Suppose it is 1800 and you have just been employed as a captain by a shipping company that regularly transports goods between London and New York. The distance between London and New York is 3,000 nautical miles. Your employer tells you that on average the trip between London and New York takes 10 weeks so the average velocity of his ships is 300nm/week. However, because the ships are powered by sail (the steam ship has yet to be invented) there is a great of variance in the velocities of his ships. In fact, your data-minded employer tells you that the velocities of his ships for trips between London and New York are normally distributed with an average velocity of 300nm/week with a standard deviation of 50nm/week.

On your first voyage, you and I agree to pass the time by estimating every week what the final velocity of our trip will be. Every week we know how many total weeks we’ve been traveling and how many nautical miles of the original 3,000 remain between us and New York. We can thus calculate what our current velocity is.For instance, after two weeks 2,500nm remain between us and New York so our current velocity is:

(3,000nm-2,500nm)/2weeks = 250nm/week.

How can we best estimate what our final velocity for the entire trip will be using only the prior distribution of average velocities (mean=300nm/week, standard deviation=50nm/week), the distance remaining, and the number of weeks traveled?

How can we estimate the variance of our final velocity for the entire trip?

Keep in mind that we will want to update our estimate every week based on the new information of how far we’ve traveled.


1 Answer 1


It looks this problem isn't completely specified. In particular, you need to distinguish within-ship variability from between-ship variability. Within-ship variability is how much a single ship's speed varies over time, whereas between-ship variability is how the relationship between time and speed varies across ships. In particular, it isn't clear to me whether "your data-minded employer" is talking about the within- or between-ship variability in the normal distribution he describes.

By the way, you keep writing "velocity", but it seems that you mean "speed", since we're not considering the ships' direction of movement here. An object's velocity comprises its speed and its direction of movement.

Edit: From your comment, it looks like the employer is only reporting the distribution of trip times (divided by the trip length, 3,000 nmi, to get average speeds), and you assume that all ships have the same distribution of trip times. It also seems that while we know that ships speed up and slow down quite a bit, not to mention changing direction, we don't have much information about the pattern of this over time. For example, it might be that if the ship is moving especially fast today, then tomorrow it's likelier to be fast than slow. But we don't have any data about this from previous voyages, so let's assume not. Then, in the mid-voyage problem, when we're estimating how fast the ship will go for the remainder of the voyage, we can ignore the ship's previous performance; we just need to know how much distance of the trip is left, and multiply that by the distribution of average speeds.

Hence, the posterior completion time for the whole voyage is simply

$T = t_0 + \frac{3,000\text{ mni} - d}{X}$

where $t_0$ is the time taken so far, $d$ is the distance traveled so far, and $X \sim \operatorname{Normal}(300\text{ nmi / week}, 50\text{ nmi / week})$. The posterior average speed is then $3,000\text{ nmi} / T$.

  • $\begingroup$ Actually I do mean velocity. A ship can get blown off course, it has to tack with the wind. The average "velocity" of all his ships is the distance between London and New York divided by the average trip time between London and New York, regardless of whether a ship was blown off course, etc. $\endgroup$ Sep 14, 2016 at 18:50
  • $\begingroup$ The "data-minded employer" is just tracking the total trip time between London and New York from which you can calculate the velocity of each trip (as discussed above). So there is no relevant difference between the speeds of different ships. $\endgroup$ Sep 14, 2016 at 18:53
  • $\begingroup$ "A ship can get blown off course, it has to tack with the wind." — But you're not tracking any of the direction changes, just the time and distance, even though it's the as-the-crow-flies distance between London and New York rather than the full distance actually traveled by the ship. A mean velocity would include a direction as well as a speed. Similarly, the "current velocity" you describe is only a speed. $\endgroup$ Sep 14, 2016 at 19:30
  • $\begingroup$ @KevinMcLaughlin See edit. $\endgroup$ Sep 14, 2016 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.