1
$\begingroup$

I'm trying to model the distribution of effects of mutations (let's call it s) in evolution but I'm stuck in generating the probability distribution function (pdf) for my model. So, my model is a mixture of 3 components. As far as I understood, if I assume independence between all components, my final pdf is a weighted sum of the pdf of each component.

My problem starts in defining the pdf of each component. I have 2 major problems. 1: In two of my components I have a function F() that relates some biochemical value (let's call it E) with the effect I want to model. I assume that E is distributed as a Gaussian (or a mixture of gaussians). My solution was to find the function G(s) = E (receive a value of s and convert in the value of E) and substitute that functions in the Gaussian pdf. Is this correct?

2: For this I have no solution yet. So I assume that s depend on 2 different variables (E and K). In my previous point I got the pdf for the cases in which only one is changing (assuming the changes are Gaussian). Know I want to build a component where I allow the 2 variables to change. So in the end I know the distribution of E and K (I assume they are independent) and the function H(E, K) = s. I can also get the function G(s, K), = E and J(s, E) = K. How can I get the pdf for this component? I found somewhere that I can use the method of the cdf (integrate the joint probability distribution for each variable), but the Gaussian pdf is not integrable!!

$\endgroup$
  • $\begingroup$ This is so vague that I am not sure what you are really asking, but it seems to focus on how to compute PDFs of transformed random variables. For information on that, please search our site. At the end, exactly what do you mean by "...the Gaussian pdf is not integrable"? By definition it is an integrable function, so this is a puzzling statement. $\endgroup$ – whuber Nov 19 '14 at 19:31
  • $\begingroup$ Sorry for the confusion. As far as I understood the result of integrating the Gaussian pdf still as an integral. After posting the question I found the transformation of random variables post that answer my first point. However, in the case I have two variables (second case) how to solve it? I didn't understood this answer $\endgroup$ – Diogo Santos Nov 19 '14 at 19:42
  • $\begingroup$ Googling keywords including "change random variable Jacobian" will give you links to the standard explanation. For another (equivalent) approach using differential forms see stats.stackexchange.com/a/101298. $\endgroup$ – whuber Nov 19 '14 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.