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Get prediction intervals of parameters on GLM Binomial models?

Terry Acree

Terry Acree, Cornell University

Posted 7 years ago

Using Binomial data we need to determine predicational intervals on x at specific probabilities (Y). Enclosed is code that produces a visual representation of example fitted data but not the + and - 95% prediction values of x. The image of the plot (copy included) indicates the two points we need to determine. Attachments: GLM Plot.png GLM_code.nb

POSTED BY: Terry Acree

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Jim Baldwin

Jim Baldwin, Retired

Posted 7 years ago

I think what you want are confidence intervals for $x$ given some probability $p$ (not prediction or predictional intervals - and I don't know what predictional intervals are). Because the model you're fitting is $$\log \left(\frac{\text{prob}}{1-\text{prob}}\right)=a+b x$$ you want confidence intervals for $x$ $$x = \frac{\log \left(\frac{\text{prob}}{1-\text{prob}}\right)-a}{b}$$ You can get the lower and upper limits for a 95% confidence interval for $x$ with the following: prob = 0.5; upper= tt /. Solve[prob == 1 - 1/(1 + Exp[parms.{1, tt} - 1.96 Sqrt[{1, tt}.cov.{1, tt}]]), tt] lower= tt /. Solve[prob == 1 - 1/(1 + Exp[parms.{1, tt} + 1.96 Sqrt[{1, tt}.cov.{1, tt}]]), tt]

POSTED BY: Jim Baldwin

Terry Acree

Terry Acree, Cornell University

Posted 7 years ago

hello Bill You hit the nail on the head I am having problems with language, as my parameter-variable faux-pax made clear. I have included a copy of the data I used to generate the plot. A glance will make it clear that it is a Binomial (n=12) data set and not Bernoulli. The column on the left is the predictor variable tested 12 times for each value of x and the column on the right is the response variable y. Or am I mistaken. { {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2.69897, 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {2., 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1.69897, 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {1., 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0.69897, 1.}, {0., 0.}, {0., 0.}, {0., 0.}, {0., 0.}, {0., 0.}, {0., 0.}, {0., 0.}, {0., 1.}, {0., 1.}, {0., 1.}, {0., 1.}, {0., 1.}, {-0.30103, 0.}, {-0.30103, 0.}, {-0.30103, 0.}, {-0.30103, 0.}, {-0.30103, 1.}, {-0.30103, 1.}, {-0.30103, 1.}, {-0.30103, 1.}, {-0.30103, 1.}, {-0.30103, 1.}, {-0.30103, 1.}, {-0.30103, 1.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1., 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 0.}, {-1.30103, 1.} }

POSTED BY: Terry Acree

Jim Baldwin

Jim Baldwin, Retired

Posted 7 years ago

I'm having a hard time understanding the seemingly inconsistent and non-standard GLM language you are using. Below I'm just attempting to state what I think you have in terms of more standard statistical language (or if you want, jargon). If you use the command `GeneralizedLinearModelFit[p, x, x, ExponentialFamily -> "Binomial"]` you have a dataset named `p` with a single predictor variable `x` such that response variable `y` has a Bernoulli distribution (equivalent to a binomial distribution with a sample size of 1) with parameter $P$ (not to be confused with the name of your dataset) equal to $$P=1-1/(1+e^{c0 +c_1 x})$$ or equivalently $$\log(P/(1-P))=c_0+c_1 x$$ where $x$ is the relative concentration of odorant $A$ with $x=A/(A+B)$. If you knew $c_0$ and $c_1$, then you could determine the value of $x$ that satisfies $$P=1-1/(1+\exp{(c0+c_1 x}))$$ But you only have a sample of binomial results from a variety of relative concentrations. To obtain confidence intervals for the above value of $x$ when $P=0.5$ you would use the commands I gave earlier. (And, sorry, using $c_0$ sometimes displays as $c0$. Don't know why.) You can probably find more concrete examples of what you want from searching for "LD50 confidence intervals".