Suppose $X$ is a continuous random variable with pdf $f$ and cdf $F$. What is $\int F(a+bx)f(x) dx$ where $a, b$ are constants?

Suppose $X$ is a continuous random variable with pdf $f$ and cdf $F$. Therefore, $F'(x) = f(x)$ wherever the cdf is continuous.

What is $\int F(a+bx)f(x) dx$ where $a, b$ are constants. Suppose $X$ is normally distributed and here's what I have so far:

Let $u = a + bx$, then $du = b dx,$

$$\begin{align*} \int_{-\infty}^{\infty} F(a+bx)f(x) dx &= \int_{-\infty}^{\infty}\frac{1}{b} F(u)f\left(\frac{u-a}{b}\right) du \end{align*}$$

It doesn't look like $u$-substitution got me anywhere? If I set $u = F(a+bx)$, then $du = bf(a + bx)dx$ which also doesn't seem to simplify things a whole lot.

Solution 1:

Developing a bit the expression the unique thing that I can see here is that

$$ \int_{\mathbb{R}}F(a+bx)f(x)\,d x=\int_{\{(x,y): y\leqslant a+bx\}}f(x)f(y)d(x,y)=\Pr [Y\leqslant a+bX] $$

where $X$ and $Y$ are independent random variables with the same distribution. I dont think we can say something more.

EDIT: ok, I didn't read that $X$ is normally distributed, then in this case we can say A LOT more, as $Z:=bX-Y$ will be normally distributed also with mean $b\mu_X-\mu_Y$ and variance $b^2\sigma ^2_X+\sigma ^2_Y$.