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I am trying to fit a model to binomially distributed data, which I have done via the maximum likelihood method. Normally I'm working with normally distributed data (or data which I can convince myself and occasionally others is normally distributed) and I'm a little confused about how to find asymmetric errors on my parameters. Usually I find uncertainties by scanning the space of the fitted parameters and finding the surface where the X^2 is one greater than the best fit value. I have found some unproven statements suggesting that this method is applicable regardless of whether one is using Gaussian maximum likelihood or not. Is this the case?

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What you're describing is the Method of Support, which relies on the asymptotic property of the likelihood ratio statistic L, namely that -2L is has an approximate chi-square distribution.

A more direct method is to use the second derivatives of the log-likelihood function to obtain the observed Fisher information matrix. The inverse of this matrix, evaluated at the point of maximum likelihood, is the estimated variance-covariance matrix for your model parameters.

I've had great success doing this in Mathematica, where the derivatives can be calculated on the fly without doing all the messy calculus by hand.

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  • $\begingroup$ Ah ok, it follows from the likelihood test, that makes me feel a little better. The error matrix approaches in the past that I have tried don't take into account any parameter non-linearities. By this I mean that if I double the level of uncertainty in my interval the error always doubles. Is this a necessary limitation of the method or have I just not been doing it the best way. Thanks either way, you answered my main question. $\endgroup$ – Bowler Dec 2 '11 at 12:29
  • $\begingroup$ +1 , but note that the information matrix approach cannot produce asymmetric confidence limits. For small-parameter problems (up to 3-4 parameters) Mathematica can graph actual contours of the confidence regions: you don't need to use this second-order approximation. $\endgroup$ – whuber Dec 2 '11 at 15:07

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