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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?

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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}

Can you simplify the equations above?

Edit:

In the second equation, I have used the fact that we are given $$Pr(Y=1|A=0)=Pr(Y=1|A=1)$$

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  • $\begingroup$ 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. $\endgroup$ – Phil Nov 2 '18 at 7:41
  • $\begingroup$ We are told $Pr(Y=1|A=0)=Pr(Y=1|A=0)$ isn't it? $\endgroup$ – Siong Thye Goh Nov 2 '18 at 7:44
  • $\begingroup$ 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. $\endgroup$ – Phil Nov 2 '18 at 7:49
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    $\begingroup$ Let me add a remark there. Thanks for the feedback. $\endgroup$ – Siong Thye Goh Nov 2 '18 at 7:50

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