Talk:Moment-generating function

Hyphentation

Spelling question: I've never (before now) seen the name spelled with a hyphen. Searches of Math Reviews (MathSciNet) and Current Index to Statistics show an overwhelming preference for no hyphen. Should the title, at least, be changed (move the article to "Moment generating function" with a redirect)? Zaslav 18:08, 8 December 2006 (UTC)

Sir Ronald Fisher always used the hyphen in "moment-generating function". This is an instance of the fact that in this era the traditional hyphenation rules are usually not followed in scholarly writing, nor in advertising or package labelling, althouth they're still generally followed in newspapers, magazines, and novels. This particular term seldom appears in novels, advertisements, etc. Personally I prefer the traditional rules because in some cases they are a very efficient disambiguating tool. Michael Hardy 20:03, 8 December 2006 (UTC)

Terms

I would like _all_ the terms such as E to be defined explicitly. Otherwise these articles are unintelligible to the casual reader. I would have thought that all terms in any formula should be defined every any article, or else reference should be made to some common form of definition of terms for that context. How about a bit more help for the randomly browsing casual student? I would like to see a recommendation in the Wikipedia "guidelines for authors" defining some kind of standard for this, otherwise it is very arbitrary which terms are defined and which are expected to be known.

Definition

the definition of the n-th moment is wrong, the last equality is identically zero, as the nth derivative of 1 evaluated at t=0 will always be zero. the evaluation bar must be placed at the end (so we know we are differentiating M_x(t) n times and evaluating it at zero).

The only thing I find here that resembles a definition of the nth moment is where it says:

the nth moment is given by

$E\left(X^{n}\right)\,$

That definition is correct.

I take the expression

$\left.{\frac {\mathrm {d} ^{n}}{\mathrm {d} t^{n}}}\right|_{t=0}M_{X}(t)$

to mean that we are first differentiating n times and then evaluating at zero. Unless you were referring to something else, your criticism seems misplaced. Michael Hardy 22:05, 8 April 2006 (UTC)

Please provide a few examples, e.g. for a Gaussian distribution.

How about adding something like this?

For the Gaussian distribution

$f(x)={\frac {1}{{\sqrt {2\pi }}\sigma }}e^{-(x-\mu )^{2}/2\sigma ^{2}},$

the moment-generating function is

$M(t)=\int _{-\infty }^{\infty }e^{tx}f(x)\,dx={\frac {1}{{\sqrt {2\pi }}\sigma }}\int _{-\infty }^{\infty }\exp \left[tx-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}\right]dx$

$={\frac {1}{{\sqrt {2\pi }}\sigma }}\int _{-\infty }^{\infty }\exp \left[{\frac {-1}{2\sigma ^{2}}}\left((x-\mu )^{2}-2\sigma ^{2}tx\right)\right]dx.$

Completing the square and simplifying, one obtains

$M(t)=\exp \left(\mu t+{\frac {1}{2}}\sigma ^{2}t^{2}\right).$

(mostly borrowed from article normal distribution.) I don't know if there's enough space for a complete derivation. The "completing the square" part is rather tedious. -- 130.94.162.64 00:37, 17 June 2006 (UTC)

I would also like to see some more in the article about some basic properties of the moment-generating function, such as convexity, non-negativity, the fact that M(0) always equals one, and also some other not-so-obvious properties (of which I lack knowledge) indicating what the mgf is used for. --130.94.162.64 00:55, 17 June 2006 (UTC)

Vector of random variables or stochastic process

We should mention the case when X is a vector of random variables or a stochastic process. Jackzhp 22:29, 3 September 2006 (UTC)

I've added a brief statement about this. (Doubtless more could be said about it.) Michael Hardy 03:48, 4 September 2006 (UTC)

Properties would be nice

There are a whole bunch of properties of MGFs that it would be nice to include -- e.g. the MGF of a linear transformation of a random variable, MGF of a sum of independent random variables, etc.

Discrete form of mgf

something should be added about the discrete form of the mgf, no? $M_{X}(s)=E(e^{sX})=\sum _{k=1}^{\infty }{p(k)e^{sk}}$ 24.136.121.150 08:37, 20 January 2007 (UTC)