Friday, April 29, 2011

Example 8.36: Quadratic equation with real roots

We often simulate data in SAS or R to confirm analytical results. For example, consider the following problem from the excellent text by Rice:

Let U1, U2, and U3 be independent random variables uniform on [0, 1]. What is the probability that the roots of the quadratic U1*x^2 + U2*x + U3 are real?

Recall that for a quadratic equation A*x^2 + B*x + C to have real roots we need the discriminant (B^2-4AC) to be non-negative.

The answer given in the second and third editions of Rice is 1/9. Here's how you might get there:

Since, B^2 > 4*A*C <=> B > 2*sqrt(A*C), we need to integrate B over the range 2*sqrt(a*c) to 1, then integrate over all possible values for A and C (each from 0 to 1).

Another answer can be found by taking y = b^2 and w = 4ac and integrating over their joint distribution (they're independent, of course). That leads to an answer of approximately 0.254. Here's how to calculate this in R:

f = function(x) {
A = x[1]; B = x[2]; C = x[3];
return(as.numeric(B^2 > 4*A*C))
}
library(cubature)
adaptIntegrate(f, c(0,0,0), c(1,1,1), tol=0.0001, max=1000000)

which generates the following output:

$integral
[1] 0.2543692

$error
[1] 0.005612558

$functionEvaluations
[1] 999999

$returnCode
[1] -1

We leave the details of the calculations aside for now, but both seem equally plausible, at first glance. A quick simulation can suggest which is correct.

For those who want more details, here's a more complete summary of this problem and solution.

SAS

Neither the SAS nor the R code is especially challenging.


data test;
do trial = 1 to 10000;
u1 = uniform(0); u2 = uniform(0); u3 = uniform(0);
res = u2**2 - 4*u1*u3;
realroot = (res ge 0);
output;
end;
run;

proc print data=test (obs=10);
run;

proc means data=test;
var realroot;
run;

Leading to the result:

The MEANS Procedure

Analysis Variable : realroot

N Mean Std Dev Minimum Maximum
-----------------------------------------------------------------
10000 0.2556000 0.4362197 0 1.0000000
-----------------------------------------------------------------


R


numsim = 10000
u1 = runif(numsim); u2 = runif(numsim); u3 = runif(numsim)
res = u2^2 - 4*u1*u3
realroots = res>=0
table(realroots)/numsim

With the result

realroots
FALSE TRUE
0.747 0.253

The simulation demonstrates both that the first solution is incorrect. Here the simulation serves as a valuable check for complicated analysis.

Insights into where the 1/9 solution fails would be welcomed in the comments.

3 comments:

Rich said...

FYI - Prof. Rice has posted an erratum for this problem (http://www.stat.berkeley.edu/~rice/Book3ed/index.html) which lists the corrected answer as

5/36 + log(2)/6

instead of the 1/9 printed on p. A34 of the book. This corrected answer agrees with the analytical result you derived in your detailed writeup.

Great example of why you should always double-check your results. Thanks for sharing!

Sarah Anoke said...

I really enjoyed the use of simulations to complement the analytical solution. Definitely something I will keep in my statistical toolbox.

Anonymous said...

I know I'm late to the game, but when I originally did the problem, I came up with the answer of 1/9. This is wrong and it's easy to see if you use the trick used in Casella and Berger exercise 4.53.

A,B,C are uniform[0,1] so -log(A), -log(B), -log(C) are exponential(1). And so, B^2 >= 4*A*C is the same as -2log(B) <= -log(4) -log(B) - log(C).

Let X = -2log(B), so X is exponential(2) and let Y = -log(B) - log(C), so Y is gamma(2,1). So now we are solving X <= -log(4) + Y. This is a simple double integral (easy b/c X and Y are independent).

You get 1/9 if you incorrectly specify the limits of integration of Y to be [0,infinity). Y has to be greater than log(4) since X <= -log(4) + Y. So I am assuming Rice did the same thing I did to get 1/9.