0

For Quasi Monte Carlo, given the standard normal distribution $$ f(x) = \frac{1}{\sqrt{2\pi}}\exp{\frac{-x^{2}}{2}} $$ over the real line ($\mathbb{R}$), how can this be transformed to be a density over $[0,1]$?

Context:
I would like to use QMC quadrature to evaluate an integral which is not over $[0,1]$. Since QMC methods involving Sobol Sequences or Lattice points are only for integrals over the hypercube, I would like to either transform my integrad (which is a more complicated form of the gaussian) into something over $[0,1]$. I would also be happy with an answer that can show how to transform points over the hypercube from Sobol sequences or Lattice methods into points over $\mathbb{R}^{n}$ that can be used for quadrature with the standard gaussian.

SpicyJalapenos
  • 177

2 Answers2

1

If random variable $X$ has a continuous distribution with positive density $f$ on $\mathbb R$, the cumulative distribution function $F(X)$ has uniform density on $[0,1]$. That is, for $0 \le t \le 1$, if $F(z) = t$, $$\mathbb P(F(X) \le t) = P(X \le z) = F(z) = t$$

Robert Israel
  • 470,583
  • For a proof here: https://math.stackexchange.com/questions/868400/showing-that-y-has-a-uniform-distribution-if-y-fx-where-f-is-the-cdf-of-contin – tintinthong Aug 22 '17 at 07:38
1

Regards @Abby. I may contribute. The original is $$ f(x)=\frac{1}{\sqrt{2 \pi}}e^{-0.5x^{2}}, \: \: -\infty < x < \infty $$ What i understood from your question is that you would like to evaluate all values of $f(x)$ by the domain $(0,1)$?

If it is, notice that $\lim_{z \rightarrow -\pi/2 } tan(z) = -\infty $ and $\lim_{z \rightarrow \pi/2} tan(z) = \infty $.

My idea is : write $x = g(z)=tan( \pi z-\frac{\pi}{2}) $, with $z \in (0,1)$. Then you would get $$ h(z)=\frac{1}{\sqrt{2 \pi}}e^{-0.5(g(z))^{2}}, \: \: 0 < z < 1 $$

Thanks.

Redsbefall
  • 5,059