That was me failing at paraphrasing, actual problem is more along the lines that given the midpoint of the rectangle is within the big circle, etc

Of course it depends on the big R. If the big R was infinite the probability would be naturally tending to 1 right???

Its just that as you go close to the edge the conditional remains 100% because its so far away from the small circle that it doesnt matter that many rectangles fall outside the big, those that fall inside will always be out of the small so its 100% even up there. But things change below r<30.2. There the conditional starts dropping below 1 and gets to zero roughly near 21+ a bit,

The point is eventually you willd or an integration of all this over the set of internal rectangles and there the big R will enter.

So what does "touching the little circle" mean, that any point on the rectangle is inside the circle, or only the midpont?

Masque, is your 91% the probability that it does NOT touch instead of that it does touch? The little circle is much smaller than than the big circle, and the long side of the rectangle is half the radius of the little circle. So there's lots of opportunity to miss the little circle. When the long side is oriented along a radius the probability should just be 20/95 = 21% by the new interpretation that only the center must be inside the big circle. If the short side is along a radius, then it's 20/99 = 20.2%.

This is a very confused question apologies in advance....

Are there any known (trivial?) techniques on (approximately) inverting a dense random matrix "perturbitavely"......

Perturbitavely means if you have an nxn matrix you can give each of the n^2 elements a random or highly correlated order weight (1,2,3....K) and only keep the 1,2,3 ect highest order terms element by element? I first thought of grahm schmiting ect into a basis that makes the order rank trivial on a term by term basis.....but we proved that can't be be done in general. I did a basic literature search and couldn't find anything (obv not my field) but it seems like an obvious problem that somebody else should have worked on, and perhaps even solved a long time ago.

Screw this, I'm tired of confusing people I will just post the actual problem word for word.


Super sorry for that confusion again... =\.

I got a hint from the TA. Find the P(hit|angle), then find p(hit) using total probability theory. And he said PDF over all angles is 1.

Yes its the probability that it doesnt have any common area ( seen as intersect) with the small circle if given that its part of the set of rectangles that are inside the large circle.

See my prior post to see why it roughly comes out as a ratio of areas when integrated over all the set. So basically some (99^2-30.2^2)/99^2 plus corrections.

I imagine a point M in the plane with (r,f) the center of the rectangle and use a rotation 0 to 2Pi of the vector say MA (A one vertex) to examine all possible orientations a rectangle with center M would happen to have. I examine of course as function of r,a,b ,r1,r2 (a,b,the sides, r1 small circle r2 big). I ask myself what fraction of those possible orientations satisfy the condition to be inside and to not intersect the samllone at the same time. This is the density function basically un-normalized.

Then all i have to do is notice spherical symmetry so this is function of r only and observe how that ratio (the conditional probability for each point M that belongs to the sets of rectangles inside the big circle) behaves. If you are away from the small circle by at least r1+(a^2+b^2)^(1/2) you are able to claim all orientations for these points are outside the small circle and inside the big one until you go up there to a bit less than r2-a say . Over there the chance a random point with r> to say about r2-b/2 (roughly of course need to be exact here but it wont affect the problem) belongs to a rectangle that is inside the big circle is less than 1 but all those that do obviously are safely away from the smallcircle anyway so the conditional is 1 there.

It drops below one only as r approaches the small circle at say r<r1-(a^2+b^2)*1/2).

So you have a matrix say 3x3 for example;

{{a11,a12,a13},{a21,a22,a23},{a31,a32,a33}} and maybe say a13,a22,a23>> everything else and you want to now consider the inverse and keep only the most significant terms expressing where possible the result as function of only the 3 big ones???

Can you give a more clear example maybe a particular 2x2 or 3x3 to get an idea what you want to do???

Cant Mathematica invert it and then look at the elements one by one and try to expand those in their most basic important significant terms plus unimportant corrections that you ignore (if its not essential to keep though because it may be).

OK, correcting the numbers from my last post: When the long side is oriented along a radius the probability should just be

(40+10)/200 = 25%.

That's because one end can cross the diameter of 40, and then move another 10 before it stops touching (like convolution). That's out of a total of 200 that the the center can move to stay inside the big circle. If the short side is along a radius, then it's

(40+2)/200 = 21%

which should be the minimum. When a diagonal of the rectangle is along the radius, that's a length of

sqrt(10^2 + 2^2) = sqrt(104)

and gives a probability of

(40 + sqrt(104)) / 200 =~ 25.1%

which should be the maximum. You have to do that for every angle of orientation of the rectangle around 2pi. Find a function for the length along the radius L(theta), and use that in the probability formula for P(hit|theta) taking into account that the center doesn't always move along a diameter of 200. Each angle has the same probability, so integrate and divide by 2pi.

Sorry should have given more info.....they are super large matrices. Standard optimized latex packages take ~minute to invert on a 20 node cluster.... We'd like to do it on millisecond scales.

