(via http://www.afc-cca.com/ page 158)
(via http://www.afc-cca.com/ page 158)
2012 will be, in several countries, a presidential election year. Some decades ago, candidates were not always supposed to spend months on some
demagogicdemocratic debates, but had time to spend on more important problems. Like mathematics… For instance, a congressman, who became the 20th president of the United States, James Garfield, gave the following proof to the Pythagorean theorem (actually, he wrote that proof five years before he become President). The legend claims that he found this proof in 1876 during a mathematics discussion with some of the members of Congress… Those were good old days, when politicians were interested in mathematics, and sciences. The proof he suggested was the following
i.e. , since
I found that nice graph in Roger Nelsen’s book. For more details, Klebe (1995), or on wikipedia. And for those who love proofs without words, look at the 96 geometrical proofs of the Pythagorean theorem mentioned on http://www.cut-the-knot.org/.
This Friday, the Second Québec-Ontario Workshop on Insurance Mathematics (WIM) will take place in Toronto, at the Fields Institute. Mathieu will give a talk on “Multivariate integer-valued autoregressive models applied to earthquake occurrences“. The paper can still be downloaded on http://arxiv.org/ and the slides can be downloaded here.
In econometrics course we always say to our students that “if you fit a linear model with no constant, then you might have trouble. For instance, you might have a negative R-squared”. So I tried to find databases on the internet such that, when we compute a linear regression, we actually obtain a negative R squared. I have generated hundreds to random databases that should exhibit such a property, in R. With no success. Perhaps to be more specific, I should explain what might happen if we do not include a constant in a linear model. Consider the following dataset, where points are on a straight line, with a negative slope, far from the origin, symmetric with respect to the first diagonal.
> x=1:3 > y=3:1 > plot(x,y)
Points are on a straight line, so it is actually possible to get a perfect linear model. But only if we integrate a constant in our model. This is related to the fact that the correlation between our two variates is -1,
> cor(x,y)  -1
The least-square program is here
i.e. the estimate of the slope is
Numerically, we obtain
> sum(x*y)/sum(x^2)  0.7142857
which is the actual slope on the illustration above. If we compute the sum of squares of errors (as a function of the slope), we have here
so the value we have computed is actually the minimum of the sum of squares of errors. But note that the sum of squares always exceeds the total sum of squares in red on the graph above
Recall that the total “coefficient of variation“, denoted , is defined as
> 1-ssr(b)/sum((y-mean(y))^2)  -2.428571
which is negative. It is also sometimes defined as “the square of the sample correlation coefficient between the outcomes and their predicted values“. Here it would be related to
> cor(b*x,y)  -1
so we would have a unit . So obviously, using the in a model without a constant would give odd results. But the weird part is that if we run that regression with R, we get
> summary(lm(y~0+x)) Call: lm(formula = y ~ 0 + x) Residuals: 1 2 3 2.2857 0.5714 -1.1429 Coefficients: Estimate Std. Error t value Pr(>|t|) x 0.7143 0.4949 1.443 0.286 Residual standard error: 1.852 on 2 degrees of freedom Multiple R-squared: 0.5102, Adjusted R-squared: 0.2653 F-statistic: 2.083 on 1 and 2 DF, p-value: 0.2857
Here, the estimation is correct. But the we obtain tells us that the model is not that bad…
So if anyone knows what R computes, I’d be glad to know. The value given by R (thanks Vincent for asking me to look carefully at the R source code) is obtained using Pythagoras’s theorem to compute the total sum of square,
So be careful, the might look good, but meaningless !