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Let $(X_n)_{n \geq 1}$ be a sequence of iid random variables where $X_i \sim \text{Uniform}(\{0, \ldots, 9\})$. Let $$Y_n = \sum_{j = 1}^n X_j10^{-j}$$ Study the convergence in distribution of $Y_n$, using the characteristic function. Recognize the limit distribution.

I think that the limit distribution is uniform over the interval $[0, 1]$, but I cannot prove it. Let $\Phi_X$ be the characteristic function of $X$. We have: $$\Phi_{X_j}(t) = \mathbb E\left[e^{iX_jt}\right] = \sum_{x = 0}^9 \frac1{10} {\left(e^{it}\right)}^x = \frac1{10}\frac{1 - e^{i10t}}{1 - e^{it}}$$

Now, since $Y_n$ is a sum of independent random variables, $$\Phi_{Y_n}(t) = \prod_{j = 1}^n \Phi_{X_j10^{-j}}(t) = \prod_{j = 1}^n \Phi_{X_j}(10^{-j}t) = \frac1{10}\prod_{j = 1}^n \frac{1 - e^{i10^{1 - j}t}}{1 - e^{i10^{-j}t}}$$

Now it remains to find the limit of $\Phi_{Y_n}(t)$ as $n \to +\infty$, but I don't know how to evaluate it unless I transform that product, which I was not able to do.

I really think that the limit is a uniform, so it should be $$\lim_{n \to +\infty} \Phi_{Y_n}(t) = \frac{e^{it} - 1}{it}$$

Any hints?

2 Answers 2

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Replace $\dfrac{1}{10}$ by $\dfrac{1}{10^n}$ here: $$\Phi_{Y_n}(t) = \prod_{j = 1}^n \Phi_{X_j10^{-j}}(t) = \prod_{j = 1}^n \Phi_{X_j}(10^{-j}t) = \frac1{10^n}\prod_{j = 1}^n \frac{1 - e^{i10^{1 - j}t}}{1 - e^{i10^{-j}t}}$$

After that simply reduce the denominator of each fraction and the numerator of the next fraction: $$ \Phi_{Y_n}(t) = \frac1{10^n} \frac{1 - e^{it}}{1 - e^{it/10}}\cdot \frac{1 - e^{it/10}}{1 - e^{it/100}}\cdot \frac{1 - e^{it/100}}{1 - e^{it/10^3}}\cdot\dots\cdot \frac{1 - e^{it/10^{n-1}}}{1 - e^{it/10^n}}= $$ $$=\frac1{10^n} \frac{1 - e^{it}}{1 - e^{it/10^n}}.$$

Then use $\lim\limits_{x\to 0}\dfrac{e^x-1}{x}=1$ and obtain $\Phi_{Y_n}(t) \to \Phi_{X}(t)=\dfrac{e^{it}-1}{it}$ as $n\to\infty$.

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    God damn it people, why are u always do that. He ask for a hint not an answer -_- – 2017-02-10
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    Ah yes that $\frac1{10}$ is a mistake. So essentially is a telescopic product. Damn, I should have thought of this. – 2017-02-10
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first of all replace

$$\Phi_{{X_j}10^{-j}}(t) = \Phi_{{X_j}}(10^{-j}t)$$

with

$$\Phi_{{X_j}10^{-j}}(t) = (\Phi_{{X_j}}(t))^{10^{-j}}$$

try again and let me know if you got it or not.

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    First of all this answer should be a comment. And then it can be proved that $$\Phi_{aX + b}(t) = e^{itb} \Phi_X(at)$$ So where's the error? – 2017-02-10
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    oh, sorry. You're right, you can write it like this. But still, don't :P. Multiplying powers of the same value is an easy task ;) – 2017-02-10