If A,B indenpendent, and P(C|A),P(C|B) >P(C), what is relation P(C|A,B) with P(C)?

Question

If A,B indenpendent, and $P(C|A),P(C|B) >P(C)$, what is relation $P(C|A,B)$ with $P(C)$?(> or <)

Answer 1

There's nothing we can say in general; it can either be $>$ or $<$ depending on the situation.

An example where $\mathbb{P}(C|A,B) < \mathbb{P}(C)$

Let $X$ and $Y$ be independent $\text{Bernoulli}(1/4)$ random variables and $Z = X+Y$.

Let $A = \{ X = 1 \}$, $B = \{Y = 1\}$, $C = \{Z= 1 \}$.

It is not too difficult to check that $A$ and $B$ are independent, and that

• $\mathbb{P}(C) = 3/8$
• $\mathbb{P}(C|A) = \mathbb{P}(C|B)=3/4$
• $\mathbb{P}(C|A,B) = 0$

So $\mathbb{P}(C|A),\mathbb{P}(C|B)>\mathbb{P}(C)$, but $\mathbb{P}(C|A,B)<\mathbb{P}(C)$.

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An example where $\mathbb{P}(C|A,B) > \mathbb{P}(C)$

Let $X,Y, Z$ be as before.

Now let $A = \{X=1\}, B= \{Y=1\}, C=\{Z \geq 1\}$.

Then $A$ and $B$ are still independent, and we have

• $\mathbb{P}(C) = 7/16$
• $\mathbb{P}(C|A) = \mathbb{P}(C|B)= 1$
• $\mathbb{P}(C|A,B) = 1$

So $\mathbb{P}(C|A),\mathbb{P}(C|B)>\mathbb{P}(C)$, and $\mathbb{P}(C|A,B)> \mathbb{P}(C)$.