Limiting a sequence of moment generating functions

Question

I was trying to solve the following problem:

Let $\{X_n\}_{n=1}^{\infty}$ be a sequence of independent random variables with the probability mass function $P\{X_n = \pm1 \} = \frac{1}{2}$, $n \in \mathbb{N}$. Let $Z_n=\sum_{j=1}^{n}{X_j/2^j}$. Show that $Z_n \xrightarrow{L} Z$, where $Z \sim U[-1, 1]$.

(From An Introduction to Probability and Statistics, V.K. Rohatgi & A. K. Md. Saleh, (c) 2015, Problems 7.5, Page 320)

Here $\xrightarrow{L}$ means convergence in law (or in distribution), and $U[-1, 1]$ is the uniform distribution on the interval $[-1, 1]$.

My approach was the following:

We need to show that $$\lim_{n\rightarrow\infty} M_{Z_n}(t) = M_{Z}(t) = \frac{e^{1 \times t} - e^{-1 \times t}}{t \times (1 - (-1))} = \frac{e^t - e^{-t}}{2t}.$$ Since $$M_{Z_n}(t) = E_{Z_n}\left(e^{tZ_n}\right) = E\left(e^{t\sum_{j = 1}^{n}{\frac{X_j}{2^j}}}\right) = E\left(\prod_{j=1}^{n}{e^{t\frac{X_j}{2^j}}} \right) = \prod_{j=1}^{n}{E_{X_j}\left(e^{t\frac{X_j}{2^j}} \right)},\\ E_{X_j}\left( e^{t \frac{X_j}{2^j}} \right) = e^{t \times \frac{-1}{2^j}} \times \frac{1}{2} + e^{t \times \frac{1}{2^j}} \times \frac{1}{2} = \frac{1}{2} \left( e^{\frac{t}{2^j}} + e^{\frac{-t}{2^j}} \right),$$ then $$M_{Z_n}(t) = \prod_{j = 1}^{n}{\frac{1}{2} \left( e^{\frac{t}{2^j}} + e^{\frac{-t}{2^j}} \right)}.$$

I cannot see how this sequence of functions converges to the required moment generating function of $U[-1,1]$.

I had many attempts, for instance using the power series representation of $e^x$ and limiting approximations, but failed in them all. After that I started thinking that perhaps I am missing knowledge of some theorems.

Any idea how to proceed?

Answer 1

$\def\e{\mathrm{e}}$Another method: Note that $a + b = \dfrac{a^2 - b^2}{a - b}$ for $a ≠ b$, then$$ M_{Z_n}(t) = \frac{1}{2^n} \prod_{k = 1}^n \left( \exp\left( \frac{t}{2^k} \right) + \exp\left( -\frac{t}{2^k} \right) \right) = \frac{1}{2^n} · \frac{\e^t - \e^{-t}}{\exp\left( \dfrac{t}{2^n} \right) - \exp\left( -\dfrac{t}{2^n} \right)}. $$ Because$$ \lim_{n → ∞} \frac{\exp\left( \dfrac{t}{2^n} \right) - \exp\left( -\dfrac{t}{2^n} \right)}{\dfrac{t}{2^n} - \left( -\dfrac{t}{2^n} \right)} = (\e^x)'\bigr|_{x = 0} = 1, $$ then $\lim\limits_{n → ∞} M_{Z_n}(t) = \dfrac{1}{2t} (\e^t - \e^{-t})$ and the result follows.

Answer 2

$\def\i{\mathrm{i}}\def\d{\mathrm{d}}$Since $X_1, X_2, \cdots$ are independent and $\displaystyle Z_n = \sum\limits_{k = 1}^n \frac{X_k}{2^k}$, then$$ φ_{Z_n}(t) = \prod_{k = 1}^n φ_{X_k}\left( \frac{t}{2^k} \right) = \prod_{k = 1}^n \cos\left( \frac{t}{2^k} \right) = \frac{\sin t}{2^n \sin\left( \dfrac{t}{2^n} \right)}, \quad \forall t \in \mathbb{R}^* $$ which implies $\displaystyle \lim_{n → ∞} φ_{Z_n}(t) = \frac{\sin t}{t}$, and the limit is the characteristic function of $Z \sim U(-1, 1)$. By the continuity theorem, $Z_n \xrightarrow{\mathrm{d}} Z$.

Answer 3

In addition to characteristic functions, one may also approach the problem via binary representation of integers, which is not as short as the answers with characteristic functions, but is quite straightforward.

Indeed, rewrite $$ Z_n = \frac{1}{2^n} \sum\limits_{j=1}^n 2^{n-j} X_j : = \frac{1}{2^n}S_n. $$ Define $\Lambda_{+} = \{ 1\leq j \leq 2^n: X_j = 1 \}$, and let $\Lambda_-$ be the complement of $\Lambda_+$ in $1\leq j\leq 2^n$. Then, $$ S_n = \sum_{j\in \Lambda_+} - \sum_{j \in \Lambda_-} = \sum_{j\in \Lambda_+} - \left( 2^n - 1 - \sum\limits_{j\in \Lambda_+} \right) = 2\sum\limits_{j \in \Lambda_+} 2^{n-j} - (2^n - 1) \tag{1}. $$ Thus, with $S_n$ we cover all integers from $-(2^n - 1) , ... 2^n - 1$ of the form $(1)$, which are precisely all the odd integers from $-(2^n - 1), ... (2^n - 1)$, $2^n$ in total. Thus, if $i \in [-(2^n - 1), ..., 2^n - 1] $ is even then $$ \mathbb{P}\left(Z_n = \frac{i}{2^n}\right) = 0 \tag{2} $$ and if $i $ is odd, then $$ \mathbb{P}\left(Z_n = \frac{i}{2^n}\right) = 2^{-n}, \tag{3} $$ since there is a single choice of index set $\Lambda_+$ in $(1)$, and hence $(3)$ follows in view of independence of $\{X_j\}$.

From $(2)$ and $(3)$ we see, by counting the number of odd integers, that for any integer $-2^{n-1} + 1 \leq i \leq 2^{n-1}$ one has $$ \mathbb{P}\left(Z_n \leq \frac{2i - 1}{2^n} \right) = \frac{1}{2} + \frac{i}{2^n}. $$

It follows that the distribution function $F_n$ of $Z_n$ coincides, on odd dyadic rationals from $[-1,1]$ (and obviously everywhere on $(-\infty, -1] \cup [1,\infty)$) with the distribution function $F$ of a random variable with the law $U[-1, 1]$. The density of dyadic rationals and right-continuity of cdf imply $F_n \to F$ everywhere on $[-1,1]$, hence the claim.