How can I determine the characteristic function of $Z = \sum_{j = 1}^{N}\frac{X_j}{10^j}$?

Question

I am currently working on an exercise regarding characteristic functions.

Consider a set of i.i.d. random variables, $\{X_1,\ldots,X_n\}$, uniformly distributed on $\{0,1,2,\ldots,9\}$. I want to determine the characteristic function of the following distribution:

$$Z = \sum_{j = 1}^{N}\frac{X_j}{10^j}$$

Additionally, I am given the following limit which is fulfilled by the characteristic function of the distribution $Z$:

$$\lim_{n \to \infty}G_Z(k) = i\frac{e^{-ik}-1}{k}$$

Here is what I have done so far:

Given the sum of iid random variables $X_n$ with characteristic function $G_X(k)$, the characteristic function of the sum is given by: $$G_S(k) = \prod_{n=1}^{N}G_X(k)$$ Now taking the characteristic function of the discrete uniform distribution from $0$ to $9$: $$G_X(k) = \frac{1-e^{10ik}}{10(1-e^{ik})}$$ Now substituting $k = \frac{k}{10^j}$ and using the aforementioned summation rule: $$G_Z(k) = \prod_{j=1}^{N}\frac{1-e^{ik10^{1-j}}}{10(1-e^{ik10^{-j}})}$$ This is where I am currently stuck. I can't seem to find a good expression for $N= 10$ nor for $N \to \infty$. I am not sure I my approach is incorrect or if I am missing an algebraic manipulation which gives me the desired result.

Edit:

Using hint and simplifying the telescoping product:

$$ \begin{align} G_Z(k) =& \ \frac{1-e^{ik}}{10(1-e^{ik10^{-1}})}\cdot \frac{1-e^{ik10^{-1}}}{10(1-e^{ik10^{-2}})} \cdot \ ... \ \cdot \frac{1-e^{ik10^{2-n}}}{10(1-e^{ik10^{1-n}})} \cdot \frac{1-e^{ik10^{1-n}}}{10(1-e^{ik10^{-n}})} \\ =& \ \frac{1-e^{ik}}{10^{n}(1-e^{ik10^{-n}})} \end{align}$$

However, I can't seem to figure out the limit:

$$\lim_{n \rightarrow \infty} \frac{1-e^{ik}}{10^{n}(1-e^{ik10^{-n}})}$$

Using L'hopitals rule yields more indeterminant forms and the expressions become extremely complicated. Is this form even correct? If yes, is there an easy or straightforward way to calculate the limit of this expression.

Answer 1

You are on the right track. Hints:

1. You have a telescoping product.

2. As $N\to\infty$ you have to evaluate an indeterminate form $\frac00$, which you can tackle using l'Hopital's rule, or the definition of the derivative.

The limiting distribution of $Z$ should be uniform$(0,1)$, since $Z$ is essentially the $N$-decimal-digit approximation of a uniform$(0,1)$ random number. So I would expect the CF of the limit to be $$G_Z(k)=\frac{e^{ik}-1}{ik}.$$

EDIT: To answer your final question, it is enough to evaluate the limit $$\lim_{t\to0} \frac{t(1-e^{ik})}{1-e^{ikt}},$$ which is more amenable to l'Hopital.

Answer 2

For evaluating the limit you can expand the $e^{ik10^{-n}}$ in the denominator as $1 + ik10^{-n}$. This will will give the required result. $\frac{1-e^{ik}}{-ik}$