# Conditional probability and independence

Suppose A and Y are discrete dichotomous variables $$(A=0,1; Y=0,1)$$

If $$Pr[Y=1|A=1] = Pr[Y=1|A=0]$$, why can we conclude that $$Pr[Y=1|A=1] = Pr[Y=1|A=0] = Pr[Y = 1],$$ without knowing beforehand whether $$A$$ and $$Y$$ are independent?

By the law of total probability, \begin{align} Pr(Y=1) &= Pr(Y=1|A=1)Pr(A=1) + Pr(Y=1|A=0)Pr(A=0) \\ &=Pr(Y=1|A=1)Pr(A=1) + Pr(Y=1|A=1)(1-Pr(A=1))\\ \end{align}
In the second equation, I have used the fact that we are given $$Pr(Y=1|A=0)=Pr(Y=1|A=1)$$
• If you, in your equation, are only using the law of total probability, and not an indepedence property, the last part of the equation should read $Pr(Y=1\mid A=0)(1-Pr(A=1))$ and not $Pr(Y=1\mid A=1)(1-Pr(A=1))$. Otherwise the equivalence does not hold. – Phil Nov 2 '18 at 7:41
• We are told $Pr(Y=1|A=0)=Pr(Y=1|A=0)$ isn't it? – Siong Thye Goh Nov 2 '18 at 7:44
• I am assuming that you mean $Pr(Y=1\mid A=1) = Pr(Y=1 \mid A=0)$. Yes, we are. But my point is that we are using external information for that part of the calculation, and not just relying on the law of total probability. I would thus add somewhere in the writing that "where the last equality holds because, here, $Pr(Y=1\mid A=1) = Pr(Y=1 \mid A=0)$" or something along those lines. – Phil Nov 2 '18 at 7:49