Wikipedia:Reference desk/Archives/Mathematics/2017 March 4

= March 4 =

Find the flaw
This appears to prove it's impossible to draw a perfect isosceles right triangle:

The ratios of the sides are 1-1-sqrt(2). Sqrt(2) is irrational. However, in the real world, all lengths are an integer multiple of the Planck length, proving that the sides must all be rational. Any flaw?? Georgia guy (talk) 00:19, 4 March 2017 (UTC)


 * False precision. In the real world, no measurement is infinitely reducible.  You can draw any arbitrarily perfect isosceles triangle given the precision of the measuring device you use to verify it.  That is, no real measuring device can tell the difference between a perfect and imperfect isosceles right triangle beyond a certain level of precision, which is good enough for the real world. -- Jayron 32 00:40, 4 March 2017 (UTC)


 * That's not what false precision is, as the article states. My bathroom scale that lists my weight to the tenth of a pound, but is only accurate to a pound, is an example of that. Then there's the joke: "He mister, how old is that dinosaur skeleton ?", "Well, it was 200 million years old when I started working at the museum, so now it would be 200 million and 7 years old".  StuRat (talk) 00:46, 4 March 2017 (UTC)


 * Every time you speak, you verify your intelligence. -- Jayron 32 02:09, 4 March 2017 (UTC)


 * The same is true of you. As the first sentence in your link states: "False precision ... occurs when numerical data are presented in a manner that implies better precision than is justified; since precision is a limit to accuracy, this often leads to overconfidence in the accuracy as well."  This applies to my examples, but not the the OP, as no numeric data has been presented, and there is no overconfidence in accuracy resulting from this.  StuRat (talk) 04:48, 4 March 2017 (UTC)


 * If you've got a ruler with markings as small as a Planck length, I'd love to see it. -- Jayron 32 05:03, 4 March 2017 (UTC)


 * You still don't get it, so I will explain further. Had somebody reported the length of the edges of a drawn triangle in units beyond what they could measure, then that would be an example of false precision.  But that has absolutely nothing to do with this question, as nobody made such a claim. StuRat (talk) 05:09, 4 March 2017 (UTC)


 * So, you mean like trying to draw a triangle and reporting it's size in units of Planck length, right? Like the OP asked?-- Jayron 32 13:22, 4 March 2017 (UTC)


 * No, they didn't ask about that at all. That would be false precision, as Planck length is LESS than the actual accuracy of a drawing, but they were complaining that Planck length is MORE than zero. StuRat (talk) 14:43, 4 March 2017 (UTC)


 * What is this level of precision; specifically its base 10 logarithm?? Has anyone proposed technology that will strengthen it?? Georgia guy (talk) 00:44, 4 March 2017 (UTC)


 * No flaw, it is impossible to draw one exactly. A bit silly to worry about it, though.  StuRat (talk) 00:46, 4 March 2017 (UTC)


 * Why do you believe "in the real world, all lengths are an integer multiple of the Planck length"? (Not that freeing yourself from that common misunderstanding will allow for construction of infinitely precise physical objects.) -- ToE 05:40, 4 March 2017 (UTC)
 * That is more of a 'fat finger' type limit and applies to the sides of the triangle as well as the hypotenuse. One can't meaningfully point more accurately than that, and one can't add up some zillions of fat finger widths to get some precise length! Dmcq (talk) 12:06, 4 March 2017 (UTC)

It's also a problem in pure mathematics. Count Iblis (talk) 19:46, 4 March 2017 (UTC)


 * No it's not. Wildberger is a known crank.  Deacon Vorbis (talk) 18:41, 5 March 2017 (UTC)


 * and Gregory Chaitin and Emile Borel, so he is in good company. Count Iblis (talk) 01:20, 6 March 2017 (UTC)


 * Nothing in the linked article supports even vaguely a statement like "[something to do with the square root of two] presents a problem in pure mathematics." And it is deeply implausible that Borel would have agreed with any such statement. --JBL (talk) 01:34, 6 March 2017 (UTC)
 * In fact, Chaitin rejects the sort of limited thinking that you appear to be advocating quite explicitly: "In spite of the fact that most individual real numbers will forever escape us, the notion of an arbitrary real has beautiful mathematical properties and is a concept that helps us to organize and understand the real world. Individual concepts in a theory do not need to have concrete meaning on their own; it is enough if the theory as a whole can be compared with the results of experiments." --JBL (talk) 01:57, 6 March 2017 (UTC)

theory of decomposition spaces
Can somebody tell me what this is and if we have an article or field about it on hand? It comes from this article: Georg Aumann. Thanks. scope_creep (talk) 12:56, 4 March 2017 (UTC)
 * Try Manifold decomposition. --RDBury (talk) 21:36, 5 March 2017 (UTC)

Psi Weights and ARIMA models
In this class lesson page, the second example asks, "Suppose that an AR(1) model is x_t=40+.6(x_t-1)+w_t." How does one find this formula from an AR model? If I have an ARIMA(1,0,1), how do I create a formula like the one on the page, in the form "x_t=..." Furthermore, how does one arrive at w_t and the variance of w_t?

