{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

note 11 - MA2216/ST2131 Probability Notes 11 Central Limit...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
MA2216/ST2131 Probability Notes 11 Central Limit Theorem Before proceeding to the central limit theorem, we will first review the topics such as moment generating functions as well as conditional expecta- tion, and then consider a few examples for reference. § 1. Moment Generating Functions. 1. Recall that, ( Cf. § 5, Notes 10), the moment generating function of a random variable X is formally defined as M X ( t ) = EE £ e tX / = X x e tx f X ( x ) , if X is discrete with p.d.f. f X ( x ); Z IR e tx f X ( x ) dx, if X is continuous with density f X ( x ). We say the moment generating function of X is well-defined if there exists an η > 0 such that M X ( t ) < for all t ( - η, η ). Obviously, the domain of M X ( · ) is the set of all real numbers such that e tX has finite expectation. A trivial observation is that M X (0) = 1. 2. Why is it called the moment generating function? Because all of the moments of X can be obtained by successively differ- entiating M X ( t ) and then evaluating the result at t = 0. To be precise, if M X ( t ) is well defined in an interval of the origin, then M ( n ) X (0) = EE [ X n ] , n 1 . As a matter of fact, if M X ( t ) is finite on - η < t < η for some positive number η (could possibly be ), then we can write M X ( t ) = X n =0 EE [ X n ] n ! t n , - η < t < η, which is the Taylor series expansion of M X ( t ) at the origin. 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3. Remark. However, not every distribution admits a finite mo- ment generating function. If it does admit one on an interval of the origin, then the distribution itself is uniquely determined by its moment generating function. Such a uniqueness result can be stated as follows: If two random variables have the same moment generating func- tion, they have the same distribution. 4. Example. Let X be a r.v. such that M X ( t ) is finite for all t . We may use the same argument as in the proof of Markov’s inequality to derive the following inequality: IP { X x } ≤ e - tx M X ( t ) , t 0 . (1 . 1) Derivation. For t 0, X x is equivalent to e tX e tx , and hence IP { X x } = IP { e tX e tx } ≤ EE £ e tX / /e tx = e - tx M X ( t ) . We are done. Furthermore, (1.1) holds for all eligible t . Thus, it follows that IP { X x } ≤ min t 0 e - tx M X ( t ) . (1 . 2) In general, the inequality in (1.2) is valid in the following form IP { X x } ≤ min t D e - tx M X ( t ) , (1 . 3) where D refers to the set of all t such that M X ( t ) exists. 2
Image of page 2
5. Let us see a simple application in the following example. Example. Let X have a gamma distribution with parameters α and λ . It is known that M X ( t ) = λ λ - t α for -∞ < t < λ . By making use of (1.3), we get IP { X 2 α/λ } ≤ min 0 t<λ e - t 2 α/λ λ λ - t α By calculus, as a function of t [0 , λ ), ψ ( t ) = e - t 2 α/λ λ λ - t α attains its minimum (2 /e ) α at t = λ/ 2. Therefore, we conclude that IP { X 2 α/λ } ≤ (2 /e ) α . (1 . 4) § 2. Conditional Expectations.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern