# Help solving the following integral with a CDF

Here is my problem

Let $H(x)=\Phi(\frac{Ax+b}{\sqrt{2}})$ where $A$, $b$ are some non-negative constant and $\Phi$ is the CDf of a standard normal distribution.

I want to find

$$\int x \, dH(x)$$

I thought maybe integration by parts which yields:

$$x H(x)-\int H(x) \, dx$$

And now I'm not sure how to proceed or if it's possible to take the integral of a CDf

-

Let us do some probability here. Let $Z$ denote a standard normal random variable, with CDF $\Phi$, and $X$ a random variable with CDF $H$. Then, for every $x$, $H(x)=\Phi((Ax+b)/\sqrt2)$ implies that $\mathrm P(X\leqslant x)=\mathrm P(Z\leqslant(Ax+b)/\sqrt2)=\mathrm P((\sqrt2Z-b)/A\leqslant x)$. That is, $X$ is distributed as $(\sqrt2Z-b)/A$. In particular, $Z$ is centered hence $$\int\limits x\mathrm dH(x)=\mathrm E(X)=\mathrm E((\sqrt2Z-b)/A)=(\sqrt2\mathrm E(Z)-b)/A=-b/A.$$

-
Why did you change your name? – Pedro Tamaroff Apr 10 '12 at 23:13
@PeterT.off Why do you ask? – Did Apr 10 '12 at 23:17
Just Curiosity. I added T.off when I saw many users named Peter, but you seem to have done the opposite. – Pedro Tamaroff Apr 10 '12 at 23:22
@PeterT.off Right, but not me‌​. :-) – Did Apr 10 '12 at 23:38
@Dider I'm considering using another name. Pierre seems a little less abundant. – Pedro Tamaroff Apr 11 '12 at 0:45

Since $H$ is an everywhere differentiable function, we have $$\int x \, dH(x) = \int x H'(x)\,dx = \int x \Phi'\left(\frac{Ax+b}{\sqrt{2}}\right) \frac{A}{\sqrt{2}}\, dx.$$ Let $u=\dfrac{Ax+b}{\sqrt{2}}$, so that $du = \dfrac{A\,dx}{\sqrt{2}}$ and $x= \dfrac{\sqrt{2}\; u - b}{A}$. Then the integral becomes $$\int \frac{\sqrt{2}\;u-b}{A} \Phi'(u) \, du.$$ Then make this into $$\frac{\sqrt{2}}{A} \int u \Phi'(u)\,du - \frac{b}{A} \int \Phi'(u)\,du.$$ The second integral above becomes $\Phi(u)+\text{constant}$. The first is $$\int u \Phi'(u)\,du = \int u e^{-u^2/2} \, du = \int e^{-w}\,dw = -e^{-w}+\text{constant}.$$ Then convert back to an expression in $u$, then back to an expression in $x$.

If you had in mind a definite integral from $-\infty$ to $\infty$, then $$\int_{-\infty}^\infty \Phi'(u)\,du = 1$$ and $$\int_{-\infty}^\infty u \Phi'(u) \, du = 0.$$ The second integral is $0$ because you're integrating an odd function over an interval that is symmetric about $0$. The first is $1$ since you're integrating a probability density function.

-