Re: Rookie questions about solving for small numbers

*To*: mathgroup at smc.vnet.net*Subject*: [mg131071] Re: Rookie questions about solving for small numbers*From*: Ray Koopman <koopman at sfu.ca>*Date*: Sun, 9 Jun 2013 04:31:09 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net

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.