How to produce synthetic (fake) data knowing the correct fit parameters?

Question

For cross-checking purposes, if I want to check that a NonlinearModelFit performed in Mathematica is producing the correct standard errors, I would need to produce simulated data based on the fitted equation with known parameters. How would I go about doing that?

Answer 1

It would be nice if next time you put the fit equation into the question. As we discussed this yesterday, I believe your question in related to the question: Why are my fitted coefficients so well-determined?

So the idea is as follows. I use R, because I don't have Mathematica. I hope you can translate the code.

First you need to define the (population) parameters of your model. These are the values, which your fit has to find later on:

##
# Define the parameters:
a   = 11
b   = 10
c   = 9
d   = 8

Next you need to set up some more parameters, which defines your data structure. E.g.

Sigma = 1  # this is the "error estimate" you got from the fit    
nSim = 400 # number of datapoints in your sample
t0   = 0   # your starting time
tEnd = 12  # your end time

Now you generate the fake data:

##
# Generate the fake data:
set.seed(3)                        # not important
e = rnorm(nSim, mean=0, sd=Sigma)  # random error. Note that R uses sigma as the second parameter. Mathematica probably uses the variance (=Sigma^2)
t = seq(t0, tEnd, length=nSim)     # time 
y = a*cos(2*pi*t) + b*sin(2*pi*t) + c*cos(4*pi*t) + d*sin(4*pi*t) # Note that I add the random error below.

##
# Check the fake data: (with and without noise)
plot(t, y+e)
plot(t, y)

Finally, you fit the fake data and evaluate the obtained error

##
# Fit your model (without intersect):
df     = data.frame(t=t, y=y+e) # here I add the random error
lm.out = lm(y~cos(2*pi*t) + sin(2*pi*t) + cos(4*pi*t) + sin(4*pi*t) -1, df)
summary(lm.out)

In order to evaluate the error of your fitted parameters, you can loop over the above stated code and check whether or not each parameter lies within the $\pm 2 \sigma$ interval 95% of the times.