# Plot explicit cdf instead of ecdf in R

I have adjusted the parameters (lambda, mu, sigma) for a mixture of two normals fitted to my data. Now I would like to plot the cdf of this model using the explicit function instead of the ecdf. Is there any way to do this or I do I have to simulate data so then I can use again ecdf?

The explicit function is something like:

ipc_values_EM\$lambda[1] * dnorm(x, ipc_values_EM\$mu[1], ipc_values_EM\$sigma[1]) + ipc_values_EM\$lambda[2] * dnorm(x, ipc_values_EM\$mu[2], ipc_values_EM\$sigma[2])


(as you can note, is the mixture of two normals different mus and different sigmas)

-

Like the title of the function ecdf() says, it is empirical and only runs on samples.

If you want the exact cdf of a Gaussian, the function you are looking for is pnorm(). Here is a demonstration.

x <- seq(from=-5, to=5, by=.1)
y <- pnorm(x)
plot(x, y, type='l')


If you replace dnorm() by pnorm() in your code, and x by the range of values you want to take the cdf over you should get the result you are looking for.

-
I tried a little different approach but with the same spirit in mind, just used the curve() function and the code is as follows: curve(ipc_values_EM\$lambda[1] * pnorm(x, ipc_values_EM\$mu[1], ipc_values_EM\$sigma[1]) + ipc_values_EM\$lambda[2] * pnorm(x, ipc_values_EM\$mu[2], ipc_values_EM\$sigma[2]), from=-0.10, to=0.07, add=TRUE, col="blue") –  natorro May 21 '12 at 10:12
You might be interested in using the distr package for plotting the theoretical distribution functions for mixture distributions. Here is a quick example:
library(distr)