Le $X$ be a standard normally distributed random variable with PDF $$f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}.$$ The Q-function is defined as $$ Q(m)=\frac{1}{\sqrt{2\pi}}\int_{m}^{\infty}e^{-\frac{x^2}{2}}dx.$$ Let $Y$ a RV with PDF $f_Y(y)$ and CDF $F_Y(y)$. Let $a$ and $b$ two constant and defin $$P=aQ(\sqrt{by}).$$ We want to avrage $P$ about $Y$ so $$Z=E_y\{aQ(\sqrt{by})\}.$$ I found the simplification in 4 steps, but i did not understand step 2 and how we go from step 2 to 3 and 4$E_y\{\}$ to $E_x$. This are the steps \begin{align} (1).Z&=E_y\{aQ(\sqrt{by})\}\\ (2).Z&=aE_y\{P(X > \sqrt{by}) \}\\ (3).Z&=aE_x\{P(y < \frac{x^2}{b})\}\\ (4).Z&=aE_x\{F_y(\frac{x^2}{b})\} \end{align}
