Let $\xi$ be a random variable with $$F_\xi = \begin{cases} e^x, & x < -1 \\ \frac{1}{8} x + \frac{1}{2}, & -1 \le x < 1 \\ 1 - e^{-x}, & x \ge 1 \end{cases}$$ and let $\eta \sim Bern(\frac{1}{2})$, such that $\xi - \eta$ and $\eta$ are independent. Find $\mathbb{E} 2^{\xi + \eta}$.
What I tried first is to get some information about $\mathrm{cov}(\xi,\: \eta) $. I got that it equals $\frac{1}{4}$. I also counted \begin{align}\mathbb{E}\xi &= \int_{-\infty}^{-1} x e^x d x + \int_{-1}^{1} x \frac{1}{8} d x + \int_{1}^{\infty} x e^{-x} d x \\[0.2cm]&=(-1) \left(\frac{1}{8}(-1) + \frac{1}{2} - e^{-1}\right ) + 1\left( 1 - e^{-1} - \frac{1}{8} + \frac{1}{2}\right) = 1 - \frac{2}{e}\end{align} and then, using that $\mathrm{cov}(\xi,\: \eta) = \frac{1}{4}$, I counted $$\mathbb{E} \xi \eta = \frac{3}{4} - \frac{1}{e}, \:\, \mathbb{E} \xi \mathbb{E} \eta = -\frac{1}{2} + \frac{1}{e}\,\, \text{ and even } \mathbb{E}(\xi - \eta)\eta = -\frac{3}{4} + \frac{1}{e}$$ However, I think I've done it only because I do not know what to do, and now I've run out of ideas! Help pls!
I think i finally got it, can you check it please? My sollution:
\begin{align}\mathbb{E} [2^{\xi + \eta}] = \mathbb{E} [2^{\xi - \eta}] \mathbb{E} [2^{2\eta}] = \mathbb{E} [\frac{2^\xi}{2^\eta}] \frac{5}{2} = \frac{5}{2} (\mathbb{E}[ 2^{\xi - \eta} I(\eta = 1)] + \mathbb{E}[2^{\xi - \eta} I(\eta = 0)] ) = \frac{5}{2} \frac{1}{2} (\mathbb{E} [2^\xi] + \mathbb{E} [2^ {\xi - 1}])\end{align}
what is trivial.
