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I am trying to solve this problem: Expected falling time of all $500$ random ants I need help with understanding why my thought process is flawed. (highlighted below)

Question: 500 ants are randomly put on a 1-foot string (independent uniform distribution for each ant between 0 and 1). Each ant randomly moves toward on end of the string (equal probability to the left or the right) at constant speed of 1 foot/minute until it falls of a t one end of the string. Also assume that the size of the ant is infinitely small. When two ants collide head-on, they both immediately change directions and keep on moving at 1 foot/min. What is the expected time for all ants to fall off the string?

I totally get that when ants collide they exchange labels, so we just consider that ants never collide and keep on walking till one end of the string.

My flawed thought process: When an ant lands on a point "x" foot away it can go either left or right (equal probability). So expected time taken by it is 0.5x + 0.5(1-x) = 0.5, which is independent of x. So all ants regardless where they land will take an expected time of 0.5min.

However the solution says that the time taken is x or 1-x and we can just solve for expected value of maximum value of 500 X's. ( because x , 1-x are symmetric.) .

What am I missing here ?

anshul aggarwal
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1 Answers1

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Both of you are correct:

If $X$ is uniformly distributed over $[0,1]$, then

  • The expected value of $X$ is $0.5$
  • The expected value of $1-X$ is also $0.5$, (and $1-X$ is also a uniform random distribution on the interval $[0,1]$, just as $X$ was)
  • The expected value of $0.5*X + 0.5*(1-X)$ is also still $0.5$. (though it is not longer a uniform distribution but a constant).

The interesting part of the qeustion is, how do you determine the expected maximum of 500 (independent) copies of $X$? For this, it is not sufficient to know the expected value of each individual $X$. The distribution is important here.

Simon
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  • probably I wrote it wrong. the 0.5 in 0.5x + 0.5(1-x) was meant to signify the probability of going right and probability of going left. I am getting the intuition that the final distribution looks different, but still unable to understand it. I understand that p.d.f. of max(Xi) is different and can take out the value 500/501. But how does that relate here? Sorry, if i am asking dumb questions – anshul aggarwal Sep 13 '21 at 21:08
  • do you mean to say that the final answer is 0.5max(Xi) + 0.5max(1-Xi) ? – anshul aggarwal Sep 13 '21 at 21:09
  • actually, thats true. The expected value of $\max(X_i)$ is the same as the expected value of $0.5 \max(X_i)+0.5\max(1-X_i)$. – Simon Sep 13 '21 at 21:11
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    Totally makes sense. I get it now. Thanks a lot for helping me out. – anshul aggarwal Sep 13 '21 at 21:16
  • you're welcome. I wasnt sure if I wrote something useful for you. Stochastics is hard to wrap ones head around sometimes ;) – Simon Sep 13 '21 at 21:17
  • haha yeah! that part where you said " max(Xi) is the same as the expected value of 0.5max(Xi)+0.5max(1−Xi)" did it for me. Actually I had solved this question around 10 days back on my own but today totally went off a different chain of thought. Just need tons of practice to weed out such thinking errors :) – anshul aggarwal Sep 13 '21 at 21:22