration by parts involving the normal probability density function (p.d.f.) and its cumulative distribution function (c.d.f.). Specifically, the term [ e^{itx} \phi(x) ]^{+ \infty}_{- \infty} is analyzed, where \phi(x) represents the normal p.d.f. The conclusion reached is that this term evaluates to zero as x approaches infinity due to the rapid decay of e^{-x^2}. This decay is sufficient to ensure that the entire expression approaches zero.

PREREQUISITES
  • Understanding of complex analysis, particularly the behavior of exponential functions.
  • Familiarity with the properties of the normal distribution, including its p.d.f. and c.d.f.
  • Knowledge of integration techniques, specifically integration by parts.
  • Basic grasp of limits and asymptotic behavior of functions as they approach infinity.
NEXT STEPS
  • Study the properties of the normal distribution, focusing on its p.d.f. and c.d.f.
  • Learn about integration by parts in the context of complex functions.
  • Explore the asymptotic behavior of exponential functions, particularly e^{-x^2}.
  • Investigate the implications of complex analysis in probability theory.
USEFUL FOR

Mathematicians, statisticians, and students studying probability theory and complex analysis will benefit from this discussion, particularly those interested in the properties of the normal distribution and integration techniques.

St41n
Messages
32
Reaction score
0
I found this passage in a book:
http://img8.imageshack.us/img8/1452/75717730.jpg
where Φ(x) is the c.d.f. of the normal distribution.
However, using integration by parts I also get this term which is not included in the passage:

[tex][ e^{itx} \phi(x) ]^{+ \infty}_{- \infty}[/tex]
where i is the imaginary unit and [tex]\phi(x)[/tex] is the normal p.d.f.
So,
[tex]e^{itx} \phi(x) = \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right)[/tex]
So, is this term:
[tex]\left[ \frac{1}{\sqrt{2\pi}} \exp \left( -0.5x^2 + itx \right) \right]^{+ \infty}_{- \infty}[/tex]
equal to zero and why? It is not clearly evident to me
 
Last edited by a moderator:
Physics news on Phys.org
Yes, as x go to infinity,
[tex]e^{-x^2}[/tex]
goes to 0 very quickly. Quickly enough so that even
[tex]e^{-x^2}e^{x}[/tex] goes to 0 as x goes to positive infinity.
 
I kept doing the same fault, but I figured it out now, thanks
 

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 27 ·
Replies
27
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K