Csdrhrt

If I only have a random number generator rand which generates a random number from a uniform distribution in [0,1], would it be possible to use some smart mathematical transformation to:

1. generate a value from a normal distribution with mean = 0 and variance = 1


2. generate a value from a poisson distribution with mean = 60

Thanks!

Write
$$
F(x) = P(X leq x)
$$
for the cumulative distribution function of the random variable you are trying to simulate. This is increasing so invertible. Take $Y = F^{-1}(U)$ where $U$ is your uniform random variable. Then
$$
P(Y leq y)= P(F_X^{-1}(U) leq y) = P(U leq F_X(y)) = F_X(y) = P(X leq y)
$$
So the distribution of $Y$ is the same as that of $X$.

If $U sim mathsf{Unif}(0,1)$ and if random variable $X$ has inverse CDF (quantile function) $F^{-1}_X(t),$ then a realization $u$ of $U$ produces a realization
$F^{-1}_X(u)$ of $X.$

For example, if $X sim mathsf{Exp}(1),$ then we have PDF $f_X(x) = e^{-x},$
CDF $F_X(x) = 1 - e^{-x},$ and quantile function $F_X^{-1}(t) = -log(1-t),$
for $x > 0, 0 < t < 1).$ Thus if you generate a sample of size 10,000 from
$U sim mathsf{Unif}(0, 1),$ then $X = -log(1 - U) sim mathsf{Exp}(1).$

This can be demonstrated in R statistical software as shown below. [In R,
runif generates a sample from a uniform distribution and dexp is an
exponential PDF.]

set.seed(917);  u = runif(10^4);  x = -log(1-u)

summary(u)

    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 

0.000243 0.280017 0.677848 0.981355 1.363933 7.717536 



hist(x, prob=T, col="skyblue2")

curve(dexp(x), 0, 8, add=T, col="red", lwd=2, n = 10001)

In the figure below, the histogram shows the 10,000 simulated values $X$
and the curve is the density of $mathsf{Exp}(1).$

This 'quantile method' works easily if the CDF of $X$ can be written in
closed form and then inverted to find the quantile function of $X.$
The idea can often be extended more widely by finding a rational approximation
of the CDF and inverting it.

In R, a random sample of size n from a standard unform
distribution can be generated with rnorm(n) which is (a few technical
details for optimization notwithstanding) essentially qnorm(runif(n)), where
qnorm uses Michael Wichura's rational approximation to the normal quantile
function. (The approximation is accurate to the degree that can be represented
by double precision arithmetic.)

set.seed(918);  z = qnorm(runif(10^5))

summary(z)

       Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 

 -4.283920 -0.673689  0.004902  0.005624  0.683117  4.532156 

An entirely different method, specific to normal distributions, is used by the
'Box-Muller transformation' which generates two standard normal variates from
two standard uniform ones. [One nice explanation is given in Wikipedia.]