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Suppose $X,Y$ are two random variables.They have MGF:$M_X(t)=\mathbb E(e^{tX}),M_Y(t)=\mathbb E(e^{tY})$

If $M_X(t)\equiv M_Y(t)$

Why can we conclude $X,Y$ have the same distribution?

(like the characteristic funtion ,could we have an equation like inversion formula?)

Lookout
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2 Answers2

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First recall that two random variables $X$ and $Y$ can have the same moments $E(X^k)=E(Y^k)$ for every nonnegative integer $k$ and yet have different CDFs (examples are in the book by Casella and Berger). However, when the MGFs are finite in an interval around $0$ and coincide, they indeed determine the CDF (see Probability and Measure by Billingsley).

Did
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From Wikipedia:

An important property of the moment-generating function is that if two distributions have the same moment-generating function, then they are identical at almost all points. That is, if for all values of t,

$M_X(t) = M_Y(t),$

then

$F_X(x) = F_Y(x)$

for all values of x (or equivalently X and Y have the same distribution).

This statement is not equivalent to "if two distributions have the same moments, then they are identical at all points", because in some cases the moments exist and yet the moment-generating function does not, because in some cases the limit

$\displaystyle \lim_{n \rightarrow \infty} \sum_{i=0}^n \frac{t^im_i}{i!}$

does not exist. This happens for the lognormal distribution.

If the moment generating function is defined on such an interval, then it uniquely determines a probability distribution.

square_one
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