Perpendicularity of random variables? I am reading Bechavod et al. (2017) [1], and at page 3 there is written:

In the example, each data point lies in $X = (X_1,X_2) = \{0, 1\}^2$
  and has two features—$X_1 = A$ is the protected attribute, and $X_2$
  is a non-protected attribute—and a label in $Y = {0, 1}$. Given
  $\epsilon \in (0,\frac{1}{4})$, we define a distribution $\mathcal{D}$
  over labelled examples as follows: 
$\mathbb{P}[Y = 1] = 0.5$ 
$\mathbb{P}[A = y|Y = y] = 1 − \epsilon $
$\mathbb{P}[X2 = y|Y = y] = 1 − 2\epsilon$
Note that $\mathcal{D}_\epsilon$ is defined s.t. $A \,\bot\, X_2|Y$.

What does it mean, mathematically, that $A \,\bot\, X_2|Y$? Aren't those random vectors?
[1] Yahav Bechavod, Katrina Ligett, "Penalizing Unfairness in Binary Classification" (link)
 A: $\bot$ usually means that the random variables in question are uncorrelated; but sometimes it means independent. Fortunately, for Bernoulli random variables, which $X_1=A$ and $X_2$ are, the two properties are the same).  So, I would interpret $A \,\bot\, X_2 \mid Y$ to mean that $X_1$ and $X_2$ are conditionally independent random variables given $Y$. 
Let us dissect the given information a little more. We are told that 
\begin{align}
P(X_1 = 1\mid Y = 1) &= 1-\varepsilon\\
P(X_2 = 1\mid Y = 1) &= 1-2\varepsilon
\end{align}
from which we can deduce that
\begin{align}
P(X_1 = 0\mid Y = 1) &= 1 - (1-\varepsilon) = \varepsilon\\
P(X_2 = 0\mid Y = 1) &= 1-(1-2\varepsilon) = 2\varepsilon
\end{align}
but we are not told the value of $P(X_1 = 1, X_2 = 1\mid Y = 1)$ explicitly; it is hidden in the $\bot$ statement which says that $X_1$ and $X_2$ are (conditionally) independent given $Y=1$, that is,
$$P(X_1 = 1, X_2 = 1\mid Y = 1) = P(X_1 = 1\mid Y = 1)P(X_2 = 1\mid Y = 1) = (1-\varepsilon)(1-2\varepsilon)$$
 Combining all this, we have that 

Conditioned on $Y=1$, $X_1$ and $X_2$ are conditionally independent Bernoulli random variables with parameters $1-\varepsilon$ and $1-2\varepsilon$ respectively.

A very similar analysis for the case when $Y=0$ gives us that

Conditioned on $Y=0$, $X_1$ and $X_2$ are conditionally independent Bernoulli random variables with parameters $\varepsilon$ and $2\varepsilon$ respectively.

We readily get that unconditionally, $X_1$ and $X_2$ are Bernoulli random variables with parameters
\begin{align}
P(X_1 = 1) &= P(X_1 = 1\mid Y = 1)P(Y=1)+P(X_1 = 1\mid Y = 1)P(Y=0)\\
&= (1-\varepsilon)\frac 12 + (\varepsilon)\frac 12\\
&= \frac 12,\\
P(X_2 = 1) &= P(X_2 = 1\mid Y = 1)P(Y=1)+P(X_2 = 1\mid Y = 1)P(Y=0)\\
&= (1-2\varepsilon)\frac 12 + (2\varepsilon)\frac 12\\
&= \frac 12
\end{align}
but they are not unconditionally independent since
$$P(X_1 = 1, X_2 = 1) = (1-\varepsilon)(1-2\varepsilon)\frac 12
+ (\varepsilon)(2\varepsilon)\frac 12\neq \frac 14 = P(X_1=1)P(X_2=1).$$
