$x_1, x_2, x_3, ..., x_i, ...$ ~ $uniform(0, 1)$

The actual random variable is the following.

$P_i = (1-x_1)(1-x_2)...(1-x_{i-1})x_i$

And the goal is proving these...

  1. $\sum_{i=1}^{n}P_i \leq 1$

  2. If i's satisfying $P_{i}=0$ exist finitely, $\lim_{n\to\infty}\sum_{i=1}^{n}P_i \to 1$

How can I do that?


2 Answers 2


To facilitate this analysis, we denote the partial sums by:

$$S_n \equiv \sum_{i=1}^n P_i \quad \quad \quad \text{for all } n \in \mathbb{N}.$$

With a bit of algebra it is simple to establish that:

$$S_n = 1- \prod_{i=1}^n (1-x_i).$$

From this form you should easily be able to establish that $0 \leqslant S_n \leqslant 1$. Assuming that the underlying random variables in your analysis are IID (you haven't specified if they are independent) then you should be able to show that $S_n \rightarrow 1$ almost surely (i.e., with probability one).


Wow, that's a beautiful exercise for me, to go back to the University times and exercise my brain :-) It needs some mathematical thinking :-)

The $\sum_{i=1}^nP_{i} \le 1$ kinda seems "obvious", but how to prove it? It took me a while:

$$\sum_{i=1}^nP_{i} = \sum_{i=1}^{n-1}P_{i} + x_{n} \prod_{i=1}^{n-1}(1 - x_{i}) \le \sum_{i=1}^{n-1}P_{i} + \prod_{i=1}^{n-1}(1 - x_{i})$$

and it can be seen that this term is exactly equal to 1. We can prove it by induction: for $n = 2$ it's certainly valid, and if we know is valid for $n - 1$, we can prove it for $n$:

$$\sum_{i=1}^{n}P_{i} + \prod_{i=1}^{n}(1 - x_{i}) = \sum_{i=1}^{n-1}P_{i} + x_{n} \prod_{i=1}^{n-1}(1 - x_{i}) + (1 - x_{n}) \prod_{i=1}^{n-1}(1 - x_{i}) =\\= \sum_{i=1}^{n-1}P_{i} + \prod_{i=1}^{n-1}(1 - x_{i}) = 1$$

As for your second point, this certainly doesn't have to be true! The limit is certainly $\le 1$ but can be whatever number between 0 and 1. For example, take $x_i = (\frac{1}{2})^{i+1}$

Where did you take this from?

  • $\begingroup$ Thanks for your answer but i doubt with your answer of second question. Even if x_i was like your defining, I don't know where the sum of "P_i" goes. Can you show me that? $\endgroup$
    – Toast
    Aug 10, 2019 at 13:43
  • $\begingroup$ And I made this question from my real world modeling problem. ;) $\endgroup$
    – Toast
    Aug 10, 2019 at 13:44
  • $\begingroup$ Re your last point: The $x_i$ are supposed to have uniform distributions. $\endgroup$
    – whuber
    Aug 10, 2019 at 15:02

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