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September 14th, 2017, 09:44 PM  #1 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0  Number of times an experiment must be repeated to ensure a range of outcomes
Let the experiment G be described by the following. Suppose I have an event B whose probability is of occurring is 1/2^13 = 1/8192. Suppose I have a random variable X measuring the number of trials necessary to observe the first success of event B occurring. B is independent, so if B fails on a previous trial, P(B) doesn't change. Now suppose I repeat the experiment G an arbitrary N number of times, recording each value of X into a table. Find the value of N such that there is exactly a 99% probability that the average value AV of all X's recorded in the table satisfies 8191 <= AV <= 8193. In other words, how many times do I have to run the experiment G to make sure that, with a 99% probability, the average of all of my outcomes is between 8191 and 8193? I honestly don't have any idea where to start with this one. I'm sure that it's related to a geometric distribution, but the question itself seems a little too out there. How would you go about solving this? 
September 15th, 2017, 05:16 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
I haven't been able to get numbers I like out of this yet but from the bit of digging I did last night it appears that the sum of $n$ iid geometric rvs is given by $P[S_n=m] = \dbinom{m1}{n1}p^n (1p)^{mn}$ where $n$ is the number of rvs being summed $p$ is the parameter for the geometric distribution $m$ is the value of the sum Now $8191 < \overline{X} < 8193 \Rightarrow 8191n < S_n < 8193n$ So it should be a simple matter of $P = \displaystyle \sum_{m=8191n}^{8193n} P[S_m] = \sum_{m=8191n}^{8193n}\dbinom{m1}{n1}p^n (1p)^{mn}$ and we want the smallest $n$ for which $P\geq 0.99$ I wasn't able to get anywhere close to $P =0.99$ and calculations were starting to get slow around $n=5000$ As with all sums of rvs this should be able to be approximated with the Normal distribution but it was getting late. I'll keep toying with this as time allows but this might get you started. Last edited by romsek; September 15th, 2017 at 05:19 AM. 
September 15th, 2017, 06:02 AM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
I'm not sure if I've done something wrong or what but I'm observing what looks like $\displaystyle \lim_{N\to \infty} P = 0.5$ I.e. you can never achieve $P>0.99$ by increasing $N$ I'll keep looking at it. 
September 15th, 2017, 05:42 PM  #4 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 
That's a very clever way of setting up the problem. I hadn't quite thought about it like that. However, I still find it odd there there's no value of N such that there's a 99% probability of the average of the random variables being in that range... after all, if you just think about it intuitively, as the number of trials increases, the average will tend toward 8192. As a result, I would imagine there would be a certain number of trials that would result in any percentage chance of being within a certain range. But I suppose intuition isn't always the best way to go about things.

September 15th, 2017, 07:53 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390  Quote:
Last edited by romsek; September 15th, 2017 at 08:46 PM.  
September 16th, 2017, 12:55 AM  #6 
Senior Member Joined: Sep 2015 From: USA Posts: 2,531 Thanks: 1390 
Ok, I used the central limit theorem and approximated the distribution of the average as a normal distribution with appropriate parameters. Once that's done the problem is simple with some software. I came up with $n=445210000$ which explains why I wasn't getting close to 99%. I wasn't trying anything near that. I leave it to you to look up how to use the central limit theorem and to reproduce the results. 
September 16th, 2017, 12:48 PM  #7 
Senior Member Joined: Oct 2015 From: Antarctica Posts: 128 Thanks: 0 
Brilliant. I'm thoroughly impressed. Great work.


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