Poisson distribution for a given mean

[Wile_E_Coyote]

Some ideas:

[Solved] Fit TH1D with Poisson Function

I think the problem is in the “x range” of your histogram. From your picture, I can see that “x_max = 10.e-6” and the Poisson distribution expects “x_max >= 1” (so, in principle you should also have “mean >= 0.5”). You could try to “scale” your “x range”: { double xmin = 0.; double xmax = 10.e-6; TF1 *f = new TF1("f", "[0] * TMath::Poisson((x/[2]), ([1]/[2]))", xmin, xmax); double constant = 160.; double mean = 1.5e-6; double xscaling = 2.e-7; f->SetParameters(constant, mean, xsc…

Fitting a Poisson Distribution to a histogram

I guess, the trick with the parameter “[2]” is to scale “x” and then also “u” (as they appear in your “definition”), so that the “mean” (and thus also the “variance”) remains simply the value of the parameter “[1]”. You could try to “scale” the “x-axis” of your histogram by the value of the parameter “[2]” (then also possibly “f1->FixParameter(2, 1.0);”) and then redo your fit: { gStyle->SetOptFit(112); Double_t xmin = 0, xmax = 10; // define the Poisson fit function TF1 *f = new TF1("…