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May 30

Calculus in a statistics class

Next semester at college, I am enrolled in a class called "Intro to Probability and Statistics". All the course description says `is: "Sample spaces; combinatorial theory; elementary probability; random variables; discrete and continuous probability distributions; moments and moment-generating functions; applications"

This makes it sound pretty basic, but 2 things give me pause. First, one of the prerequisites to this particular course is multivariable calculus. Second, while browsing some of the student feedback on the course, several warned that your integration skills have to be sharp to do well in this class.

Fortunately, I did well in my multivariable calculus class and have a lot of practice doing integration problems, but I'm still curious: what concepts taught in this type of class are likely to require us to use integration? I might want to read up on these subjects just so I have a better idea what I'm getting into.--Captain Breakfast (talk) 03:02, 30 May 2015 (UTC)[reply]

See probability density function, which is integrated into the cumulative distribution function, and some of the other articles that are linked to about continuous distributions. I will comment that, if you know calculus, continuous distribution functions are generally easier to work with than discrete distribution functions. If you don't know calculus, the opposite is true. Robert McClenon (talk) 03:13, 30 May 2015 (UTC)[reply]

I'll add that (as with most integration), the challenge with integrating a density function is determining the limits of integration. Be sure you know how to set up and solve double integrals over general regions (e.g. not just over a rectangle): you need this to derive the joint density function of a function of two continuous variables (something like "If X has density function f, and Y has density function g, and Z = 2X - 3Y, find the density function of Z). You will probably also have to solve for one of the limits of integration, given an integral and it's solution: this is how you find quantiles. Don't worry about trigonometric functions, though: they won't appear (integration by parts will, though, particularly for expected value problems). Good luck! OldTimeNESter (talk) 19:12, 2 June 2015 (UTC)[reply]

To be clear, when I say "general regions," I'm still referring to integrals defined with rectangular Euclidean coordinates (e.g. not parametric functions, or vector fields). OldTimeNESter (talk) 19:12, 2 June 2015 (UTC)[reply]

Thanks for the feedback! I actually love doing integration over triangles, trapezoids, teardrops, or and areas defined by diagonal lines, parabolas, etc. My favorite kind of problem was doing double integrals where you had to switch the order of integration in order to solve. I also enjoy doing integration by parts, but I'm relieved I won't to have to mess with trigonometric stuff--that frequently messes me up.

Also, now that you mention, I am now recalling a lecture in my multivariable calc where my professor solved an integral where the answer was the normal distribution probability density: i.e., 1/(2pi)*e^((-x^2)/2). I don't remember what the integration problem was that had that answer, but I assume this is the type of stuff we will be doing in my upcoming stats class.--Captain Breakfast (talk) 07:33, 6 June 2015 (UTC)[reply]

How to calculate the length, width, height of a rectangular cuboid given two opposite corners

How do you calculate the length, width, and height of a rectangular cuboid when given the coordinates of two diagonally opposite corners? I came across some code[1] that does exactly this:

   viewer->addCube (min_point_AABB.x, max_point_AABB.x, min_point_AABB.y, max_point_AABB.y, min_point_AABB.z, max_point_AABB.z, 1.0, 1.0, 0.0, "AABB");

I didn't even think this was possible before, since so little information is given. My other car is a cadr (talk) 10:04, 30 May 2015 (UTC)[reply]

I think the code contains the implicit assumption that the sides of the cuboid are parallel to the axes of a given co-ordinate system, which is enough additional information to solve the problem. Without that assumption, a cuboid with a given space diagonal can have any volume from as close to zero as desired, up to a cube with the given diagonal; it can also be in any orientation, and have any one side of any length from zero to the length of the diagonal (a choice which will constrain the other side lengths). AlexTiefling (talk) 10:09, 30 May 2015 (UTC)[reply]

Sorry, my bad, it's just rectangular cuboids instead of general cuboids. I've updated the question. That should make it doable, no? My other car is a cadr (talk) 10:19, 30 May 2015 (UTC)[reply]

How the length, width and height match the x, y and z directions is rather arbitrary, but the length in each direction is simply the smaller coordinate value subtracted from the larger one, i.e. xmax - xmin, ymax-ymin and zmax-zmin.→109.145.13.182 (talk) 11:57, 30 May 2015 (UTC)[reply]

