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I am given the MGF of IID RVs as $ e^{\lambda(e^t-1)}$ and am supposed to find the MGF of their sample mean of n such RVs. I am arriving at

$M_n(t) = e^{n\lambda(e^{t/n} - 1)}$

This does appear to be a valid MGF, however, there is another part to the question where it asks for the $\lim_{n \to \infty} M_n(t)$

I am confused if it is proper to have the MGF equal to 1 for ${n \to \infty}$.

Thank You

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    $1$ is MGF of r.v. which is equal to zero a.s. But this is clearly not the case. Use Taylor series expansion $e^{t/n}=1+\frac{t}{n}+o(\frac{t}{n})$ and find proper limit of $n\lambda(e^{t/n}-1)$. – 2017-02-13
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    thank you so much NCh. So, $/lim_{n\to\infty} M_n(t)=e^{\lambda t}$. What kind of distribution does it imply? – 2017-02-13
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    MGF of $X$ is $Ee^{tX}$. It is equal to $e^{\lambda t}$ iff $X=\lambda$ a.s. So, the limiting distribution is degenerate at $\lambda$. – 2017-02-13
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    Oh I see. Thank you. When will i be as good at maths – 2017-02-15

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