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I have question about example illustration Convergence of random variables in probability but not almost surely.

Suppose a person takes a bow and starts shooting arrows at a target. Let $X_n$ be his score in $n$-th shot. Initially he will be very likely to score zeros, but as the time goes and his archery skill increases, he will become more and more likely to hit the bullseye and score $10$ points. After years of practice the probability that he hit anything but $10$ will be getting increasingly smaller and smaller and will converge to $0$. Thus, the sequence $X_n$ converges in probability to $X = 10$.

Note that $X_n$ does not converge almost surely however. No matter how professional the archer becomes, there will always be a small probability of making an error. Thus the sequence $(X_n)$ will never turn stationary: there will always be non-perfect scores in it, even if they are becoming increasingly less frequent.

https://en.wikipedia.org/w/index.php?title=Convergence_of_random_variables&diff=879355996&oldid=879219919

For me this example is true. But not for 69.181.249.190. Why? You have any idea?

69.181.249.190 talk‎ 36,722 bytes -957‎ →‎Convergence in probability: removed false archer example

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  • $\begingroup$ It is hard to read but that example looks pretty crappy, it has no probability model but makes sweeping claims about convergence. Those claims cannot be made without a probability model! $\endgroup$ – Michael Jun 2 at 13:37
  • $\begingroup$ If $P[X_n=10]=1-1/2^n$ then $X_n$ converges to 10 both in probability and with prob 1 since with prob 1, all but a finite number of values will be 10. By Borel-Cantelli. This contradicts the crappy archer claims. $\endgroup$ – Michael Jun 2 at 13:46
  • $\begingroup$ Oh, I remember seeing this example on this page a few years ago when I was still studying. At the time I definitely cocked an eyebrow at it but didn't want to remove it in case I was just missing something. Glad it's gone to be honest, for essentially the same reasons as Michael. At the very least it needed more detail. $\endgroup$ – Jack M Jun 2 at 20:12
  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, use MathJax. $\endgroup$ – dantopa Jun 3 at 5:08
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The archer example incorrectly suggests that if there is always a small-but-positive probability of missing the target, then the archer misses infinitely often.

Counter-example: Consider $\{X_n\}_{n=1}^{\infty}$ with: $$X_n=\left\{\begin{array}{} 10 & \mbox{ with prob $1-1/2^n$}\\ 0& \mbox{ else (miss target)} \end{array}\right.$$ We don’t care about independence but you can assume the variables are mutually independent if you want. We observe $$\sum_{n=1}^{\infty}P[X_n\neq 10]<\infty$$ So by the Borel-Cantelli lemma we conclude that (with prob 1) we miss only finitely often. So $X_n\rightarrow10$ both in probability and with prob 1.


On the other hand, if the variables are mutually independent with $$P[X_n=10]=1-1/n, P[X_n=0]=1/n$$ then we can conclude (either by Borel-Cantelli or by direct calculation) that we miss infinitely often with prob 1. This is a classic example where $X_n\rightarrow10$ in probability but not with prob 1.

Overall, it depends on the rate at which the miss probabilities converge to zero.

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    $\begingroup$ Yes, it seems intuitive that if there's always a non-zero chance of missing, then with probability one we will miss infinitely many times by some sort of "whatever can happen does happen" law, but this turns out to be false. $\endgroup$ – Jack M Jun 2 at 20:20

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