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Let $(X_n)$ be a sequence of independent random variables, which have Laplace-distribution on the set $\{0,..,9\}$. $Y_n:= \sum_{i=1}^n X_i \frac {1}{10^i}$ for all $n$.

now I want to show, that $Y_n$ converges in distribution against a $U(0,1)$-distributed random variable

I would be thankful for any help

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    (1) the [Laplace distribution](https://en.wikipedia.org/wiki/Laplace_distribution) usually describes a continuous distribution on the whole real line – 2017-01-09
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    (2) You presumably intend a uniform discrete distribution from $\{0,1,2,3,4,5,6,7,8,9\}$ as without the $0$ you will not get your result – 2017-01-09
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    your are right, sorry! the 0 has to be included – 2017-01-10

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Consider some $a \in [0, 1)$. We want to show that $\lim_{n \rightarrow \infty} \mathbb{P}[Y_n \le a] = a$. We will denote $\lfloor a \rfloor_n$ to be $a$ floored to the nearest $n$th digit in its decimal representation. It is clear then that $\mathbb{P}[Y_n \le a] = \mathbb{P}[Y_n \le \lfloor a \rfloor_n]$. Let us write the decimal representation of $\lfloor a \rfloor_n$ as $0.a_1a_2\ldots a_n00000\ldots$ Since each of the $X_i$ are i.i.d. and discrete uniform, it is clear that each $Y_n$, written as, $0.x_1x_2\ldots x_n 0000 \ldots$ with $x_j \in \{0, \ldots, 9\}$ is equally likely. Therefore, $\mathbb{P}[Y_n \le \lfloor a \rfloor_n]$ is precisely the number of integers less than or equal to $a_1a_2\ldots a_n$, which is $\lfloor a \rfloor_n$. Therefore, $$ \mathbb{P}[Y_n \le a] = \lfloor a \rfloor_n$$ Taking limits on both sides we have, $$ \lim_{n \rightarrow \infty} \mathbb{P}[Y_n \le a] = \lim_{n \rightarrow \infty} \lfloor a \rfloor_n = a$$

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    I´m not really sure why this answeres my question correctly. Where did you use the definition of Yn and why is this showing convergence in distribution? – 2017-01-10
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    $Y_n$ is simply a decimal such that the $j \le n$ digit of $Y_n$ is equal to $X_j$, which is uniform on $\{0, 1, \ldots, 9\}$. The definition of convergence in distribution is that the CDF of $Y_n$ converges pointwise at all points of continuity (which for the uniform 0-1 case, is just proving that $\lim_{n \rightarrow \infty} \mathbb{P}[Y_n \le a] = a$ for all $a \in [0, 1]$). – 2017-01-10