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I'm looking for a way of integrating the following formula where ppf() is the percentile point function for the standard normal distribution, cdf() is its inverse, and A is a constant:

\begin{equation} \int_{0}^{1} cdf(ppf(x)-A)dx \end{equation}

I can do it with a monte carlo technique but I'm hoping there's a faster way! Thanks in advance if anyone can help...

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I have not. If that's the suggestion that makes the most sense, I'll try it. I need to do this in a software module -- if anyone could make a suggestion about a C package (preferably with Python wrapper) to do that, that would be a help! –  garyrob Mar 26 at 18:46
    
Ah, I see it's there in scipy, thanks! –  garyrob Mar 26 at 19:15

2 Answers 2

up vote 7 down vote accepted

Since the percentile point function is the inverse of the standard normal CDF $\Phi(\cdot)$, we can write the desired integral as $$\begin{align} \int_0^1 \Phi\left(\Phi^{-1}(x) - A\right)\, \mathrm dx &= \int_{-\infty}^\infty \Phi(y-A)\phi(y)\,\mathrm dy\\ &= \int_{-\infty}^\infty \left[\int_{-\infty}^{y-A} \phi(z)\,\mathrm dz\right]\phi(y)\,\mathrm dy\\ &= \int_{-\infty}^\infty \int_{-\infty}^{y-A} f_{Y,Z}(y,z)\,\mathrm dz \, \mathrm dy\\ &= P\{Z \leq Y-A\}\\ &= P\{Y-Z \geq A\}\\ &= 1 - \Phi\left(\frac{A}{\sqrt{2}}\right) \end{align}$$ where we have

  • used $\phi(\cdot)$ to denote the standard normal pdf

  • substituted $x = \Phi(y)$, $\mathrm dx = \phi(y)\,\mathrm dy$, $x=0 \to y = -\infty$, $x=1 \to y = \infty$ as in whuber's answer,

  • replaced $\Phi(y-A)$ by its definition as the integral of $\phi(\cdot)$

  • recognized the integrand as the joint density of two independent standard normal random variables $Y$ and $Z$

  • recognized that the double integral gives $P\{Z \leq Y-A\}$

  • recognized that $Y-Z$ is a zero-mean normal random variable with variance $2$

The final result is the same as that given in whuber's answer.

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(+1) Although the mathematics is essentially the same as in my answer, the focus on expressions that have probabilistic meaning provides a welcome clarity and simplicity. Note that only the very last equality made any assumption about the distribution apart from its continuity. –  whuber Mar 26 at 21:17
    
Awesome! Thanks so much! I had no idea it could come out to such a simple final calc. I'm going to mark @Dilip's as the answer because his explanation is so good, but equal thanks to @whuber! –  garyrob Mar 26 at 23:12

Let $f$ be the standard normal PDF and $F$ the CDF. Substituting $x=F(y)$ gives

$$g(a) = \int_0^1 F(F^{-1}(x)-a)dx = \int_{-\infty}^\infty F(y-a)f(y)dy.$$

The derivative of this expression with respect to $a$ can be found by differentiating under the integral sign, whence

$$\frac{dg(a)}{da} = -\int_{-\infty}^\infty f(y-a)f(y)dy= -\int_{-\infty}^\infty f(a-y)f(y)dy$$

because $f$ is an even function. The integral is the formula for the PDF of the sum of two standard Normal variables, which will therefore have mean $0$ and standard deviation $\sqrt{2}$.

Integrating with respect to $a$ to reverse the differentiation demonstrates that $g(a)$ is the negative of the CDF of a Normal$(0, \sqrt{2})$ variable, up to an additive constant of integration. Since the limiting value of $g$ as $a\to\infty$ is obviously $0$ and the limiting value of the CDF is $1$, the constant of integration must equal $1$. Therefore

$$g(a) = 1 - F\left(\frac{a}{\sqrt{2}}\right).$$

Consequently, any method to compute a Normal CDF will do the job.


There is a simple graphical interpretation of this result based on the probability integral transform. Recall that the CDF $F_X$ of an absolutely continuous distribution re-expresses the variable $X$ as a value $F_X(X)$ which has a uniform distribution. Moreover, $F$ is invertible with inverse $F^{-1}$. Assuming $\xi$ and $\eta$ are independently distributed with distribution $F$, consider the event

$$A = \{(\xi, \eta)\ |\ \xi-\eta\gt a\}.$$

This is depicted by the region beneath the surface in the left-hand plot and by the colored region in the middle plot, both shown on $(\xi,\eta)$ axes:

Figures

The value $a=-3/2$ is illustrated.

When $\xi$ is re-expressed as $x = F(\xi)$ and $\eta$ as $y=F(\eta)$, $A$ can be written as

$$A = \{(x, y)\ |\ F^{-1}(x) - F^{-1}(y)\gt a\}.$$

This is the shaded region on the right hand plot, which is shown in the re-expressed $(x,y)$ coordinates. To show more clearly the re-expression, I have labeled the axes in this plot with the corresponding $(\xi, \eta)$ values, so that the grid lines in the middle plot match the grid lines in this plot. The simultaneous re-expression of $\xi$ and $\eta$ has turned the boundary of $A$, which was a line $\xi-\eta=a$ on the left, into a curvilinear boundary.

The key point is that the variable density shown in the left and middle plots (the joint density of $\xi$ and $\eta$) becomes uniform in the right plot. This reduces questions of finding probabilities--which involve integrating the joint PDF over $A$--to those of finding areas.

Solving for $y$ shows us that $A$ is the region under the graph

$$y = F(F^{-1}(x) - a)$$

which extends only from $x=0$ through $x=1$. That is, when both $\xi$ and $\eta$ have densities given by $F$,

$${\Pr}_X(\xi-\eta\gt a)={\Pr}_X(A) = \int_0^1 F(F^{-1}(x) - a)dx = g(a).$$

This perfectly general result, when applied to a standard normal distribution, easily produces the earlier answer.

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