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

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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 $[{x^{-1}+x^{-1/2}}]^{-2}\neq {(x^{-1})^{-2}+(x^{-1/2})^{-2}}$. 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 $1+t=625t^{2}$. 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

Group W (3 voters)	A > B > C > D
Group X (3 voters)	D > A > B > C
Group Y (1 voter)	C > D > A > B
Group Z (1 voter)	B > C > D > A

Sorted tally

A (7)	>	B (1)
A (6)	>	C (2)
B (6)	>	C (2)
C (5)	>	D (3)
D (5)	>	A (3)
B (4)	==	D (4)

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)