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can anyone help me to inform the name of any book where I can get the following theorem, or give some detailed hint to solve this one:

Let $X$ and $Y$ be two random variables, if the moment generating functions of $X$ and $Y$ are equal then the probability distributions of $X$ and $Y$ will be same. In other words, if $E(e^{tX})=E(e^{tY})$ for all $t$ real, then $X$ and $Y$ have the same distribution.

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elinor, I see you have asked 12 questions so far without upvoting or accepting any of the answers you have received. Just so you know, it is considered polite on this site both to upvote answers you find helpful and to formally accept the answer to each question you consider the best. In addition to being polite, doing so is a way to express appreciation for the help others are giving you. –  Mike Spivey Mar 7 '11 at 17:04
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4 Answers

One can look at the contrapositive: $$F_{X}(x) \neq F_{Y}(y) \implies M_{X}(t) \neq M_{Y}(t)$$

I think it would be easier to look at characteristic functions (i.e. $\varphi_{X}(t) = E[e^{it}X]$).

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sorry,could you explain a little more?I am not able to follow how to proceed then –  elinor Mar 7 '11 at 7:42
    
PEV: I would appreciate if you could make precise the ways in which your post is relevant to the question asked. To me, the fact that you did not answer elinor's request for more explanations is on a par with elinor's complete absence of upvotes mentioned by Mike Spivey. –  Did May 28 '11 at 19:03
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An even stronger result is Levy's continuity theorem. Your question is easier, look it up here. If your variable is continuous, then the characteristic function is the Fourier transform of the probability density function, so all you need to do is run the inverse Fourier transform. For the general case, see the reference.

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This chapter (which I found immediately by googling for 'moment generating function proof') contains a proof in the discrete case and a sketch of a proof in the continuous case. It also says that the theorem holds only under the hypothesis that the random variables have finite range.

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Nice reference. There is an ambiguity in your last sentence: in fact, the theorem holds at least under a finite range hypothesis, and effectively in a strictly wider context, see en.wikipedia.org/wiki/Carleman%27s_condition for a start. –  Did May 28 '11 at 19:07
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The idea is to reduce the problem to the well-known case of characteristic functions (cf. Yuval's or PEV's answer). For a complete proof (for moment-generating functions), see Theorem 4.9 here. Further, see Theorem 8.1 here, and property (M1) here.

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