Simple theoretical car question.

[JoshuaD]

I had an argument with a friend about this situation, and I was unable to convince him. Please post what you think the answer is and why so I can link him to this thread.

You're about to leave your house and go to the gas station. On the way there you've got to make a few left turns that are subject to bad traffic and there are a few traffic lights.

If you leave 5 seconds earlier than you were initially going to, how much earlier will you get there on average?

[fiskebent]

5 seconds

[ADDboy]

The mean is 5 seconds, but the median difference in arrival time is smaller and possibly 0.

To see that the mean is 5 seconds, let T_1 be the travel time if we leave now, and T_2 the travel time if we leave in 5 seconds. These are clearly far from independent, but for expectation what matters is that they have the same distribution and thus E{T_1-T_2)=ET_1-ET_2=0. Since the expected difference in travel time is 0, the expected difference in arrival time is 5 seconds.

There are two reasons why the median is almost 0. The first is to think about your trip--it seems likely that the car leaving 5 seconds later will catch up to the ghost car that left on time, go through each light at the same time, and get to the gas station at the same time. Occasionally, leaving 5 seconds earlier will get you through an intersection at the last possible moment, thus saving a huge chunk of time. The fact that these two effects cancel each other out and the mean difference in arrival time is 5 seconds implies that saving a big chunk of time is a rare event, hence the median event is that the 5 seconds doesn't end up making a difference.

A more formal way of saying that is to realize that the distribution of the difference cannot be negative, but has a long right tail, hence the mean is noticeably larger than the median.

[richard kurtz]

no way to tell in fact it could take you more time depending how the lights are set up.

[AaronBrown]

I agree with ADDBoy. Unless there is some difference between the two times, the expected arrival difference should be five seconds. I also agree that the average is probably composed of a lot of zeros with a few one minute or longer differences.

An example of a difference beween the two times that could change the answer is if you normally leave when you hear a noon air raid siren, and that allows you to arrive at your first traffic light just as it's turning green. If you leave five seconds earlier, you save no time, you just have to wait five seconds at that light. That's pretty unrealistic for a five second delay, but if you said five hours it's a different matter.

[Pog0]

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no way to tell in fact it could take you more time depending how the lights are set up.

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Can you give an example?

[fiskebent]

Let's say that there are no lights, no turns and no traffic. Then you'll always arrive 5 seconds earlier. The average is 5 seconds and there's no variance.

Then let's say that there's a really weird traffic light right in front the gas station that is only green at noon and red the rest of the day. So no matter when you leave, you'll arrive at noon. Let's then divide the day into 17280 5-second intervals. Normally it won't matter that you leave 5 seconds earlier. But there's one 5-second interval where if you'll be at the light just before noon and you'll arrive a day (86400 seconds) earlier.
This averages to 86400 / 17280 = 5 seconds earlier.
So the average is still 5 seconds but with large variance. Depending on how the lights and turns are set up, the variance will be large or small.

[SheetWise]

[ QUOTE ]
[ QUOTE ]
no way to tell in fact it could take you more time depending how the lights are set up.

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Can you give an example?

[/ QUOTE ]

I raised an eyebrow at that one as well. But if you include all of the unknowns -- getting stuck behind a bus, or behind someone who needs to make a left turn mid-block, etc., it could be true. One of the most frustrating parts of getting stuck in traffic are those situations where you get locked in and get to watch all of the traffic from behind you advance in other lanes. But I'm still not sure how the lights alone could account for a reversal ...

[ohnonotthat]

I am WAY out of my element here but it seems the keyword is - [on] AVERAGE

Given this stipulation I am unable to see how a 5 second delay in the time of departure could result in anything other than a 5 second AVERAGE delay in arrival.

A situation could easily exist whereby the TYPICAL benefit is greater than 5 seconds (arrive much sooner by virtue of leaving only a little sooner); conversely it is possible that the TYPICAL result could be a cost of greater than 5 seconds (arrive later, in some cases much later, in spite of having left earlier) but I cannot see how the average would be anything other than static - leave "X" seconds sooner, arrive "X" seconds sooner.

[StinkWater]

it is much less than five seconds.

if the question was phrased such that the trips began at random times and one began 5 seconds before the selected random time for that specific trip, then the answer would be 5 seconds.

since the trips begin 5 seconds from each other, the lights and traffic correlate.