In R language version 4.4.1 (the version should not matter, just for the sake of discussion), we write the code :
set.seed(1234)
x <- 5
y <- rnorm(1, mean = x, sd = 0.1)
We will be able to print the number y. Suppose that there is a second person, he knows the number y, that we are using seed 1234 and that he knows the number is generated using this code. The only thing he does not know is the number x. Can he work out that x=5?
0
It looks like the source code for rnorm() roughly defines it as:
rnorm(1, mean, sd) = mean + sd * norm_rand()
or using your variable names y = x + 0.1*norm_rand()
So to reverse it, I’d think the following would work:
set.seed(1234)
x <- y - 0.1*norm_rand()
If norm_rand() isn’t an exposed function, I think you could replace it with rnorm(1,0,1)
I’m not an ‘R’ coder, and haven’t tried this myself, but it looks straightforward.
2
If we knew x was an integer, we could try to estimate how much the normal distribution is shifted by subtracting a normal distribution with mean 0 and same sd from y to see how much it is shifted.
> set.seed(1234)
> x <- 5
> y <- rnorm(1, mean=x, sd=0.1)
> f <- (y, sd) {y - rnorm(1L, 0, sd)}
> est <- replicate(10, f(y, 0.1))
> summary(est)
Min. 1st Qu. Median Mean 3rd Qu. Max.
4.771 4.840 4.930 4.910 4.937 5.114
This would strongly suggest x == 5.
The problem is that this only works for x integer or when SD is very low and would fail dramatically if it was higher:
> set.seed(1234)
> x <- 5
> y <- rnorm(1, mean=x, sd=2)
> f <- (y, sd) {y - rnorm(1L, 0, 2)}
> est <- replicate(1e3, f(y, 2))
> summary(est)
Min. 1st Qu. Median Mean 3rd Qu. Max.
-3.806 1.354 2.665 2.639 3.932 9.378
You probably had a low SD in mind when you did your thought experiment?