Probability question: show that $\Pr[X = k] = x_k$, where $X = \sum\limits_{k = 0}^\infty k I_A(u)$ and $u \sim \text{Unif}[0,1]$

Question

Let $\{x_k| k \geq 0, 0 \leq x_k \leq 1\}$ be a sequence of numbers such that $\sum\limits_{k = 0}^\infty x_k = 1$

Let $u \sim \text{Unif}[0,1]$ be a uniformly distributed random variable. Let $$X = \sum\limits_{k = 0}^\infty k I_{A_k}(u)$$ where $I_{A_k}$ is the indicator function (i.e., $I_{A_k}(u) = 1$ if $u \in A_k$) and $A_k$ is the set defined by $A_k = (\sum\limits_{i = 0}^{k-1} x_i, \sum\limits_{i = 0}^k x_i]$.

I wish to show that,

$$\Pr[X = k] = x_k$$

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There seems to be a lot of moving parts to this question. An initial attempt would look like this:

$$\Pr[X = k]= \Pr[\sum\limits_{k = 0}^\infty k I_{A_k}(u) = k] = \Pr[ I_{A_1}(u) + 2I_{A_2}(u) + 3 I_{A_3}(u) + \ldots = k]$$

It just doesn't seem to give me anything. Can someone please offer assistance?

Answer 1

Hint: The key insight is that the event $\{ U\in A_k\} $ is the same as the event $\{X=k\}$. To prove this, use the fact that the $\{A_k\}$ are disjoint and their union is the unit interval.

1. For the forward inclusion, suppose that $U\in A_k$. Then $I_{A_k}(U)=1$ and (by disjointness) $I_{A_j}(U)=0$ for all $j\ne k$, so that $X=k$.

2. Conversely, suppose $U\not\in A_k$. Then there exists $i\ne k$ such that $U\in A_i$, so that $I_{A_i}(U)=1$ while (by disjointness) $I_{A_j}(U)=0$ for all $j\ne i$, whence $X=i$ and therefore $X\ne k$.

To finish off: what's the probability that $U\in A_k$?

(Strictly speaking, the union of the $\{A_k\}$ is the half-open interval $(0,1]$, so to be picky we've shown $\{U\in A_k\}\subset \{X=k\} \subset \{U\in A_k\}\cup \{U= 0\}$. But the event $\{U=0\}$ has probability zero.)