Find the mean of $\text{max}_{t\in[0,1]}\{|At+B|\}$ with $A,B\sim N(0,\sigma^2)$ I have the following exercise that I am having trouble solving:
Let $A,B\sim N(0,\sigma^2)$. Find the mean of the following random variables:
a) $\text{max}_{t\in[0,1]}\{At+B\}$
b) $\text{max}_{t\in[0,1]}\{|At+B|\}$
I may have found a way of solving (a): we write
$$Y=\text{max}_{t\in[0,1]}\{At+B\}=(A+B)\unicode{x1D7D9}_{\{A\geq0\}}+B\unicode{x1D7D9}_{\{A< 0\}}.$$
Then
\begin{align} \mathbb{E}[Y] & = \mathbb{E}[A\unicode{x1D7D9}_{\{A\geq0\}}] + \mathbb{E}[B] \\
 & = \int_0^\infty \frac{x}{\sqrt{2\pi\sigma^2}}\exp{(-\frac{x^2}{2\sigma^2})} dx \\
 & = \frac{-\sigma^2}{\sqrt{2\pi\sigma^2}}\exp{(-\frac{x^2}{2\sigma^2})}|_0^\infty \\
 & = \frac{\sigma}{\sqrt{2\pi}}
\end{align}
Now splitting the variable up in two cases $A<0$ and $A\geq 0$ using indicator variables is doable, but how to approach (b)? Splitting up seems like an incredibly convoluted approach, that will yield integrals that are very difficult to solve.
How do I solve (b) and is there an easier way to solve (a)?
 A: Draw a picture with the values of $Y$:

Then integrate each of the 4 areas and add them up:
$$m_1=\int _0^{\infty }\int _{-\frac{a}{2}}^{\infty }\frac{(a+b) e^{-\frac{a^2}{2 \sigma ^2}} e^{-\frac{b^2}{2 \sigma ^2}}}{\left(\sqrt{2 \pi } \sigma \right) \left(\sqrt{2 \pi } \sigma \right)}dbda=\frac{\left(3 \sqrt{5}+5\right) \sigma }{10 \sqrt{2 \pi }}$$
$$m_2=\int _{-\infty }^0\int _{-\infty }^{-\frac{a}{2}}-\frac{(a+b) e^{-\frac{a^2}{2 \sigma ^2}} e^{-\frac{b^2}{2 \sigma ^2}}}{\left(\sqrt{2 \pi } \sigma \right) \left(\sqrt{2 \pi } \sigma \right)}dbda=\frac{\left(3 \sqrt{5}+5\right) \sigma }{10 \sqrt{2 \pi }}$$
$$m_3=\int _{-\infty }^0\int _{-\frac{a}{2}}^{\infty }\frac{b e^{-\frac{a^2}{2 \sigma ^2}} e^{-\frac{b^2}{2 \sigma ^2}}}{\left(\sqrt{2 \pi } \sigma \right) \left(\sqrt{2 \pi } \sigma \right)}dbda=\frac{\sigma }{\sqrt{10 \pi }}$$
$$m_4=\int _0^{\infty }\int _{-\infty }^{-\frac{a}{2}}-\frac{b e^{-\frac{a^2}{2 \sigma ^2}} e^{-\frac{b^2}{2 \sigma ^2}}}{\left(\sqrt{2 \pi } \sigma \right) \left(\sqrt{2 \pi } \sigma \right)}dbda=\frac{\sigma }{\sqrt{10 \pi }}$$
$$m_1+m_2+m_3+m_4=\frac{\left(\sqrt{5}+1\right) \sigma }{\sqrt{2 \pi }}$$
