Let , , and be events. If $(|∩)=(|)$ , then is said to screen off from . Suppose that $(∩)>0$. Show that screening off is equivalent to saying that $$(∩|)=(|)(|).$$ What does this latter equation say in terms of independence?
I am struggling to find a way to formulate an answer to this.
Essentially what I am thinking is something along the lines of:
As $$ P(E|C) = \frac{P(E∩C)}{ P(C)} $$ And $$ P(A∩C) = P(A|C)P(C) $$ Then something along the lines of: $$ \frac{P(E∩(P(A∩C)P(C))) }{P(A∩C)P(C)} = \frac{P(E∩C) }{P(C)} $$
That is a very rough idea. I am not sure if that is even remotely close to what I need to be doing. Any suggestions would be great. Ideally I would like to figure this out myself but a nudge in the right direction would be great.
The later equation indicates that event E and A are independent of one another as knowing the outcome of event A does not interfere with the outcome of event E.
Your attempt is close. Applying the definition of conditional probability,
$$\begin{align*} P(E\mid A\cap C) &= P(E\mid C)\\ \frac{P(E\cap A \cap C)}{P(A\cap C)} &= P(E\mid C)\\ \frac{P(A\cap E\mid C)P(C)}{P(A\mid C)P(C)} &= P(E\mid C)\\ P(A\cap E\mid C) &= P(A\mid C)P(E\mid C) \end{align*}$$
Compare these, with or without condition:
$$\begin{align*} P(E\mid A\cap C) &= \frac{P(E\cap A\mid C)}{P(A\mid C)} &&(P(C)\ne 0)\\ P(E\mid A) &= \frac{P(E\cap A)}{P(A)} \end{align*}$$
Compare these, for 1) independent $A$ and $E$ given $C$, and 2) independent $X$ and $Y$:
$$\begin{align*} P(A \cap E\mid C) &= P(A\mid C)P(E\mid C)\\ P(X\cap Y) &= P(X)P(Y) \end{align*}$$