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March 20

Normal Distribution

If I have a sample of size n drawn from a normal distribution with mean μ and variance σ², what is the probability density function of the value of the k^th percentile P_k? — Preceding unsigned comment added by 123.136.64.14 (talk) 07:31, 20 March 2013 (UTC)[reply]

See Order statistic. Bo Jacoby (talk) 14:22, 20 March 2013 (UTC).[reply]

Integral of 1/x

I was taught that the integral of 1/x is ln|x|+C but does this actually mean that the integral of dx/x is ln|x| + C? Clover345 (talk) 16:01, 20 March 2013 (UTC)[reply]

Yes. Because 1 * dx = dx. Obviously. — 79.113.214.121 (talk) 16:30, 20 March 2013 (UTC)[reply]

No. It means

\int {1 \over x}\,dx=\ln \left|x\right|+C.

We talk about the integral of a function (meaning the indefinite integral in this case), not the integral of dx times a function. The dx goes with the integral and indicates that the integral is over x. RockMagnetist (talk) 20:42, 20 March 2013 (UTC)[reply]

\int {1 \over x}dx=\int {dx \over x}

— 79.113.214.121 (talk) 20:56, 20 March 2013 (UTC)[reply]

While the two statements are equivalent, I would say that the second is more accurate: dx/x is a differential form and so can be integrated. The bare statement "integral of 1/x" is meaningless without saying what measure the integral is taken with respect to. Sławomir Biały (talk) 21:54, 20 March 2013 (UTC)[reply]

This isn't really what you asked, but it's probably worth pointing out that the absolute-value sign in the expression

\ln |x|+C

is kind of a cheat. If you're not careful, you might think it means that, say, you can write

\int _{-2}^{2}{\frac {dx}{x}}=\ln |x|{\bigg |}_{-2}^{2}=0

. But you can't — that definite integral is not defined. The version with the absolute value works for intervals completely contained within the negative reals or completely contained within the positive reals, but it doesn't work for intervals containing zero. --Trovatore (talk) 22:05, 20 March 2013 (UTC)[reply]

While the Lebesgue and improper Riemann integrals of

{\frac {1}{x}}

are not defined over intervals containing zero, the formula still gives the correct Cauchy principal value for such integrals. In that sense, the formula

\int _{-2}^{2}{\frac {dx}{x}}=0

is true (but you have changed the meaning of the

\int

). —Kusma (t·c) 10:43, 21 March 2013 (UTC)[reply]

Yeah, but Cauchy principal values are mostly a curiosity. They're not very robust under stuff you'd like to be able to do. In my view they're ordinarily best avoided unless you have a special problem with a clear rationale as to why that value is relevant. --Trovatore (talk) 14:58, 21 March 2013 (UTC)[reply]

Yes the principal value is a far more useful idea, but then you're dealing with complex numbers. Any idea who first called it that anyone? Cauchy didn't also do this did he? Dmcq (talk) 15:54, 21 March 2013 (UTC)[reply]

I really don't see any very direct connection between those two notions. The article you linked to seems to be just talking about making a branch cut so that you have a single-valued analytic function on a slightly reduced domain. Cauchy principal values, on the other hand, are ways of assigning a value to integrals that don't converge. They're similar to the various ways of assigning values to the sum of a divergent series, and have the same sorts of problems. --Trovatore (talk) 16:18, 21 March 2013 (UTC)[reply]

I'm just saying that for complex numbers you've got Cauchy's integral theorem and the constant is fixed within a branch, the integral is given by the difference in the principal values of the integral at the two points, in this case by the values of log z as given in the principal value article. It is true though that one can always go around 0 a few times and so not stay with a branch. Dmcq (talk) 17:53, 21 March 2013 (UTC)[reply]

Oh, true — I thought you were trying to find a commonality between the two senses of "principal value". If you take one of the two direct paths from −2 to 2, the integral will be either −iπ or iπ, depending on which one you take. --Trovatore (talk) 19:13, 21 March 2013 (UTC)[reply]

I’d take issue with the claim that Cauchy principal values are mostly a curiosity. They come up naturally in Fourier analysis, for one thing.—Emil J. 13:27, 26 March 2013 (UTC)[reply]