What does it mean to take $\| \cdot \|_{L1}$ of Bernoulli r.v.?

Question

What does it mean to take $\| \cdot \|_{L1}$ of Bernoulli r.v.?

I saw an example that said that if $\epsilon_k$ is Bernoulli taking values $1$ and $-1$, then $\| \epsilon_k \|_{L1}=1$ for all $k$.

Answer 1

$$\Vert X \Vert_{L_1}:= \int_\Omega |X| d \mathbb{P}= \mathbb{E}|X|$$

Similarly for $p \geq 1$,

$$\Vert X \Vert_{L_p}:= \left(\int_\Omega |X|^p d\mathbb{P}\right)^{1/p}= \mathbb{E}[|X|^p]^{1/p}$$

Thus if $X$ takes the value $-1$ with probability $p$ and $X$ takes the value $1$ with probability $1-p$, we have

$$\Vert X \Vert_{L_1} = \mathbb{E}|X| = |-1| \mathbb{P}\{X=-1\} + 1\mathbb{P}\{X=1\} = 1.p + 1.(1-p) = 1$$

Answer 2

The $L^1$ norm of a random variable $X$ defined on a probability space $(\Omega, P)$ in general is the integral $\int_{\Omega} |X(\omega)|dP$. For discrete random variables, it is the sum $$\sum_{\omega \in \Omega} |X(\omega)|P(X=\omega).$$ Thus, in the case $X$ takes the values $\pm 1$, we have $$||X||_{L^1}= |-1| \cdot P(X=-1)+ |1| \cdot P(X=1) = 1.$$