If $X_n\thicksim\text{Unif}\left\{\frac{1}{n},\frac{2}{n},\dots,\frac{n-1}{n},1\right\}$ then $X_n\overset{d}\to Z$ where $Z\thicksim\text{Unif}[0,1]$

Question

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 $X_n\overset{d}\longrightarrow Z.$

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What do you think about my proof? Any feedback is most welcomed and appreciated.
Thank you for your time.

Answer 1

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 $\mu$.

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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.

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About notation:

$\nu f :=\int f\,d\mu=\mathbb{E}_{\nu}[f(X)]$ is expectation under the probability measure $\nu$

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Answer 2

I also agree the MGFs are overkill here. For part b we need to show:

$$\left |\mathbb{P}(X_n\leq t) - \mathbb{P}(Z\leq t) \right | \rightarrow 0$$

Rewriting:

$$\left |\frac{\left\lfloor nt \right\rfloor}{n} - t \right | = \left |\frac{\left\lfloor nt \right\rfloor}{n} - \frac{nt}{n} \right |$$ which is at most $1/n$ which goes to $0$ as desired.

Your proof for both parts looks correct, but you are already pretty close after showing the MGF in part b. Note that you want to show:

$$\left |\frac{e^t-1}{t} - \frac{e^t-1}{n(1-e^{-t/n})}\right | \rightarrow 0$$

So all that's needed to show is that $n(1-e^{-t/n}) \rightarrow t$. We can show this by applying the monotone convergence theorem. Observe that $n(1-e^{-t/n})$ is increasing and bounded above by $t$ (this bound is tight):

$$(1-e^{-t/n}) \leq t/n$$

so we have:

$$n(1-e^{-t/n}) \rightarrow t$$

which leads to the desired convergence.

Answer 3

a very fast way to prove this Convergence in law is to observe that

$$ F_{X_{n}}(t) = \begin{cases} 0, & \text{if $t<\frac{1}{n}$} \\ t, & \text{if $t=\{\frac{1}{n},\frac{2}{n},...,\frac{n-1}{n}\}$} \\ 1, & \text{if $t \geq \frac{n}{n}$} \end{cases}$$

Thus the limit when $n\rightarrow +\infty$ is

$$ F_{X}(t) = \begin{cases} 0, & \text{if $t<0$} \\ t, & \text{if $t \in [0;1)$} \\ 1, & \text{if $t \geq 1$} \end{cases}$$

Similar example, if you wanna try, is when $X_n$ is discrete uniform on the set $\{0;\frac{1}{n},\frac{2}{n},...,1\}$