![]() You clearly will have to establish a scaling factor between intensity and fluorophore density. I interpret your last posting to mean that you are essentially trying to determine the statistics of the number of fluorophores per pixel. WaveMetrics, has beat me to it in replying, but I will add my two cents. Since they are continuous distributions, fitting will be more straightforward. Your picture looks very much like a Gamma distribution, which could be fit using a user-define fitting function that wraps our StatsGammaPDF or StatsGammaCDF functions. It seem like a Poisson curve with peak at such large numbers is more nearly Gaussian. I wonder if you need some scaling factor, or an X offset. As stated before, you will need to be careful that the histogram bin values fall on integer values, or StatsPoissonXXX will give bad values. ![]() Igor has both StatsPoissonCDF and StatsPoissonPDF that can be wrapped in a user-defined fitting function. That makes more sense, although when you publish you may need to convince a reviewer that the image doesn't have camera artifacts like systematic variations in response across the image. I am used to Matlab and there you can specify an array containing the bins - have to find the equivalent now in Igor :) or at least a way to see the scale (so far I can only change the scale, but can only guess how the automatically generated histogram-wave is scaled.Īh, I see- the pixels in your image constitute 1 million samples judged to be independent samples of the same underlying value, so they also are a good estimate of the sampling error. My fitting problem is probably related to making the histogram in a good way (choose bins so that I have integers, as you pointed out). However, the best estimator should actually be the mean value (I indicated it with the black line) - but it does not look like that to me -> I might have a wrong understanding of the maths. Attached are images of the histogram and the cumulative histogram. I look at the histogram of the image and want to find EXP(I) by fitting a Poisson PDF of CDF. To determine EXP(I), I want to treat the 1024x1024 pixels as 1mio independent measurements of EXP(I). Thus, every pixel should have the same intensity, more precisely they should be Poisson-distributed around the same expectation value EXP(I). To clarify the problem: I take a confocal image of a fluorophore solution. If I have mis-interpreted your question, please provide more details. I used a limited number of "measurement" samples to demonstrate that the fit is not exactly along the analytic result, and that the histogram can deviate a lot from the ideal case. I attach a figure showing an analytic Poisson PDF with mean value 6, the histogram of 200 data variates simulated using that mean, and the curve fit of the Histogram data to the above function. ![]() Do not use the /CUM flag on the Histogram. ![]() You have to be careful that the fitting takes place only on the integer values of the data and of the Histogram. Return exp (-w ) * (w ^x ) / factorial ( x ) ![]() CurveFitDialog/ Independent Variables 1 CurveFitDialog/ f(x) = exp(-mu)*(mu^x) / factorial(x) CurveFitDialog/ make the function less convenient to work with in the Curve Fitting dialog. CurveFitDialog/ These comments were created by the Curve Fitting dialog.
0 Comments
Leave a Reply. |