The Problem: Let $Z\thicksim\text{Unif}[0,1]$.
$\textbf{(a)}$ Find the moment generating function of $M_Z(t)$ of $Z$.
$\textbf{(b)}$ For $n\in\mathbb N$, let $X_n$ be a uniform random variable on the set $\left\{\frac{1}{n},\frac{2}{n},\dots,\frac{n-1}{n},1\right\}$, by which we mean that
$P\left(X_n=\frac{k}{n}\right)=\frac{1}{n}$ for each $k\in\{1,2,\dots,n\}$. Prove a limit in distribution $X_n\overset{d}\longrightarrow X$ using moment generating functions and identify the limit.
My Attempt: (a) We have for $t\ne0$,
$$M_Z(t)=E\left[e^{tZ}\right]=\int_0^1 e^{tz}\,dz=\frac{e^t-1}{t}.$$
If $t=0$ then we see from the integral above that $M_Z(t)=1.$
(b) Observe that
\begin{align}M_{X_n}(t)&=E\left[e^{tX_n}\right]=\sum_{k=1}^{n}\frac{e^{tk/n}}{n}\\
&=\frac{1}{n}\sum_{k=1}^n \left(e^{t/n}\right)^k=\frac{1}{n}\cdot\frac{e^{t/n}-e^{t(n+1)/n}}{1-e^{t/n}}\\
&=\frac{1/n(1-e^t)}{e^{-t/n}-1},
\end{align}
for $t\ne0$ and $M_{X_n}(0)=1.$ For $t\in(-1,1)\setminus\{0\}$,
$$M_{X_n}(t)=\frac{1/n(1-e^t)}{e^{-t/n}-1}\longrightarrow\frac{0}{0}\quad\text{as }n\to\infty.$$
So we may apply L'Hopital's rule to the function
$$f(x)=\frac{1/x(1-e^t)}{e^{-t/x}-1}\quad\text{for }x\in\mathbb R\setminus\{0\}.$$
Hence,
$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{e^t-1}{te^{-t/n}}=\frac{e^t-1}{t},$$
whence for all $t\in(-1,1)$,
$$\lim_{n\to\infty}M_{X_n}(t)=\frac{e^t-1}{t}.$$
Since $M_Z(t)$ is finite in the interval $(-1,1)$ and the limit above holds for all $t\in(-1,1)$, the continuity theorem for moment generating functions implies that $X_n\overset{d}\longrightarrow Z.$
What do you think about my proof? Any feedback is most welcomed and appreciated.
Thank you for your time.
Best Answer
You are on the right track. There is another more direct method.
Let $\mu_n$ the uniform distribution on $\{\frac{1}{n},\ldots,\frac{n-1}{n},n\}$ and $\mu$ the uniform distribution over $[0,1]$.
(a) Let $g_t(x)=e^{tx}$. Then $$\mathbb{E}_{\mu_n}[e^{tZ}]=\mu_n g_y=\frac{1}{n}\sum^n_{j=1}e^{t\tfrac{j}{n}}$$ which is the a Riemann sum $g_t$ over $[0,1]$ with the equally spaced partition. $$\mathbb{E}_{\mu}[e^{tZ}]=\mu g_t =\int^1_0 e^{tx}\,dx$$
For any bounded continuous function $f$ $$ \mu_n f =\frac{1}{n}\sum^n_{j=1}f\big(\tfrac{j}{k}\big)\xrightarrow{n\rightarrow\infty}\int^1_0f(x)\,dx = \mu f$$
for $f$ is Integrable and $\mu_n f$ is a convergent sequence of Riemann sums of the integral $\int^1_0f$.
Then, by definition of convergence in distribution, $\mu_n$ converges to $\mu$.
One can use in particular $f_t(x)=e^{i\pi tx}$ to get $\hat{\mu_n}(t)=\mu f_t\xrightarrow{n\rightarrow\infty}\mu f_t=\hat{\mu}(t)$. the get a proof along the lines of your attempt.
About notation:
$\nu f :=\int f\,d\mu=\mathbb{E}_{\nu}[f(X)]$ is expectation under the probability measure $\nu$