# Homework Help: The Waiting Time Paradox

1. Jan 29, 2008

### e(ho0n3

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.

2. Jan 29, 2008

### Dick

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.

3. Jan 29, 2008

### e(ho0n3

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.

4. Jan 29, 2008

### Dick

I would say 'expected waiting time' and 'average waiting time' are the same thing. Am I wrong?

5. Jan 29, 2008

### e(ho0n3

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?

6. Jan 29, 2008

### Dick

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.

7. Jan 30, 2008

### e(ho0n3

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".

8. Jan 30, 2008

### EnumaElish

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.

Last edited: Jan 30, 2008
9. Jan 30, 2008

### Dick

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.

10. Jan 30, 2008

### e(ho0n3

No I'm not.

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.

11. Jan 30, 2008

### Hurkyl

Staff Emeritus

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.

12. Jan 30, 2008

### e(ho0n3

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.

13. Jan 30, 2008

### Dick

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.

Last edited: Jan 30, 2008
14. Jan 30, 2008

### EnumaElish

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).

Last edited: Jan 30, 2008
15. Jan 31, 2008

### e(ho0n3

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.