Wikipedia:Reference desk/Archives/Mathematics/2009 June 27

= June 27 =

sample path
The article Dirichlet process begins "In probability theory, a Dirichlet process over a set S equipped with a suitable sigma-algebra is a stochastic process whose sample path is a probability distribution on S." I think I sort of understand this, but to be sure, what is a sample path? There is currently no article about it. Thanks. 208.70.31.206 (talk) 09:42, 27 June 2009 (UTC)

Algebraic equation
Hi, just for fun I was doing some algebraic math. (It is a few years ago since I graduated from the (Swedish) gymnasium.) But I've encountered this problem which drives me crazy. Any help on this would be much appreciated.

$$\frac{1}{x}+\frac{1}{\sqrt{x}}=625$$

I will give my attempt at a solution, but it seems I'm wrong somehow. Please point out how and why I'm wrong:

$$\frac{1}{x}+\frac{1}{\sqrt{x}}=625\Rightarrow{x^{-1}+x^{{(1/2)}^{-1}}}=625$$

$${x^{-1}+x^{{(-1/2)}^{-1}}}=625\Rightarrow{x^{-1}+x^{-1/2}}=625$$

$${x^{-1}+x^{-1/2}}=625\Rightarrow{(x^{-1})^{-2}+(x^{-1/2})^{-2}}=625^{-2}$$

$${(x^{-1})^{-2}+(x^{-1/2})^{-2}}=625^{-2}\Rightarrow{x^2}+x=625^{-2}$$

$${x^2}+x=625^{-2}\Rightarrow{x^2}+x-\frac{1}{625^2}=0$$

So, $$x=-\frac{1}{2}\pm\sqrt{(\frac{1}{2})^2+\frac{1}{625^2}}$$

But this gives me that $$x_1\approx{0.000002559}$$; and $$x_2\approx{-1.00000256}$$. But my book tells me that the answer should be $$x=\frac{1}{625}=0.0016$$ I can clearly see the logic in this, but I can not figure out how to solve it in steps. Why is my method wrong, and what should I have done instead? The chapter (in the book) is about substituting varibales (e.g. $$x^2:=\!\,t$$), and I can not figure out how I could apply that method in this equation. Instead, I thought this method above would work as well. Apparently not... //Misopogon 85.229.101.18 (talk) 14:44, 27 June 2009 (UTC)


 * I assume that's a typo in the second line, where you have x raised to -1/2 raised to -1.
 * Your mistake is where you tried to raise both sides to the -2 power, but $$[{x^{-1}+x^{-1/2}}]^{-2} \ne {(x^{-1})^{-2}+(x^{-1/2})^{-2}}$$. A better approach would be to, with the first equation, multiply through by x, getting $$1 + \sqrt{x}=625x$$, and solve for $$\sqrt{x}$$, which then leads to x. If it helps you to visually see what you're solving for, you can substitute $$t = \sqrt{x}$$ to get $$1 + t = 625 t^2$$. Then don't forget to check your solution with the original equation. --COVIZAPIBETEFOKY (talk) 15:06, 27 June 2009 (UTC)


 * BTW, the book must have made a mistake (it's not unheard of for that to happen), because $$x=\frac{1}{625}$$ does not solve the original equation. --COVIZAPIBETEFOKY (talk) 15:11, 27 June 2009 (UTC)
 * Thank you for the help. Oops... The second line contains a typo, yes. And oops again... The math book isn't wrong, but instead I'm wrong. It should be $$x=\frac{1}{650}$$, not $$x=\frac{1}{625}$$! Just a mistake, although a huge one! So the book's answer really does make sense... But typos aside, you're absolutely right. Now I get it. Thanks! 85.229.101.18 (talk) 15:41, 27 June 2009 (UTC)
 * Uhmm...actually, the book really does state the $$x=\frac{1}{625}$$. Which does, in fact, solve the equation. I'm a little tired right now, hence all the mistakes. But the answer really solves the equation (I tried it with my calculator.) 85.229.101.18 (talk) 15:47, 27 June 2009 (UTC)
 * Probably the original question should have had the sum equal to 650 and then x=1/625 would be right. Dmcq (talk) 16:35, 27 June 2009 (UTC)
 * Dmcq, you beat me right to it. --COVIZAPIBETEFOKY (talk) 16:39, 27 June 2009 (UTC)
 * Oh! I'm really, really sorry for that! Well, I got the solution method I was looking for, so it didn't really matter. But a little embarassing that I made so many mistakes! Well, as I said, I'm blaming my tiredness... //Misopogon 85.229.101.18 (talk) 16:45, 27 June 2009 (UTC)
 * Don't worry about it; it happens to the best of us. --COVIZAPIBETEFOKY (talk) 01:02, 28 June 2009 (UTC)


