Suppose $X$ and $Y$ are two random variables such that all of their moments exist and $E[X^r]=E[Y^r]$ for all integers $r=0,1,2...$
If $X$ and $Y$ have bounded support, then this implies that they have the same distribution.
When the support is not bounded, equality of moments does not imply that they have the same distribution. Here, the classic example is the log normal and the perturbation of it. The mgf does not exist for the log normal.
My question is: if the moment generating functions for $X$ and $Y$ exist, and all the moments of $X$ and $Y$ are the same, does this imply that the they are equal in distribution (i.e. that the mgf are the same)?
My answer is (not very rigorous): $$E[e^{tX}]=E[1+tX + \frac{t^2}{2}X^2+...]$$ $$E[e^{tY}]=E[1+tY + \frac{t^2}{2}Y^2+...]$$ so that if $E[X^r]=E[Y^r]$, then indeed the moment generating functions have to be the same around 0. Is this correct?
Provided that this is true, then it seems that the values of the derivative at 0 is determining the function in a neighborhood of 0? Why is this?