The exact weighting shouldn't matter (we want to be able to go it for general weightings) but I don't think there is an obvious way to generalize from small matrices ....but that could be wrong. We have a semi rigorous definition for "close enough" to the inverse that's passed a lot of tests....and matrix multiplication is obv very cheap to check.

Reposting my prior post to make a few corrections.

Yes its the probability that it doesnt have any common area ( seen as intersect) with the small circle if given that its part of the set of rectangles that are inside the large circle.

See my prior post to see why it roughly comes out as a ratio of areas when integrated over all the set. So basically some (99^2-30.2^2)/99^2 plus corrections.

I imagine a point M in the plane with (r,f) the center of the rectangle and use a rotation 0 to 2Pi of the vector say MA (A one vertex) to examine all possible orientations a rectangle with center M would happen to have. I examine of course as function of r,a,b,r1,r2 (a<b the sides, r1 small circle r2 big). I ask myself what fraction of those possible orientations satisfy the condition to be inside and to not intersect the small one at the same time. This is the density function basically un-normalized for each point (r,f).

Then all i have to do is notice spherical symmetry so this density is function of r only and observe how that density=ratio (the conditional probability for each point M that belongs to the sets of rectangles inside the big circle) behaves. If you are away from the small circle by at least r1+(a^2+b^2)^(1/2)~30.2 you are able to claim all orientations for these points are outside the small circle and inside the big one until you go up there to a bit less than r2-a say . Over there near the circumference of the big circle the chance a random point with r> r2-(a^2+b^2)^(1/2) belongs to a rectangle that is inside the big circle is less than 1 but all those that do obviously are safely away from the small circle anyway so the conditional is still 1 there too. Eventually the chance to be inside drops to 0 at about r2-a-tiny bit (a the small side)

So just study how that conditional probability for each point evolves with r and that way you have assigned a function a density to each point. Then you integrate over all points of the plane. Since its 100>>20 here its easy to see the small edge effects are not going to alter the basic area ratio i claimed above although to be rigorous one needs to take care of the radii below r1+(a+b^2)^*(1/2) where the conditional drops below 1 and of course to also calculate accurately the upper r near r2 that you can no longer have rectangles satisfy the requirement to be inside.

Yes i was assuming it was fully inside however the density function is 1 even above the last 100% inside radius which is r=r2-(a^2+b^2)^(1/2)/2=100-10.2/2=94.9.

So basically above 94.9 all the way to a bit less than 99 those points that represent the center of a rectangle are not all corresponding to rectangles that are fully inside but those that do are safely away from r1 to still give a density of 1.

This is why i basically say its an area thing. Why arent you also considering areas?

If i imagine that only the center is inside after seeing his problem as worded then i suppose i could say its not going to significantly change that much still around 90% or so.


edit

Correction the area ratio is more like; (due to his claim only the center needs be inside)

(100^2-25.1^2)/100^2+corrections=93.7%+corrections

ie 6.3% plus corrections to hit any part of the city.


Basically let me restate what the density does for his re-adjusted problem;

From r=21+ to 25.1 it is 0 rising to 1 eventually.From r=25.1 all the way to r=100 it remains 1. So we have to integrate over all those points inside the big circle and properly weight them with the above density hence an area thing.

We also need to be rigorous in this calculation to capture the effects properly (corrections to the above) but in my opinion its kind of ridiculous to be rigorous on a problem that makes such crude assumptions like tornado being a rectangle, city being circular etc. we need to know how rigorous the class wants us to be here is it a geometric problem or a real life estimation that doesnt have to be nitty.

The center will always fall on some diameter, and any diameter is equally likely. So given a particular diameter, I'm just rotating the rectangle around it's center point. If I do that, the center should always have a range of 200 along that diameter (contrary to my last edit). The min probability of touching should occur when the short side is along the diameter, and the max should occur when the diagonal is along the diameter.

My P(hit|theta) won't be (L(theta)+40)/200 for all theta, only when a side or a diagonal is along a diameter. Otherwise, the center still moves 200, but the L+40 isn't right because one side will hit before the length L hits. I still think the min probability should occur when the short side is along the diameter, and the max probability occurs when the diagonal is along the diameter. I see what you're saying you'd think it would depend on the circle areas. I seem to have something that depends on the circle diameters.

Do you agree that all rectangle center points with (polar coordinates) r>25.1 and r<100 are both inside the big one and outside the small one? In other words if you randomly select a point in this sector it fails to hit the city. The 25.1 is 20+(2^2+10^2)^(1/2)/2.

So why isnt then the chance to miss the city simply the relative area ie (100^2-25.1^2)/(100^2)=93.7%. Plus we have corrections because if we go below 25.1 there are orientations that safely avoid the city all the way to about 21 but those basically correspond to density functions between 1 and 0 so they are the correction term that is the core rigorous calculation i havent done yet since i view it as nitty although for the real geometric problem needs to be done. Basically all that is needed is to model the density as function of r by finding the range of angles that produce safe rectangles between 21 and 25.1. Then we can do a more rigorous integration but since that area is only (25^2-21^2)/100^2=0.0184 in relative terms and say has avg density about 0.75 (higher near25 is not only more area but also closer to 1) for example it is not going to significantly change the overall estimate from say 92-93%.

