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Re: Rookie questions about solving for small numbers
It's better to compute Erfc[a]-Erfc[b] as Erf[a,b], especially when the answer is small. When w is quadratic in x then Pr(w > 1) will always be either Pr(x < x1) + Pr(x > x2) or Pr(x1 < x < x2), but will Pr(w > 1) always have one of those forms when you include higher-order terms? In other words, does w = 1 always have either two or no real roots? ----- Samuel Mark Young <sy81 at sussex.ac.uk> wrote: > Thank you for the advice. What I need is something that will do that automatically (without, for example, plotting a log-log graph to find approximate solutions) - so that I can quickly run this for a large range of values of f. I also need it to work for higher order equations (i.e. past 4th order, the equation cannot be solved symbolically). > > Essentially, I'm solving for a taylor series type expansion, where a variable (call it w) is expanded in terms of a gaussian variable x. i.e. > w=x+f*(x^2-s^2)+g*x^3+h*(x^4 - 3*s^4)+... (Equation 1). > > And then solving Pr(w>1)=e (Equation 2) for s. The range of values of x which give w>1 depends on the coefficients f, g, h, etc, as well as s (the standard deviation of the Gaussian variable x). As seen for the quadratic case, x + f*(x^2 - s^2), if f is negative we have: > Pr(w>1)=Pr(x>x1)-Pr(x>x2) > I've written the RHS that way for 2 reasons: it was easier to write the code that way, and this gives a sum of 2 ERFCs to solve for. > > The code I included in the original email works for the majority of cases - but the problems I run into are the same as those which occur for just the quadratic case: > > 1) If the highest order term is even, then typically for negative values of the coefficients there are no solutions to Equation 1 unless s becomes large. For example, there may be no solutions unless s>0.7, so to solve I use a temporary value for s, say stemp=0.8, and I use NSolve to find the solutions for x. I then use find root to find a value for s by solving Equation 2 - which can come out as, for example, 0.1. However, s=0.1 would have meant there were no solutions for w=1. At the moment, my best method is to use a trial and error approach until stemp matches s to within 3sf (an approximation to 3sf is fine). > > This doesn't take too long to do once, around 10s or so on my laptop, but I want the code to run for multiple values of f, g, h etc, so even running for only 10 values of each coefficient quickly means I have to run this thousands of times. Ideally, I'd like Mathematica to have a streamlined way of doing the trial and error, or a way for FindRoot to solve the system of equations simultaneously (so that I don't need to use a temporary value for s to solve equation 1). > > 2) Underflow occurs a lot - you seem to have got around that using WorkingPrecision? Where do I need to place that in the code? > > > Many thanks, > Sam > ________________________________________ > From: Ray Koopman [koopman at sfu.ca] > Sent: 08 June 2013 08:29 > To: Samuel Young > Cc: mathgroup at smc.vnet.net > Subject: RE: Re: Rookie questions about solving for small numbers > > Writing s instead of sigma, and letting e denote the probability > that is variously specified as 10^-5 or 10^-20, we have > > Pr[x + f*(x^2 - s^2) > 1] = e, > > where x is Normal with mean 0 and standard deviation s. Rewrite that as > > Pr[z + f*s*(z^2 - 1) > 1/s] = e, where z is standard normal. > > If f = 0 then we have Pr[z > 1/s] = e. > > Plot log e as a function of log s to identify approximate solutions, > then use FindRoot. > > f = 0, e = 10^-5 gives s -> .234473 > f = 0, e = 10^-20 gives s -> .107964 > > If f > 0 then z + f*s*(z^2 - 1) = 1/s has two real roots, say z1 < z2. > The coefficient of z is > 0, so we want Pr[z < z1] + Pr[z > z2]. > Again do a log-log plot, then use FindRoot. > > f = .5, e = 10^-5 gives s -> .173683 > f = .5, e = 10^-20 gives s -> .0792307 > > If -1/4 < f < 0 then the roots are real for all s > 0. > The coefficient of z^2 is < 0, so we want Pr[z1 < z < z2]. > > f = -.125, e = 10^-5 gives s -> .271650 > f = -.125, e = 10^-20 gives s -> .126184 > > If f < -1/4 then the roots are real only for s > sqrt[-1-4f]/(-2f). > f = -1/2 requires s > 1. > > f = -1/2, e = 10^-5, WorkingPrecision->30 gives s -> > 1.000000000213493355543893763569933088415336334927`30. > > f = -1/2, e = 10^-20, WorkingPrecision->60 gives s -> > 1.0000000000000000000000000000000000000002134933555514410720359161572336783883`60. > > In general, s -> smin as e -> 0 with f < -1/4. > How much precision do you need? Would an approximation suffice? > > ----- Samuel Mark Young <sy81 at sussex.ac.uk> wrote: >> Yes that's correct. The code I've included does this fine for positive values of f, but can't handle negative values very well. I also run into problems when I want to solve for Pr[x + f*(x^2 - sigma^2) > 1] = 10^-20 instead. >> >> Sam >> ________________________________________ >> From: Ray Koopman [koopman at sfu.ca] >> Sent: 07 June 2013 04:19 >> To: mathgroup at smc.vnet.net >> Subject: Re: Rookie questions about solving for small numbers >> >> It's still not clear exactly what you want. Is this it? >> >> X is a random variable whose distribution is normal with mean 0 >> and standard deviation sigma. Given a constant f, solve for sigma >> such that Pr[x + f*(x^2 - sigma^2) > 1] = 10^-5.