Timeline for Probability of collision (two bivariate normal distributions)
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Nov 14, 2012 at 21:37 | vote | accept | mmacx | ||
| Nov 14, 2012 at 17:16 | answer | added | whuber♦ | timeline score: 7 | |
| Nov 14, 2012 at 15:02 | comment | added | mmacx | @whuber That is exactly what I meant, lets forget the shooter analogy. The data that I have is 2 pairs of x,y coordinates that mark the estimated positions of 2 ships. Also, positional errors are bivariate normal with 95% probability of ship's actual position being within 1 mile of the expected position. | |
| Nov 14, 2012 at 13:30 | comment | added | whuber♦ | That is my point: because the positions are not known, your shooting analogy is incorrect. But there remains potential for confusion here, because it still is not apparent exactly what data you have and what you really aim to do with it. My understanding still is that you have used position measurements for both ships to estimate the ship positions and that you wish to find a probability distribution for the error in the associated estimate of the inter-ship distance, assuming that that positional errors are bivariate normal. If this is not what you mean, then please modify the question. | |
| Nov 14, 2012 at 0:17 | comment | added | mmacx | @whuber Ships' positions are not known, ships' expected positions are known (nominal positions or means or centers of the error circles, I don't know how to express myself more clearly, English is my second language). This expected position is calculated from position given by the ship's positioning system which is then extrapolated into probable trajectory. For the sake of simplicity, lets ignore how the center of the error circle is calculated, lets just focus on the fact that the ship is expected to be within 1 mile of the center with 95% probability. | |
| Nov 14, 2012 at 0:02 | comment | added | whuber♦ | Please explain to me, then, why the positioning systems are needed at all when you are willing to assume the ships' positions are known? Of what value is that information? It seems supernumerary at best. | |
| Nov 13, 2012 at 23:35 | comment | added | mmacx | @whuber In the shooter analogy, I said that the 'points at which he aimed each time are known' not the actual points of bullet impact. If we know the points at which he aimed, and we know how accurate he is, we should be able to calculate (with a degree of certainty) how far apart the bullet holes should be. Ships' expected positions are known (aiming points), probability distribution is known (shooter's accuracy), probability of a certain distance between ships (distance between bullet holes) is to be calculated. I fail to see the difference. I'm truly sorry for the confusion I caused. | |
| Nov 13, 2012 at 22:38 | comment | added | whuber♦ | What you just said in your comment is not what you wrote in your question. The analogy to the question is that you observe where the bullets hit and you want to know the chances that the shooters were aiming at points within distance $x$ of each other. This is an "inverse probability" problem whose solution requires you to assume some prior knowledge--in the form of probability distributions--concerning where those shooters were aiming. After all, if you know the ships' positions with perfect accuracy, why ask probability questions about their positions? | |
| Nov 13, 2012 at 22:19 | comment | added | mmacx | @whuber I am not sure if this info helps or not. The problem is that I do not understand completely the following part from your comment: 'you need to update a prior distribution...'. I read up on Bayes' Theorem but I can't find the application here. This problem is analogous to the probability that the sharpshooter will shoot 2 bullets within 5 inches of each other if the points at which he aimed each time are known, and his shots are 95% accurate within 1 inch of that aiming point. Hope this doesn't complicate things any further. | |
| Nov 13, 2012 at 22:08 | comment | added | mmacx | Okay, I'll try to give more information. Given information above, if there is probability >1% that the ships are less than 5 miles apart, a maneuver must be made to assure adequate separation. An outside observer can predict trajectories of each ship with confidence of 95% that the ships will be in their expected positions (as mentioned, that probability is based on bivariate normal distribution). The observer must decide whether to order the separation maneuver or not. This model is a decision support system which alerts the observer to order the maneuver before the loss of separation happens | |
| Nov 13, 2012 at 20:45 | comment | added | whuber♦ | Sorry about the vagueness: I was thinking of "mutual distance" as a random variable describing the actual positions. My concern is that you might be asking for something you can't get: to derive a probability distribution for the actual ship-to-ship distance based on measurements, you need to update a prior distribution using Bayes' Theorem. If that's not how you're thinking about this problem, you might actually be asking how to compute confidence or prediction intervals. I'm trying to elicit enough information from you to clarify this point, because it may be important. | |
| Nov 13, 2012 at 20:42 | comment | added | mmacx | @whuber I understood your comment, and I agree that the correlation of the position errors should be investigated more thoroughly, however, that is too much to cope with for me right now. I would be more than satisfied if I could find a solution for this simplified version of the problem. When you say 'mutual distance', do you think about distance between their expected or actual positions? Because, their expected positions (means?) are known. I guess I misunderstood something again :). Thanks for the patience. | |
| Nov 12, 2012 at 17:29 | comment | added | whuber♦ | My earlier comment still stands: combining correlated sources of error may produce correlated estimates of position, which will affect the calculations. Although in practice the magnitude of the resulting error may be inconsequential, some analysis of it ought to be offered to justify the solution. Incidentally, this question has no unique answer unless you provide a prior distribution for the ship locations (or at least their mutual distance): that may be why a solution has been eluding you. | |
| Nov 7, 2012 at 10:45 | comment | added | mmacx | I see, do you have any idea what model could be more suitable? | |
| Nov 7, 2012 at 8:20 | comment | added | image_doctor | I don't think the model of a circular region centered on a point is correct here, as that gives the probability that the two ships are both within a fixed distance of that point. There are many cases when neither ship will be in that region, yet both ships may be close to one another. | |
| Nov 6, 2012 at 0:16 | answer | added | image_doctor | timeline score: 2 | |
| Nov 5, 2012 at 20:42 | comment | added | mmacx | To whuber: The position is calculated form multiple sources, but the basis is GPS. However, I'm not interested in position errors of individual navigation systems because the ship's position error is a combination of multiple factors which are combined into single normal distribution. By law, the onboard systems are guaranteed to maintain this level of accuracy. To image_doctor: I am interested in general solution. I'm new here so can anyone explain how can I see one comment that was here before but is now missing, it was by @Erik, thanks | |
| Nov 5, 2012 at 18:20 | comment | added | image_doctor | Did I miss the nominal positions of each ship, or do you want a general solution for any pair of positions ? | |
| Nov 5, 2012 at 15:39 | comment | added | whuber♦ | What positioning systems are the ships using? In some cases (such as GPS), the errors may be correlated, which will affect the calculations. | |
| Nov 5, 2012 at 13:49 | answer | added | ziggystar | timeline score: 1 | |
| Nov 5, 2012 at 11:05 | review | Close votes | |||
| Nov 5, 2012 at 15:37 | |||||
| Nov 5, 2012 at 10:40 | review | First posts | |||
| Nov 5, 2012 at 10:47 | |||||
| Nov 5, 2012 at 10:20 | history | asked | mmacx | CC BY-SA 3.0 |