What is the solution to the Waiting Time Paradox?

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[SOLVED] The Waiting Time Paradox

Suppose there's a bus stop where the interarrival time T between buses is exponentially distributed with E[T] = 1. I randomly arrive at this bus stop and because of the memoryless property of the exponential distribution, my average waiting time is E[W] = 1.

The paradox here is that E[W] = E[T] where "common sense" would suggest that E[W] < E[T]. The paradox is explained by noting that I'm more likely to arrive during a large interarrival time than I am during a short one.

This however doesn't make any sense to me because P{0 <= T <= 1} > P{T > 1}, i.e. T is more likely to be small.
 
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Sure, your probability of waiting less than T=1 is greater than 1/2. But the waiting time is a weighted average of the probabilities of the waiting times. So a waiting time of T=2 counts twice as much in the average as a waiting time of T=1.
 
Would you elaborate on "the waiting time is a weighted average of the probabilities of the waiting times". I don't understand. You seem to be referring to the expected waiting time.
 
I would say 'expected waiting time' and 'average waiting time' are the same thing. Am I wrong?
 
No, you're not wrong. What I need to see is a calculation that demonstrates that the probability of arriving at an interval greater than 1 is greater than 50 %. The calculation of the expected or average waiting time says nothing about this. Or does it?
 
It does. To get the probability you integrate the probability density. To get the waiting expectation value time you integrate T times the probability density. That weights longer times even with lower probability versus shorter times with higher probabilities.
 
Sorry. I'm still unconvinced because I still think that P{T >1} < P{T <= 1} implies that I'm more likely to arrive during a short interval, even if "T=2 counts twice as much in the average as a waiting time [sic] of T=1".
 
You are comparing mean (E[T]) to median (P{T >1} vs. P{T <= 1}).

Mean = 1 but median = Log 2. The latter implies P{T > Log 2 } = P{T <= Log 2}.

The calculation of E(T) does say something about this. If you know E(T) = u then you know median = u Log 2.
 
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e(ho0n3 said:
Sorry. I'm still unconvinced because I still think that P{T >1} < P{T <= 1} implies that I'm more likely to arrive during a short interval, even if "T=2 counts twice as much in the average as a waiting time [sic] of T=1".

It IS TRUE that the odds of it arriving in T<1 are better than 1/2. So? That doesn't mean E[W]<1. I'm losing track of exactly what your question is.
 
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You are comparing mean (E[T]) to median (P{T >1} vs. P{T <= 1}).

No I'm not.

It IS TRUE that the odds of it arriving in T<1 are better than 1/2. So? That doesn't mean E[W]<1. I'm losing track of exactly what your question is.

Yes. There seems to be a misunderstanding. Going back to my first post, why am I more likely to arrive in a T with T > 1 (large interarrival time) than with T < 1 (small interarrival time)? I'm looking for a probability calculation that explicitly demonstrates this. So far, I've calculated the contrary.
 
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e(ho0n3 said:
Yes. There seems to be a misunderstanding. Going back to my first post, why am I more likely to arrive in a T with T > 1 (large interarrival time) than with T < 1 (small interarrival time)? I'm looking for a probability calculation that explicitly demonstrates this. So far, I've calculated the contrary.
Could you show your calculation?

By the way, T is a random variable describing "the time between successive arrivals"; it is not a random variable describing "the amount of time between the arrival before I got here and the arrival after I got here". I will call this latter variable X. It seems pretty clear to me that X is the sum of two independent random variables with the same distribution as T.
 
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Yes, T is the time between arrivals (of the buses at the bus stop). That is clear from the problem statement (cf. first post).

My calculation: P{T > 1} = 1 - P{T <= 1} = 1 - (1 - e^-1) = e^-1 < 1/2. My interpretation: Large interarrival times are less likely than small ones. Conclusion: The explanation of the paradox in my first post is bogus.
 
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e(ho0n3 said:
Yes. There seems to be a misunderstanding. Going back to my first post, why am I more likely to arrive in a T with T > 1 (large interarrival time) than with T < 1 (small interarrival time)? I'm looking for a probability calculation that explicitly demonstrates this. So far, I've calculated the contrary.

You AREN'T more likely to wait T>1 than not. Nobody is claiming that. The 'paradox' is talking about AVERAGE. 1/2, 1/2, 1/2, 1/2, 8. Pick a number. The odds are 80% that it's less than one. The average is 2. Which is greater than 1.
 
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Say I make 100 draws from ExpDist[1]; 63 of them turn out to be < 1. The expected value (average) conditional on T < 1 is a = 0.42. And the expected value (average) conditional on T > 1 is A = 2.

While it is true that I have obtained more short draws than long, the time spanned by the short ones is 0.42 * 63 = 26.5 < 2 * 37 = 74 = time spanned by the long draws. If my arrivals (as opposed to buses' arrivals) are uniformly spaced over time, then I am about 3 times as likely to arrive during a long stretch than a short one (74/26.5 is approximately 3).
 
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  • #15
EnumaElish said:
Say I make 100 draws from ExpDist[1]; 63 of them turn out to be < 1. The expected value (average) conditional on T < 1 is a = 0.42. And the expected value (average) conditional on T > 1 is A = 2.

While it is true that I have obtained more short draws than long, the time spanned by the short ones is 0.42 * 63 = 26.5 < 2 * 37 = 74 = time spanned by the long draws. If my arrivals (as opposed to buses' arrivals) are uniformly spaced over time, then I am about 3 times as likely to arrive during a long stretch than a short one (74/26.5 is approximately 3).

Aha! This is exactly the type of explanation I was looking for. I was going to write that I didn't know how to compute the aggregate times for T < 1 and T > 1 and that perhaps this was the source of my troubles. Indeed, it was. I have never dealt with conditional expected values before. It makes perfect sense now. Thanks a lot.
 
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