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I have two infinite sets of events $A$ and $B$ with the following probabilities:

  • $P(A_n)=\frac{2}{6n-1}$
  • $P(B_n)=\frac{2}{6n+1}$

Note that I have divided them into two sets only because it is easier to depict the probabilities; other than that, these two sets are essentially "indistinguishable".

I want to prove that the probability that any one of the events will take place is smaller than $1$.

I've figured I can use Inclusion-Exclusion principle, but the problem, of course, is that I have an infinite amount of events so I'm not quite sure how to accomplish that.

I tested it using the following Python script, so I believe that the probability of one of the events taking place is indeed smaller than $1$:

from decimal import Decimal
from decimal import getcontext

getcontext().prec = 100

def getBits(num):
    bit = 0
    bits = []
    while num > 0:
        if num & 1:
            bits.append(bit)
        bit += 1
        num >>= 1
    return bits

def prod(arr):
    res = Decimal(1)
    for val in arr:
        res *= val
    return res

SIZE = 20

sums = [Decimal(0) for k in range(SIZE)]

A = [Decimal(2)/(6*k-1) for k in range(1,SIZE)]
B = [Decimal(2)/(6*k+1) for k in range(1,SIZE)]
probabilities = [p for pair in zip(A,B) for p in pair]

for n in range(1,1<<SIZE):
    bits = getBits(n)
    sums[len(bits)-1] += prod([probabilities[bit] for bit in bits])

print(sum([sums[k]*(-1)**k for k in range(SIZE)]))

After a minute or two, the printout shows a probability of approximately $89\%$.

Does anyone see a way for me to achieve my purpose mathematically?

Again - my primary objective is to prove that the probability of one of the events taking place is smaller than $1$. I don't mind calculating the exact value of this probability along the way, but if there's another way which only proves that it is smaller than $1$ then I'm fine with that.

Thank you very much!

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  • $\begingroup$ So you want to prove that, for all $n \in \Bbb Z^+$, that $P(A_n)$ and $P(B_n)$ (individually) are always less than $1$? Just want to clarify. $\endgroup$ – Eevee Trainer May 11 at 6:53
  • $\begingroup$ @EeveeTrainer: No, that one is obvious from the definition. $\endgroup$ – goodvibration May 11 at 6:54
  • $\begingroup$ @EeveeTrainer: I want to prove that there is a chance that none of the events will take place (or equivalently, that the probability that either one of them will take place is smaller than $1$). $\endgroup$ – goodvibration May 11 at 6:56
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Without further information about the relation between any two of $\{A_n\}_{n\ge 1} \cup \{B_n\}_{n\ge 1}$, like independence or exclusivity, you cannot prove what you want to prove, as there are counterexamples:

Take a uniformly random continuous variable $u$ on $[0,1]$. With $0 \le a \le b \le1$, the probability for the event "$u \in [a,b]$" is $b-a$.

Define

$$a_0:=0, \text{ and }a_{n}:=a_{n-1}+\frac2{6n-1} \text{ for }n \ge 1.$$

This means

$$a_k=\sum_{n=1}^k\frac2{6n-1}.$$

Now that sum in the above forumula is (asymptotically, for large $n$) equivalent to the harmomic series, so it diverges. That means at some point $k_1$ we get $a_{k_1} \ge 1$ for the first time. According to Wolfram Alpha we have $k_1=8$, but the exact value is unimportant.

Now define your events:

$$A_n=u \in [a_{n-1},a_n], \text{ for }n=1,2,\ldots,k_1-1$$

and

$$A_{k_1} = u \in [1-\frac2{6k_1-1},1]$$

We have $P(A_n)=\frac2{6n-1}$ for $n=1,2,\ldots,k_1$. Also note that by definition of $k_1$, we have $1-\frac2{6k_1-1} \ge a_{k_1-1}$.

In the end that means that

$P(A_1 \cup A_2 \cup \ldots A_{k_1-1} \cup A_{k_1})=1$, because the intervals used to define each $A_n$ cover the complete interval $[0,1]$.


In your python code (I have no knowledge of python) it looks like you are multiplying probabilites, which suggests that there is some kind of independence. If you can specify more about that, maybe what you want to prove is correct. But as my example proves, if you can't assume anything, your desired conclusion is incorrect.

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  • $\begingroup$ Thank you. I will rephrase the question then. $\endgroup$ – goodvibration May 13 at 7:09
  • $\begingroup$ @goodvibration Make a new question, if possible, and add a link to it in this question's text. $\endgroup$ – Ingix May 13 at 8:37

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