1
$\begingroup$

How can I sample from a mixture distribution in particular a mixture of Normal distributions and Exponential distribution in R using composition method?

For instance if I want to sample from: $0.3\textrm{Exp}(1)+0.5\textrm N(0,1)+0.2\textrm N(4,1)$.

The algorithm should be following:

  1. Generate random number $N$ with distribution $\left \{ \frac{a_n}{n+1} \right \}$
  2. Generate random number $X$ with probability distribution $g_N(x)$. Using inverse transformation we get: $Y\sim U(0,1) \Rightarrow Y^{\frac{1}{N+1}}\sim g_N$.

I do not know understand how to get $N.$

$\endgroup$
4
  • 1
    $\begingroup$ @Xi'an yes it is, I updated my post $\endgroup$
    – Tegig
    Nov 25, 2022 at 16:18
  • $\begingroup$ What is $a_n$ In 1.? $\endgroup$
    – utobi
    Nov 25, 2022 at 16:38
  • $\begingroup$ The inverse transform generation is not what you wrote.(*Hint: It is called the inverse cdf transform.) $\endgroup$
    – Xi'an
    Nov 25, 2022 at 17:07
  • $\begingroup$ R code to generate samples from a mixture of Normals appears as the rmix function in my post at stats.stackexchange.com/a/428083/919. Here it is in its entirety: rmix <- function(n, mu, sigma, p) { matrix(rnorm(length(mu)*n, mu, sigma), ncol=n)[ cbind(sample.int(length(mu), n, replace=TRUE, prob=p), 1:n)] } It is readily modified to generate from a mixture of any distributions. $\endgroup$
    – whuber
    Nov 25, 2022 at 17:20

1 Answer 1

2
$\begingroup$

Let me rephrase the problem as follows:

Question In order to sample $X$ from $$0.3\,\mathcal Exp(1)+0.5\, \mathcal N(0,1)+0.2\,\mathcal N(4,1)\tag{1}$$

a. Write $\text{Prob}(X\le x)$ as $$0.3\,\text{Prob}(X_1\le x)+0.5\,\text{Prob}(X_2\le x)+0.2\,\,\text{Prob}(X_3\le x)$$ and specify the distributions of the three random variables $X_1,X_2,X_3$

b. Identify an integer-valued random variable $Z$ such that

  1. $Z\in\{1,2,3\}$ with probability one
  2. $X|Z\sim\begin{cases}X_1 &\text{if }Z=1\\X_2 &\text{if }Z=2\\X_3&\text{if }Z=3\\\end{cases}$
  3. $X$ is marginally distributed as (1).

c. Conclude with a generation of $X$ based on the joint generation of $(Z,X)$.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you I understood it completely wrong $\endgroup$
    – Tegig
    Nov 25, 2022 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.