Show that $ \frac{X_{n}}{n} \stackrel{\mathrm{d}}{\longrightarrow} X $ for $ n \rightarrow \infty $

Question

Let $ X_{n} \sim \mathrm{U}\{1, \ldots, n\} $ and $ X \sim \mathrm{U}(0,1) $. Show that $ \frac{X_{n}}{n} \stackrel{\mathrm{d}}{\longrightarrow} X $ for $ n \rightarrow \infty $.

Attempt/Idea:

To show that $ \frac{X_n}{n} $ converges in distribution to $ X $ as $ n \rightarrow \infty $, we need to prove the convergence of their distribution functions.

First, let's consider the distribution function of $ \frac{X_n}{n} $. Let $ t $ be any point in $ (0,1) $. The distribution function of $ \frac{X_n}{n} $ is defined as:

$ F_{\frac{X_n}{n}}(t) = P\left(\frac{X_n}{n} \leq t\right) $

Since $ X_n $ is a discrete uniform random variable on $ \{1, \ldots, n\} $, we have:

$ P\left(\frac{X_n}{n} \leq t\right) = P(X_n \leq nt) $

As the values of $ X_n $ are integers, we can express the probability as the number of possible values of $ X_n $ that are less than or equal to $ nt $ divided by the total number of possible values of $ X_n $:

$ P(X_n \leq nt) = \frac{\text{{number of integers in }} \{1, \ldots, n\} \text{{ that are less than or equal to }} nt}{n} $

Since the random variable $ X_n $ is uniformly distributed, all values in $ \{1, \ldots, n\} $ are equally likely. Therefore, the number of integers in $ \{1, \ldots, n\} $ that are less than or equal to $ nt $ is equal to $ \lfloor nt \rfloor $, where $ \lfloor \cdot \rfloor $ denotes the floor function:

$ \frac{\text{{number of integers in }} \{1, \ldots, n\} \text{{ that are less than or equal to }} nt}{n} = \frac{\lfloor nt \rfloor}{n} $

Thus, the distribution function of $ \frac{X_n}{n} $ is given by:

$ F_{\frac{X_n}{n}}(t) = \frac{\lfloor nt \rfloor}{n} $

Now, let's consider the distribution function of $ X $, which is given by:

$ F_X(t) = P(X \leq t) $

Since $ X $ is a continuous uniform random variable on $ (0,1) $, the distribution function is given by:

$ F_X(t) = t $

To show that $ \frac{X_n}{n} $ converges in distribution to $ X $, we need to demonstrate the convergence of the distribution functions $ F_{\frac{X_n}{n}}(t) $ and $ F_X(t) $ for all $ t $ in $ (0,1) $:

$ \lim_{{n \to \infty}} F_{\frac{X_n}{n}}(t) = \lim_{{n \to \infty}} \frac{\lfloor nt \rfloor}{n} = \lim_{{n \to \infty}} \frac{nt – \{ nt \}}{n} = t $

Since the fractional part $ \{ nt \} $ lies between $ 0 $ and $ 1 $, we have $ \lim_{{n \to \infty}} \frac{\{ nt \}}{n} = 0 $. Thus, we obtain:

$ \lim_{{n \to \infty}} \frac{\lfloor nt \rfloor}{n} = \lim_{{n \to \infty}} \frac{nt – \{ nt \}}{n} = t $

As $t $ is arbitrary in $ (0,1) $, the distribution function of $ \frac{X_n}{n} $ converges to the distribution function of $ X $. Therefore, $ \frac{X_n}{n} $ converges in distribution to $ X $ as $ n \rightarrow \infty $.

Does this argumentation make sense and is there maybe another way to show the converges in distribution?

Answer 1

A shorter approach. The measures $\mu_n=P\circ (X_n/n)^{-1},\mu=P\circ X^{-1}$ are probability measures on $([0,1],\mathscr{B}[0,1])$. Let $f \in C_b([0,1])$ (continuous and bounded). We get $$E[f(X_n/n)]=\frac{1}{n}\sum_{1\leq k \leq n}f(k/n)\to \int_0^1f(x)dx=E[f(X)]$$ We conclude.