What happens to your method if you take for example the very small rectangle case of 1x1 that starts looking like a point? Shouldnt we both then be converging to something like;

~(100^2-21^2)/100^2~95.6%???
What would your method give then???

I believe that the difference between our solutions represents a paradox similar to Bertrand's paradox which lies at that the heart of probability theory. This may even be the point of the problem.


Yes I agree. Now do you agree that each of the diameters are equally likely to contain the center of the rectangle? Then do you agree that all points along any diameter are equally likely to be the center? Then we only need to consider a single diameter, and whenever r > 25.1 the city will be missed, and the probability of this is 74.9/100.

The problem told us that each point of the circle will be chosen by the tornado with equal probability. But what does this mean? To you it means that equal areas are equally likely. But I have described a different experiment for choosing a random point by choosing first a random diameter, say by choosing a random theta, and then choosing a random point on that diameter uniformly distributed. Each of these can be simulated by different simulations, and each should give our respective answers. This is similar to the Bertrand paradox in which there are 3 ways choose a random chord, and they each give very different answers. None are to be preferred a priori in that case, and I think not in this case either.

The way i see it is very simple and i believe the correct without any chance of alternative interpretation. I assume the probability the center is at a point x,y within error dx,dy is the same for all points of the plane target area A and equal to dx*dy/A. If i asked you to take random points in the plane inside a unit square what would you do??? x= rnd*1, y=rnd*1 right? Uniform distribution.

How about if the area was a fat W. What diameter will you see there?

With your approach you arrive at the funny result that if the "tornado" is a tiny dust particle (or a small 1mm radius "bomb" say) that can fall anywhere within the big circle with the same probability you will tell me ~20/100.Basically you will tell me a tiny circle gets 20% of the hits when it has only 4% of the total target area? How can such approach make sense?

By the way if one asked me what is the avg distance from the center if i throw random dots in a circle the answer is clearly

Integral[r*(dr*r*df) with r in (0,R) and f in(0,2Pi)]/Integral[(dr*r*df) with r in (0,R) and f in(0,2Pi)]=1/3*R^3*2*Pi/(Pi*R^2)=2/3R.

The way you see it though the answer would be R/2 since any point in the radius can happen with equal density from 0 to R. (that of course is true only in 1 dimension)

That's an assumption that wasn't specified by the problem. It may make sense for tornadoes, or for darts, I don't know. Maybe the tornado chooses a random trajectory into the city, and then makes a random hop along that trajectory. Then my answer is right. The problem just said all points are equally likely, and they are in my scenario. They all have probability 0. That's an assumption that wasn't specified. We don't have enough information if this is purely a probability problem. The same thing happened with your problem of finding the distance between 2 points chosen at random on a unit circle. If you recall, checktheriver got a different answer, and that answer corresponded to choosing a random radius, and then a random point on that radius, then constructing the perpendicular through that point to choose the 2 points.

Areas of r*dr*dtheta would be the element of areas with equal probability in my scenario, and these areas are larger near the outside of the circle than near the inside, so areas near the outside would have lower probability than equal areas near the inside. One thing that works that way is a disk drive. As the disk spins and the head makes a random access, sectors on the outside can have a lower probability because there are more of them. In fact, a difference between mac and pc drives, at least at one time, was that one encoded data to be read at constant angular velocity while the other encoded it to be read with constant linear velocity. Tornadoes also move in circular motions analogous to a spinning disk drive, and they make hops analogous to the movement of r/w heads.

I think I am really close but need a little push.... sigh I suck at this stuff :/

For g<s, t> = < (st+1)/(st-1), (s-t) / (st-1), (s+t) / (st-1) >, show that g<s_o, t> and g<s, t_o) where s_o and t_o are constants.

I started by saying (st+1)/(st-1) = 1 + 2/ (s_ot-1), and let u = 1/(s_ot-1). So the first component can be represented as 1 + 2u.

I am just having a bit of trouble rewriting the next components to get it in the form ct + d

Shae, you said "show that" but didn't say what to show...

Oh blah, show that ... are subsets of lines.

g(s,t_0) takes the form (k1*f(s),k2*f(s),k3*f(s)).

Is there an eloquent way to write it as I did with the first component?

Oh, I read your post and you wanted help with writing eg. (s0 - t)/(s0*t - 1) in the form xu + y? (where u = 1/(s0*t - 1) ) ?

This is essentially polynomial division. You have a polynomial in t of degree 1 in the numerator, and a polynomial in t of degree 1 in the denominator. For practice, try dividing (3-t)/(3t - 1). If you can do that, you can probably divide (s0 - t)/(s0*t - 1).

I'm unsure how to deal with the remainder. So I have 1/t - 1/s_o + [(1/t)-1/s_o] / (s_o*t-1)

Should I define u to be the same thing, and then this becomes (1/t - 1/s_o) + (1/t - 1/s_o)*u ?