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In my book "Mathematical Statistics with Applications" by Dennis Wackerly it's stated that the moment generating function exists if

The moment-generating function m(t) for a random variable Y is defined to be m(t) = E($e^{tY}$ ). We say that a moment-generating function for Y exists if there exists a positive constant b such that m(t) is finite for |t| ≤ b.

What does it mean having a constant be such that the absolute value of the argument is less or equals to B?

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  • $\begingroup$ 1. You should not change the case from $b$ to $B$; stick with the same notation as the thing you quote. 2. If there's any positive value for which the statement is true, "$b$" can stand for it in the statement. $\endgroup$ – Glen_b Jul 14 at 0:48
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You probably need to brush up on your $\epsilon-\delta$ definitions! First an example: The mgf of the exponential distribution with rate $\lambda>0$ is $m(t)=\lambda / (\lambda-t)$. By analyzing that function as in high school math, you will see it is undefined (division by zero) for $t=\lambda$, and negative for $t>\lambda$. Since mgf's can never be negative, that cannot be right, and by looking at the actual evaluation of the integral defining the mgf, you will see that the integral diverges for such $t$, so to get the expression given, one must assume $t<\lambda$.

So, by choosing $b=\lambda$, there is a positive constant $b$ such that $m(t)$ exists (and is finite) for $|t|<b$. (In this case it actually exists for $t<b$, but that is not necessary.) Hoping this example clears your confusion.

See also Existence of the moment generating function and variance for good explanations of basic facts about the mgf.

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    $\begingroup$ Thank you. Yes, it helped a lot. $\endgroup$ – Maria Lavrovskaya Jul 13 at 12:27

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