Generate random distribution of truncated distribution

Imagine that you have estimated a sex ratio being 0.8 SE 0.2 by maximum likelihood. If you want make an histogram showing this distribution, the first idea is to do:

> sr <- rnorm(100000, mean=0.8, sd=0.2)
> sd(sr)
[1] 0.2001526
> hist(sr)
> sum(sr>1)
[1] 15680
> sum(sr<0)
[1] 3

The sd of sr is right 0.2 but clearly, it is not correct !

Let do a truncation:
> sr <- sr[(sr>0) & (sr<1)]
> hist(sr)
> length(sr)
[1] 84317
> sd(sr)
[1] 0.1591352

The histogram is correct but the standard deviation is less than 0.2 due to the truncation.

> sfinal <- NULL
> x <- seq(from=1E-6, to=1, length.out = 100)
> for (sigma in x)
+   sfinal <- c(sfinal, abs(sd(sin(rnorm(100000, mean=2*asin(sqrt(0.8)), sd=sigma)/2)^2) - 0.2))
> plot(x, sfinal, type="l")
> sigma <- x[which.min(sfinal)]
> distribution_sr <- sin(rnorm(100000, mean=2*asin(sqrt(0.8)), sd=sigma)/2)^2
> hist(distribution_sr, xlim=c(0, 1))

Now the distribution has a mean of 0.75 and a sd of 0.2. I am not sure that we can do better.

> mean(distribution_sr)
[1] 0.7564771
> sd(distribution_sr)
[1] 0.2000063

It is important to recognize that there is an infinite number of solutions that will give the same mean and sd.

So in conclusion, it is not correct to represent graphically sex ratio as mean ± 2 SD !


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