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Let $X_i, i\ge1$ be i.i.d random variables where $P(X_i = 0) =P(X_i = 1) = 1/2$.

Sequence 1): $\sum_1^n{X_i2^{-i}}$. What is its distribution and show that this sequence converges as $n\to\infty$. What's the distribution of the limit?

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    Any thoughts on the problem would be really appreciated. – 2017-01-25
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    @saz For the first sequence, was wondering whether the following can be applied: Because we have $\sum_1^n {1/2^n}$ = 1, then we can apply the strong law of large numbers where $S_n \to X_1 +X_2 +...+X_n \to {nE(X_i)}$ where ${E(X_i)}$ = 1/2 – 2017-01-25

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It is easy to see that the outcomes of the partial sums are convergent pointwise for they are bounded from above by one and are increasing.

For the first series

The sequence of partial sums converges in distribution to the uniform distribution over the unit interval. Think of halving the unit interval, say three times. We'll have eight intervals of equal length. The probability that the infinite sequence falls in one of the eight intervals is the same. The landing interval depends only on the first three bits...

You can repeat this argumentation for any number of halving.

In the second case the outcomes of the infinite series belong to the Cantor set. The partial sums seem to converge in distribution to the uniform distribution over the Cantor set.

The distribution function to which the partial sums' distribution tends to is the Cantor function.

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    Thank you! However, I am still confused as how to algebraically show the convergence of both series? – 2017-01-25
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    For the first case: if you do $n$ halves then you will have $2^n$ intervals and a step function as the distribution function. This function step by step increases from $\frac1{2^n}$ to $1$ in steps of $\frac{2^n}$. This step function converges in every point of $[0,1]$ to the function $y=x$ if $n\to\infty$. As far as the second case. Sorry, I cannot do it analytically. – 2017-01-25