...or, alternatively,(|x₁-x₂|, |y₁-y₂|, |z₁-z₂|). StuRat (talk) 12:13, 30 May 2015 (UTC)[reply]

No, I was assuming a rectangular cuboid. There's still no guarantee that the cuboid is aligned with the axes. AlexTiefling (talk) 21:28, 30 May 2015 (UTC)[reply]

Agreed. If we don't assume the edges are parallel to the axes, then we would need to know what they are parallel to, in order to solve it. (An interesting problem would be if you also knew the volume, and from that and the opposite corner points, had to figure out the dimensions of a rectangular cuboid.) StuRat (talk) 21:59, 30 May 2015 (UTC)[reply]

If x²+y²+z² = L² where L is the length of the diagonal, then there exists a rectangular cuboid with length = x, width = y, and height = z. Bo Jacoby (talk) 22:17, 30 May 2015 (UTC).[reply]

Not just one. There should be an infinite number of solutions, if the orientation is not specified (although if they specify that x, y, and z must be integers, then there may only be one solution, excluding the trivial duplicate solutions where x, y, and z are swapped around). StuRat (talk) 15:20, 2 June 2015 (UTC)[reply]

Surely, pick x² ≤ L². Pick y² ≤ L²–x². Let z² = L²–x²–y². Bo Jacoby (talk) 13:48, 6 June 2015 (UTC).[reply]

Rotations in four-dimensional space using quaternions

Does anyone know of a good, preferably free, introduction to this topic for a non-mathematician? Most of the material I've come across is a bit too formal for me.--Leon (talk) 10:33, 30 May 2015 (UTC)[reply]

Chapter 11 (Hypercomplex numbers) of Penrose's The Road to Reality has a couple of sections on this. Be warned though, while the book may be intended for non-mathematicians, even mathematicians may find it pretty heavy going. --RDBury (talk) 13:13, 30 May 2015 (UTC)[reply]

Our article Rotations in 4-dimensional Euclidean space isn't bad and includes a short section on quaternion representation of 4D rotations. --Mark viking (talk) 17:21, 30 May 2015 (UTC)[reply]

You might want to also look at use of geometric algebra as a mathematically similar approach, as being more intuitive than the use of quaternions. I'm surprised that the article Rotations in 4-dimensional Euclidean space does not mention this. —Quondum 17:45, 30 May 2015 (UTC)[reply]

Name of a general "orthogonal group"

We could define (in a broad sense) the orthogonal group of a vector space V over a field K with an associated symmetric bilinear form B to mean the group of all linear transformations that preserve the form, namely all T for which B(Tv, Tv) = B(v, v) for all v in V. It appears to be a fairly normal to define orthogonality in terms if any reflexive bilinear form (and, I suspect, any bilinear form in general). We already have a name for a group that preserves a symmetric bilinear form. My question is: what would be call the group that preserves an alternating bilinear form, and what would we call the group that preserves a bilinear form in general? Does the latter even make sense? —Quondum 17:31, 30 May 2015 (UTC)[reply]

The group associated to an alternating bilinear form is called the symplectic group. Sławomir Biały (talk) 18:04, 30 May 2015 (UTC)[reply]

Ah, thanks, I read that article too cursorily, and missed that statement. Which leaves the question for an arbitrary bilinear form (neither symmetric nor alternating in the general position). —Quondum 19:41, 30 May 2015 (UTC)[reply]

An automorphism of a general bilinear form preserves the alternating and symmetric parts. In general, the group should decompose into irreducible factors consisting of unitary groups, orthogonal groups and symmetric groups. Sławomir Biały (talk) 21:51, 30 May 2015 (UTC)[reply]

Okay, thanks, that is valuable info, which makes a nice overall picture, and gives me a few areas of study: unitary groups and decomposition of groups. —Quondum 22:26, 30 May 2015 (UTC)[reply]

Classical group summarizes this concisely (without going much into general fields or rings). YohanN7 (talk) 11:44, 1 June 2015 (UTC)[reply]

Thanks, I've made a note to include this article when I review this. —Quondum 18:57, 1 June 2015 (UTC)[reply]