How to integrate cdf(ppf(x)-A) for standard normal ppf and cdf 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...
 A: 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.
A: 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:

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.
