How do I convert a uniform value in [0,1) to a standard normal (Gaussian) distribution value?

Question

I have uniform value in [0,1). I'd like to transform it into a standard normal distribution value, in a deterministic fashion.

What I'm confused about with the Box-Muller transform is that it takes two uniform values in [0, 1), and transform them into two normal random values.

However, I only have one uniform value. How do I apply Box-Muller over a single value?

Answer 1

The Box-Muller method is commonly used. It's simple to implement. And if you need several values, you can use it to produce normal samples two at a time. Otherwise, you could just discard one of the values and pretend you never created it.

George Marsaglia's Ziggurat method is more efficient than Box-Muller but more complicated.

Answer 2

Use the inverse transform method.

Answer 3

Given one uniform value in [0,1) you can use alternate digits to get two uniform values. Or alternate bits.

Some other methods to generate standard gaussians are here: http://en.wikipedia.org/wiki/Normal_distribution#Generating_values_from_normal_distribution

Answer 4

Although your setup is in the interval $[0,1)$, I will ignore the left endpoint and work with $(0,1)$. Recall the cumulative distribution function for the normal distribution: $$ \Phi(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^t e^{-(1/2)x^2}dx. $$ This function is an increasing map from ${\mathbb R}$ onto $(0,1)$ and its value at $t$ gives the probability that a normal random variable has value less than or equal to $t$. Let $g \colon (0,1) \rightarrow {\mathbb R}$ be its inverse, i.e., $g(y)$ is the unique solution $t$ to $\Phi(t) = y$. That is, $$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{g(y)} e^{-(1/2)x^2}dx = y. $$ Note the input to $g$ is a number in $(0,1)$ and the output is a real number. That's the kind of setup you're looking for.

Claim: If $X$ is a uniform random variable on $(0,1)$ then $g(X)$ is a normally distributed random variable on the real line.

Proof: For $a < b$ in ${\mathbb R}$, we want to show the probability $a \leq g(X) \leq b$ is $$ \frac{1}{\sqrt{2\pi}}\int_{a}^b e^{-(1/2)x^2}dx = \Phi(b) - \Phi(a). $$ Since $g$ and $\Phi$ are inverses of each other, the condition $a \leq g(X) \leq b$ is the same as $\Phi(a) \leq X \leq \Phi(b)$, which is a condition in $(0,1)$ and your hypothesis about $X$ being uniform in $(0,1)$ is that the probability of that is $\Phi(b)- \Phi(a)$. Since $$ \Phi(b) - \Phi(a) = \frac{1}{\sqrt{2\pi}}\int_{a}^b e^{-(1/2)x^2}dx, $$ we've got a Gaussian distribution. (That you want $X$ to be uniform made this a lot easier to describe.)

Quite generally, if you want to model a probability distribution on the real line with density function $f(x)$ by sampling a uniform random variable $X$ on $(0,1)$, you can use the function $g(X)$, where $g$ is the inverse of the cumulative distribution function $F(t) = \int_{-\infty}^t f(x)dx$.

Answer 5

The best way to obtain the inversion from U[0, 1] to Normal distribution is by using an algorithm presented in a famous short paper of Moro (1995). Moro presented a hybrid algorithm: he uses the Beasley & Springer algorithm for the central part of the Normal distribution and another algorithm for the tails of the distribution.

He modeled the distribution tails using truncated Chebyschev series..... .................. http://marcoagd.usuarios.rdc.puc-rio.br/quasi_mc.html