 * The most straight forward substitution would probably be to set $$t = \frac{1}{\sqrt{x}}$$. This immediately gives you a quadratic in a nice form:
 * t2 + t - 650 = 0
 * which you can plug into the quadratic formula or just notice from inspection that it factors into
 * (t + 26)(t - 25) = 0
 * Rckrone (talk) 07:03, 3 July 2009 (UTC)

Resolving ties in Ranked Pairs
I'm in the middle of building a web application that shows the outcomes of voter preferences under different electoral systems. I understand what Ranked Pairs does in the general case, but there are some edge cases that need to be considered


 * Voter preferences


 * Sorted tally


 * Locking pairs

This is where I get confused. We can lock A>B immediately. We can lock A>C and B>C at the same time (same number of votes, no cycle created). What do we do when we get to C>D and D>A? Individually they're fine, but including them both results in a cycle. Including neither isn't an option either as it leaves D dangling out on the graph with no edges. -- 08:22, 27 June 2009 User:BradBeattie


 * This is actually more of a math question. AnonMoos (talk) 15:08, 27 June 2009 (UTC)


 * Hrm... Doing this scenario with the Schulze Method results in a tie between A and D. The same result could be obtained by not adding C>D and D>A (general rule: don't add them because doing so would create a cycle). This is fine because D ends up tied with the winner from A, B and C (which is A, which is the same result that Schulze gives us). Brad Beattie (talk) 16:40, 27 June 2009 (UTC)


 * Our Ranked Pairs article links to this article which suggests a pre-determined tie-breaking ranking (chosen randomly, for example). I don't know how common that is in actuality. (Of course, with significantly more than 8 voters, this scenario becomes much less likely.) — JAO • T • C 16:51, 27 June 2009 (UTC)


 * Yeah, I was the one that added that link yesterday. The difficulty is that in a plurality voting system, we'd say that two candidates with the same number of votes are tied. Likewise, ranked pairs could also have that accomodation if neither C>D nor D>A are included. In this sense, it kind of becomes a reverse iteration of the Schulze method, yeah? Anywho, I agree that it's statistically unlikely. Brad Beattie (talk) 22:04, 27 June 2009 (UTC)

Measurable functions on [0,1]
I wonder about this. Take a measurable function f :[0,1]→R. Can we write it as a composition f(x)=g(h(x)), where g :[0,1]→R is a monotone non-decreasing function and h :[0,1]→[0,1] is a Lebesgue measure preserving map? I think the function g is the distribution function of f and is unique up to a countable set; I think there exists also such an h but it's a bit more tricky.. Is there something about it in a wiki article?--78.13.140.73 (talk) 20:25, 27 June 2009 (UTC)
 * There is such a g, what you're talking about is really just something like the symmetrically decreasing rearrangement of f (Lieb and Loss has a discussion of this). The business of symmetrically rearranging f means that the level sets of f are also rearranged, in a measure-preserving way. It's not so clear to me whether this rearrangement represents a measurable transformation over the entire sigma-algebra, however. Just my 2 cents. Hopefully somebody here will have a clearer answer. Ray  Talk 20:47, 27 June 2009 (UTC)


 * This I think should do, e.g. for $$f:[0,1]\to[0,1]$$
 * $$g(x):=\min\{y \in [0,1]\, : \Big|f^{-1}[0,y]\Big| \geq x \},$$
 * $$h(x):=\Big|f^{-1}[0,f(x)[\Big| + \Big|[0,x]\cap f^{-1}(f(x))\Big|$$
 * where $$|S|$$ denotes Lebesgue measure; but you better check it. --pma (talk) 16:55, 29 June 2009 (UTC)