new order

The latest riddle from The Riddler was both straightforward: given four iid Normal variates, X¹,X²,X³,X⁴, what is the probability that X¹+X²<X³+X⁴ given that X¹<X³ ? The answer is ¾ and it actually does not depend on the distribution of the variates. The bonus question is however much harder: what is this probability when there are 2N iid Normal variates?

I posted the question on math.stackexchange, then on X validated, but received no hint at a possible simplification of the probability. And then erased the questions. Given the shape of the domain where the bivariate Normal density is integrated, it sounds most likely there is no closed-form expression. (None was proposed by the Riddler.) The probability decreases roughly in N³ when computing this probability by simulation and fitting a regression.

> summary(lm(log(p)~log(r)))

Residuals:
      Min        1Q    Median        3Q       Max 
-0.013283 -0.010362 -0.000606  0.007835  0.039915 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) -0.111235   0.008577  -12.97 4.11e-13 ***
log(r)      -0.311361   0.003212  -96.94  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.01226 on 27 degrees of freedom
Multiple R-squared:  0.9971,	Adjusted R-squared:  0.997 
F-statistic:  9397 on 1 and 27 DF,  p-value: < 2.2e-16

2 Responses to “new order”

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