# Rejection sampling from a normal distribution

I am running a Monte-Carlo simulation and I sample from various normal distributions. I was just wondering, is there a way by which I can increase the probability of selecting a point from the tails (i.e., $[-5\sigma,-3\sigma]$ and $[3\sigma,5\sigma]$) as compared to the probability associated with the interval $(-3\sigma, 3\sigma)$?

Edit:

• @gung Could you give me any insights? Thanks. Mar 16, 2014 at 19:29
• As it stands your question is unclear. If you arbitrarily increase the probability of selecting from the tails compared to the center, you're no longer sampling from a normal with variance $\sigma$ (and if that's okay, it's easy - sample from something with higher probability in the tails -- like a normal with large variance). What are you trying to achieve, exactly? Are you trying to do some kind of importance sampling? Mar 16, 2014 at 23:37
• @Glen_b Thanks. It isn't importance sampling really. I am essentially keen on obtaining a larger percentage of values from the tails as I want to see how these will effect the final output of my Monte-Carlo Simulation. However, the samples I am obtaining are from parameters which have their variabilities modelled through Normal Distributions Mar 17, 2014 at 9:00
• Not certain I correctly understood your meaning, but if you're simulating a normal location-mixture of normals, that would be a normal with a larger variance. Mar 17, 2014 at 9:20
• It seems like not enough detail is in the framing of your question. I would not have guessed from the way your question was asked that this was the kind of thing you were after. It feels like something relevant is missing. Mar 17, 2014 at 13:32

Well, after discussion in comments, I think it might be clear enough to hazard an answer to the question.

It seems that a symmetric-about-0 location-mixture of two normals is required, such that the probability of being in (0, 3) is about 0.15 and in (3, 5) is about 0.35 (and the same on the other side of 0).

This can be done; we'll do the positive component and the negative one will simply be the same but with $$-\mu$$ in place of $$\mu$$. The positive component should therefore have approximately 0.3 chance of being in (0, 3) about about 0.7 chance of being in (3, 5), since these probabilities will be halved when we select each half with probability 0.5.

Since almost all the probability for this positive part must lie in (0,5), and $$\mu$$ must exceed 3, $$\sigma$$ should probably be less than 1 (so that not too much probability is above 5).

Like so:

For a given $$\sigma$$ we need $$P(X<3) = 0.3$$, so $$P(\frac{(X-\mu)}{\sigma}<\frac{(3-\mu)}{\sigma})=0.3)$$, or $$\frac{(3-\mu)}{\sigma}=\Phi^{-1}(0.3)$$ where $$\Phi^{-1}$$ is the inverse cdf of the standard normal. Hence $$\mu = 3-\sigma\, \Phi^{-1}(0.3)$$.

Calculating in R:

> sig=c(0.6,0.8,1.0); data.frame( sigma = sig, mu = 3-sig*qnorm(0.3) )
sigma       mu
1   0.6 3.314640
2   0.8 3.419520
3   1.0 3.524401


Check how much probability is above 5 for the $$\sigma=1$$ case:

> pnorm(5,3.524,1,lower.tail=FALSE)
[1] 0.06997195


That's perhaps a little high, we only get (0.7-0.07)/2 = 0.315 in (3,5). Checking $$\sigma=0.8$$:

 pnorm(5,3.4195,0.8,lower.tail=FALSE)
[1] 0.02409863


That looks reasonably good, almost 0.34 in (3,5). (The value for $$\sigma=0.6$$ is 0.349.)

You can use $$\mu = 3-\sigma\, \Phi^{-1}(0.3)$$ with whatever value of $$\sigma$$ you prefer, or you can manipulate the equation so that $$\mu$$ is given and $$\sigma$$ is calculated.