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Moment generating function for a random variable X is given by M(t)=$E[e^{tX}]$ for some −h < t < h.

Why does −h < t < h ?.

I am not able to understand the importance and meaning of parameter t in the definition of moment generating function. What does it signify?

I guess understanding this below question might help me in understanding the concept of parameter t in moment generating function. So please explain that also.

Let X be a random variable with mgf M(t), −h < t < h.

Prove that

(a) $P(X ≥ a) ≤ e^{−at}M(t) , 0 < t < h$

and that

(b) $P(X ≤ a) ≤ e^{−at}M(t) , − h < t < 0 $

(I am able to derive part(a) using markov's inequality but I dont understand for (a) to be true , why $ 0 < t < h$ ?)

Ayush
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1 Answers1

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There are at least two reasons for this.

  1. Many r.v. Do not have an expected value because their PDFs are too large at infinity. In particular $E(e^X)$ may not be defined

  2. All the useful information (namely the moments) is in the the derivatives at zero, and those exist as soon as you have a function defined in a neighborhood of zero.

Also, note that $M$ is a version of the Laplace transform of the pdf. (It is bilateral though) which exists typically for $t$ with a small real part.

GReyes
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  • thank you

    I have proved (a) and (b) which dont have answer yet:-

    (a) $P[X\geq a]$

    $=P[tX \geq ta]$ -> t is positive

    $=P[e^{tX} \geq e^{at}]$

    $ \leq \frac{E[e^{tX}]}{e^{at}}$ -> markov's inequality

    (b) $P[X\gt a]$

    $=P[tX \lt ta]$ ->t is negative

    $=P[e^{tX} \lt e^{at}]$

    $=1-P[e^{tX} \geq e^{at}]$

    $ \geq 1- \frac{E[e^{tX}]}{e^{at}}$ -> by markov's inequality

    $P[X\gt a] \geq 1-\frac{E[e^{tX}]}{e^{at}}$

    $=>\frac{E[e^{tX}]}{e^{at}} \geq 1-P[X\gt a]$

    $=>\frac{E[e^{tX}]}{e^{at}} \geq P[X\leq a]$

     – Ayush Aug 20 '21 at 12:52
  • Seems correct to me… – GReyes Aug 20 '21 at 19:38