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In many applications, for example Monte Carlo methods, we require the inverse CDF of a standard normal random variable. But the CDF:

$$ \Phi(x)= \int_{-\infty}^ x \frac{1}{\sqrt{2\pi}} e^{-t^2/2} dt $$

cannot be obtained in closed form. Is there a way to define its inverse, say $\Phi^{-1}(y)$, then? Any help is greatly appreciated. Thank you.

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Since $\Phi(x)$ is strictly increasing, its inverse is well defined.

In general for a regular function $f$, in order to find x such that $f(x) = y$, there are plenty of numerical algorithms, such as Newton-Raphson method.

And for this particular case, we have approximative formula with good precision, such as this one: http://www.johndcook.com/normal_cdf_inverse.html

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  • $\begingroup$ I'll give it a look, thanks. $\endgroup$ – JohnK Dec 1 '13 at 14:48

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