My goal is construct the prediction intervals, but I am new to ARIMA models, so I might have missed some of the fundamentals in my learning. I appreciate any guidance in this regard. Schyler ( exquirere bonum ipsum ) 22:05, 4 March 2017 (UTC)


 * The general form of the equation might be just assumed. The specific numerical values of the parameters are estimated from a data set. We have pretty good articles: Autoregressive model and Autoregressive integrated moving average. If reading them leaves you with more specific questions, I'll try to answer them. Loraof (talk) 00:21, 5 March 2017 (UTC)


 * Also Order of integration. Loraof (talk) 00:44, 5 March 2017 (UTC)

Okay, I think it's identifying the lag operator is the fundamental I'm missing. How do I find the lag operator of a time series? Schyler ( exquirere bonum ipsum ) 01:09, 5 March 2017 (UTC)


 * I assume you mean how do you find the maximum lag (the lag operator L or sometimes B is just a function that moves your focus back one period: so Lxt = xt–1, and L2xt = xt–2). For the choice of a maximum lag, see Autoregressive model and the wikilink therein. Loraof (talk) 01:19, 5 March 2017 (UTC)


 * Okay, I don't think that's it, then. I'm asking when I don't know what I don't know. So,



\left( 1 - \sum_{i=1}^p \phi_i L^i\right) (1-L)^d X_t = \delta + \left( 1 + \sum_{i=1}^q \theta_i L^i \right) \varepsilon_t. \, $$

defines ARIMA (p, d, q). My ARIMA is of order (1,0,1), so that means my model is formulated by



\left( 1 - \sum_{i=1}^1 \phi_i L^i\right) (1-L)^0 X_t = \delta + \left( 1 + \sum_{i=1}^1 \theta_i L^i \right) \varepsilon_t. \, $$

Correct? But then how do I identify phi, L, delta, and theta? I appreciate your consideration of this problem I am having. Schyler ( exquirere bonum ipsum ) 01:50, 5 March 2017 (UTC)


 * The symbol L is not something to be identified; it is the name of a function. So $$L^i\varepsilon_t$$ means $$\varepsilon_{t-i}.$$ Thus your last equation can be written equivalently (noting that $$(1-L)^0 =1$$) as


 * $$X_t-\phi_1X_{t-1}=\delta+\varepsilon_t+\theta_1\varepsilon_{t-1}.$$


 * You need to estimate $$\phi_1, \,\,\, \delta, \,\,$$ and $$\theta_1.$$ For the AR parameter $$\phi_1$$ by itself you would go by the lengthy section Autoregressive model. For the MA parameter $$\theta_1$$ we have the very short section Moving-average model. For your case of a combined AR and MA model (called an ARMA model) all we have is Autoregressive integrated moving average. Maybe the manual for one of the software packages would tell you how it is estimated, or maybe you would be satisfied to just tell the package to give you the results. All I can remember about it is that ordinary least squares can be used for AR models but not for MA models, which typically use maximum likelihood. Loraof (talk) 03:33, 5 March 2017 (UTC)


 * Also Autoregressive moving average model. Loraof (talk) 04:23, 5 March 2017 (UTC)


 * Here, I'll take a leap of faith. Using MLE, I can identify the coefficients of an ARMA (1,1) model. If the coefficients of my model at ar=0.35 and ma=0.7, then I can graph said model of a time series with the equation $$X_t = 0.35X_{t-1}+0.7\varepsilon_{t-1}$$. Yes? Also, psi weights are given by $$\psi_j=\phi^j$$. Therefore, the first two psi weights of my equation are $$\psi_1=0.35, \psi_2=0.1225$$. Is my faith rewarded? Finally, I'm still unclear as to how to get the $$\hat{\sigma}_{w}^2$$. I really appreciate the detailed guidance here. It's a difficult topic for me. Schyler  ( exquirere bonum ipsum ) 19:23, 5 March 2017 (UTC)


 * With an ARMA(1, 1) model and with AR coefficient 0.35 and MA coefficient 0.7, the equation would be $$X_t = 0.35X_{t-1}+0.7\varepsilon_{t-1}+\varepsilon_t$$ (don't forget the current error term). (Note that $$\phi$$ and $$\psi$$ are two different alternative notations for the same thing, so I don't know what you mean by $$\psi_j=\phi^j.$$) Here your AR order (longest lag) is 1, so there is no such thing as $$\psi_2.$$ I don't know how an estimate of the variance of the error term is obtained. Loraof (talk) 20:43, 5 March 2017 (UTC)


 * They say repeatedly that $$E(\sigma^2)=0$$. I understand why that's true, but I'm not sure if that might be helpful. Furthermore, there is a given, that the standard error of $$(x_{n+2}^n-x_{n+2})=\sqrt{\hat{\sigma}_w^2(1+\psi_1^2)}$$.


 * This thread is approaching a singularity. I am greatful for your help in this. By the way, this is for a paper I am preparing on an educational intervention. Schyler  ( exquirere bonum ipsum ) 01:41, 6 March 2017 (UTC)


 * Two quick points: (1) Note that their notation $$x_{n+m}^n$$ is defined in their notation section at the start as the m-period ahead forecast from time n (and not as x raised to a power, which is what it looks like). (2) They don't say $$E(\sigma^2)=0, $$ which would be impossible since the variance is positive. Instead, they say that $$E(w)=0.$$ Loraof (talk) 02:08, 6 March 2017 (UTC)