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$a)$ Consider an uniform discrete distribution ( $X$ ~ uniform{$0,...,n-1$} ) and find the moment-generating function.

$b)$ Now Consider $Y$ ~ uniform{$\frac{0}{n},...,\frac{n-1}{n}$} and find the moment-generating function $m_n(t)$. Calculate $lim_{n \rightarrow \infty} m_n(t). $

$a)$ : $ m[n] (t) = \mathbb{E}(e^{tX}) = \sum_{k=0}^{n-1} e^{tk} P(X=k) = \sum_{k=0}^{n-1} e^{tk} \frac{1}{n} = \frac{1}{n} \sum_{k=0}^{n-1} (e^{t})^{k} = \frac{1}{n} \frac{1-e^{tn}}{1-e^t} $ for $t \neq 0$. for $t=0$ we would have $1$. So far so good.

but how can I solve $b)$ ? I mean it looks like $a)$. But there's a difference, I think.

Thank you.

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Since $Y = X/n$, we can compute $$ m_Y(t) = \mathbb{E}(e^{tX/n}) = m_X(t/n).$$ From your computation, this is 1 if $t=0$, and $$ \frac{1}{n} \frac{1 - e^t}{1-e^{t/n}}$$ otherwise. The limit of this expression as $n\rightarrow \infty$, is $$ \frac{1}{t} (e^t - 1).$$

Note that this limit is the moment generating function of the uniform distribution on $[0,1]$, which is what $Y$ is approximating.

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