# Conditional probabilities for discrete random variables

Let $X$ and $Y$ be two (dependent) non-negative discrete random variables. Assume that these variables can take the following values: $0$, $1$ and $2$.

I have some questions regarding conditional probabilities:

1) Is the following relation correct: $$P(X>0 \mid Y \ge 0)=P(X>0 \mid Y=0)+P(X>0 \mid Y=1)+P(X>0 \mid Y=2)$$ If it is not the case, how can we express $P(X>0 \mid Y \ge 0)$ ?

2) Can we claim that $P(X>0 \mid Y \ge 0) \ge P(X>0 \mid Y > 0)$ ? if so, why ?

3) For general non-negative random variables $X$, $Y$, $Z$ (discrete or continuous), what is the relation between $P(X>0 \mid Y \ge 0, Z<t)$ and $P(X>0 \mid Y \ge 0, Z \ge t)$, for some constant $t >0$.
In other words, what is the relation between $P( Y \ge 0, Z<t \mid X>0 )$ and $P( Y \ge 0, Z \ge t \mid X >0 )$ ?

• do you mean non-negative (instead of positive)? Apr 9 '17 at 14:16
• @drhab you are correct. I have edited the question. Thank you.
– din
Apr 9 '17 at 14:18

$\mathsf{Q1}.$ No. It is not correct. In fact, since $Y$ takes values from $\{0, 1, 2\}$, $Y \geq 0$ gives no information for $X$. Therefore, $$\Pr(X > 0 \mid Y \geq 0) = \Pr(X > 0)$$ However, we have by law of total probability $$\Pr( X>0, Y \geq 0) = \Pr(X > 0, Y = 0) + \Pr(X > 0, Y = 1) + \Pr(X > 0, Y = 2)$$ but $\Pr(X > 0, Y \geq 0)$ is different from $\Pr(X > 0 \mid Y \geq 0)$.
$\mathsf{Q2}.$ No. It depends. An example would be $X = Y$ and $$X = \begin{cases} 0 &\text{with probability } 1 / 3 \\ 1 & \text{with probability } 1/ 3 \\ 2 & \text{with probability } 1/ 3 \end{cases}$$ In this example, $$\Pr( X> 0 \mid Y \geq 0) = \frac{2}{3} \quad\text{and}\quad\Pr(X > 0 \mid Y > 0) = 1$$
$\mathsf{Q3}.$ We have $$\Pr(Y \geq 0, Z < t \mid X > 0) + \Pr(Y \geq 0, Z \geq t \mid X > 0) = \Pr(Y \geq 0 \mid X > 0)$$