Motivation of Moment Generating Functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:22:59Z http://mathoverflow.net/feeds/question/34070 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34070/motivation-of-moment-generating-functions Motivation of Moment Generating Functions student 2010-08-01T00:02:07Z 2010-08-01T14:39:54Z <p>What is the motivation of defining the mmoment generating function of a random variable $X$ as: $E[e^{tX}]$? I know that one can obtain the mean, second moment, etc.. after computing it. But what was the intuition in using $e^{x}$? Is it because its one-to-one and always increasing?</p> http://mathoverflow.net/questions/34070/motivation-of-moment-generating-functions/34071#34071 Answer by Yuval Filmus for Motivation of Moment Generating Functions Yuval Filmus 2010-08-01T00:15:25Z 2010-08-01T00:15:25Z <p>If X and Y are independent then $E[e^{t(X+Y)}] = E[e^{tX}] E[e^{tY}]$, so convolution corresponds to multiplication of the mgf's. Another reason: the moment generating function is actually a Fourier transform.</p> <p>Now suppose $X_i$ are i.i.d. with zero mean, and define $Y_n = \sum_{i=1}^n X_i/\sqrt{n}$. Define $\phi(t) = E[e^{tX_1}]$. Then $E[e^{tY_n}] = \phi(t/\sqrt{n})^n$. Under reasonable assumptions, $\phi(t) = 1 + V[X_1]t^2/2 + O(t^3)$, and so $E[e^{tY_n}] = (1 + V[X_1]t^2/2n + O(t^3)/n^{1.5})^n \longrightarrow e^{V[X_1]t^2/2}$, and we get the central limit theorem (by continuity of the Fourier transform). </p> http://mathoverflow.net/questions/34070/motivation-of-moment-generating-functions/34072#34072 Answer by Gerald Edgar for Motivation of Moment Generating Functions Gerald Edgar 2010-08-01T00:22:31Z 2010-08-01T00:22:31Z <p>Take the definition of "generating function" for a sequence. Do it where the sequence is the sequence consisting of the moments of $X$. That's it. </p> http://mathoverflow.net/questions/34070/motivation-of-moment-generating-functions/34104#34104 Answer by Mark Meckes for Motivation of Moment Generating Functions Mark Meckes 2010-08-01T11:53:27Z 2010-08-01T13:52:29Z <p>As you suggest, the fact that $e^x$ is increasing is a useful property here. In particular, that lets you in some cases to apply Markov's/Chebychev's inequality to the random variable $e^{tX}$ in order to get exponentially decaying bounds on the tails of $X$; see e.g. <a href="http://en.wikipedia.org/wiki/Chernoff_bound" rel="nofollow">Chernoff bounds</a>. In principle any other positive increasing function could be used in the same way, but $e^x$ is a particularly useful choice because it is especially well suited to studying sums of independent random variables, as noted already in Yuval's answer.</p> http://mathoverflow.net/questions/34070/motivation-of-moment-generating-functions/34112#34112 Answer by Chris Godsil for Motivation of Moment Generating Functions Chris Godsil 2010-08-01T14:39:54Z 2010-08-01T14:39:54Z <p>The goal is to to put all the moments in one package. Since $$e^{tx} = \sum \frac{x^n}{n!} t^n$$ the coefficients of $t^n$ in $E(e^{tx})$ are (scaled) moments. In other contexts we can use $$(1-xt)^{-1} = \sum x^n t^n$$ in place of $e^{tx}$. This gives more or less what engineers call the "z-transform" and in combinatorics it is known as "ordinary generating function". Using the exponential has the happy advantage that convolution of random variables translates to product of moment generating functions.</p>