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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)

$\xrightarrow{L}$ 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 $\lim\limits_{n\rightarrow\infty} M_{Z_n}(t) = M_{Z}(t) = \frac{e^{1 \times t} - e^{-1 \times t}}{t \times (1 - (-1))} = \frac{e^t - e^{-t}}{2t}$.

$$M_{Z_n}(t) = E_{Z_n}\left(e^{tZ_n}\right) = E\left(e^{t\sum_{j = 1}^{n}{\frac{X_j}{2^j}}}\right) = E\left(\prod_{j=1}^{n}{e^{t\frac{X_j}{2^j}}} \right) = \prod_{j=1}^{n}{E_{X_j}\left(e^{t\frac{X_j}{2^j}} \right)}$$

$$ E_{X_j}\left( e^{t \frac{X_j}{2^j}} \right) = e^{t \times \frac{-1}{2^j}} \times \frac{1}{2} + e^{t \times \frac{1}{2^j}} \times \frac{1}{2} = \frac{1}{2} \left( e^{\frac{t}{2^j}} + e^{\frac{-t}{2^j}} \right) $$

Hence,

$$ M_{Z_n}(t) = \prod_{j = 1}^{n}{\frac{1}{2} \left( e^{\frac{t}{2^j}} + e^{\frac{-t}{2^j}} \right)} $$

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'm missing knowledge of some theorem.

Any idea how to proceed?

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migrated from mathoverflow.net Jan 5 at 15:15

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