Recent content by imsolost

This is my problem : it all starts with some very basic linear least-square to find a single parameter 'b' : Y = b X + e where X,Y are observations and 'e' is a mean-zero error term. I use it, i find 'b', done. But I need also to calculate uncertainties on 'b'. The so-called "model uncertainty"...

I do agree about what you say about "goodness" of the fit, and indeed as u guessed, this is a study that will be published so i need to make sure people will "adhere" to it. And i haven't find a paper where people faced the same thing. I already saw about "total" least square but i have no idea...

Hello all, I am a bit lost with a problem and my reasoning and would like to hear your thoughts about it. Problem is the following : For a set of data yi, I need to find the best value for Am given : "i" is an indice {1, 2, 3, ... up to something like 20}. "m" stands for "mass" (Am is a...

Thank you for these answers ! Yes, this is what I was saying in the last part of my post. About using more than 15 points, yes I can do that but this doesn't give me more data. It just approximates the integral better, so the error is on the 4 unknown parameters should be smaller. The problem...

I have the following expression : $$ y_{E} = \int_{0}^{\infty} 0.5 * [E_{1}(µ(E)*r) - E_{1}(\frac{µ(E)*r}{cos \alpha})] * f(r) dr $$ where : - $y_{E}$ has been measured for some E (something like 5 different $E_{i}$, to give you an idea) - µ(E) is retrieved from a table in the litterature...

I followed your suggestion and rewrote the code with more intermediate calculation subfunctions. It worked since problem seems to be gone. Thank you all guys for this help.

Thank you for your answer but i don't get what you mean. I want indeed to use the first value stored in the vector mubeton. So mubeton(1) seems right to me.

Code is the following : %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% xiter=[0.7 20 10]; mubeton=[0.100 0.80 0.70 0.60 0.50 0.25]; muair=[0.05 0.042 0.04 0.032 0.021 0.011]; zl=[5 30 70 100]; fnorm1 = @(d,a0,dmax) (a0*(1-exp(-(0.5*log(d./dmax)).^2))+exp(-(0.5*log(d./dmax)).^2)); fnorm2 = @(d,dmax,DRL)...

No, I don't know its value.

Using your notation (x,y), then your "y" is my ##y_{E,L}##. Your "x" is my couple of variables (E, L). Basically, during my experiment, I changed L (I used 3 different values of L) and E (i used 5 different values of E) for a total of 3x15 data points ##y_{E,L}##. I guess this should be enough...

The problem is the following : I have some measured data's obtained when measuring a physical process. Let's call these : yE,L where E and L are 2 physical parameters of the experiment (an energy and a length). I also know that : $$\frac{y_{E,L}}{\sum_{k=1} ^{k=100} {\epsilon_{E,L} (k * v) *...

I'll take more time into reading your 2 last posts with attention but I can already answer your first question. The measurand is not really a concentration. It's a count rate which relates to a radioactive activity per unit mass (see here for a quick summary). Basically for a given number of...

Okay so I just read Tashi's very interesting post and I think we definitely are getting close to the core of my problem and the mistakes I'm probably making here. Honestly I think I was doing both at the same time (conventional stats and bayesian ones) and that's probably not appropriate. So...

Each sample doesn't change as a function of time : ρi is a constant over time. As stated above, it's just that the longer the measurement, the more accurate it gets. The estimate µi of the real value ρi gets closer to ρi. (see my post above with the distribution function f(ρi I µi). Now, is it...

Please note that µi/t = σi2 (the square disappeared from your expression). Now, I'll try to answer to your remark. The real value of the concentration in the sample i, let's call it ρi, is unknown but considering we measured µi, we assume that ρi is somewhere around µi and µi is the best...