# Simulating rnorm() using runif()

I am trying to 'simulate' rnorm() using only runif().

I don't know if I should do:

sqrt(-2*log(U1))*cos(U2)


or

sqrt(-2*log(U1))*sin(U2)


Where U1 is a runif(0,1) and U2 runif(0,6.28)

I do not know if I should do it using cos or sin, or is it that I need to sample from one and the other consecutively? What is the mathematical logic behind it?

• You will want to investigate the Probability Integral Transform. – StatsStudent Mar 14 at 15:28
• Because $\cos(U_2)$ is the horizontal coordinate of a point uniformly distributed on the unit circle and $\sin(U_2)$ is its vertical coordinate, notice that the symmetry of the circle under a switch of coordinates implies $\cos(U_2)$ and $\sin(U_2)$ are identically distributed. Thus, your two methods are equivalent (assuming, anyway, that you replace "6.28" by $2\pi$). – whuber Mar 14 at 18:44

It looks like you are trying to use the Box-Muller transform.

The method is to use $$U_1, U_2$$ which are independently distribution Uniform(0,1). Then

$$Z_0 = \sqrt{-2 \ln(U_1)}\cos(2 \pi U_2)$$

and

$$Z_1 = \sqrt{-2 \ln(U_1)} \sin(2 \pi U_2)$$

are a pair of independent $$N(0,1)$$.

The derivation comes from using polar coordinates, but to be honest the exact steps escape me at the moment (see link if you are interested).

Let $$N_1,N_2$$ be independent $$N(0,1)$$ random variates. $$(N_1,N_2)$$ defines a point in Cartesian coordinates. We transform to polar by

$$N_1 = R \cos \theta$$ $$N_2 = R \sin \theta$$

This transformation is one to one and has continuous derivatives. So we can derive the joint distribution $$f_{R,\theta}(r,\theta)$$ by using

$$f_{R,\theta}(r,\theta) = f_{N_1,N_2}(n_1,n_2)|J^{-1}|$$

where $$J$$ is the Jacobian of the transformation and

$$J^{-1}(r,\theta )= {\begin{bmatrix}{\dfrac {\partial n_1}{\partial r}}{\dfrac {\partial n_1}{\partial \theta }}\\{\dfrac {\partial n_2}{\partial r}}{\dfrac {\partial n_2}{\partial \theta }}\end{bmatrix}} = \begin{bmatrix}\cos \theta & -r\sin \theta \\\sin \theta & r\cos \theta \end{bmatrix}$$

Hence $$|J^{-1}| = r \cos^2 \theta + r \sin^2 \theta = r$$

and

$$f_{R,\theta}(r,\theta) = r \frac{1}{2\pi} e^{-\frac{1}{2}(n_1^2+n_2^2)} = \boxed{\frac{r}{2 \pi}e^{-\frac{1}{2}(r^2)}} \qquad 0 \leq \theta \leq 2 \pi, 0 \leq r \leq \infty \$$

And we see $$\theta \sim unif[0,2\pi]$$ (since $$f_{\Theta}(\theta) = \frac{1}{2\pi}$$ ) and $$R^2 \sim \exp(1/2)$$

Hence, to simulate $$\Theta$$ we simply take $$2 \pi U_2$$ where $$U_2 \sim unif[0,1]$$ and to simulate $$R$$ we can take $$-2\ln U_1$$ where $$U_1 \sim unif[0,1]$$. ( to see why find the density of $$X=-\ln U, U \sim unif[0,1]$$).

So, Box and Muller simply inverted $$N_1= R \cos \theta$$, $$N_2= R \sin \theta$$ and moved from $$(R,\Theta)$$ to $$(N_1,N_2)$$ by simulating $$\Theta$$ from $$2 \pi U_2$$, and an independent $$R$$ from $$\sqrt{- 2 \ln U_1}$$

explicitly

$$Z_0 = \sqrt{-2 \ln(U_1)}\cos(2 \pi U_2)$$

$$Z_1 = \sqrt{-2 \ln(U_1)} \sin(2 \pi U_2)$$

And that is that is the mathematical logic behind it.

p.s. As I don't know whether it is clear through the mathematical justification (I believe it is, but you may have been lost in details), note you need a pair of drawings from the uniform to get a pair of normals